Glencoe Solution Manual
Glencoe Solution Manual
Glencoe Solution Manual
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Copyright © by The McGraw-Hill Companies, Inc. All rights reserved. Except aspermitted under the United States Copyright Act, no part of this publication may bereproduced or distributed in any form or by any means, or stored in a database orretrieval system, without prior permission of the publisher.Send all inquiries to:<strong>Glencoe</strong>/McGraw-Hill8787 Orion PlaceColumbus, OH 43240-4027ISBN: 0-02-834177-5Printed in the United States of America.4 5 6 7 8 9 10 024 08 07 06 05 04
Chapter 1 Linear Relations and Functions1-1Pages 8–91.Relations and Functionsx y4 26 10 58 42 24 02. Sample answer:yOCheck for Understandingxx yy1 32 23 1Ox4 05 16 27 3Equation: y x 46. {(3, 4), (0, 0), (3,4), (6, 8)}; D {3, 0, 3, 6};R {8, 4, 0, 4}7. {(6, 1), (4, 0), (2, 4), (1, 3), (4, 3)};D {6, 4, 2, 1, 4}; R {4, 0, 1, 3}84y42O244 8x3. Determine whether a vertical line can be drawnthrough the graph so that it passes through morethan one point on the graph. Since it does, thegraph does not represent a function.4. Keisha is correct. Since a function can beexpressed as a set of ordered pairs, a function isalways a relation. However, in a function, there isexactly one y-value for each x-value. Not allrelations have this constraint.5. Table: Graph:8.9.x y4 73 42 11 20 51 82 113 144 17x y1 52 53 54 55 56 57 58 54yOy1284O42 4x10. {3, 0, 1, 2}; {6, 0, 2, 4}; yes; Each member of thedomain is matched with exactly one member ofthe range.11. {3, 3, 6}; {6, 2, 0, 4}; no; 6 is matched withtwo members of the range.12a. domain: all reals; range: all reals12b. Yes; the graph passes vertical line test.13. f(3) 4(3) 3 (3) 2 5(3)108 9 15 or 8414. g(m 1) 2(m 1) 2 4(m 1) 2 2(m 2 2m 1) 4m 4 2 2m 2 4m 2 4m 4 2 2m 215. x 1 0x 1The domain excludes numbers less than 1.The domain is {x⏐x 1}.16a. {(83, 240), (81, 220), (82, 245), (78, 200),(83, 255), (73, 200), (80, 215), (77, 210), (78, 190),(73, 180), (86, 300), (77, 220), (82, 260)}; {73, 77,78, 80, 81, 82, 83, 86}; {180, 190, 200, 210, 215,220, 240, 245, 255, 260, 300}x1 Chapter 1
16b.Weight(lb)30028026024022020018016c. No; a vertical line at x 77, x 78, x 82, orx 83 would pass through two points.Pages 10–12 Exercises17. Table Graph:x y1 32 63 94 125 156 187 218 249 27Equation: y 3x18. Table:O 70 72 74 76 78 80 82 84 86Height (in.)y2418126O2 4 6 8 10xx y6 115 104 93 82 71 6Equation: y x 519. Table: Graph:x yy4 43 52 61 70 81 92 103 114 12OEquation: y 8 x20. {(5, 5), (3, 3), (1, 1), (1, 1)};D {5, 3, 1, 1}; R {5, 3, 1, 1}21. {(10, 0), (5, 0), {0, 0), (5, 0)};D {10, 5, 0, 5}; R {0}yOxx22. {(4, 0), (5, 1), (8, 0), (13, 1)};D {4, 5, 8, 13}; R {0, 1}23. {(3, 2), (1, 1), (0, 0), (1, 1)};D {3, 1, 0, 1}; R {2, 0, 1}24. {(5, 5), (3, 3), (1, 1), (2, 2), (4, 4)};D {5, 3, 1, 2, 4}; R {4, 2, 1, 3, 5}25. {(3, 4), (3, 2), (3, 0), (3, 1), (3, 3)}; D {3};R {4, 2, 0, 1, 3}26.yx yO4 93 82 71 60 51 427.28.29.30.x y1 12 23 34 45 56 6x y5 54 43 32 21 10 01 1x y1 02 33 64 95 12x y11 311 342O–2–4yOyOyyO4 8 12xxxxxChapter 1 2
31.x y4 24 2y51a. x 1Ox32. {4, 5, 6}; {4}; yes; Each x-value is paired withexactly one y-value.33. {1}; {6, 2, 0, 4}; no; The x-value 1 is paired withmore than one y-value.34. {0, 1, 4); {2, 1, 0, 1, 2}; no; The x-values 1 and 4are paired with more than one y-value.35. {0, 2, 5}; {8, 2, 0, 2, 8}; no; The x-values 2 and 5are paired with more than one y-value.36. {1.1, 0.4, 0.1}; {2, 1}; yes; Each x-value ispaired with exactly one y-value.37. {9, 2, 8, 9}; {3, 0, 8}; yes; Each x-value is pairedwith exactly one y-value.38. domain: all reals; range: all reals; Not a functionbecause it fails the vertical line test.39. domain: {3, 2, 1, 1, 2, 3}; range: {1, 1, 2, 3};Afunction because each x-value is paired withexactly one y-value.40. domain: {x⏐8 x 8}; range: {y⏐8 y 8};Not a function because it fails the vertical linetest.41. f(3) 2(3) 3 6 3 or 942. g(2) 5(2) 2 3(2) 2 20 6 2 or 12143. h(0.5) 0 .5 244. j(2a) 1 4(2a) 3 1 4(8a 3 ) 1 32a 345. f(n 1) 2(n 1) 2 (n 1) 9 2(n 2 2n 1) n 1 9 2n 2 4n 2 n 1 9 2n 2 5n 1246. g(b 2 1) 3 ( b2 1)5 ( b2 1) 3 b 26 1b2 or 2 b26 b 247. f(5m) ⏐(5m) 2 13⏐ ⏐25m 2 13⏐48. x 2 5 0x 2 5x 5; x 549. x 2 9 0x 2 93 x 3; x 3 or x 350. x 2 7 0x 2 77 x 7; x 7 or x 751b. x 551c. x 2, 252a. {(13,264, 4184), (27,954, 4412), (21,484, 6366),(23,117, 3912), (16,849, 2415), (19,563, 5982),(17,284, 6949)}; {13,264, 16,849, 17,284, 19,563,21,484, 23,117, 27,954}; {2415, 3912, 4184, 4412,5982, 6366, 6949}52b.7NumberAttending(thousands)52c. Yes; no member of the domain is paired withmore than one member of the range.153. x 2m 1, so x 2 m.Substitute x 12for m in f(2m 1) to solve for f(x),24m 3 36m 2 26m165432O12 16 20 24 28Number Applied(thousands)11 24 x 2 3 36 x 2 2 26 x 2x 24 3 3x 2 3x 1 2x 1 36 x2 4 26 x 1 82 3x 3 9x 2 9x 3 9x 2 18x 9 13x 13 3x 3 4x 754a. t(500) 95 0.005(500) 92.5°F54b. t(750) 95 0.005(750) 91.25°F54c. t(1000) 95 0.005(1000) 90°F54d. t(5000) 95 0.005(5000) 70°F3 Chapter 1
54e. t(30,000) 95 0.005(30,000)55°F55a. d(0.05) 299,792,458(0.05) 14,989,622.9 md(0.02) 299,792,458(0.2) 59,958,491.6 md(1.4) 299,792,458(1.4) 419,709,441.2 md(5.9) 299,792,458(5.9) 1,768,775,502 m55b. d(0.008) 299,792,458(0.08) 23,983,396.64 m) 156. P(4) (1)(2 3 1P(5) (2)(3 ) 11 7P(6) (3)(1 ) 17 4 7 57. 7 2 (3 2 4 2 ) 49 (9 16) 49 25 or 24The correct choice is B.1-2Composition of Functions3.4. Sample answer: The (sum/difference/product/quotient) of the function values is the functionvalues of the (sum/difference/product/quotient)of the functions.5. Sample answer: For functions f(x) and g(x),(f g)(x) f(x) g(x); (f g)(x) f(x) g(x);(f g)(x) f(x) g(x); and f g(x) f( x)g(x), g(x) 0Page 131.2.Graphing Calculator ExplorationPage 17 Check for Understanding1. Sample answer: f(x) 2x 1 and g(x) x 6;Sample explanation: Factor 2x 2 11x 6.2. Iteration is composing a function on itself byevaluating the function for a value and thenevaluating the function on that function value.3. No; [f g](x) is the function f(x) performed on g(x)and [g f ](x) is the function g(x) performed on f(x).See students’ counter examples.4. Sample answer: Composition of functions isperforming one function after another. Aneveryday example is putting on socks and thenputting shoes on top of the socks. Buying an itemon sale is an example of when a composition offunctions is used in a real-world situation.5. f(x) g(x) 3x 2 4x 5 2x 9 3x 2 6x 4f(x) g(x) 3x 2 4x 5 (2x 9) 3x 2 2x 14f(x) g(x) (3x 2 4x 5)(2x 9) 6x 3 35x 2 26x 45f f( x)(x) g g( x) 3x2 4x 52x 9, x 9 2 6. [f g](x) f(g(x)) f(3 x) 2(3 x) 5 2x 11[g f ](x) g(f(x)) g(2x 5) 3 (2x 5) 2x 8Chapter 1 4
7. [f g](x) f(g(x)) f(x 2 2x) 2(x 2 2x) 3 2x 2 4x 3[g f ](x) g(f(x)) g(2x 3) (2x 3) 2 2(2x 3) (4x 2 12x 9) 4x 6 4x 2 16x 158. Domain of f(x): x 1Domain of g(x): all realsg(x) 1x 3 1x 2Domain of [f g](x) is x 2.9. x 1 f(x 0 ) f(2) 2(2) 1 or 5x 2 f(x 1 ) f(5) 2(5) 1 or 11x 3 f(x 2 ) f(11) 2(11) 1 or 235, 11, 2310a. [K C](F) K(C(F)) K 5 9 (F 32) 5 9 (F 32) 273.1510b. K(40) 5 9 (40 32) 273.1540 273.15 or 233.15K(12) 5 9 (12 32) 273.1524.44 273.15 or 248.71K(0) 5 9 (0 32) 273.1517.78 273.15 or 255.37K(32) 5 9 (32 32) 273.15 0 273.15 or 273.15K(212) 5 9 (212 32) 273.15 100 273.15 or 373.15Pages 17–19 Exercises11. f(x) g(x) x 2 2x x 9 x 2 x 9f(x) g(x) x 2 2x (x 9) x 2 3x 9f(x) g(x) (x 2 2x)(x 9) x 3 7x 2 18x f g(x) x2 2xx ,9x 9x12. f(x) g(x) x x 2 11x x 1 (x2 1)(x 1)x 1x x 1 x3 x2 x 1x 1 x3 x 2 1x ,1x 1xf(x) g(x) x (x 2 1)1x x 1 (x2 1)(x 1)x 1x x 1 x3 x2 x 1x 1x 3 x 2 2x 1 x 1, x 1xf(x) g(x) x (x 2 1) f g(x) 1 x(x 1)(x 1)x 1 x 2 x, x 1xx 1 x 2 1x 1 x 1 x 2 1x x 3 x 2 ,x 1x 1 or 1313. f(x) g(x) x x 2 5x73 x 7 (x2 5x)(x 7)x 73 x3 7x 2 5x 2 35x x 7x 7x 3 2x 2 35x 3 x 7, x 73f(x) g(x) x (x 2 5x)73 x 7 (x2 5x)(x 7)x 73 x3 7x 2 5x 2 35x x 7x 7 x 7 , x 73f(x) g(x) x 7 (x 2 5x)2 15x, x 7 f g(x) 3 1 x 7 x 2 5x3x 3 2x 2 35x, x 5, 0, 72x14. f(x) g(x) x 3 x 5 (x 3)(x 5) 2xx 5 x 5 x2 2x 15 2xx 5 x 5 x2 15x ,5x 52x 5 (x 3)(x 5) 2xx 5 x 5 x2 2x 15 2xx 5 x 5 x2 4x 15x ,5x 52x 5 2x 2 6xx ,5x 5f(x) g(x) x 3 xf(x) g(x) (x 3) x g 3x x 73x 7x 2 5xf (x) x 32xx 5x 3 2x 2 35x 3 5 x 3 x 2x x2 2x 15, x 0 or 52x5 Chapter 1
15. [f g](x) f(g(x)) f(x 4) (x 4) 2 9 x 2 8x 16 9 x 2 8x 7[g f ](x) g(f(x)) g(x 2 9) x 2 9 4 x 2 516. [f g](x) f(g(x)) f(x 6) 1 2 (x 6) 7 1 2 x 3 7 1 2 x 4[g f ](x) g(f(x)) g( 1 2 x 7) 1 2 x 7 6 1 2 x 117. [f g](x) f(g(x)) f(3x 2 ) 3x 2 4[g f ](x) g(f(x)) g(x 4) 3(x 4) 2 3(x 2 8x 16) 3x 2 24x 4818. [f g](x) f(g(x)) f(5x 2 ) (5x 2 ) 2 1 25x 4 1[g f ](x) g(f(x)) g(x 2 1) 5(x 2 1) 2 5(x 4 2x 2 1) 5x 4 10x 2 519. [f g](x) f(g(x)) f(x 3 x 2 1) 2(x 3 x 2 1) 2x 3 2x 2 2[g f ](x) g(f(x)) g(2x) (2x) 3 (2x) 2 1 8x 3 4x 2 120. [f g](x) f(g(x)) f(x 2 5x 6) 1 x 2 5x 6 x 2 5x 7[g f ](x) g(f(x)) g(1 x) (x 1) 2 5(x 1) 6 x 2 2x 1 5x 5 6 x 2 7x 1221. [f g](x) f(g(x)) f 1x 1 1 x 1 11 x 1 x 1x 1x x ,1x 1[g f ](x) g(f(x)) g(x 1)1 x 1 1 1 x , x 022. Domain of f(x): all realsDomain of g(x): all realsDomain of [f g](x): all reals23. Domain of f(x): x 0Domain of g(x): all realsg(x) 07 x 07 xDomain of [f g](x) is x 7.24. Domain of f(x): x 2Domain of g(x): x 0g(x) 2 1 4 x 21 8x 1 8 xDomain of [f g](x) is x 1 8 , x 0.25. x 1 f(x 0 ) f(2) 9 2 or 7x 2 f(x 1 ) f(7) 9 7 or 2x 3 f(x 2 ) f(2) 9 2 or 77, 2, 726. x 1 f(x 0 ) f(1) (1) 2 1 or 2x 2 f(x 1 ) f(2) (2) 2 1 or 5x 3 f(x 2 ) f(5) (5) 2 1 or 262, 5, 2627. x 1 f(x 0 ) f(1) 1(3 1) or 2x 2 f(x 1 ) f(2) 2(3 2) or 2x 3 f(x 2 ) f(2) 2(3 2) or 22, 2, 2Chapter 1 6
28. $43.98 $38.59 $31.99 $114.56Let x the original price of the clothes, or$114.56.Let T(x) 1.0825x. (The cost with 8.25% tax rate)Let S(x) 0.75x. (The cost with 25% discount)The cost of clothes is [T S](x).[T S](x) T(S(x)) T(0.75x) T(0.75(114.56)) T(85.92) 1.0825(85.92) 93.0084Yes; the total with the discount and tax is $93.01.29. Yes; If f(x) and g(x) are both lines, they can berepresented as f(x) m 1 x b 1 and g(x) m 2 x b 2 . Then [f g](x) m 1 (m 2 x b 2 ) b 1 m 1 m 2 x m 1 b 2 b 1Since m 1 and m 2 are constants, m 1 m 2 is aconstant. Similarly, m 1 , b 2 , and b 1 are constants,so m 1 b 2 b 1 is a constant. Thus, [f g](x) is alinear function if f(x) and g(x) are both linear.30a. W n W p W f F p d F f d d(F p F f )30b. W n d(F p F f ) 50(95 55) 2000 J31a. h[f(x)], because you must subtract beforefiguring the bonus31b. h[f(x)] h[f(400,000)] h(400,000 275,000) h(125,000) 0.03(125,000) $375032. (f g)(x) f(g(x)) f(1 x 2 ) x2 (x 2 1)1 x 2 x 2(1 x 2 ) 1So, f(x) x 1 and f 1 2 1 2 1 1 2 .33a. v(p) 7 p47 33b. r(v) 0.84v33c. r(p) r(v(p)) r 7 p47 0.84 7 p47 5. 84 8p7or 1 47p1 175(423.18)17533d. r(423.18) 147 1 $52.94r(225.64) 147 (2251175 $28.23r(797.05) 147 (7971175 $99.72.64).05)34a. I prt 5000(0.08)(1) 400I prt 5400(0.08)(1) 432I prt 5832(0.08)(1) 466.56I prt 6298.56(0.08)(1) 503.88I prt 6802.44(0.08)(1) 544.20(year, interest): (1, $400), (2,$432), (3, $466.56),(4, $503.88), (5, $544.20)34b. {1, 2, 3, 4, 5}; {$400, $432, $466.56, $503.88,$544.20}34c. Yes; for each element of the domain there isexactly one corresponding element of the range.35. {(1, 8), (0, 4), (2, 6), (5, 9)}; D {1, 0, 2, 5};R {9, 6, 4, 8}36. D {1, 2, 3, 4}; R {5, 6, 7, 8}; Yes, every elementin the domain is paired with exactly one elementof the range.37. g(4) ( 4)3 54(38.4) 6 4 516 5911 16or 3 1 639. f(n 1) 2(n 1) 2 (n 1) 9 2(n 2 2n 1) n 1 9 2n 2 5n 12The correct choice is C.1-3x y2 61 30 01 32 63 9Graphing Linear EquationsPage 23 Check for Understanding1. m represents the slope of the graph and brepresents the y-intercept2. 7; the line intercepts the x-axis at (7, 0)3. Sample answer: Graph the y-intercept at (0, 2).Then move down 4 units and right 1 unit to grapha second point. Draw a line to connect the points.yOx7 Chapter 1
4. Sample answer: Both graphs are lines. Both lineshave a y-intercept of 8. The graph of y 5x 8slopes upward as you move from left to right onthe graph and the graph of y 5y 8 slopesdownward as you move from left to right on thegraph.5. 3x 4(0) 2 0 3(0) 4y 2 03x 2 0 4y 2 03x 24y 2x 2 3 y 1 2 y9. 1 2 x 6 012 1 2 x 6x 12812844 O4yx2( ,03)O1(0, )2x10. Since m 0 and b 19, this function has nox-intercept, and therefore no zeros.y246. x 2(0) 5 0 0 2y 5 0x 5 0 2y 5 0x 5 2y 5y 5 2 yO(0, 52)x(5, 0)1684 2 O 2 4x11a. (38.500, 173), (44.125, 188)188 17311b. m 44.125 38.50015 5. 625 or about 2.66711c. For each 1 centimeter increase in the length of aman’s tibia, there is an 2.667-centimeterincrease in the man’s height.7. The y-intercept is 7. Graph (0, 7).The slope is 1.y(0, 7)(1, 8)Pages 24–25 Exercises12. The y-intercept is 9. The slope is 4.yOxy 4x 9O x8. The y-intercept is 5. Graph (0, 5).The slope is 0.y13. The y-intercept is 3. The slope is 0.yy 3(0, 5)OxOxChapter 1 8
14. 2x 3y 15 03y 2x 15y 2 3 x 5The y-intercept is 5. The slope is 2 3 .y19. 2x y 0y 2xThe y-intercept is 0. The slope is 2.y2x 3y 15 0O2x y 0x15. x 4 0x 4There is no slope. The x-intercept is 4.yO16. The y-intercept is 1. The slope is 6.yOx 4 0xx20. The y-intercept is 4. The slope is 2 3 .yO2y 3 x 421. The y-intercept is 150. The slope is 25.yy 25x 15010050xy 6x 164 2 O 2x50O x17. The y-intercept is 5. The slope is 2.y22. 2x 5y 85y 2x 8y 2 5 x 8 5 The y-intercept is 8 5 . The slope is 2 5 .yy 5 2xO2x 5y 8xOx18. y 8 0y 8The y-intercept is 8. The slope is 0.yOx23. 3x y 7y 3x 7y 3x 7The y-intercept is 7. The slope is 3.yOxy 8 03x y 79 Chapter 1
24. 9x 5 09x 5x 5 9 The y-intercept is 5.25. 4x 12 04x 12x 3The y-intercept is 12.26. 3x 1 03x 1x 1 3 The y-intercept is 1.27. 14x 0x 0The slope is 14.Of(x) 9x 5( , 0)59f(x)f(x)Of(x) 4x 12f(x) 3x 1( )1,03f(x)2814(0, 0)Of(x)(3, 0)Oxxxf(x) 14xx33a. (1.0, 12.0), (10.0, 8.4)m 8 . 4 12.010.0 1.0 3 .69or 0.4(0.4) 0.4 ohms12 v33b. 0.4 1.0 25.02 v240.4 1 9.6 12 vv 2.4 volts9 734. m 274 3 2 7 1 9 a (4) 2 7 8 a 435a. (86.85, 90), (126.85, 100)100 – 90m 126.85 – 86.85 1 042(a 4) 562a 8 562a 48a 240 or 1 4 35b. For each 1 degree increase in the temperature,there is a 1 4 -pascal increase in the pressure.35c. P1008028. None; since m 0 and b 12, this function hasno x-intercepts, and therefore no zeros.f(x)84O x29. 5x 8 05x 8x 8 5 The y-intercept is 8.30. 5x 2 05x 2x 2 5 f(x) 1231. The y-intercept is 3. The slope is 3 2 . 3 2 x 3 0 3 2 x 3x 232. Sample answer: f(x) 5; f(x) 0f(x)f(x) 5x 822 O468yO( )8, 052 x3y x 32(2, 0)x604020O20 40 60 8036. No; the product of two positives is positive, so forthe product of the slopes to be 1, one of theslopes must be positive and the other must benegative.37a. 10,440 290t 0290t 10,440t 36The software has no monetary value after36 months.37b. 290; For every 1-month change in the numberof months, there is a $290 decrease in the valueof the software.37c. v(t)(0, 10,440)10,0008000600040002000O(36, 0)T8 16 24 32tChapter 1 10
38. Afunction with a slope of 0 has no zeros if itsy-intercept is not 0; a function with a slope of 0has an infinite number of zeros if its y-intercept is0; a function with any slope other than 0 hasexactly 1 zero.39a. (56, 50), (76, 67.2)m 6 7.2 5076 56 1 7.220or 0.8639b. 1805(0.86) $1552.3039c. 1 MPC 1 0.86 0.1439d. 1805(0.14) $252.7040. (f g)(x) 2x x 2 4 or x 2 2x 4(f g)(x) 2x (x 2 4)x 2 2x 441a. 1 0.12 0.88d(p) 0.88p41b. r(d) d 10041c. r(d(p)) r(0.88p) 0.88p 10041d. r(799.99) 0.88(799.99) 100 603.9912 or about $603.99r(999.99) 0.88(999.99) 100 779.9912 or about $779.99r(1499.99) 0.88(1499.99) 100 1219.9912 or about $1219.9942. [f g](3) f(g(3)) f(3 2) f(5) (5) 2 4(5) 5 25 20 5 or 50[g f ](3) g( f(3)) g((3) 2 4(3) 5) g(9 12 5) g(26) 26 2 or 2443. f(9) 4 6(9) (9) 3 4 54 729 or 67144. No; the graph fails the vertical line test.45.x y3 142 131 120 1146. Let s sum.s 4 15s 60The correct choice is D.{(3, 14), (2, 13), (1, 12),(0, 11)}, yes1-3BGraphing Calculator Exploration:Analyzing Families of LinearGraphsPage 261. See students’ graphs. All of the graphs are lineswith y-intercept at (0, 2). Each line has adifferent slope.2. A line parallel to the ones graphed in the Exampleand passing through (0, 2).3. See students’ sketches. Sample answer: Thegraphs of lines with the same value of m areparallel. The graphs of lines with the same valuefor b have the same y-intercept.1-4Writing Linear EquationsPage 29 Check for Understanding1. slope and y-intercept; slope and any point; twopoints2. Sample answer:Use point-slopeUse slope-interceptform:form.y y 1 m(x x 1 ) y mx by (4) 1 4 (x 3) 4 1 4 (3) by 4 1 4 x 3 4 1 94 bx 4y 19 0Substitute the slopeand intercept into thegeneral form.y 1 4 x 1 94Write in standard form.x 4y 19 03. 55 represents the hourly rate and 49 representsthe fee for coming to the house.4. m 3 00 6y 1 2 x 3 36or 1 2 5. Sample answer: When given the slope and they-intercept, use slope-intercept form. When giventhe slope and a point, use point-slope form. Whengiven two points, find the slope, then use pointslopeform.6. y mx b → y 1 4 x 107. y 2 4(x 3) 8. m 9 27 5y 2 4x 12 7 2 y 4x 10y 2 7 2 (x 5)9. y 2 0(x (9))y 2 0y 2y 2 7 2 x 3 5 2y 7 2 x 3 1210a. y 5.9x 211 Chapter 1
27b. Using sample answer from part a,10b. y 5.9(7) 2 5 8 41.3 2 or 43.3 in.y 9 7 (19) 1 6710c. Sample answer: No; the grass could not supportit own weight if it grew that tall. 17 17 1 67or 18 77or about 26.7 mpg27c. Sample answer: The estimate is close but notexact since only two points were used to writePages 30–31 Exercisesthe equation.11. y mx b → y 5x 212. y 5 8(x (7))28a. See students’ work.28b. Sample answer: Only two points were used toy 5 8x 56make the prediction equation, so many points lieoff of the line.y 8x 6113. y mx b → y 3 29. Yes; the slope of the line through (5, 9) and (3, 3)4 x3 914. y mx b → y 12x 1 is 2 3 5or 3 4 . The slope of the line through6 3(3, 3), and (1, 6) is 15. y 5 6(x 4) 16. x 121 (3)or 3 4 . Since these twolines would have the same slope and would sharey 5 6x 24a point, their equations would be the same. Thus,y 6x 19they are the same line and all three points are9 517. m 8 1y 5 4 9 (x 1)collinear.4 9y 5 4 9 x 4 9 30. 3x 2y 5 0yy 4 9 x 4 92y 3x 5918. (8, 0), (0, 5) y 0 5 y 8 3(x (8))2 x 5 3x 2y 5 02 5 0m 0 (8)y 5 8 O xx 51 119. m 3 8y 1 0(x 8)0y 1 0 11 or 0y 131a. (1995, 70,583), (1997, 82,805)20. x 4 21. x 022. y 0 0.25(x 24)y 0.25x 623. y (4) 1 2 (x (2))y 4 1 2 x 1 1 2 x y 5 0x 2y 10 0x 2y 103 024. m 1 ( 2) 331y 0 1(x 2)y x 2x y 225a. t 2 x 207000004 700000025b. t 2 14,49 2 2 3.747 or 5.747; about 5.7 weeks26. Ax By C 0By Ax CAy x B C B A; m B27a. Sample answer: using (20, 28) and (27, 37),m 3 7 2827 20y 28 9 7 (x 20) 9 7 y 28 9 7 x 18 07y 9 7 x 1 67m 82, 805 70,5831997–1999 12, 2222 or 6111; $6111 billion31b. The rate is the slope.32. g[f(2)] g(f(2)) g((2) 3 ) g(8) 3(8) or 2433. (f g)(x) f(x) g(x) x 3 (x 2 3x 7) x 5 3x 4 7x 3f (x)x3 x, where g(x) 02 34. gx y4 163 92 43x 735. x 1 y y yThe correct choice is A.{(4, 16), (3, 9), (2, 4)}, yes1 y 2y 1 y 2 y2 1y 2 y2 1y 1 2 y2 12yChapter 1 12
Page 31 Mid-Chapter Quiz1. {2, 2, 4}, {8, 3, 3, 7}; No, 2 in the domain ispaired with more than one element of the range.2. f(4) 7 4 2 7 16 or 94. Let x original price of jacketLet T(x) 1.055x. (The cost with 5.5% tax rate)Let S(x) 0.67x. (The cost with 33% discount)The cost of the jacket is [T S](x).[T S](x) T(S(x)) T(0.67x) 1.055(0.67x)The amount paid was $49.95.45.95 1.055(0.67x)43.55 0.67x65 x; $655. [f g)(x) f(g(x) f(x 1)1 x 1 1 1 x , x 0[g f ](x) g(f(x))1 g( x )11 x 1 11 x 1 x 1x 1x x ,1x 16. 2x 4y 84y 2x 8y 1 2 x 27. 3x 2y 3 2 x y8. 5x 3 05x 3x 3 5 33. g(n 2) n 2 1O3x 2y 5 2 3 5 8,186,453 6,478,2162000 – 19901,708,237 10or about 170,823.73 n 19. m 8 7y 5 3 5 (x 2)y 5 3 5 x 6 5 3 5 x y 1 95 03x 5y 19 010a. (1990, 6,478,216), (2000, 8,186,453)m yOyx2x 4y 8x10b. y – 6,478,216 170,823.7(x – 1990)y – 6,478,216 170,823.7x – 339,939,163y 170,823.7x – 333,460,9471-5Writing Equations of Parallel andPerpendicular LinesPages 35–36 Check for Understanding1. If A, B, and C are the same or the ratios of the Asand the Bs and the Cs are proportional, then thelines are coinciding. If A and B are the same andC is different, or the ratios of the As and the Bsare proportional, but the ratio of the Cs is not,then the lines are parallel.2. They have no slope.3. 4x 3y 19 0y 4 3 x 1 93 4 3 ; 3 4 4. All vertical lines have undefined slope and onlyhorizontal lines are perpendicular to them. Theslope of a horizontal line is 0.5. none of these 6. perpendicular7. y x 6 x y 8 0y x 8parallel8. y 2x 8 4x 2y 16 0y 2x 8coinciding9. y 9 5(x 5)y 9 5x 255x y 16 010. 6x 5y 24y 6 5 x 2 45y (5) 5 6 (x (10))y 5 5 6 x 2 536y 30 5x 505x 6y 80 011. m of EF: m 8 44 3m of EH: m 2 46 3 4 1 or 4 2 3 68 4 2 3 m of GH: m 2 6 7m of FG: m 6 74 or 4parallelogram 1Pages 36–37 Exercises12. y 5x 18 2x 10y 10 0y 1 5 x 1slopes are opposite reciprocals; perpendicular13 Chapter 1
13. y 7x 5 0 y 7x 9 0y 7x 5 y 7x 9same slopes, different y-intercepts; parallel14. different slopes, not reciprocals; none of these15. horizontal line, vertical line; perpendicular16. y 4x 3 4.8x 1.2y 3.6y 4x 3same slopes, same y-intercepts; coinciding17. 4x 6y 11 3x 2y 9y 2 3 x 1 61y 3 2 x 9 2 Slopes are opposite reciprocals; perpendicular.18. y 3x 2 3x y 2y 3x 2different slopes, not reciprocals; none of these19. 5x 9y 14 y 5 9 x 1 49y 5 9 x 1 49same slopes, same y-intercepts; coinciding20. y 4x 2 0 y 4x 1 0y 4x 2 y 4x 1same slopes, different y-intercepts; parallel21. None of these; the slopes are not the same noropposite reciprocals.22. y (8) 2(x 0)y 8 2x2x y 8 0423. m ( 9)or 4 9 y (15) 4 9 (x 12)y 15 4 9 x 1 639y 135 4x 484x 9y 183 024. y (11) 0(x 4)y 11 025. y (3) 1 5 (x 0)y 3 1 5 x5y 15 xx 5y 15 0626. m ( 1)or 6; perpendicular slope is 1 6 y (2) 1 6 (x 7)y 2 1 6 x 7 6 6y 12 x 7x 6y 5 027. x 12 is a vertical line; perpendicular slope is 0.y (13) 0(x 6)y 13 028a. 5y 4x 1044x 5y 10 0 m ( 5)or 4 5 y 8 4 5 (x (15))y 8 4 5 x 125y 40 4x 604x 5y 100 028b. perpendicular slope: 5 4 y 8 5 4 (x (15))y 8 5 4 x 7 544y 32 5x 755x 4y 43 029a. 8x 14y 3 0 kx 7y 10 08m ( 14)or 4 7 km ( 7)or k 7 4 7 k 7 → k 429b. k 7 7 4 4k 49k 4 9430a. Sample answer: y 1 0, x 1 030b. Sample answer: x 7 0, x 9 031. altitude from A to BC:m of BC 5 (5)10 4 0 6 or 0m of altitude is undefined; x 7altitude from B to AC:m of AC 5 104 715 or 5 3m of altitude 1 5 y (5) 1 5 (x 10)y 5 1 5 x 25y 25 x 10x 5y 15 0altitude from C to AB:m of AB 5 1010 715 or 5 3m of altitude 1 5 y (5) 1 5 (x 4)y 5 1 5 x 4 5 5y 25 x 4x 5y 29 032. We are given y m 1 x b 1 and y m 2 x b 2 withm 1 m 2 and b 1 b 2 . Assume that the linesintersect at point (x 1 , y 1 ). Then y 1 m 1 x 1 b 1and y 1 m 2 x 1 b 2 . Substitute m 1 x 1 b 1 for y 1 iny 1 m 2 x 1 b 2 . Then m 1 x 1 b 1 m 2 x 1 b 2 .Since m 1 m 2 , substitute m 1 for m 2 . The result ism 1 x 1 b 1 m 1 x 1 b 2 . Subtract m 1 x 1 from eachside to find b 1 b 2 . However, this contradicts thegiven information that b 1 b 2 . Thus, theChapter 1 14
assumption is incorrect and the lines do not shareany points.33a. Let x regular espressos.Let y large espressos.216x 162y 783 248x 186y 914y 4 3 x 2 96y 4 3 x 4 5793No; the lines that represent the situation do notcoincide.33b. Let x regular espressos.Let y large espressos.216x 162y 783 344x 258y 1247y 4 3 x 2 96y 4 3 x 2 96Yes; the lines that represent the situationcoincide.34a. (15, 1939.20), (16, 1943.09)1943.09 – 1939.20m 16 – 1589 or 3.89 3. 1y 1943.09 3.89(x 16)y 1943.09 3.89x 62.24y 3.89x 1880.85(16, 1943.09), (17, 1976.76)1976.76 – 1943.09m 17 – 16 33 .671or 33.67y 1976.76 33.67(x 17)y 1976.76 33.67x 572.39y 33.67x 1404.37(17, 1976.76), (18, 1962.44)m 1962.44 1976.7618 1714.32 1or –14.32y 1962.44 –14.32(x 18)y 1962.44 –14.32x 257.76y –14.32 2220.2(18, 1962.44), (19, 1940.47)m 1940.47 1962.4419 18 –21 .971or –21.97y 1940.47 21.97(x 19)y 1940.47 21.97x 417.43y 21.97x 2357.934b. parallel lines or the same line; no34c. y 3.89(22) 1880.85 1966.43y 33.67(22) 1404.37 2145.11y 14.32(22) 2220.2 1905.16y 21.97(22) 2357.9 1874.56No; the equations take only one pair of days intoaccount.35. y 5 2(x 1)y 5 2x 2y 2x 736a. (40, 295), (80, 565)m 56 5 29580 40 2 7040or 6.75y 565 6.75(x 80)y 565 6.75x 540y 6.75x 2536b. $6.75 36c. $2537. 3x 2y 6 0yy 3 2 x 33x 2y 6 038. [g h](x) g(h(x)) g(x 2 ) x 2 139. Sample answer: {(2, 4), (2, 4), (1, 2), (1, 2),(0, 0)}; because the x-values 1 and 2 are pairedwith more than one y-value.40. 2x y 12 x 2y 6y 2x 12 x 2(2x 12) 6x 4x 24 63x 30x 102(10) y 12y 82x 2y 2(10) 2(8) 20 (16) or 41-6Modeling Real-World Data withLinear FunctionsPages 41–42 Check for Understanding1. the rate of change2. Choose two ordered pairs of data and find theequation of the line that contains their graphs. Finda median-fit line by separating the data into threesets and using the medians to locate a line. Use agraphing calculator to find a regression equation.3. Sample answer: age of a car and its value4a.Personal Consumption on Durable Goods3500300025002000Dollars150010005000O1995 1997 1999 2001 2003Yearx15 Chapter 1
4b. Sample answer: using (1995, 2294) and(2002, 3158)m 86 47or 123.4y 3158 123.4(x 2002)y 123.4x 243,888.84c. y 132.8x 262,621.2; r 0.984d. y 132.8(2010) 262,621.2 4306.8$4306.80; yes, the correlation coefficient shows astrong correlation.5a.Students per Computer2520Average 151050’89 ’91 ’93 ’95 ’97 ’99 ’01Year5b. Sample answer: using (1997, 6.1) and (2001, 4.9)4.9– 6.1m 20 01– 1997 –1 .24or 0.3y 6.1 0.3(x 1997)y 0.3x 605.25c. y 1.61x 3231.43; r 0.975d. 1 1.61x 3231.433230.43 1.61x2006.47 x2006; yes, the number of students per computer isdecreasing steadily.Pages 42–446a.Wins400200ExercisesAll-Time NFL Coaching Victories00 10 20 30 40Years6b. Sample answer: using (18, 170) and (40, 324)324 – 170 40 – 18 1 5422or 7m 7a.3158 – 22942002 – 1995y 324 7(x 40)y 7x 446c. y 7.57x 33.38; r 0.886d. y 7.57(16) 33.38 154.5155; yes, r is fairly close to 1. (Actual data is159.)Dollars(thousands)7b. Sample answer: using (1995, 23,255) and(2002, 30,832)30,832 – 23,2552002 – 1995 75 777m y 23,255 1082.43(x 1995)y 1082.43x 2,136,192.857c. y 1164.11x 2,299,128.75; r 0.997d. y 1164.11(2010) – 2,299,128.75 40,732.35$40,732.35; yes, r shows a strong relationship.8a.Car Weight and MileageAverageMileage or 1082.43806040203530252015000Personal Income1995 1997 1999 2001 2003Year15 20 25 30 35Weight (hundreds of pounds)8b. Sample answer: using (17.5, 65.4) and(35.0, 27.7)m 2 7.7 65.435.0 17.5 37.717.5or 2.15y 65.4 2.15(x 17.5)y 2.15x 103.08c. y 1.72x 87.59; r 0.778d. y 1.72(45.0) 87.59y 10.1910.19; no, r doesn’t show a particularly strongrelationship.9a.Acorn Size and RangeRange(hundredsof km 2 )30,00020,00010,00000 2 4 6 8 10 12Acorn Size (cm 3 )Chapter 1 16
9b. Sample answer: using (0.3, 233) and (3.4, 7900) 12b. Sample answer: a medication that proves to help37.6 – 26.2m y 0.6x 1029.42003 – 1990 1 1.4No; the lines do not coincide.13or 0.8814b. Let x computers.m 7 900 233delay the progress of a disease; because any3.4 0.3 7 positive correlation is better than none or a6673.1or 2473.23negative correlation.y 7900 2473.23(x 3.4)y 2473.23x 508.9712c. Sample answer: comparing a dosage of medicineto the growth factor of cancer cells; because thegreater the dosage the fewer cells that are9c. y 885.82 6973.14; r 0.38produced.9d. The correlation value does not show a strong or13. Men’s Median Salary Women’s Median Salarymoderate relationship.LinRegLinReg10a.Working Womeny ax by ax b40a 885.2867133 a 625.04195835b –1,742,768.136 b –1,234,368.06130r .9716662959 r .986950900925The rate of growth, which is the slope of thePercent in20graphs of the regression equations, for the womenManagementis less than that of the men’s rate of growth. If15that trend continues, the men’s median salary will10always be more than the women’s.514a. Let x computers.0Let y printers.’86 ’88 ’90 ’92 ’94 ’96 ’98 ’00 ’02 ’0424x 40y 38,736Yeary 0.6x 968.410b. Sample answer: using (1990, 26.2) and(2003, 37.6)30x 50y 51,470y 26.2 0.88(x 1990)y 0.88x 1725Let y printers.24x 40y 38,736y 0.6x 968.410c. y 0.84x 1648.27; r 0.98430x 50y 48,42010d. y 0.84(2008) 1648.27 or 38.45y 0.6x 968.438.45%; yes, r is very close to 1.Yes; the lines coincide.11a.World Population15. y (4) 6(x (3))7000y 4 6x 1860006x y 22 05000yMillions16a.4000of People30002520 y 0.82x 24200015100010050 1000 2000YearO 1 2 3 4 5x11b. Sample answer: using (1, 200) and (2000, 6050)6050 – 200 2000 – 1 5 8501999m or 2.93y 200 2.93(x 1)y 2.93x 197.0711c. y 1.65x 289.00; r 0.5611d. y 1.65(2010) 289.00 3027.53028 million; no, the correlation value is notshowing a very strong relationship.12a. Sample answer: the space shuttle; becauseanything less than perfect could endanger thelives of the astronauts.16b. $24 billion16c. If the nation had no disposable income, personalconsumption expenditures would be $24 billion.For each 1 billion increase in disposable income,there is a 0.82 billion dollar increase in personalconsumption expenditures.17. [f g](x) f(g(x)) f(x 1) (x 1) 3 x 3 3x 2 3x 1[g f ](x) g(f(x)) g(x 3 ) x 3 118. Yes; each domain value is paired with exactly onerange value.17 Chapter 1
19. The y-intercept is 1.The slope is 3 (move down 3 and right 1).The correct choice is C.Pages 48–49 Check for Understanding1. f(x) x if x 0x if x 02. reals, even integersx 2 if x 03. f(x) x if 0 x 4x 2 if x 44. Alex is correct because he is applying thedefinition of a function.5. y6.1-7OPiecewise Functionsyx8.x f(x)1 40 31 22 13 04 19. greatest integer function; h is hours, c(h) is the50h if [[h]] hcost, c(h) 50[[h 1]] if [[h]] hx f(x)0 x 1 501 x 2 1002 x 3 1503 x 4 200O 210. long term lot:2(6) 3(1) 12 3 or 15shuttle facility:4(3) 12shuttle facilityy400300200100yOx4 6 8 10xPages 49–5111. yExercisesOx7.xf(x)3 x 2 32 x 1 21 x 0 10 x 1 01 x 2 12 x 3 23 x 4 34 x 5 4y12.Ox f(x)1 43 25 07 29 4xyOxOx13.x f(x)2 x 1 01 x 0 10 x 1 21 x 2 32 x 3 4yOxChapter 1 18
14.xf(x)y18.xf(x)5 73 31.5 00 32 71 x 2 3 3 2 3 x 1 3 2 1 3 x 0 1Ox0 x 1 3 015.xf(x) 1 3 x 2 3 12 x 1 31 x 0 20 x 1 11 x 2 02 x 3 1y 2 3 x 1 2y42–1 O 1–2x–4Ox19. 20.yy16. ARTyOxOx21.xf(x)Ox5 x 4 2 5 17.x f(x)0 61 42 23 04 2y4 x 33 x 2 1 2 2 3 2 x 1 11 x 0 21 x 2 22 x 3 13 x 4 2 3 4 x 5 1 2 5 x 6 2 5 yOx21O1x219 Chapter 1
22.x f(x)5 63 01 60 91 63 05 623. Step; t is the time in hours, c(t) isthe cost in dollars,⎛ 6 if t 1⎢⎢ 2 10 if c(t) 1 ⎨ 2 t 1⎢⎢ 16 if 1 t 2⎝ 24 if 2 t 24d(t)2416yOx26c.Shipping(dollars)O 25 50 75 100 125 150Value of Order(dollars)27. If n is any integer, then all ordered pairs (x, y)where x and y are both in the interval [n, n 1)are solutions.28a. absolute value 28b. d(t) ⏐65 t⏐28c. d(t)806040d(t) |65 t|820O 2 4 6 8 10 12 14 16 18 20 22 24 t24. Greatest integer; w is the weight in ounces, c(w) isthe cost in dollars,0.37 0.23(w 1) if [[w]] wc(w) 0.37 0.23[[w]] if [[w]] w25. Absolute value; w is the weight in pounds, d(w) isthe discrepancy, d(w) ⏐1 w⏐26a. stepxf(x)0 x 1 0.371 0.371 x 2 0.602 0.602 x 3 0.833 0.83x f(x)0 11 02 13 226b. v is the value of the order, s(v) is the shipping,⎧⎪s(v) ⎨⎪⎩O3.50 if 0.00 v 25.005.95 if 25.01 v 75.007.95 if 75.01 v 125.009.95 if 125.01 vc(w)1.00.80.60.40.2Od(w)2 4ww28d. d(63) ⏐65 63⏐ or 2d(28) ⏐65 28⏐ or 37 2 372 19.5 heating degree days29a. step⎧5% if x $10,000⎪29b. t(x) ⎨7.5% if $10,000 x $30,000⎪⎩9.3% if x $30,00029c.O 20TaxRate(percent)y105040 60 80 t1020 30Income(thousands of dollars)29d. 9.5%30. No; the functions are the same if x is positive.However, if x is negative, the functions yielddifferent values. For example, [g f ](1.5) 1and [f g](1.5) 1; [g f ](1.5) 2 and[f g](1.5) 1.xChapter 1 20
31a.Percent who usepublic transportation60504030201031b. Sample answer: using (3,183,088, 53.4) and(362,777, 3.3)3.3 53.4m 362,777 3,183,08850.1 2 ,820,311 or 0.0000178Public Transport0 0 5 10 15 20 25 30 35Working Population(hundreds of thousands)y 53.4 0.0000178(x 3,183,088)y 0.0000178x 3.2631c. y 0.0000136x 4.55, r 0.6831d. y 0.0000136(307,679) 4.55y 8.738.73%; No, the actual value is 22%.32. y 2 2(x 4)y 2 2x 82x y 6 033a. (39, 29), (32, 15) 33b. m 1 5 232 399 147 or 233c. The average number of points scored eachminute.34. p(x) (r c)(x) (400x 0.2x 2 ) (0.1x 200) 399.9x 0.2x 2 20035. Let x the original price, or $59.99.Let T(x) 1.065x. (The cost with 6.5% tax rate)Let S(x) 0.75x. (The cost with 25% discount)[T S](x) (T(S(x)) (T(0.75x)) (T(0.75(59.99)) (T(44.9925) 1.065(44.9925) $47.9236. {7, 2, 0, 4, 9}; {2, 0, 2, 3, 11}; Yes; no elementof the domain is paired with more than oneelement of the range.37. 5 6 12 10,883,911,6805 6 12 2,176,782,341So, 5 6 12 is not greater than 5 6 12 .The correct choice is A.Graphing Linear InequalitiesPages 54–55 Check for Understanding1. y 2x 62. Graph the lines 3 2x y and 2x y 7. Thegraph of 3 2x y is solid and the graph of2x y 7 is dashed. Test points to determinewhich region(s) should be shaded. Then shade thecorrect region(s).3. Sample answer: The boundaries separate theplane into regions. Each point in a region eitherdoes or does not satisfy the inequality. Using atest point allows you to determine whether all ofthe points in a region satisfy the inequality.4. y5. ART6.7.1-8OOx y ! 4O3x y ^ 6yy7 ! x y ^ 9yxxxy ! |x 3|Ox21 Chapter 1
8a. c(m) 45 0.4m8b. c(m)7014.y605040c(m) ^ 45 0.4mO–4 ^ x y ^ 5x30201015. y16.y 6 |x|y2 ^ x 2y ^ 4O 20 40 60 80 100 m8c. Sample answer: (0, 45), (10, 49), (20, 50)OxOxPages 55–569.yExercises17.yy 6 |x| 4y < 3OxOx10.y18.yy ! |2x 3|Oxx – y > – 5Ox11.y19.y8 ^ 2x y ! 62x 4y 6 7OxOx12. y13.y20.y|x 3| ! y 1y ! 2x 1y 6 2 195x 5OxOxOxChapter 1 22
21.y26b.yx 6 0y 6 04.2524.254.248OxO7.998 8 8.002x22.y1086422344321O2468 10x27a. 0.6(220 a) r 0.9(220 a)27b. r2000.6 (220 a) ^ r ^ 0.9 (220 a)10023a. 8x 10y 8(60)8x 10y 48023b. y80604020O20 40 60 8023c. Sample answer: (0, 48) (60, 0), (45, 6)23d. Sample answer: using complex computerprograms and systems of inequalities.24. y|y| 6 xOx25a. points in the first and third quadrants25b. If x and y satisfy the inequality, then eitherx 0 and y 0 or x 0 and y 0. If x 0 andy 0, then ⏐x⏐ x and ⏐y⏐ y. Thus,⏐x⏐ ⏐y⏐ x y. Since x y is positive,⏐x y⏐ x y.If x 0 and y 0, then ⏐x⏐ x and ⏐y⏐ y.Then ⏐x⏐ ⏐y⏐ x (y) or (x y). Sinceboth x and y are negative, (x y) is negative,and ⏐x y⏐ (x y).126a. ⏐8 x⏐ 5 00 ; ⏐4 1 4 1 y⏐ 5xO27522016511055O11020 30 40 50 60 7028a. step28b. Let c(h) represent the cost for h hours.55h if [[h]] hThen c(h) 55[[h 1]] if [[h]] h28c. y2 3 4 5Hours29a. 3x y 10y 3x 10y (2) 3(x 0)3x y 2 029b. perpendicular slope: 1 3 xy (2) 1 3 (x 0)y 2 1 3 x3y 6 xx 3y 6 030. m 7 45 1y 7 3 4 (x 5) 3 4 y 3 4 x 3 1 4 31a. (0, 23), (16, 48);m 4 8 2316 0 2 516 or 1.562531b. the average change in the temperature per hour32. 95 9 4 65618 59,049 8 52, 4888 or 6561 or 9 4The correct choice is E.a00 23 Chapter 1
Chapter 1 Study Guide and AssessmentPage 57 Understanding the Vocabulary1. c 2. f3. d 4. g5. i 6. a7. h 8. j9. e 10. bPages 58–60 Skills and Concepts11. f(4) 5(4) 10 20 10 or 1012. g(2) 7 (2) 2 7 4 or 313. f(3) 4(3) 2 4(3) 9 36 12 9 or 5714. h(0.2) 6 2(0.2) 3 6 0.016 or 5.98415. g 1 3 20. (f g)(x) f(x) g(x) 4 x 2 3x or 4 3x x 2(f g)(x) f(x) g(x) 4 x 2 (3x) 4 3x x 2(f g)(x) f(x) g(x) (4 x 2 )(3x) 12x 3x 3f (x) f( x) g g( x) 4 x 23x, x 021. (f g)(x) f(x) g(x) x 2 7x 12 x 4 x 2 8x 16(f g)(x) f(x) g(x) x 2 7x 12 (x 4) x 2 6x 8(f g)(x) f(x) g(x) (x 2 7x 12)(x 4) x 3 11x 2 40x 48f (x) f( x) g2 g ( x)5 1 3 x2 7x 12x 42 (x 4)(x 3)or 6 5 5 x 43 g ( x) x 2 1x 1 (x 1)(x 1)x 1 12x 8x) x 1, x 1( x)23. (f g)(x) f(x) g(x) 6x 4 x 2 42 4x x 4 x3 8x 2 16x 4x 4 x 2 4 4x x 16. k(4c) (4c) 2 2(4c) 4 16c 2 8c 417. f(m 1) ⏐(m 1) 2 3(m 1)⏐ ⏐m 2 2m 1 3m 3⏐ ⏐m 2 5m 4⏐18. (f g)(x) f(x) g(x) 6x 4 2 6x 2(f g)(x) f(x) g(x) 6x 4 (2) 6x 6(f g)(x) f(x) g(x) (6x 4)(2) gf (x) gf( 3x 219. (f g)(x) f(x) g(x) x 2 4x x 2 x 2 5x 2(f g)(x) f(x) g(x) x 2 4x (x 2) x 2 3x 2(f g)(x) f(x) g(x) (x 2 4x)(x 2) x 3 2x 2 8x g( x)f (x) f g()x2 4x, x 2 x x 2 x 3, x 422. (f g)(x) f(x) g(x) x 2 1 x 1 x 2 x(f g)(x) f(x) g(x) x 2 1 (x 1) x 2 x 2(f g)(x) f(x) g(x) (x 2 1)(x 1) x 3 x 2 x 1f (x) f( x) g(f g)(x) f(x) g(x) x 2 4 4x x ,4x 4 4 x3 8x 2 16x 4x 4 x 2 4 4x x ,4x 4(f g)(x) f(x) g(x) (x 2 4 4x) x 4 4x, x 4f (x) f( x) g g( x)x 2 4x x 3 8x 2 16xor , x 444x 4Chapter 1 24
24. [f g](x) f(g(x)) f(2x) (2x) 2 4 4x 2 4[g f ](x) g(f(x)) g(x 2 4) 2(x 2 4) 2x 2 825. [f g](x) f(g(x)) f(3x 2 ) 0.5(3x 2 ) 5 1.5x 2 5[g f ](x) g(f(x)) g(0.5x 5) 3(0.5x 5) 2 0.75x 2 15x 7526. [f g](x) f(g(x)) f(3x) 2(3x) 2 6 18x 2 6[g f ](x) g(f(x)) g(2x 2 6) 3(2x 2 6) 6x 2 1827. [f g](x) f(g(x)) f(x 2 x 1) 6 (x 2 x 1) x 2 x 7[g f ](x) g(f(x)) g(6 x) (6 x) 2 (6 x) 1 x 2 11x 3128. [f g](x) f(g(x)) f(x 1) (x 1) 2 5 x 2 2x 4[g f ](x) g( f(x)) g(x 2 5) x 2 5 1 x 2 429. [f g](x) f(g(x)) f(2x 2 10) 3 (2x 2 10)2x 2 7[g f ](x) g(f(x)) g(3 x) 2(3 x) 2 10 2x 2 12x 2830. Domain of f(x): x 16Domain of g(x): all realsg(x) 165 x 16x 11Domain of [f g](x) is x 11.31. The y-intercept is 6. The slope is 3.y6 O 6 12x32. The y-intercept is 8. The slope is 5.–418126Oy 3x 6y8 y = 8 – 5x433. y 15 0y 15The y-intercept is 15. The slope is 0.y10 O434. 0 2x y 7y 2x 7The y-intercept is 7. The slope is 2.35. The y-intercept is 0. The slope is 2.x20 y 15 010yOyO10xxy 2x 7y 2xx36. The y-intercept is 2. The slope is 8.yOy 8x 2x25 Chapter 1
37. 7x 2y 5y7x 2y 5O38. The y-intercept is 6. The slope is 1 4 .y4 O 2 414y x 6y 7 2 x 5 2 The y-intercept is 5 2 . The slope is 7 2 .39. y 2x 3 40. y x 141. y 2 1 2 (x (5))y 2 1 2 x 5 2 y 1 2 x 9 2 5 242. m 2 (4) 3 6 or 1 2 y 5 1 2 (x 2)y 5 1 2 x 1y 1 2 x 443. (1, 0), (0, 4)m 4 00 1 41or 4y (4) 4(x 0)y 4 4xy 4x 444. y 1 45. y 046. y 0 0.1(x 1)y 0.1x 0.147. y 1 1(x 1)y 1 x 1x y 048. y 6 1 3 (x (1))y 6 1 3 x 1 3 3y 18 x 1x 3y 19 049. m 2 1 or 2y 2 2(x (3))y 2 2x 62x y 4 0xx50. y (8) 1 2 (x 4)y 8 1 2 x 22y 16 x 4x 2y 20 0451. m ( 2)or 2, perpendicular slope is 1 2 y 4 1 2 (x 1)y 4 1 2 x 1 2 2y 8 x 1x 2y 9 052. x 8 is a vertical line; perpendicular slope is 0.y (6) 0(x 4)y 6 053a.Overseas VisitorsVisitors(millions)2821147053b. Sample answer: using (1994, 18,458) and(2000, 25,975)25,975 – 18,458m 2000 – 1994 75 176 or 1252.81994 1996 1998 2000Yeary 25,975 1252.8(x 2000)y 1252.8x 2,479,62553c. y 1115.9x 2,205,568; r 0.944127574453d. y 1115.9(2008) 2,205,568 35,159.235,159,200 visitors; Sample answer: This is agood prediction, because the r value indicates astrong relationship.54. f(x)OxChapter 1 26
55.h(x)60. ARTyx ^ 5OxOx61.y56.xf(x)2 x 1 11 x 0 00 x 1 11 x 2 22 x 3 362.x y ^ 1Oyxf(x)2y x ^ 4f(x) x 1OxOx63. y64.yy 3x 1 257.58.x f(x)2 81 40 01 42 8x f(x)2 61 40 21 42 6g(x)Og(x) |4x|k(x)Oxx65.66.O xy ^ |x |yy ! |x| 2O xyOy ! |x 2|xOx59.yy 1 4Ox27 Chapter 1
Page 61 Applications and Problem Solving2. Since this is a multiple-choice question, you can67a. d 1 2 (20)(1) 2try each choice. Choice A, 16, is not divisible by 12,so eliminate it. Choice B, 24, is divisible by both 8 10and 12. Choice C, 48, is also divisible by both 8d 1 2 (20)(2) 2and 12. Choice D, 96, is also divisible by both 8 40and 12. It cannot be determined from thed 1 2 (20)(3) 2information given. The correct choice is E.3. Write the mixed numbers as fractions. 90d 1 4 12 (20)(4) 23 1 332 3 5 1 35Remember that dividing by a fraction is 160equivalent to multiplying by its reciprocald 1 2 (20)(5) 24 1 1 33 2503 1 3 53 1 3 5 3 1 32 3 5 510 m, 40 m, 90 m, 160 m, 250 mThe correct choice is B.67b. Yes; each element of the domain is paired withexactly one element of the range.4. Since this is a multiple-choice question, try eachchoice to see if it answers the question. Start with68a. (1999, 500) and (2004, 636)10, because it is easy to calculate with tens. If 10636 – 500m 2004 – 1999 adult tickets are sold, then 20 student tickets 13 6must be sold. Check to see if the total sales5or 27.2; about $27.2 billionexceeds $90.68b. y 500 27.2(x 1999)Students sales Adult sales $90y 27.2x 53,872.820($2.00) 10($5.00) 40 50 $9069. y 0.284x 12.964; The correlation isSo 10 is too low a number for adult tickets. Thismoderately negative, so the regression line iseliminates answer choices A, B, C, and D. Checksomewhat representative of the data.choice E. Eleven is the minimum number of adulttickets.19($2.00) 11($5.00) 38 55 $93Page 61 Open-Ended Assessment1. Possible answer: f(x) 4x 4, g(x) x 2 The correct choice is E.;[f g](x) f(g(x)) 4(x 2 ) 4 4x 2 45. Recall the definition of absolute value: the numberof units a number is from zero on the number line.2a. No; Possible explanation: If the lines have theSimplify the expression by writing it without thesame x-intercept, then they either intersect onabsolute value symbols.the x-axis or they are the same line. In eithercase, they cannot be parallel.⏐7⏐ 72b. Yes; Possible explanation: If the lines have the⏐7⏐ 7same x-intercept, they can intersect on the⏐7⏐⏐5⏐3⏐4⏐ 7 5 12 24x-axis. If they have slopes that are negativeThe correct choice is A.reciprocals, then they are perpendicular.6. Write each part of the expression without⎧4 if x 4exponents.3a. y ⎨⎩ 2x 4 if x 4(4) 2 16⎧x 1 if x 13b. y (2) 4 1 2 4 ⎨116⎩ 3x 1 if x 1116 1 6 3 4 16 1 316 16 1 316Chapter 1SAT & ACT PreparationPage 65 SAT and ACT Practice1. Prime factorization of a number is a product ofprime numbers that equals the number. ChoicesA, B, and E contain numbers that are not prime.Choice C does not equal 54. Choice D, 3 3 3 2, is the prime factorization of 54. The correctchoice is D.The correct choice is A.7. Use your calculator. First find the total amountper year by adding.$12.90 $16.00 $18.00 $21.90 $68.80Then find one half of this, which is the amountpaid in equal payments.$68.80 2 $34.40Then divide this amount by 4 to get each of 4monthly payments.$34.40 4 $8.60.The correct choice is A.Chapter 1 28
8. First combine the numbers inside the square rootsymbol. Then find the square root of the result.64 36 100 10The correct choice is A.9. 60 2 2 3 5 2 2 3 5The number of distinct prime factors of 60 is 3.The correct choice is C.10. First find the number of fish that are not tetras. 1 8 (24) or 3 are tetras. 24 3 or 21 are nottetras. Then 2 3 of these are guppies. 2 3 (21) 14The answer is 14.29 Chapter 1
Chapter 2 Systems of Linear Equations and Inequalities2-1Solving Systems of Equations inTwo VariablesPage 69 Graphing Calculator Exploration1. (1, 480)2. 3x 4y 320y 3 4→ x 805x 2y 340 y 5 2 x 170(76.923077, 22.30769)3. accurate to a maximum of 8 digits4. 5x 7y 70y 5 7→ x 1010x 14y 120 y 5 7 x 6 07Inconsistent; error message occurs.5. See students’ systems and graphs; any point inTRACE mode will be the intersection point sincethe two lines intersect everywhere.Page 70 Check for Understanding1. Sample answer:4x 7x 21y 2x 1The substitution method is usually easier to usewhenever one or both of the equations are alreadysolved for one variable in terms of the other.2. Sample answer: Madison might consider whetherthe large down-payment would strap herfinancially; if she wants to buy the car at the endof the lease, then she might also consider whichlease would offer the best buyout.3. Sample answer: consistent systems of equationshave at least one solution. A consistent,independent system has exactly one solution; aconsistent, dependent system has an infinitenumber of solutions. An inconsistent system hasno solution. See students’ work for examples andsolutions.4. 2y 3x 6y 3 2→ x 34y 16 6x y 3 2 x 4Inconsistent; Sample answer: The graphs of theequations are lines with slope 3 2 , but eachequation has a different y-intercept. Therefore,the graphs of the two equations do not intersectand the system has no solution.5. y(1, 3)y 5x 2(1, 3)y 2x 56. yy x 2 no solutionOy x 57. 7x y 9 7x y 95x y 15 7(2) y 912x 24 y 5x 2 (2, 5)8. 3x 4y 13x 4y 1→2(6x 2y) 2(3) 12x 4y 615x 5x 1 3 3x 4y 13 1 3 4y 14y 2y 1 2 1 3 , 1 2 9. 30 1 3 x 3 2 y 30(4) 10x 45y 120→2(5x 4y) 2(14) 10x 8y 2837y 148y 45x 4y 145x 4(4) 145x 30x 6 (6, 4)10. Let b represent the number of baseball racks andk represent the number of karate-belt racks.b 6k b 6k3b 5k 46,000 6(2000)3(6k) 5k 46,000 12,00023k 46,000k 200012,000 baseball, 2000 karatePages 71–72 Exercises11. x 3y 18y 1 3→ x 6x 2y 7y 1 2 x 7 2 consistent and independent12. y 0.5xy 0.5x→2y x 4 y 1 2 x 2inconsistent13. 35x 40y 55→ y 7 8 x 1 1 87x 8y 11 y 7 8 x 1 81consistent and dependentOxChapter 2 30
14. y15.y 45x 4O(5, 0)x 5(5, 0) (4, 3)16. y17. yy x 2xO y 3x 10x(2, 4)yOx 4y 3 (4, 3)Oy 13x 56y 13x(2, 4) no solution18. y19. yy x 2y 14x 3(0, 3)xy x 2(0, 2)O12Oy x 3(0, 2) (0, 3)20. 3x 8y 10→ y 3 8 x 5 2 16x 32y 75 y 1 2 x 7 532 Consistent and independent; if each equation iswritten in slope-intercept form, they havedifferent slopes, which means they will intersectat some point.21. 3(5x y) 3(16) 15x 3y 482x 3y 3→2x 3y 317x 51x 35x y 165(3) y 16y 1y 1 (3, 1)22. 3x 5y 83x 5y 83(x 2y) 3(1) →3x 6y 311y 11x 2y 1 y 1x 2(1) 1x 1 (1, 1)23. x 4.5 y y 6 xx 4.5 6 x y 6 5.252x 10.5 y 0.75x 5.25 (5.25, 0.75)24. 5(2x 3y) 5(3) 10x 15y 1512x 15y 4→12x 15y 422x 112x 3y 3x 1 2 2 1 2 3y 33y 2y 2 3 1 2 , 2 3 32xxx25. 2(3x 10y 5) 6x 20y 103(2x 7y 24)→6x 21y 7241y 822x 7y 24 y 22x 7(2) 242x 10x 5 (5, 2)26. 2x y 7 x 2y 82(2y 8) y 7 x 2(3) 83y 9 x 2y 3 (2, 3)27. 3(2x 5y) 3(4)6x 15y 122(3x 6y) 2(5)→6x 12y 103y 22x 5y 4y 2 3 2x 5 2 3 42x 2 3 x 1 3 1 3 , 2 3 28. 5 3 5 x 1 6 y 5(1)→ 3x 5 6 y 5 1 5 x 5 6 y 1 1 5 x 5 6 y 11 1 5 x 5 6 y 11 1 5 (5) 5 6 y 11 1 65x 16x 5 5 6 y 10y 12 (5, 12)29. 7(4x 5y) 7(8) 28x 35y 565(3x 7y) 5(10)→15x 35y 5043x 664x 5y 8 x 4 364 4 3 5y 85y 3 2043y 6 4330. 2(3x y) 2(9)4x 2y 84 6 4,3 6 4436x 2y 18→4x 2y 82x 103x y 9x 53(5) y 9y 6y 6 (5, 6)31. Sample answer: Elimination could be consideredeasiest since the first equation multiplied by 2added to the second equation eliminates b;substitution could also be considered easiest sincethe first equation can be written as a b, makingsubstitution very easy.a b 0a ba b 0a (3) 0a 3 (3, 3)32a. B3a 2b 153(b) 2b 155b 15b 331 Chapter 2
32b. S 4V 0 S V 30,000S 4V 4V V 30,0005V 30,000V 6000S 4V 0S 4(6000) 0S 24,000 0S 24,000Spartans: 24,000; visitors: 600033a. Let b represent the base and represent the leg.Perimeter of first triangle: b 2 20Perimeter of second triangle: 6 b 20b 2 20b 20 26 b 20 b 2 206 20 2 20 b 2(6) 20 6 b 8 66, 6, 8; 6, 6, 833b. isosceles34. y (3) 4(x 4)y 4x 19y (3) 1 4 (x 4)y 1 4 x 235a. Let x represent the number of refills. Then,x 1 number of drinks purchased.C 2.95 0.50xC 0.85 0.85xC 2.95 0.50x0.85 0.85x 2.95 0.50x0.35x 2.1x 6x 1 7 C 2.95 0.50xC 2.95 0.50(7) or 5.95(7, 5.95)35b. If you drink 7 servings of soft drink, the price foreach option is the same. If you drink fewer than7 servings of soft drink during that week, thedisposable cup price is better. If you drink morethan 7 servings of soft drink, the refillable mugprice is better. See students’ choices.35c. Over a year’s time, the refillable mug would bemore economical.36a. a b d e 36b. a b d e , c f36c. a b d e , c f37. Let x represent the full incentive.Let y represent the value of the computer.x 516 y38. Let x represent the number of people in linebehind you. 200 x represents the number infront of you. Let represent the whole line.200 x 1 x 3x200 x 1 x 3x 3x201 x 3(201)603 people 60339. y40.41. y 6 2(x 0)y 2x 642. $12,50043. [f g](x) f(g(x)) f(x 2) 3(x 2) 5 3x 144. {18}, {3, 3}; no, because there are two rangevalues paired with a single domain value.25 5 5 45. 5 1 1The correct choice is A.2-22x 7 yOx f(x)2 11 10 31 12 1xf(x)Solving Systems of Equations inThree VariablesPage 76 Check for Understanding1. Solving a system of three equations involveseliminating one variable to form two systems oftwo equations. Then solving is the same.Of(x) 2|x| 3x 3 4.5 x 264 y 3 .5 x4 264 y 3 .5 (5164 y) 264 y451.5 0.875y 264 yy $1500Chapter 2 32
2. The solution would be an equation in twovariables. Sample example: the system 2x 4y 6z 12, x 2y 3z 6, and 3x 5y 6z 27 has a solution of all values of x and y thatsatisfy 5x y 39.2x 4y 6z 12 2(x 2y 3z) 2(6)3x 5y 6z 27 2x 4y 6z 125x y 39 ↓2x 4y 6z 122x 4y 6z 120 0all reals3. Sample answer: Use one equation to eliminate oneof the variables from the other two equations.Then eliminate one of the remaining variablesfrom the resulting equations. Solve for a variableand substitute to find the values of the othervariables.4. 4(4x 2y z) 4(7) 2(4x 2y z) 2(7)2x 2y 4z 4 x 3y 2z 8↓↓16x 8y 4z 28 8x 4y 2z 142x 2y 4z 4 x 3y 2z 818x 10y 24 9x 7y 618x 10y 2418x 10y 242(9x 7y) 2(6)→18x 14y 124y 12y 39x 7y 6 4x 2y z 79x 7(3) 6 4(3) 2(3) z 7x 3 z 1(3, 3, 1)5. 2(x y z) 2(7) x 2y 3z 12x 2y 3z 12 3x 2y 7z 30↓ 2x 4z 182x 2y 2z 14x 2y 3z 12x 5z 22(x 5z) 2(2)2x 10z 4→2x 4z 18 2x 14z 1814z 14z 1x 5z 2 x y z 7x 5(1) 2 7 y 1 7x 7 y 1 (7, 1, 1)6. 2(2x 2y 3z) 2(2x 3y 7z)2(6)2(1)4x 3y 2z 0 4x 3y 2z 0↓↓4x 4y 6z 12 4x 6y 14z 24x 3y 2z 0 4x 3y 2z 0y 4z 12 3y 12z 23(y 4z) 3(12)3y 12z 36→3y 12z 2 3y 12z 20 38no solution7. 75 1 2 a(1) 2 v 0 (1) s 0 0.5a v 0 s 075 1 2 a(2.5) 2 v 0 (2.5) s 0 3.125a 2.5v 0 s 03 1 2 a(4) 2 v 0 (4) s 0 8a 4v 0 s 02.5(75) 2.5(0.5a v 0 s 0 )75 3.125a 2.5v 0 s 0↓187.5 1.25a 2.5v 0 2.5s 075 3.125a 2.5v 0 s 0112.5 1.875a 1.5s 04(75) 4(0.5a v 0 s 0 )3 8a 4v 0 s 0↓300 2a 4v 0 4s 03 8a 4v 0 s 0297 6a 3s 02(112.5) 2(1.875a 1.5s 0 )297 6a 3s 0↓225 3.75a 3s 0297 6a 3s 072 2.25a32 a297 6a 3s 0 3 8a 4v 0 s 0297 6(32) 3s 0 3 8(32) 4v 0 s 035 s 0 56 v 0acceleration 32 ft/s 2 , initial velocity: 56 ft/s,initial height: 35 ftPages 76–77 Exercises8. 3(5x 3y z) 3(11)x 2y 3z 5↓15x 9y 3z 33x 2y 3z 516x 11y 282(5x 3y z) 2(11)3x 2y 2z 13↓10x 6y 2z 223x 2y 2z 137x 4y 94(16x 11y) 4(28) →64x 44y 11211(7x 4y) 11(9) 77x 44y 9913x 13x 116x 11y 28 x 2y 3z 516(1) 11y 28 1 2(4) 3z 5y 4 z 4(1, 4, 4)33 Chapter 2
9. 7(x 3y 2z) 7(16) x 3y 2z 167x 5y z 0 x 6y 2z 18↓ 3y 2z 27x 21y 14z 1127x 5y z 026y 15z 11215(3y z) 15(2)45y 15z 3026y 15z 112→26y 15z 11271y 142y 23y z 2x 6y z 183(2) z 2x 6(2) 4 18z 4x 2(2, 2, 4)10. 2(2x y 2z) 2(11)x 2y 9z 13↓4x 2y 4z 22x 2y 9z 133x 5z 353(x 3z) 3(7)3x 9z 21→3x 5z 35 3x 5z 3514z 14z 1x 3z 7 2x y 2z 11x 3(1) 7 2(10) y 2(1) 11x 10y 7(10, 7, 1)11. 3(x 3y 2z) 3(8) 5(x 3y 2z) 5(8)3x 5y z 9 5x 6y 3z 15↓↓3x 9y 6z 24 5x 15y 10z 403x 5y z 9 5x 6y 3z 154y 7z 33 9y 13z 559(4y 7z) 9(33)36y 63z 297→4(9y 13z) 4(55)36y 52z 22011z 77z 74y 7z 33x 3y 2z 84y 7(7) 33x 3(4) 2(7) 8y 4x 6(6, 4, 7)12. 8x y z 4y z 58x y 91(8x y) 1(9)8x y 911x y 15→11x y 153x 6x 28x z 4 y z 58(2) z 4 y 12 5z 12y 7(2, 7, 12)13. 3(x y z) 3(3) 2(x y z) 2(3)4x 3y 2z 12 2x 2y 2z 5↓↓3x 3y 3z 9 2x 2y 2z 64x 3y 2z 12 2x 2y 2z 57x z 21 0 11no solution14. 3(36x 15y 50z) 3(10)5(54x 5y 30z) 5(160)↓108x 45y 150z 30270x 25y 150z 800162x 20y 77081(2x 25y) 81(40) 162x 2025y 3240162x 20y 770→162x 20y 7702005y 4010y 22x 25y 4036x 15y 50z 102x 25(2) 40 36(5) 15(2) 50z 10x 5 z 4(5, 2, 4)15. 4(x 3y z) 4(54)4x 2y 3z 32↓4x 12y 4z 2164x 2y 3z 3210y 3z 1845(2y 8z) 5(78)10y 40z 39010y z 184→10y z 18441z 574z 142y 8z 78 x 3y z 542y 8(14) 78 x 3(17) 14 54y 17 x 11(11, 17, 14)16. 1.8x 1.2y z 0.71.2y z 0.71.8x 1.2y 03(1.8x 1.2y) 3(0) 5.4x 3.6y 01.2(1.5x 3y) 1.2(3)→1.8x 3.6y 3.67.2x 3.6x 0.51.5x 3y 3 1.8x z 0.71.5(0.5) 3y 3 1.8(0.5) z 0.7y 0.75 z 0.2(0.5, 0.75, 0.2)17. y x 2z z 1 2xy (y 14) 2z7 1 2x7 z 4 xx y 144 y 1410 y (4, 10, 7)Chapter 2 34
18. 5 2 3 4 x 1 6 y 1 3 z 5 2 (12) 1 8 x 2 3 y 5 6 z 8↓5x 1y 5 6 z 30 1 582 1 8 x 2 3 y 5 6 z 2x 1 4 y838 7 4 3 4 x 1 6 y 1 3 z 7 4 (12) 2 13 1 63 1x6 5 8 7y 1z225↓1 7 7x6 2y4 1z2 21x 5 8 7y 1z 25 9 8 11 x y 1 2 4 1962(38) 9 6 →8 9x 6y4 17 81 y 4 9 8 11 x y 1 4 2x 1 4 y 19 9 8 x 1122x 1 4 y 382 2 0319213y 20 8y 24 3 4 x 1 6 y 1 3 z 122x 1 4 (24) 38 3 4 (16) 1 6 (24) 1 3 z 12x 16 z 12(16, 24, 12)19. Let x represent amount in International Fund, yrepresent amount in Fixed Assets Fund, and zrepresent amount in company stock.x y z 2000x 2z0.045x 0.026y 0.002z 58x y z 20002z y z 2000y 3z 2000y 2000 3z0.045x 0.026y 0.002z 580.045(2z) 0.026y 0.002z 580.026y 0.088z 580.026y 0.088z 580.026(2000 3z) 0.088z 58z 60020c. Sample answer: x y z 6;2x y 2z 8; x 2y 3z 221. 124 1 2 a(1) 2 v 0 (1) s 0272 1 2 a(3) 2 v 0 (3) s 082 1 2 a(8) 2 v 0 (8) s 0↓124 1 2 a v 0 s 0272 9 2 a 3v 0 s 082 32a 8v 0 s 01(124) 1 1 2 a v 0 s 0 272 9 2 a 3v 0 s 0↓124 1 2 a v 0 s 0272 9 2 a 3v 0 s 0148 4a 2v 01(124) 1 1 2 a v 0 s 0 82 32a 8v 0 s 0↓124 1 2 a v 0 s 082 32a 8v 0 s 042 31.5a 7v 07(148) 7(4a 2v 0 )2(42) 2(31.5a 7v 0 )↓1036 28a 14v 084 63a 14v 01120 35a32 a148 4a 2v 0 124 1 2 a v 0 s 0148 4(32) 2v 0 124 1 2 (32) 138 s 0138 v 0 2 s 0(32, 138, 2)22a. Sample answer: A system has no solution whenyou reach a contradiction, such as 1 0, as yousolve.22b. Sample answer: A system has an infinite numberof solutions when you reduce the system to twoequivalent equations such as x y 1 and2x 2y 2.x 2z x y z 2000x 2(600) 1200 y 600 2000x 1200 y 200International Fund $1200; Fixed Assets Fund $200; company stock $60020a. Sample answer: x y z 15;2x z 1; 2y z 720b. Sample answer: 4x y z 12;4x y z 10; 5y z 935 Chapter 2
23. x yz 2y xz 2z xy 2x yz 2→x yz 2y xz 2 y xz 2(x y) (yz xz) 0(x y) z(x y) 0(1 z)(x y) 01 z 0 or x y 0z 1 x yy xz 2→y xz 2z xy 2 z xy 2( y z) (xz xy) 0( y z) x(y z) 0(1 x)(y z) 01 x 0 or y z 0x 1 y zIf z 1 and x 1, 1 1y 2 and y 1.If z 1 and y z, x 1 1 2 and x 1.If x y and x 1, 1 1z 2 and z 1.If x y and y z, x y z.x x x 2x x 2 2x 2 x 2 0(x 2)(x 1) 0x 2 0 or x 1 0x 2 2 x 1If x 2, y 2 and z 2.The answers are (1, 1, 1) and (2, 2, 2).24. 3x 4y 375 → 3x 4y 3752(5x 2y) 2(345)10x 4y 6907x 315x 455x 2y 3455(45) 2y 345y 60 (45, 60)25. yOy 13 x 2x26.y(1, 3)AOAB: d (1 3) 2 (1)) (2 2 5BC: d (2 1)) ( 2 (6 2) 2 5CD: d (6 ) 2 2 (3 6) 2 5AD: d (6 ) 3 2 (31)) ( 2 51 3AB: m 2 ( 1)BC: m 2 6 4 3 (1) 2 3 4 AB BC CD AD 5 units; ABCD isrhombus. Slope of AB 4 3 and slope of BC 3 4 ,so AB ⊥ BC. A rhombus with a right angle is asquare.27a. (20, 3000), (60, 5000)m 500 0 300060 20 20 0040 or 50y 5000 50(x 60)y 50x 2000C(x) 50x 200027b. $2000; $5027c. c(x)4Cost($1000)321OD(3, 6)C(6, 2)B(2, 1)c(x) 50x 20001 2 3 4 5 6 7 8 9 xTelevisions Produced28. A s 2 A r 22 s 2 (2) 22 s 2The correct choice is C.x2-3Modeling Real-World Data withMatricesPages 82–83 Check for Understanding1. Sample answer:film pain reliever blow dryer(24 exp.) (100 ct)Atlanta $4.03 $6.78 $18.98 ⎤Los Angeles $4.21 $7.41 $20.49Mexico City $3.97 $7.43 $32.25⎥Tokyo $7.08 $36.57 $63.71 ⎦2. 2 43. The sum of two matrices exists if the matriceshave the same dimensions.⎤⎥⎦Chapter 2 36
4. Anthony is correct. A third order matrix has 3rows and 3 columns. This matrix has 4 rows and 3columns.5. 2y x 3 2y y 5 3x y 5 y 2x y 5x 2 5x 7 (7, 2)6. 18 4x y 24 12y24 12y 2 y18 4x y18 4x 25 x (5, 2)7. 16 4x 16 4x0 y 4 x2x 8 y (4, 0)8. X Z 4 (1) 1 3 2 0 6 (2) 3 4 2 49. impossible1 4 3 10 (2) 2 65 2 2 810. Z X 11. 4X 4(4) 4(1)4(2) 4(6) 16 4 8 2412. impossible4 113. YX [0 3] 2 6 [0(4) (3)(2) 0(1) (3)(6)]or [6 18]14.Budget Viewers($ million) (million)soft-drink 40.1 78.6 ⎤package delivery 22.9 21.9 ⎥telecommunications 154.9 88.9 ⎦Pages 83–86 Exercises15. y 2x 1 y 2(y 5) 1x y 5 y 11x y 5x 11 5x 6 (6, 11)16. 9 x 2y 13 4x 113 4x 1 3 x9 x 2y9 3 2y3 y (3, 3)17. 4x 15 x 4x 15 x5 2y x 5⎤⎥⎦18. x 2y y 2(2y) 6y 2x 6 y 2x 2yx 2(2)x 4 (4, 2)19. 27 3y 27 3y8 5x 3y 9 y8 5x 3y8 5x 3(9)7 x (7, 9)20. 4x 3y 11x y 14x 3y 114x 3y 113(x y) 1→3x 3y 37x 14x 2x y 12 y 1y 1 (2, 1)21. 2x 10 2x 10 y 3xy 3x x 5 y 3(5)y 15y 15(5, 15)22. 12 6x 12 6x 2 y 12 y 1 2 x 1 y12y 10 x (2, 1)23. x y 0y y 23 2y x6 4 2x6 4 2x x y 01 x 1 y 0y 1 (1, 1)24. x 2 1 2x y 55 y xy 4 2y 4 2 x y 5y 6 x 6 5x 1 (1, 6)25. 3x y 1 15 64 3z 6z3x y3x 3y 3 15 6 129z6z3x y3x 1512 6z3y 3 69z 3x y3x 15 12 6z 3y 3 6x 5 2 z y 3(5, 3, 2)5 2y2.5 y (5, 2.5)37 Chapter 2
26. 2 w 5 x z 16 43y 8 6 2x 8z2w 10 2x 2z 16 4 6y 16 6 2x 8z2w 10 16 2w 10 16 6y 66y 6 w 3 y 12x 2z 416 2x 8z2x 2z 4 2x 2z 42x 8z 16 2x 2(2) 410z 20 x 0z 2 (3, 0, 1, 2)27. A B 5 3 7 56 (1) 1 88 12 28. impossible29. impossible790 4 1 (2) 2 330. D C 2 5 3 0 0 (1) 4 9 4 0 2 14 5 3 1 13134131. B A B (A)3 5 5 71 8 6 13 (5) 5 (7)1 6 8 (1)32. C D C (D)4 2 3 05 0 1 29 0 1 4533. 4(0)4D 4(2) 4(4)08 16 or 2 2 5 71 2 3 04 24 0 2 (1) 3 (2)5 2 0 (3) 1 09 (4) 0 (4) 1 24 3 1or 7 3 1 4121644(1)4(3)4(4)80834(2)4(0)4(2)34. 2F 2(6) 2(1) 2(0)2(1) 2(4) 2(0) 12 0 22 8 035. F E F (E)6 1 01 4 014 3 2 2 3 536. E F E (F)8 4 23 1 514 3 2 2 3 5 8 4 231 6 1 01 4 0 537. 5A 5(5) 5(7)5(6) 5(1) 25 3530538. BA 3 5 5 7 1 8 6 13(5) 5(6)1(5) 8(6)15 26or 5339. impossible4 2 340. FC 6 1 0 5 0 11 4 016(2) (1)(0) 0(0)1(2) 4(0) 0(0)6(3) (1)(1) 0(1)1(3) 4(1) 0(1) 29 12 1724 2 1 0 1 28 4 22 3 03 1 54 4 28(0) (4)(2) 2(4)3(0) 1(2) (5)(4)8(1) (4)(3) 2(4)3(1) 1(3) (5)(4)8(2) (4)(0) 2(2)3(2) 1(0) (5)(2) 16 4 122214 165 7 5 76 1 6 15(5) 7(6) 5(7) 7(1)6(5) 1(6) 6(7) 1(1) or17 4236410 1 26 1 02 3 01 4 04 4 2 6(0) (1)(2) 0(4)1(0) 4(2) 0(4)6(1) (1)(3) 0(4)1(1) 4(3) 0(4)6(2) (1)(0) 0(2)1(2) 4(0) 0(2) 2 9 12813 2E FD 8 2 4 (9) 2 (12)3 (8) 1 13 5 2 10 13 10514 342. AA 3(7) 5(1)1(7) 8(1) 9 0 6(4) (1)(5) 0(9)1(4) 4(5) 0(9)41. ED 43. FD 1Chapter 2 38
44. 3AB 3 5 7 3 56 1 1 8 (3)(5) 3(7) 3 5(3)(6) 3(1) 1815 21 3 5 18 3 1 815(3) (21)(1) 18(3) (3)(1)15(5) (21)(8)18(5) (3)(8) 24 243 57 6645. (BA)E 3 5 5 7 8 4 2 1 8 6 13(5) 5(6)1(5) 8(6)8 4 231 515 26 8 453 1 3 115(8) 26(3) 53(8) 1(3)15(2) 26(5)53(2) 1(5) 42 1603 13(7) 5(1)1(7) 8(1) 2 5515(4) 26(1)53(4) 1(1)86421213 11146. F 2EC F (2EC)4 2 38 4 25 0 13 1 59 0 12(8) 2(4) 2(2)2(3)2(1) 2(5) 4 2 35 0 1 9 0 14 2 316 8 45 0 16 2 109 0 116(4) 8(5) (4)(9)6(4) (2)(5) 10(9)16(2) 8(0) (4)(0)6(2) (2)(0) 10(0)16(3) 8(1) (4)(1)6(3) (2)(1) 10(1) 60 32 60 56 12 6F (2EC) 6 (60) 1 32 1 56 4 1266 31 60 57 16 62EC 2 0 (60)0 (6) 247. 3XY 3 3 3 68 4 42 653(2) 3(4)3(8) 3(4)3(2) 3(6)6 12324 1256 1846(3) 12(5) 24(3) (12)(5)6(3) 18(5)6(6) 12(2)24(6) (12)(2)6(6) 18(2)78122 3 3 6 3 6 1272301209016872544226(3) 12(4)24(3) (12)(4)6(3) 18(4)1 7 4 53 21 12(1) 2(7) (3)(4) 3(5)2(3) 2(2) (3)(1) 3(1)2 14 12 156 4 332 12 14 (15)6 (3) 4 3 14 29 3 749. Sample answer:1996 2000 200618 to 24 ⎡ 8485 8526 8695 ⎤25 to 34 ⎢ 10,102 9316 9078 ⎥35 to 44 ⎢⎥⎢8766 9036 8433⎥45 to 54 ⎢ 6045 6921 7900 ⎥55 to 64 ⎢ 2444 2741 3521 ⎥⎢65 and older⎥⎣ 2381 2440 2572⎦18 24 19 26 31 24 16 24 1750a. 22 28 216 6 6 12 9 712 2 4 17 4 624 19 26 31 2424 17 22 28 216 6 12 9 748. 2K 3J 2 (3) 1826612or2 418 (26)16 (22)6 (12)12 (17)86657432 1724 (31)24 (28)6 (9)2 (4)54124619 (24)17 (21)6 (7)4 (6)TV Radio RecordingClassical 8 7 5⎡⎤Jazz 6 4 4Opera 6 3 1Musicals ⎢ 5 2 2 ⎥⎣⎦50b. classical performances on TV39 Chapter 2
2 351a. a b 2 34 5 c d 4 52(a) 3(c) 2(b) 3(d) 2 3 4(a) (5)(c)2a 3c 2 2b 3d 34a 5c 44b 5d 52(2a 3c) 2(2) →4a 6c 44a 5c 4 4a 5c 4c 04a 5c 44a 5(0) 4a 12(2b 3d) 2(3)4b 5d 54(b) (5)(d)→4b 6d 64b 5d 5d 14b 5d 54b 5(1) 5b 051b. a matrix equal to the original matrix552a. [42 59 21 18]52b. ⎡33.81 30.94 27.25 ⎤⎢⎥⎢15.06 13.25 8.75 ⎥[42 59 21 18] ⎢⎥⎢54 54 46.44⎥⎢⎥⎢52.06 44.69 34.38 ⎥⎣⎦4233.81 5915.06 21(54) 1852.064230.94 5913.25 21(54) 1844.694227.25 598.75 2146.44 1834.38 [4379.64 4019.65 3254.83]July, $4379.64; Aug, $4019.65; Sep, $3254.8353. The numbers in the first row are the triangularnumbers. If you look at the diagonals in thematrix, the triangular numbers are the endnumbers. To find the diagonal that contains 2001,find the smallest triangular number that isgreater than or equal to 2001. The formula for thenth triangular number is n(n 1)2. Solve n(n 1)22001. The solution is 63. So the 63rd entry in thefirst row is 63(63 1)2 2016. Since 2016 2001 15, we must count 15 places backward along thediagonal to locate 2001 in the matrix. Thismovement takes us from the position (row,column) (1, 63) to (1 15, 63 15) (16, 48).54a.A B C DA ⎡ 0 1 0 0⎤B 1 0 1 1C 0 1 0 2D ⎢ 0 1 2 0 ⎥⎣⎦54b. No; since the matrix shows the number of nodesand the numbers of edges between each pair ofnodes, only equivalent graphs will have thesame matrix.455. 2x 6y 8z 52x 9y 12z 515y 20z 102(2x 9y 12z) 2(5) → 4x 18y 24z 104x 6y 4z 3 4x 6y 4z 324y 20z 1315y 20z 10 → 15y 20z 101(24y 20z) 1(13)24y 20z 139y 3y 1 3 15y 20z 10 2x 6y 8z 515 1 3 20z 10 2x 6 1 3 8 1 4 5z 1 4 x 1 2 1 2 , 1 3 , 1 4 56. 4x 2y 7→12x 6y 21consistent and dependent57. y58.O6 3x y 12x f(x)2 81 50 21 52 8y 2x 7 2 y 2x 7 2 59. Sample answer: using (60, 83) and (10, 65),m 6 5 8310 60 1850or 0.36y 65 0.36(x 10)y 0.36x 61.460. m 7 45 1y 7 3 4 (x 5) 3 4 y 3 4 x 3 1 4 61. f(x) 5x 30 5x 3 3 5 x62. [f g](x) f(x) g(x) 2 5 x(40x 10) 16x 2 4x63. f(x) 4 6x x 3f(14) 4 6(14) (14) 32656xf(x)Of(x) |3x| 2xChapter 2 40
64. 2x 3x 3 x22d. Let A , B ,2(2x 3) x(3 x)4x 6 3x x 2c 11 c 12x 2 x 6 0c 21 c 22(x 3)(x 2) 0(AB)C ax 3 0 or x 2 011 b 11 a 12 b 21 a 11 b 12 a 12 b 22 c11 c 12a 21 b 11 a 22 b 21 a 21 b 12 a 22 b 22 c 21 c 22x 3 x 2The correct choice is A. a 11 b 11 c 11 a 11 b 12 c 21 a 12 b 21 c 11 a 21 b 22 c 21a 21 b 11 c 11 a 21 b 12 c 21 a 22 b 21 c 11 a 22 b 22 c 21a 11 b 11 c 12 a 11 b 12 c 22 a 12 b 21 c 12 a 12 b 22 c 22a 21 b 11 c 12 a 21 b 12 c 22 a 22 b 21 c 12 a 22 b 22 c 22A(BC) a11 a 12 b11 c 11 b 12 c 21 b 11 c 12 b 12 c 22a 21 a 22 b 21 c 11 b 22 c 21 b 21 c 12 b 22 c 222a. Let A a a11 12 and B b b11 12a a b b .21 2221 22A B a a11 12 b b11 12and C bBA 11 a 11 b 12 a 21b 21 a 11 b 22 a 21b 11 a 12 b 12 a 22b 21 a 12 b 22 a 22 .Page 86 Graphing Calculator Exploration1. All of the properties except for the CommutativeProperty of Multiplication hold true. Whenmultiplying matrices, the order of themultiplication produces different results. a 11 b 11 c 11 a 11 b 12 c 21 a 12 b 21 c 11 a 21 b 22 c 21a 21 b 11 c 11 a 21 b 12 c 21 a 22 b 21 c 11 a 22 b 22 c 21However, in addition of matrices, order is nota 11 b 11 c 12 a 11 b 12 c 22 a 12 b 21 c 12 a 12 b 22 c 22important.a 21 b 11 c 12 a 21 b 12 c 22 a 22 b 21 c 12 a 22 b 22 c 22Therefore (AB)C A(BC).3. All properties except the Commutative Property ofa 21 a 22 b 21 b Multiplication will hold for square matrices. A22proof similar to the ones in Exercises 2a-2d can bea b a b11 11 12 12a 21 b 21 a 22 b 22used to verify this conjecture.b a b a 11 11 12 12b 21 a 21 b 22 a 22b b 11 21 a a11 12b 21 b 12 a 21 a 222-4A Transformation Matrices2b. Let A a a11 12 , B b b11 12a a b b ,Page 8721 2221 221. The new figure is a 90° counterclockwise rotationand C c c11 12of LMN.c 21 c 222. The new figure is an 180° counterclockwisea a(A B) C b b11 11 12 12 c c11 12rotation of LMN.a 21 b 21 a 22 b 22 c 21 c 22(a 11 b 11 ) c 11(a 21 b 21 ) c 21(a 12 b 12 ) c 12(a 22 b 22 ) c 22a11 (b 11 c 11 )a 21 (b 21 c 21 )a 12 (b 12 c 12 )a 22 (b 22 c 22 ) a a11 12 b11 c 11a 21 a 22 b 21 c 21b 12 c 12b 22 c 22 A (B C)3. The new figure is an 270° counterclockwise2c. Let A a a11 12 and B b b11 12a a b b .rotation of LMN.21 2321 22aAB 11 b 11 a 12 b 21a 21 b 11 a 22 b 21a 11 b 12 a 12 b 22a 21 b 12 a 22 b 23Thus, AB BA, since a 12 b 21 b 12 a 21 .a 11a 21a 12a 22b 11b 21b 12b 2241 Chapter 2
4. See students’ work for graphs. Multiplying avertex matrix by0 1 results in a vertex1 0matrix for a figure that is a 90° counterclockwiserotation of the original figure. Multiplying avertex matrix by 1 0results in a vertex0 1matrix for a figure that is a 180° counterclockwiserotation of the original figure. Multiplying avertex matrix by 0 1results in a vertex1 0matrix for a figure that is a 270° counterclockwiserotation of the original figure.2-4Modeling Motion with MatricesPages 92–93 Check for Understanding1. Translation, reflection, rotation, dilation;translations do not affect the shape, size, ororientation of figures; reflections and rotations donot change the shape or size of figures; dilationsdo not change the shape, but do change the size offigures.2. 90° counterclockwise (360 90)° or 270°clockwise; 180° counterclockwise (360 180)°or 180° clockwise; 270° counterclockwise (360 270)° or 90° clockwise.3. Sample answer: the first row of the reflectionmatrix affects the x-coordinates and the secondrow affects the y-coordinates. A reflection overthe x-axis changes (x, y) 0 (x, y), so the firstrow needs to be [1 0] so the x is unchangedand the second row needs to be [0 1] so they-coordinates are the opposite. Similar reasoningcan be used for a reflection over the y-axis, whichchanges (x, y)to(x, y)andareflection over the liney x,which interchanges the values for x and y.4a. 6 4b. 2 4c. 3 4d. 45. 1.52 1 0 3 1.5 0 5 3 2 7.5 4.5 3J(3, 7.5), K(1.5, 4.5), L(0, 3)JJyKK6. 1 3 3 1 1 1 1 1 3 2 2 2 21 1 3 3A(2, 1), B(2, 1), C(2, 3), D(2, 3)yABBAOC xDDC7. 1 0 1 4 3 00 1 2 1 2 11 4 3 0 2 1 2 1A(1, 2), B(4, 1), C(3, 2), D(0, 1)yA ABBOxD DCC8. 0 1 3 1 1 2 4 2 130121423P(2, 3), Q(4, 1), R(2, 1)yQPOQRxRP 9.1 0 1 0 1 00 1 0 1 011 0 6 3 1 6 3 1 0 1 4 2 2 4 2L(6, 4), M(3, 2), N(1, 2)L yM N2221122LLxLO xNM10a. 10b.x 3 Landingy 4Ball3E4NChapter 2 42
Pages 93–96Exercises11. 3 1 1 5 3 3 15 1 4 1 3 12 3A(3, 3), B(3, 12), C(15, 3)yB121086 B4 AAC2 C2 4 6 8 101214 x08593212. 3 4 0 46 2 7415 9 4X(0, 6), Y3 3 4 , 6 3 4 , Z2 1 4 , 1 1 2 Y13. 2 3 2 1 4 6 4 2 8 0 2 3 2 0 4 6 4P(6, 0), Q(4, 4), R(2, 6), S (8,4)PYZQZPQyOyO14a. 21 0 1 0 2 0 2 00 1 0 1 0 2 0 23 2 0 2 0 6 0 6 0 RR0 2 0 2 0 6 0 6A(2, 0), B(0, 2), C(2, 0), D(0, 2); A(6, 0),B(0, 6), C(6, 0), D(0, 6)yXXBxS 3 2 Sx14b. 31 0 1 0 3 0 3 0 010132 3 0 3 0 6 0 6 0 0 3 0 3 0 6 0 6A(3, 0), B(0, 3), O(3, 0), D(0, 3); A(6, 0),B(0, 6), C(6, 0), D(0, 6)14c. The final results are the same image.2 1 3 3 3 3 10 5 1 2 2 2 2W(1, 2), X(4, 3), Y(6, 3)y X15. 4 6 WOWY16. 0 1 4 3 2 2 2 20 5 7 2 11 1 1 2 1 2 1OAPyOAPRAQyDRBBBDDXQCxxY0C3C 1 4 6 1Q(2, 1), P(1, 4), Q(2, 6), R(1, 1)x0333 1 5 117. 3 3 3 3 0 4 8 4 1513C(0, 5), D(4, 9), E(8, 5), F(4, 1)44445951AA ADBBC CDCxyCDDEDCOFExF43 Chapter 2
18a. 4 0 2 6 6 6 2 6 4 131222F(2, 1), G(6, 1), H(4, 3)18a-b.yG12 6 418b. 1 1 1 1 5 3 1 1 3 5 5 5 4 6F (1, 4), G(5, 6), H(3, 2)18c. translation of 5 units left and 3 units up1 019. 1 0 2 1 0 20 1 2 4 3 2 4 3A(1, 2), B(0, 4), C(2, 3)20. 1 0 2 6 3 1 2 6 3 1 0 1 4 2 4 2 4 2 4D(2, 4), E(6, 2), F(3, 4), G(1, 2)EAGyBOA21.0 1 1 3 1 2 2 1 5 4 1 0 2 1 5 4 1 3H(2, 1), I(1, 3), J(5, 1), K(4, 2)IBFHHHCCyDGyJHxOIF FOGODGKFHKxJFxEx1132220 122. 1 2 3 1 2 1 1 0 1 2 1L(1, 1), M(2, 2), N(1, 3)23. 1 0 0 4 4 0 0 4 4 0 0 1 0 0 4 4 0 0 4O(0, 0), P(4, 0), Q(4, 4), R(0, 4)0 124. 1 3 5 4 2 1LO LPQy0 2 1 2 4 41 3 5 4 2S(2, 1), T(1, 3), U(2, 5), V(4, 4),W(4, 2)UVWVaN25a. Let R x-axisc da b 1 2 1 1 2 1c d 3 1 3 31 3a 3b 2a b a 3b 1 2 1 c 3dMbMy2NRO ORU12c dxPST O xSWTyQ24c 3d43a 3b 1c 3d 32a b 22c d 1a 3b 1c 3d 3Thus, a 1, b 0, c 0, and d 1. Bysubstitution, R x-axis 1 00 1 .x123143Chapter 2 44
acbda b 1c d 3a 3b 2a bc 3d 2c d25b. Let R y-axis . 2 1 1 2 11 3 3 1 3a 3b 1 2 1 c 3d31a 3b 1 2a b 2 a 3b 1c 3d 3 2c d 1 c 3d 3Thus, a 1, b 0, c 0, and d 1. Bysubstitution, R y-axis 1 00 1 .a bc da b 1 2 1 3 1 3c d 3 1 3 1 2 1a 3b 2a b a 3b 3 1 3c 3d 2c d c 3d 1 2 125c. Let R y x .3 a 3b 3 2a b 1 a 3b 3c 3d 1 2c d 2 c 3d 1Thus, a 0, b 1, c 1, and d 0. Bysubstitution, R y x 0 11 0 .a bc da b 1 2 1 3 1 3c d 3 1 3 1 2 1a 3b 2a b a 3b 3 1 3c 3d 2c d c 3d 1 2 125d. Let Rot 90 a 3b 3 2a b 1 a 3b 3c 3d 1 2c d 2 c 3d 1Thus, a 0, b 1, c 1, and d 0. Bysubstitution, Rot 90 0 1 1 0a bc da b 1 2 1 1 2 1c d 3 1 3 31 3a 3b 2a b a 3b 1 2 1c 3d 2c d c 3d 31 325e. Let Rot 180 .a 3b 1 2a b 2 a 3b 1c 3d 3 2c d 1 c 3d 3Thus, a 1, b 0, c 0, and d 1. Bysubstitution, Rot 180 1 00 1 .a bc d25f. Let Rot 270 .a b 1 2 1 3 1 3c d 3 1 3 12 1a 3b 2a b a 3b 3 1 3 c 3d2c dc 3d1a 3b 3c 3d 12a b 12c d 2a 3b 3c 3d 1Thus, a 0, b 1, c 1, and d 0. Bysubstitution, Rot 270 0 11 0 .2126.1 0 6 3 1 6 3 10 1 4 2 2 42 26 3 1 2 2 2 4 1 1 4 2 2 5 5J(4, 1), K(1, 3), L(1, 7)J27. 1 0 6 3 1 6 3 10 1 4 2 2 42 21 0 6 3 1 6 3 1 0 1 4 2 2J(6, 4), K(3, 2), L(1, 2)JJ54121 028. 6 3 1 6 3 10 1 4 2 2 42 20 1 6 3 1 4 2 2 1 0 4 2 2 6J(4, 6), K(2, 3), L(2, 1)JJKKKKKLLyLLyOLyKLOLOLLKKKJ29a. The bishop moves along a diagonal until itencounters the edge of the board or anotherpiece. The line along which it moves changesvertically and horizontally by 1 unit witheach square moved, so the translationmatrices are scalars. Sample matrices arec1 1 , c 1 1 , c 1 11 1 1 1 1 1 , andxJxJJx3132745 Chapter 2
c 1 1,where c is the number of squares1 1moved.29b. The knight moves in combinations of 2 vertical-1horizontal or 1 vertical-2 horizontal squares.These can be either up or down, left or right.Sample matrices are1 122 ,1 1 , 1 1 , 1 1 , 2 22 2 2 2 2 2 11 ,2 2 , 2 2 , and 2 2 .1 1 1 1 1 129c. The king can move 1 unit in any direction. Thematrices describing this are1 100 ,1 1 , 0 0 , 0 0 , 1 1 , 1 10 0 1 1 1 1 1 1 11 ,1 1 , and 1 1 11 .a b30. Consider .c dDilation with scale factor 1a b1 a bc d cdRotation of 180°1 0 a b a b0 1 c d cdThe vertex matrices for the images of a dilationwith scale factor 1 and a rotation of 180° are thesame, so the images are the same.31. (0, 125); (125, 0), (0, 125), (125, 0)32. Sample answer: There is no single matrix toachieve this. You could reflect over the x-axis andthen translate 2(4) or 8 units upward.33. See students’ work; the repeated dilations animatethe growth of something from small to large,similar to a lens zooming into the origin.34. 1 4 212 0 334a. Sample answer: the figure would be enlargeddisproportionally.3 y034b. 1 1 4 2 3 3 12 60 62B 1 2 0 3 24 0 64 B2C CA A2 4 6 8 1012 x4 D6D34c. See students’ work; the figure appears as ifblown out of proportion.35.3 8 1 5 3 1 8 52 4 2 8 2 (2) 4 8 4 13 4 1236. x y z 1.8 x 2y 4.6y z 5.6 x 2y 3.8x 2y 3.8 2x 0.8x 0.4x 2y 3.8y z 5.637. Let x represent hardback books and y representpaperback books.4x 7y 5.753x 5y 4.253(4x 7y) 3(5.75)→4(3x 5y) 4(4.25)4x 7y 5.754x 7(0.25) 5.75x 1hardbacks $1, paperbacks $0.2538. x y 3yy x 3(2, 4)(3, 2)39. y 1 4(x 2))y 1 4x 84x y 9 040. y 6 2(x 1)y 2x 441. (f g)(x) f(x) g(x) x 3 (x 2 3x 7) x 5 3x 4 7x 3f f( x)(x) g g( x)x3 x 2 3x 712x 21y 17.2512x 20y 17y 0.2542. 2x y 122x y 12→2(x 2y) 2(6) 2x 4y 123y 24y 82x y 122x (8) 12x 102x 2y 2(10) 2(8) 4The correct choice is B.O (0, 0)Page 96 Mid-Chapter Quiz1. 1 2 1x 5y 17y →1x0 3 2 5 3x 2y 18 y 3 2 x 9yO3x 2y 1812x 5y 17(4, 3)x(4, 2)(3, 2)xx y 3Chapter 2 46
2. 4x y 8 6x 2y 9y 8 4x6x 2(8 4x) 9x 1 2 4x y 84 1 2 y 8 1 2 , 6y 63. Let x represent trucks and y represent cars.x 4y6x 5y 29,000 x 4y6(4y) 5y 29,000 x 4(1000)y 1000 40004000 trucks, 1000 cars4. 2x y 4z 133x y 2z 15x 2z 122(3x y 2z) 2(1)→6x 2y 4z 24x 2y z 19 4x 2y z 1910x 3z 172(5x 2z) 2(12)→10x 4z 2410x 3z 17 10x 3z 177z 7z 15x 2z 12 2x y 4z 135x 2(1) 12 2(2) y 4(1) 13x 2 y 5(2, 5, 1)5. x y 1 x y 12x y 2 1 3 y 13x 1 y 1 1 3 x 1 3 4x y z 84 1 3 1 1 3 z 8z 8 1 3 , 1 1 3 , 86. y 3 x y 3 xy 2x2x 3 xx 3y 2xy 2(3) 6 (3, 6)3 (2) 5 8 7 67. A B 1 5 0 (9) 4 10 1 1134 9 148. impossible9. B 3A B (3A)3A 33 5 710 43(3) 3(5) 3(7)3(1)3(0) 3(4) or9 15 21 3 0 122 8 6B (3A) 9 15 215 9 10 3 0 122 (9) 8 (15) 6 21 5 3 9 0 10 (12) 11 7 278 9 210. The result is the original figure. The originalfigure is represented bya b cde f . Thereflection over the x-axis is found by1 0 a b c a b c0 1 d e f de f . Thereflection of the image over the x-axis is found by1 0 a b c a b c . The matrix0 1 d e f d e ffor the final image is the same as that of theoriginal figure.2-5Determinants and MultiplicativeInverses of MatricesPage 102 Check for Understanding1. Sample answer: a matrix with a nonzerodeterminant2. Sample answer:3 2 0is not a square43 5 2 5matrix. 1 1 also has no determinant. 0 91 0 0 0013. Sample answer:0 00 0 1 00 0 0 14. Sample answer: The system has a solution ifad bc 0, since you can use the inverse of thematrixa c to find the solution.b d421312 2615 324 1 05 15 12 10 715 1 5 110 7 2 7 2 10 4(95) 1(33) 0(20)4136 4 13 3 0 30 3 30 0 9 0 9 09 0 0 6(0) 4(27) (1)(27)1352 35 71 7 32 95 24 66 9does not exist5. 4(3) (2)(1) or 106. 12(32) (15)(26) or 67. 4 1 0 5 158. 6 4 (1) 0 3 9. 2(7) 5(3) or 2910. 4(9) 6(6) or 047 Chapter 2
4 1 23 5 y 2420. 4 2 1 (1) 0 1 (2) 0 20 2 11 3 2 3 215 42 1 3 113 5 43 5 3 5 4(5) 1(2) 2(4) 2611.5 4 x 3 11315 43 5 5 4 5 4 x3 1 1, 113 1 291 3 535Chapter 2 48y 113x y 5 4 3 1 1113 1 29136 312. x 635 9 y 851 9 3 1 3 9 6 3 5 956 939553 624 63 1 9 3 3 9 6 3 x 19 35 6 5 9 y 35 6 85x 8y 5(8, 5)13. Let x represent the amount of metal with 55%aluminum content, and let y represent the amountof metal with 80% aluminum content.x y 200.55x 0.8y 0.7(x y)0.55x 0.8y 0.7x 0.7y0.15x 0.1y 015x 10y 0x y 2015x 10y 01 1 1 15 10 21510 1 215→1 1 x 20 10 1 1 1 x1515111510151015y10 215x 8 111015y11020 y 128 kg of the metal with 55% aluminum and 12 kg ofthe metal with 80% aluminumPages 102–1053 42 54 10 19 1212 162 32 113 75 86 50 8Exercises14. 3(5) 2(4) or 715. 4(1) 0(1) or 416. 9(16) 12(12) or 017. 2(1) (2)(3) or 418. 13(8) (5)(7) or 6919. 6(8) 0(5) or 4802 1 321. 3 0 2 18390 2 0 2 (1) 3 2 3 3 03 0 1 2(6) 1(2) 3(9)3735 7 3 772 4 1 44 8(6) 9(19) 3(11)22. 8 9 3 3 53 5 12904 6 72 4 33 2 41 1 11 1 1 4(2) 6(7) 7(5) 125 36 1531 12 217 15 90112323. 4 6 4 7 3 2 24. 25 12 2 36 31 2 15 31 1215 9 17 9 17 15 25(78) 36(313) 15(669)31831.5 3.6 2.34.3 0.5 2.21.6 8.2 6.6 1.5 0.5 2.2 (36) 4.3 2.28.2 6.6 1.6 6.623 4.3 0.51.6 8.2 1.5(14.74) 3.6(31.9) 2.3(36.06) 175.6680 1 42 3 3 3 3 23 2 33 4 8 4 8 38 3 4 0(17) 1(12) 4(25) 1122 32 21 2 31 02 22 01 0does not exist4 21 2 1 6 2 2146 76 71 7 78 46 625. 26. 0 1 (4) 27. 2(2) (2)(3) or 1028. 2(0) 1(0) or 029. 4(2) 1(2) or 630. 6(7) (6)(7) or 844 68 12does not exist31. 4(12) 8(6) or 0111
9271336362732. 9(36) 27(13) or 27 21713 933. 3 4 1 8 3 4 1 2 5 1 8 or 15 1 2 51 8 3 4 1 2 4 134. x 11 2 y 71 2 1 1 9 2 1 4 1 1 2141 1 9 2 1 4 1 x 1 9 2 1 11 4 1 2 y 1 4 7x 1(1, 3)35.9 6 x 4 6 y 121 6 6 1 67 8 9 6 4 6 71849y446 9 6 6 9 6 x 1 6 6 7 8 4 9 4 6 y 4 9 12x 0 (0, 2)36.1 5 x 263 2 y 411 2 5 1 2 51 7 1 5 3 2 11731y332 2 5 1 5 x 1 2 5 263 1 3 2 y 1 73 1 41x 9 (9, 7)4 837. x 73 3 y 01 3 8 316 3 8 4 8 3 3 3163338444383y3xy 31 7 71, 2 1 2 147x 71y2 17238.3 5 x 245 4 y 31 4 5 317 317 3 5 5 453 645335 384 7 4 5 3 5 x 1 4 5 3 7 245 3 5 4 y 5 3 3x 3 (3, 3)y309 339. x 15 1 y 11 1 3 1 6 1 3 9 3 5 1595 1 6 1 3 9 3 x 1 6 1 3 15 1 3 , 2 3 951y9x y 53 2 3 x 440. 1 2 2 y 02 1 1 z 1 1 4 1 10 3 2 39 x 3 3 3 1 2 2 y5 1 8 2 1 1 z 1 4 1 10 49 3 3 3 05 1 8 1x 1 9y zx 1 69y 9z 12x 2y 3 1 2 3 , 1, 4 3 z 1 3 2 3 4(4) 1(0) (10)(1)3(4) 3(0) (3)(1)5(4) 1(0) 8(1) 4 3 6 5 3 x 941. 9 2 1 y 53 1 1 z 1 1 1 2 1 6 5 39 x 12 15 21 9 2 1 y15 21 33 3 1 1 z 1 1 2 1 99 12 15 21 515 21 33 1x 1 9y zx 1 29 y 12z 3x y z 2 9 , 4 3 , 1 3 42. 21643. 30,143 2 9 4 3 1 3 1(9) (2)(5) 1(1)12(9) (15)(5) 21(1)15(9) 21(5) (33)(1)9149 Chapter 2
52. Let x represent Jessi’s first test score, and let yrepresent Jessi’s second test score.x y 179→x y 179y x 7x y 71 1 x 1791 1 y 71 1 1 1 2 11 1 11 1 1 1 1 1 2 1 1 1 1 x 1 2 1 1 1791 1 1 1 y 1 1 7x 86 y 93first test: 86, second test: 938 4 0 4 3 3 3 35 1 5 9 4 4 4 48 (3) 4 (3) 0 (3) 4 (3) 5 4 1 4 5 4 9 45 1 3 195 9 13H(5, 9), I(1, 5), J(3, 9), K(1, 13)53. 54. 3 4 6 2 18 744 0 3 055. x 3y 2z 6→x 3y 2z 62(4x y z) 2(8) 8x 2y 2z 169x y 224(4x y z) 4(8)→7x 5y 4z 1016x 4y 4z 327x 5y 4z 109x y 221(9x y) 1(22)→9x y 229x y 22 9x y 220 0infinitely many solutions56.xf(x)g(x)6 x 5 25 x 4 0O4 x 3 23 x 2 42 x 1 6 g(x) 2x 557. y 5 1 2 (x 2)y 5 1 2 x 12y 10 x 2x 2y 8 05 1 2158. m 3 2 or 2(y 5) 2(x 1) or (y 3) 2(x 2)y 5 2(x 1)y 5 2x 2y 2x 7159a. 1 2or approximately 0.0833159b. 1 2 1x8 18 12x1.5 x; 1.5 ftx60. [f g](x) f(g(x)) f(x 1) (x 1) 2 3(x 1) 2 x 2 2x 1 3x 3 2 x 2 x[g f ](x) g(f(x)) g(x 2 3x 2) x 2 3x 2 1 x 2 3x 161. No, more than one element of the range is pairedwith the same element of the domain.62. The radius of circle E is 3, so the circumference is2(3) or about 18.85. The diagonal of the square6has length 6, so each side has length 6 2 2 232. The perimeter of the square is 4(32) or122 16.92.The difference between the circumference of thecircle and the perimeter of the square isapproximately 18.85 16.92 1.93.The correct choice is B.2-5BGraphing Calculator Exploration:Augmented Matrices andReduced Row-Echelon FormPage 1062 1 2 : 7 1 0 0 : 11 2 5 : 1 0 1 0 : 54 1 1 : 1 0 0 1 : 2(1, 5, 2)1 1 1 : 6 1 0 0 : 72. 2 3 4 : 3 , 0 1 0 : 1 ,4 8 4 : 12 0 0 1 : 2(7, 1, 2)1. , ,1 1 1 1 : 0 1 0 0 0 : 12 1 1 1 : 13.0 1 0 0 : 11 1 1 1 : 0 0 0 1 0 : 2,0 2 1 0 : 0 0 0 0 1 : 2(1, 1, 2, 2)4. Exercise 1: x 1, y 5, z 2; Exercise 2:x 7, y 1, z 2; Exercise 3: w 1, x 1,y 2, z 2; They are the solutions for thesystem.5. The calculator would show the first part of thenumber and follow it by ....2-6Solving Systems of LinearInequalitiesPage 109–110 Check for Understanding1a. the sum of twice the width and twice the height1b. Sample answer: skis, fishing rods2. Tomas is correct. There are functions in which thecoordinates of more than one vertex will yield thesame value for the function.51 Chapter 2
3. You might expect five vertices; however, if theequations were dependent or if they did notintersect to form the sides of a convex polygon,there would be fewer vertices.4. y5.O(1, 3)(1, 0)yx 2y 41(33 , 13 )x y 3O(0, 4)x(7, 0)(7, 0.5)(1, 0), (1, 3), (0, 4), (7, 0.5), (7, 0)6. yx8. Let x represent the number of greeting cards sold,and let y represent the income in dollars.x 0y 0y 0.45x 1500y 1.70xy600040002000Oy 1.70xPages 110–1119. yy 0.45x 1500(1200, 2040)2000 4000 6000 xat least 1200 cardsy x 1Exercises(0, 2)(4, 3)O(1,0)xO1( 23 , 13 )xy x 125, 67. yf(x, y) 4x 3yf(0, 2) 4(0) 3(2) or 6f(4, 3) 4(4) 3(3) or 25f2 1 3 , 1 3 42 1 3 3 1 3 or 8 1 3 (7, 7)10. y3y 3x 32(0, 1) ( 2 3 , 1)1 y 1y 3x11 O 1 2 x(3, 5)(5, 3)11.yxOf(x, y) 3x 4yf(3, 5) 3(3) 4(5) or 11f(7, 7) 3(7) 4(7) or 7f(5, 3) 3(5) 4(3) or 33, 112x 5y 25O5x 7y 14y 3x 2x12. yes, it is true for both inequalities:y 1 3 x 5 y 2x 12 ? 1 3 (3) 5 2 ? 2(3) 12 4 true 2 7 trueChapter 2 52
13.14.y 0.5x 1O 3 5 , 3 1 5 , 2 5 , 1 1 5 , 1 3 5 , 1 5 (0, 0), (0, 3), (3, 3)15. y(2, 5) y 5 0(1, 5)y x 7(2, 5), (7, 0), (4, 0), (1, 5)16. f(x, y) 8x yf(0, 0) 8(0) 0 or 0f(4, 0) 8(4) 0 or 32f(3, 5) 8(3) 5 or 29f(0, 5) 8(0) 5 or 532, 017. y x 5yy 2x 2( 0, 2)2( 25 , 4 5 , 1 1) ( 5)5y 0(3, 3)x 3 0Ox y5x 3y 20Oxy 3x 2y(0, 0)2x 5y 10Ox(0, 3)(4, 0) (7, 0)x(5, 4)(0, 2) (5, 2) y 2 x18.x 2 2(0, 0)Of(x, y) y xf(0, 0) 0 0 or 0f(0, 4) 4 0 or 4f(2, 0) 0 2 or 24, 219. y 4x 5y 10(5, 6) y 6O(0, 2)f(x, y) x yf(0, 2) 0 2 or 2f(5, 6) 5 6 or 11f(10, 6) 10 6 or 1616, 220. y(0, 4)y(0, 4)(2, 0) xy 0y 2x 42x 5y 10x 0 x y 4(0, 1) (3, 1)y 1O(10, 6)f(x, y) 4x 2y 7f(0, 1) 4(0) 2(1) 7 or 9f(0, 4) 4(0) 2(4) 7 or 15f(3, 1) 4(3) 2(1) 7 or 2121, 921.yy 4x 6xxf(x, y) 3x yf(0, 2) 3(0) 2 or 2f(5, 4) 3(5) 4 or 19f(5, 2) 3(5) 2 or 1719, 2x 4y 7(1, 2)(2, 2)Ox 6y 10(3, 1)x(4, 1)2x y 7f(x, y) 2x yf(2, 2) 2(2) (2) or 0f(1, 2) 2(1) 2 or 4f(3, 1) 2(3) 1 or 5f(4, 1) 2(4) (1) or 99, 453 Chapter 2
22.y(1, 4)O2x y 2 16(1, 8) (3, 8)(3, 2)y 5 xf(x, y) 2x y 5f(1, 4) 2(1) 4 5 or 7f(1, 8) 2(1) 8 5 or 11f(3, 8) 2(3) 8 5 or 7f(6, 2) 2(6) 2 5 or 5f(3, 2) 2(3) 2 5 or 111, 523. x 4, x 4, y 4, y 424. Sample answer: y 3, x 4, 4x 3y 1225a. 3y 2x 11 3y 2x 11y 0 3(0) 2x 115 1 2 x 5 1 2 , 0y 2x 13 y 2x 13y 0 0 2x 13y 16 xy 2x 13y 8(6, 2)y 2x6 1 2 x 6 1 2 , 0y 16 x2x 13 16 xx 2 93y 16 xy 16 2 93 1 93 2 93, 1 93 y 16 x 2y 172y 17 2(16 x) 17x 7 1 2 2y 17y 8 1 2 7 1 2 , 8 1 2 2y 17 2y 17y 3x 1y 8 1 2 y 3x 18 1 2 3x 12 1 2 x 2 1 2 , 8 1 2 y 7 2xy 7 2xy 3x 13x 1 7 2xx 6 5 y 7 2xy 7 2 6 5 2 35 6 5 , 2 53y 2x 11 3y 2x 11y 7 2x 3(7 2x) 2x 112 1 2 xy 7 2xy 7 22 1 2 23 2 1 2 , 225b. f(x, y) 5x 6yf5 1 2 , 0 55 1 2 6(0) or 27 1 2 f6 1 2 , 0 56 1 2 6(0) or 32 1 2 f 2 93, 1 939 5 2 3 6 1 39 or 86 1 3 f7 1 2 , 8 1 2 57 1 2 68 1 2 or 88 1 2 f2 1 2 , 8 1 2 52 1 2 68 1 2 or 63 1 2 f 6 5 , 2 35 5 6 5 6 2 35 or 33 1 5 f2 1 2 , 2 52 1 2 6(2) or 24 1 2 max at 7 1 2 , 8 1 2 88 1 2 ; min at 2 1 2 , 2 24 1 2 26. x y 2002x y 300x 0y 0300200x 0100(0, 0)Oy(0, 200)2x y 300(100, 100)x y 200y 0 200 300(150, 0)xf(x, y) $6.00x $4.80yf(0, 0) $6.00(0) $4.80(0) or 0f(0, 200) $6.00(0) $4.80(200) or $960f(100, 100) $6.00(100) $4.80(100) or $1080f(150, 0) $6.00(150) $4.80(0) or $900$108027a. Let x represent the Main Street site, and let yrepresent the High Street site.x 20y 2010x 20y 1200yx 20O(20, 50 )10x 20y 1200(20, 20 )(80, 20 )y 2027b. f(x, y) 30x 40y27c. f(x, y) 30x 40yf(20, 20) 30(20) 40(20) or 1400f(20, 50) 30(20) 40(50) or 2600f(80, 20) 30(80) 40(20) or 320080 ft 2 at the Main St. site and 20 ft 2 at the HighSt. site27d. Main Street: $1200 $10 120 ft 2120 30 3600 customersHigh Street: $1200 20 60 ft 260 40 240 customersThe maximum number of customers can bereached by renting 120 ft 2 at Main St.xChapter 2 54
28a. 3 is $3 profit on each batch of garlic dressing and2 is $2 profit on each batch of raspberrydressing.28b. 2x 3y 182x y 10x 0y 030.31.2 13 22 13 2y 1 7 Oy 2x 8d80604020O321201 2 340608032. {16}, {4, 4}; no, two y-values for one x-value33. w x y z4 15w x y z 602-7(0, 6)x 0(0, 0)Oy2x y 10y 0(3, 4)2x 3y 18(5, 0)xf(x, y) 3x 2yf(0, 0) 3(0) 2(0) or 0f(0, 6) 3(0) 2(6) or 12f(3, 4) 3(3) 2(4) or 17f(5, 0) 3(5) 2(0) or 153 batches garlic dressing, 4 batches raspberrydressing29. 2(2) (3)(1) or 7Linear ProgrammingxpPages 115–116 Check for Understanding1. Sample answer: These inequalities are usuallyincluded because in real life, you cannot make lessthan 0 of something.2. Sample answer: In an infeasible problem, theregion defined by the constraints contains nopoints. An unbounded region contains an infinitenumber of points.3. Sample answer: First define variables. Then writethe constraints as a system of inequalities. Graphthe system and find the coordinates of the verticesof the polygon formed. Then write an expression tobe maximized or minimized. Finally, substitutevalues from the coordinates of the vertices into theexpression and select the greatest or least result.4.y8765 0.5x 1.5y 7432 3x 9y 21 O2 2 4 6 8 1012145a. 25x 50y 42005b. 3x 5y 4805c.16014012010080604020y(0,0)O 2060 100x(0, 84)3x 5y 48025x 50y 4200(120, 24)(160,0)x1405d. P(x, y) 5x 8y5e. P(x, y) 5x 8yP(0, 0) 5(0) 8(0) or 0P(0, 84) 5(0) 8(84) or 672P(120, 24) 5(120) 8(24) or 792P(160, 0) 5(160) 8(0) or 800160 small packages, 0 large packages5f. $8005g. No; if revenue is maximized, the company willnot deliver any large packages, and customerswith large packages to ship will probably chooseanother carrier for all of their business.6. Let x the number of brochures.Let y the number of fliers.3x 2y 600x 50y 150y300200100Ox 50(50, 225)(100, 150)C(x, y) 8x 4yC(50, 150) 8(50) 4(150) or 1000¢C(50, 225) 8(50) 4(225) or 1300¢C(100, 150) 8(100) 4(150) or 1400¢50 brochures, 150 fliersy 150(50, 150)3x 2y 600100 200 300 x55 Chapter 2
7. Let x the number of Explorers.Let y the number of Grande Expeditions.x y 375y2x 3y 450400x 0300y 0x 0 x y 375200(0, 150) 2x 3y 450100(0, 0) (225, 0)R(x, y) 250x 350yR(0, 0) 250(0) 350(0) or 0R(0, 150) 250(0) 350(150) or 52,500R(225, 0) 250(225) 350(0) or 56,250225 Explorers, 0 Grande Expeditions8. Let x the number of loaves of light whole wheat.Let y the number of loaves of regular wholewheat.y2x 3y 9060x 2y 8040 x 2y 80x 0(0, 30)2x 3y 90y 020x 0(0, 0) (45, 0) xO 20 40 60 80y 020P(x, y) 1x 1.50yP(0, 0) 1(0) 1.50(0) or 0P(0, 30) 1(0) 1.50(30) or 45P(45, 0) 1(45) 1.50(0) or 45alternate optimal solutionsPages 116–118 Exercises9. yinfeasibleOy 65x 3y 15O 100 200 300y 010. 56yunbounded4840 2x y 48322416 x 2y 428xO 8 162432404856xx11.yO( 3, 3)4(3, 0)x 4y 3(4, 3)(4, 0) x4x 3y 12f(x, y) 3 3yf 3 4 , 3 3 3(3) or 12f(4, 3) 3 3(3) or 12f(4, 0) 3 3(0) or 3f(3, 0) 3 3(0) or 3alternate optimal solutions12a. Let g the number of cups of Good Start foodand s the number of cups of Sirius food.0.84g 0.56s 1.5412b. 0.21g 0.49s 0.5612c. s321O(0, 2.75)0.81y 0.56 s 1.540.21g 0.49 s 0.56(1.5, 0.5)1 212d. C(g, s) 36g 22s12e. C(g, s) 36g 22sC(0, 2.75) 36(0) 22(2.75) or 60.5C(1.5, 0.5) 36(1.5) 22(0.5) or 65C(2.66, 0) 36(2.66) 22(0) or 95.760 cups of Good Start and 2.75 cups of Sirius12f. 60.5¢13a. Let d the number of day-shift workersand n the number of night-shift workers.d 5n 6d n 1413b. dd 5(6, 8)(2.67, 0)d n 14(9, 5)n 6gO13c. $5.50 4 $7.50 4 $52$7.50 8 $60C(n, d) 52d 60n13d. C(n, d) 52d 60nC(6, 8) 52(8) 60(6) or 776C(9, 5) 52(5) 60(9) or 8008 day-shift and 6 night-shift workersnChapter 2 56
13e. $77614a. Let x the number of acres of corn.Let y the number of acres of soybeans.x y 180yx 40200 x 40y 20x 2yx y 180100x 2y(40, 20)(120, 60)(160, 20)y 20 xO 100 200P(x, y) 150x 250yP(40, 20) 150(40) 250(20) or 11,000P(120, 60) 150(120) 250(60) or 33,000P(160, 20) 150(160) 250(20) or 29,000120 acres of corn, 60 acres of soybeans14b. $33,00015. Let x the questions from section I.Let y the questions from section II.6x 15y 90yy 2x 012x 08 (0, 6)6x 15y 904(10, 2)(0, 2) y 2 xO 4 8 12 16S(x, y) 10x 15yS(0, 2) 10(0) 15(2) or 30S(0, 6) 10(0) 15(6) or 90S(10, 2) 10(10) 15(2) or 13010 section I questions, 2 section II questions16. Let x the number of food containers.Let y the number of drink containers.x y 1200yx 300y 4501200(300, 900)800 x y 1200(300, 450) (750, 450)400y 450x 300O 400 800 1200 xP(x, y) 17.50x 20yP(300, 450) 17.50(300) 20(450) or 14,250P(300, 900) 17.50(300) 20(900) or 23,250P(750, 450) 17.50(750) 20(450) or 22,125$23,25017. Let x amount to deposit at First Bank.Let y amount to deposit at City Bank.x y 11,0000 x 75001000 y 7000y12,000x 7500x y 11,0008000(4000, 7000) y 7000(0, 7000)4000(7500, 3500)(0, 1000)(7500, 1000)O 4000 8000 12,000xy 1000I(x, y) 0.06x 0.065yI(0, 1,000) 0.06(0) 0.065(1,000) or 65I(0, 7000) 0.06(0) 0.065(7,000) or 455I(4000, 7000) 0.06(4000) 0.065(7000) or 695I(7500, 3500) 0.06(7500) 0.065(3500) or 677.5I(7500, 1000) 0.06(7500) 0.065(1000) or 515$4000 in First Bank, $7000 in City Bank18. Let x the number of nurses.Let y the number of nurse’s aides.x y 50yx y 20y 1260x 2yx y 5040 x 2yx y 20 (33.3, 16.7)20y 12(24, 12) (38, 12)O 20 40 60 xC(x, y) 35,000x 18,000yC(24,12) 35,000(24) 18,000(12) or1,056,000C(33.3, 16.7) 35,000(33.3) 18,000(16.7) or1,466,100C(38, 12) 35,000(38) 18,000(12) or1,546,00024 nurses, 12 nurse’s aides19. Let x units of snack-size bags.Let y units of family-size bags.x y 2400yx 600x 600y 9002400(600, 1800)1600 x y 2400(1500, 900)800y 900(600, 900)O 800 1600 2400 xP(x, y) 12x 18yP(600, 900) 12(600) 18(900) or 23,400P(600, 1800) 12(600) 18(1800) or 39,600P(1500, 900) 12(1500) 18(900) or 34,200600 units of snack-size, 1800 units of family-size57 Chapter 2
20. Let x batches of soap.Let y batches of shampoo.12x 6y 4820x 8y 76x 0y 0Oy(0, 8) 20x 8y 7612x 6y 48x 0(0, 0)Oy128(0, 6)4x 0P(x, y) 40x 40yP(0, 0) 40(0) 40(0) or 0P(0, 6) 40(0) 40(6) or 240P(4, 5) 40(4) 40(5) or 360P(9, 0) 40(9) 40(0) or 360alternate optimal solutions22. y E(0, 14)2x 6y 8412D(12, 10)F(0, 11)8C(12, 9)8x 3y 334 2x 3y 3A(0, 1) B(3, 3) x 1248 12y 0xy 10y 0 Oy 0O 4(0, 0)y604020(3, 2)(3.8, 0)3 batches of soap and 2 batches of shampoo21. Let x the number of small monitors.Let y the number of large monitors.x 2y 16x y 9x 4y 24x 0y 0Area of trapezoid ACDE 1 2 (12)(13 1) 84Area of ABF 1 2 (10)(3) 15Area of shaded origin 84 15 69 square units23a. Let x oil changes.Let y tune-ups.x 250 y 1030x 60y 30(60)P(x, y) 12x 20yP(25, 0) 12(25) 20(0) or 300P(25, 10) 12(25) 20(10) or 500P(40, 10) 12(40) 20(10) or 680P(60, 0) 12(60) 20(0) or 720$720xx y 9x 2y 16x 4y 24(4, 5)(9, 0)8 12y 0x 25x30x 60y 1800(25, 10)(40, 10)(60, 0)20 40 60 x(25, 0)23b. Sample answer: Spend more than 30 hours perweek on these services.24.y20y x 5 10 y x 5(0, 5)20 10 O 10 20x10(15, 10)y 5 520 (15, 10)f(x, y) 1 3 x 1 2 yf(15, 10) 1 3 (15) 1 2 (10) or 0f(0, 5) 1 3 (0) 1 2 (5) or 2 1 2 f(15, 10) 1 3 (15) 1 2 (10) or 10minimum: 2 1 2 , maximum: 1025. 4x y 6 4x y 6x 2y 12 4(2y 12) y 68y 48 y 6y 6x 12y 12x 2(6) 12 or 0 (0, 6)26.x yy1 90 61 32 03 327. Sample answer: C $13.65 $0.15(n 30);C $13.65 $0.15(n 30)C $13.65 $0.15(42 30) $15.4528. 2x 3x 3 x22(2x 3) x(3 x)4x 6 3x x 2x 2 x 6 0(x 3)(x 2) 0x 3 0 or x 2 0x 3 x 2The correct choice is A.y 3|x 2|Chapter 2 Study Guide and AssessmentPage 119 Understanding and Using the Vocabulary1. translation 2. added3. determinant 4. inconsistent5. scalar multiplication 6. equal matrices7. polygonal convex set 8. reflections9. element 10. multipliedOxChapter 2 58
Pages 120–122 Skills and Concepts11. 2y 4x y x 22(x 2) 4x y 2 22x 4 4xy 4x 2(2, 4)12. 6y x 0 y x 56(x 5) x 0 y 6 56x 30 x 0 y 1x 6(6, 1)13. 3y x 1 2x 5yx 1 3y 2(1 3y) 5y2 6y 5y 121 y2x 5y2x 5 121 x 151 151 , 121 14. 2y 15x 4 y 6x 12(6x 1) 15x 4 y 6(2) 112x 2 15x 4 y 13 (2, 13)x 215. 5(3x 2y) 5(1) 15x 10y 52(2x 5y) 2(12)→4x 10y 2419x 19x 12x 5y 122(1) 5y 12y 2 (1, 2)16. x 5y 20.5x 5y 20.53y x 13.5→x 3y 13.58y 34x 5y 20.5 y 4.25x 5(4.25) 20.5x 0.75 (0.75, 4.25)17. 3(x 2y 3z) 3(2) 3(x 4y 3z) 3(14)3x 5y 4z 0 3x 5y 4z 0↓↓3x 6y 9z 6 3x 12y 9z 423x 5y 4z 0 3x 5y 4z 0y 5z 6 7y 13z 427(y 5z) 7(16)→7y 35z 427y 13z 42 7y 13z 4248z 0z 0y 5z 6 x 2y 3z 2y 5(0) 6 x 2(6) 3(0) 2y 6x 10(10, 6, 0)18. x 2y 6z 4 2(x 2y 6z) 2(4)x y 2z 3 2x 3y 4z 53y 4z 7↓2x 4y 12z 182x 3y 4z 157y 16z 134(3y 4z) 4(7)→12y 16z 287y 16z 13 7y 16z 135y 15y 33y 4z 7 x y 2z 33(3) 4z 7 x 3 2(0.5) 3z 0.5x 1(1, 3, 0.5)19. x 2y z 7 2(3x y z) 2(2)3x y z 2 2x 3y 2z 74x y 9 ↓6x 2y 2z 42x 3y 2z 78x 5y 115(4x y) 5(9) → 20x 5y 458x 5y 11 8x 5y 1128x 56x 24x y 9 x 2y z 74(2) y 9 2 2(1) z 7y 1 z 3(2, 1, 3)20. A B 7 (3) 8 (5) 0 2 4 (2) 4 3 2 621. B A 3 7 5 82 0 2 (4) 10 13 2 222. 3B 3(3) 3(5)3(2) 3(2) 9 15 6 623. 4C 4(2)4(5)8 24. AB 207 8 3 50 4 2 27(3) 8(2)0(3) (4)(2)or5 518825. impossible26. 4A 4B 4A (4B)7(5) 8(2)0(2) (4)(2) 4A 4(7) 4(8) 4B 4(3) 4(5)4(0) 4(4)4(2) 4(2) 28 32 12 200 168 84A (4B) 28 12 32 200 (8) 16 840 52 8859 Chapter 2
27. impossible28. 4 2 5 3 3 3 7 1 2 3 1 3 4 4 4A(7, 1), B(1, 3), C(2, 7)A yBOxABC13732.1 0 4 2 5 4 2 50 1 3 1 3 31 30 1 4 2 5 3 1 3 1 0 3 1 3 4A(3, 4), B(1, 2), C(3, 5)C yABBOx2529. 1 0 2 1 0 1 2 1 0 1 0 1 3 2 4 2 3 2W(2, 3), X(1, 2), Y(0, 4), Z(1, 2)yYWX ZO xXZWY1 030. 2 2 1 1 2 2 1 1 0 1 3 5 5 3 3 5D(2, 3), E(2, 5), F(1, 5), G(1, 3)yE FG DO xD GF E3 1 131. 0.5 1.5 0.5 0.5C4 2 1 2 1P(6, 8), Q(2, 4), R(2, 2)4y QR2QRx2 O 2 4 624 P68P0.545234 4 433. 5 5 5 1 1 13 6 6 6 33 54 78 46 33 1 45 2 67 3 42 63 4 7 4 3(10) 1(62) 4(29) 24334. 3(7) (4)(5) or 135. 8(3) (6)(4) or 036. 3 (1) 5 6 4 5 25 0 437. 7 3 1 2 2 63 12 6 2 6 5(16) 0(44) 4(20) 16038. no, not a square matrix 5 0 7 1 (4) 7 33839. 3(5) (1)(8) or 231 51 5 82 31 35 210 4no inverse3 51 4 1 7 4 51340. 5(4) 10(2) or 041. 3(4) 1(5) or 73 277542. 3(7) 5(2) or 115 111 2 32 56 11 1 53 26 22 41 2no inverse43. 2(1) 6(5) or 3244. 2(2) (1)(4) or 0AC7233233Chapter 2 60
2 545. x 11 3 y 21 3 5 1 3 52 5 1 111 3 5 2 5 x 1 3 5 11 2 1 3 y 1 2 2x 13 (13, 5)46.3 2 x 36 4 y 61 4 2 214 43 2 21446236 436 21(1, 0)6243xyxy2y 464610 512 3233 547. x 12 4 y 21 4 5 1 2 4 5 3 5 2 42322 3 3 1 2 4 5 3 5 x 1 2 4 5 12 3 2 4 y 2 3 2x 7 (7, 4)8.82.92.74.6148. 1 48 4.6 2.7 2.9 8.8 481.31x 1O(1, 2)y 2y 10 2x(1, 5)y x 6(4, 2)(6, 2)xy .3148.82.98.8 y8.82.95.76.62.7 4.6 8.8 2.7 4.6 2.7 x2.9 4.6 2.9 1 2.7 48 .31 8.44.6 74.61x (5.7, 6.6)49. yf(x, y) 2x 3yf(1, 2) 2(1) 3(2) or 4f(1, 5) 2(1) 3(5) or 17f(4, 2) 2(4) 3(2) or 14f(6, 2) 2(6) 3(2) or 617, 4y650. yx y 1112 (0, 9)x 6(4, 7)82y x 18(0, 4)(6, 5)4y 4 (6, 4)Of(x, y) 3x 2y 1f(0, 4) 3(0) 2(4) 1 or 9f(0, 9) 3(0) 2(9) 1 or 19f(4, 7) 3(4) 2(7) 1 or 27f(6, 5) 3(6) 2(5) 1 or 29f(6, 4) 3(6) 2(4) 1 or 2729, 951. Let x gallons in the truck.Let y gallons in the motorcycle.x y 28y0 x 2530 x y 280 y 6x 2520m(x, y) 22x 42ym(0, 0) 22(0) 42(0) or 0m(0, 6) 22(0) 42(6) or 252m(22, 6) 22(22) 42(6) or 746m(25, 3) 22(25) 42(3) or 676m(25, 0) 22(25) 42(0) or 55022 gallons in the truck and 6 gallons in themotorcyclePage 12348 1210 (0, 6)(22, 6)y 6(25, 3)y 0 O 10 20 30 x(0, 0) (25, 0)Applications and Problem Solving2 5 5 5 2(5) 5(3) 5(1) 3052. 8 2 3 3 8(5) 2(3) 3(1) 496 4 1 1 6(5) 4(3) 1(1)Broadman 30; Girard 49; Niles 43x4361 Chapter 2
53. Let x the shortest side.Let y the middle-length side.Let z the longest side.x y z 83 x y z 83z 3x x y 3x 83z 1 2 (x y) 17 4x y 83z 1 2 (x y) 173x 1 2 (x y) 175x y 344 1511 4 1 5 1 x 83y 341 1 1 9 1 1 55 1 9 1 1 4 1 x 1 9 1 1 83 5451x 13y 31z 3xz 3(13)z 3913 in., 31 in., 39 in.54a. Let x number of Voyagers.Let y number of Explorers.5x 6y 240y 3x 2y 1203x 2y 120605x 18y 5405x 6y 240x 0(18, 25)40(30, 15)y 0(0, 30) 5x 18y 54020y 0(40, 0)O 20 40 60 x(0, 0)P(x, y) 2.40x 5.00yP(0, 0) 240(0) 5.00(0) or 0P(0, 30) 2.40(0) 5.00(30) or 150P(18, 25) 2.40(18) 5.00(25) or 168.20P(30, 15) 2.40(30) 5.00(15) or 147P(40, 0) 2.40(40) 5.00(0) or 9618 Voyagers and 25 Explorers54b. $168.20Page 123 Open-Ended Assessment1a. A(2, 2), B(1, 2), C(2, 1), and D(3, 0)Sample answer: Two consecutive 90° rotations isthe same as one 180° rotation. An additional180° rotation will return the image to its originalposition.1b. Two consecutive 90° rotations is the same as one180° rotation.2. No; such a coefficient matrix will not have aninverse. Consider the matrix equation2 4 x 12 . The coefficient 2 44 8 y 24 48 has adeterminant of 0, so it has no inverse.4y 54434Chapter 2 SAT & ACT PreparationPage 125 SAT and ACT Practice1. Translate the information from words into anequation. Then solve the equation for x. Use thecorrect order of operations.(1 2)(2 3)(3 4) 1 2 (20 x)(3)(5)(7) 1 2 (20 x)105 1 2 (20 x)210 20 xx 190The correct choice is D.2. First convert the numbers to improper fractions.6 2 545 1 3 6 1 4 1 3Express both fractions with a commondenominator. Then subtract.5 1 3 6 1 4 1 36 2 54 6 412 7 512 11 1 2The correct choice is A.3. You can solve this problem by writing algebraicexpressions.Amount of root beer at start: xAmount poured into each glass: yNumber of glasses: zTotal amount poured out: yzAmount remaining: x yzThe correct choice is D.4. 2⏐x 2⏐ 1 52⏐x 2⏐ 6⏐x 2⏐ 3x 2 3 or x 2 3x 5 or xor x 1The correct choice is D.5. The total amount charged is $113. Of that, $75 isfor the first 30 minutes. The rest (113 $75 $38) is the cost of the additional minutes. At $2per minutes, $38 represents 19 minutes. (19 $2 $38). The plumber worked 30 minutes plus 19minutes, for a total of 49 minutes.The correct choice is C.Chapter 2 62
6. Start by simplifying the fraction expression on theright side of the equation. 2 x5 x 2 5 2 5 2 x5 x 4 5 To finish solving the equation, treat it as aproportion and write the cross products. 2 x5 x 4 5 5(2 x) 4(5 x)10 5x 20 4xx 10The correct choice is E.7. Notice that the question asks what must be true.There are two ways to solve this problem. Thefirst is by choosing specific integers that meet thecriteria and finding their sums.I. 2 3 5, 3 4 7Choose consecutive integers where the firstone is even and where the first one is odd. Ineither case, the result is odd. So statement I istrue. Eliminate answer choice B.II. 2 3 4 9, 3 4 5 12One sum is odd, and the other is even. Sostatement II is not always true. Eliminateanswer choices B, C, and E.III. 2 3 4 9, 3 4 5 12,10 11 12 33Statement III is true for these examples andseems to be true in general. Eliminate answerchoice A.Another method is to use algebra. Representconsecutive integers by n and n 1. Representeven integers by 2k, and odd integers by 2k 1.I. n (n 1) 2n 12n 1 is odd for any value of n. So statementI is always true.II. n (n 1) (n 2) 3n 3If n is even, then 3n 3 3(2k) 3 is odd.If n is odd, then 3n 3 3(2k 1) 3 6k 3 3 6k 6 is even. So statement IIis not always true.III. By the same reasoning as in II, the sum is amultiple of 3, so statement III is always true.n (n 1) (n 2) 3n 3 3(n 1)The correct choice is D.8. Start by representing the relationships that aregiven in the problem. Let P represent the numberof pennies; N the number of nickels; D the numberof dimes; and Q the number of quarters. He hastwice as many pennies as nickels.P 2NSimilarly, N 2D and D 2Q. You know he hasat least one quarter. Since you need to find theleast amount of money he could have, he musthave exactly one quarter.Since he has 1 quarter, he must have 2 dimes,because D 2Q. Since he has 2 dimes, he musthave 4 nickels. Since he has 4 nickels, he musthave 8 pennies.Now calculate the total amount of money.1 quarter $0.252 dimes $0.204 nickels $0.208 pennies $0.08The total amount is $0.73. The correct choice is D. 3 2 9. 3 2 4 9 1 1 3 2 2 3 2 9 4 28 2 3 The correct choice is C.10. There are two products, CDs and tapes. You needto find the number of tapes sold. You also haveinformation about the total sales and CD sales.You might want to arrange the information in atable. Let t be the number of tapes sold.Price each Number TotalSold SalesCDs 40 $480TapesTotal $600You can calculate the price of each CD. Since40 CDs sold for $480, each CD must cost $12($480 40 $12). You know that the price of aCD is three times the price of a tape. So a CDcosts 1 3 of $12 or $4. You can calculate the totalsales of CDs by subtracting $480 from $600 to get$120.tPrice each Number TotalSold SalesCDs $12 40 $480Tapes $4 t $120Total $600Now you can find t using an equation.4t 120t 30Thirty cassette tapes were sold. The answer is 30.63 Chapter 2
Chapter 3 The Nature of Graphs3-1Symmetry and CoordinateGraphsPage 133 Graphing Calculator Exploration1. f(x) f(x) 2. f(x) f(x)3. even; odd4. f(x) x 8 3x 4 2x 2 2f(x) (x) 8 3(x) 4 2(x) 2 2 x 8 3x 4 2x 2 2 f(x)f(x) x 7 4x 5 x 3f(x) (x) 7 4(x) 5 (x) 3x 7 4x 5 x 3(x 7 4x 5 x 3 )f(x)5. First find a few points of the graph in either thefirst or fourth quadrants. For an even function, afew other points of the graph are found by usingthe same y-values as those points, but with oppositex-coordinates. For an odd function, a few otherpoints are found by using the opposite of both thex- and y-coordinates as those original points.6. By setting the INDPNT menu option in TBLSETto ASK instead of AUTO, you can then go toTABLE and input x-values and determine theircorresponding y-values on the graph. By inputtingseveral sets of opposite pairs, you can observewhether f(x) f(x), f(x) f(x), or neither ofthese relationships is apparent.Pages 133–134 Check for Understanding1. The graph of y x 2 12 is an even function.The graph of xy 6isanoddfunction. The graphsof x y 2 4 and 17x 2 16xy 17y 2 225 areneither.2. The graph of an odd function is symmetric withrespect to the origin. Therefore, rotating the graph180° will have no effect on its appearance. Seestudent’s work for example.3a. Sample answer: y 0, x 0, y x, y x3b. infinitely many3c. point symmetry about the origin4. Substitute (a, b) into the equation. Substitute(b, a) into the equation. Check to see whetherboth substitutions result in equivalent equations.5. AliciaGraphically: If a graph has origin symmetry, thenany portion of the graph in Quadrant I has animage in Quadrant III. If the graph is thensymmetric with respect to the y-axis, the portionin Quadrants I and II have reflections inQuadrants II and IV, respectively. Therefore, anypiece in Quadrant I has a reflection in QuadrantIV and the same is true for Quadrants II and III.Therefore, the graph is symmetric with respect tothe x-axis.Algebraically: Substituting (x, y) into theequation followed by substituting (x, y) is thesame as substituting (x, y).6. f(x) x 6 9xf(x) (x) 6 9(x) f(x) (x 6 9x)f(x) x 6 9xf(x) x 6 9xno17. f(x) 5 x x 191f(x) 5( x) (x) 19 f(x) 15x x 19 1f(x) 5 x x 19 1f(x) 5 x x 19yes8. 6x 2 y 1 → 6a 2 b 1x-axis 6a 2 (b) 16a 2 b 1 noy-axis 6(a) 2 b 16a 2 b 1 yesy x 6(b) 2 a 16b 2 a 1 noy x 6(b) 2 (a) 16b 2 a 1 noy-axis9. x 3 y 3 4 → a 3 b 3 4x-axis a 3 (b) 3 4a 3 b 3 4 noy-axis (a) 3 b 3 4a 3 b 3 4 noy x (b) 3 (a) 3 4a 3 b 3 4 yesy x (b) 3 (a) 3 4a 3 b 3 4 noy x10.y(2.5, 3) (2.5, 3)(4, 2) (4, 2)(1, 2) (1, 2)Ox11. y 2 x 2 → b 2 a2x-axisb 2 a2 noy-axis b 2 a) (2b 2 a2 yesy(1,1)2(0√2)(1,1)1(√2,0)(√2,0)21 O 1122 xChapter 3 64
18. f(x) 7x 5 8x12. ⏐y⏐ x 3 → ⏐b⏐ a 3311311y y y x(b)(a) 55 5ab 5 yesx-axis ⏐b⏐ a 3f(x) 7(x) 5 8(x)⏐b⏐ a 3 yesf(x) 7x 5 8xy-axis ⏐b⏐ (a) 3f(x) (7x 5 8x)⏐b⏐ a 3f(x) 7x 5 8xnoyesy19. f(x) 1 x x 100f(x) 1 ( (x) 100 f(x) x) 1 x x 100 (1,1)f(x) 1 x x 100 f(x) 1 x x 100O x(1,1)no20. yes;g(x) x2 1xx213. x-intercept: 2 5 y 2g(x) (x ) 2 1Replace x with x.9 1(x)x2 2 5 0 2g(x) x2 1 (x) 2 x 2x9 1g(x) x2 x2 1x Determine the opposite 2 5 1of the function.x 2 25g(x) g(x)x 5 (5, 0) 21. xy 5 → ab 5other points:x-axisa(b) 5when x 6when x 6ab 5 3 625 y 29 1 3 625 y 2ab 5 no9 1y-axis(a)b 5 y 2 11 9 2 5 y 2 11 ab 59 2 5ab 5 noy 2 9 925 y 2 9 925 y x(b)(a) 5ab 5 yes311311311y x and y x6, , 6, , 6, 555 22. x y 2 1 → a b 2 1x-axis a (b) 2 1Pages 134–136 Exercisesa b 2 1 yesy-axis (a) b 2 114. f(x) 3xa b 2 1 nof(x) 3(x)f(x) (3x)y x (b) (a) 2 1f(x) 3xf(x) 3xb a 2 1 noyesy x (b) (a) 2 115. f(x) x 3 b a 1 1 nof(x) (x) 3 1 f(x) (x 3 x-axis 1)f(x) x 3 1 f(x) x 3 23. y 8x → b 8a 1x-axis(b) 8ano16. f(x) 5x 2 b 8a no 6x 9y-axisb 8(a)f(x) 5(x) 2 6(x) 9b 8a nof(x) 5x 2 6x 9y x(a) 8(b)f(x) (5x 2 6x 9)a 8b nof(x) 5x 2 6x 9y x(a) 8(b)noa 8b no17. f(x) 14 x 7f(x) 4(f(x) 1x) 714x 7 none of these1f(x) 4 x 7 1f(x) 4 x 7yes65 Chapter 3
24. y 1 x 2 → b 1 a 2x-axis (b) a12b a12 noy-axis b 1 ( a) 2b a12 yesy x (a) 1(b ) 2a 1b 2 noy x (a) 1( b) 2a 1b 2 no; y-axis25. x 2 y 2 4 → a 2 b 2 4x-axis a 2 (b) 2 4a 2 b 2 4 yesy-axis (a) 2 b 2 4a 2 b 2 4 yesy x (b) 2 (a) 2 4a 2 b 2 4 yesy x (b) 2 (a) 2 4a 2 b 2 4 yesall26. y 2 4 x 2 4 → b 2 4 a 2 499x-axis (b) 2 4 a 2 49b 2 4 a 2 49yesy-axis b 2 4( a) 2 49b 2 4 a 2 49yesy x (a) 2 4( b) 2 49a 2 4 b 2 49noy x (a) 2 4( b) 2 49a 2 4 b 2 49nox-axis and y-axis27. x 2 1y 2 → a 2 1b 2x-axis a 2 1( b) 2a 2 1b 2y-axis (a) 2 1 b 2a 2 1b 2y x (b) 2 1(a ) 2b 2 a12a 2 1b 2y x (b) 2 1( a) 2b 2 a12yesyesyesa 2 1b 2 yesx-axis, y-axis, y x, y x28.29.(2, 1)yO(1, 2)(4, 4)(2, 1)(1, 2)(4, 4)(4, 4) y (4, 4)(1, 2) (1, 2)(2, 1)(2, 3)yOO30. Sample answer:yOyO(1, 2)(4, 4)(2, 1)(2, 1)31. y 2 x 2 → b 2 a 2x-axis (b) 2 a 2b 2 a 2y-axis b 2 (a) 2xxxxxb 2 a 2yes; both32. ⏐x⏐ 3y → ⏐a⏐ 3bx-axis⏐a⏐ 3(b)⏐a⏐ 3b noy-axis⏐(a)⏐ 3b⏐a⏐ 3b yesy-axisChapter 3 66
33. y 2 3x 0 → b 2 3a 0x-axis (b) 2 3a 0b 2 3a 0y-axis b 2 3(a) 0b 2 3a 0x-axisyOxyesno37. ⏐y⏐ x 3 x → ⏐b⏐ a 3 ax-axis⏐(b)⏐ a 3 a⏐b⏐ a 3 a yesy-axis⏐b⏐ (a) 3 (a)⏐b⏐ a 3 a nox-axisThe equation ⏐y⏐ x 3 x is symmetric about thex-axis.y11O1x34. ⏐y⏐ 2x 2 → ⏐b⏐ 2a 2x-axis ⏐(b)⏐ 2a 2⏐b⏐ 2a 2 yesy-axis ⏐b⏐ 2(a) 2yO⏐b⏐ 2a 2yes; both35. x 12 8y 2 → a 12 8b 2x-axisa 12 8(b) a 12 8b 2yesy-axis(a) 12 8b 2a 12 8b 2yes; bothyO36. ⏐y⏐ xy → ⏐b⏐ abx-axis⏐(b)⏐ a(b)⏐b⏐ ab noy-axis⏐b⏐ (a)b⏐b⏐ ab noneithery1(1, 0)O1(1, 0)xxx2 y2 138a. x 8 0 1 → a 8x-axis a 2838b.1y-axis ( a) 28origin ( a) 28yOx2 b2 1 0 ( b) 210 a 2 b28 1 0b2 1 0 a 2 b28 1 0 ( b) 210 a 2 b28 1 0 1 1 1 1 1 1yesyes 1 yesx- and y-axis symmetry38c. (2, 5), (2, 5), (2, 5)39. Sample answer: y 040. Sample answer:y x 1 y x 1 y x 1y x 1 y 2x 4 y 2x 4y 2x 4 y 2x 4y212 1 O 1 212x67 Chapter 3
y2x241. 1 2 1 6 ( 6) 2 x212 1 6x23 1 6x2 1 6 1 1 1 2x 2 32x 42(42, 6) or (42, 6)42. No; if an odd function has a y-intercept, then itmust be the origin. If it were not, say it were(0, 1), then the graph would have to contain (1, 0).This would cause the relation to fail the verticalline test and would therefore not be a function.But, not all odd functions have a y-intercept.Consider the graph of y 1 x .yO1y x43. Let x number of bicycles.Let y number of tricycles.3x 4y 4505x 2y 400x 0y 0xy150100x 050(0, 0)O5x 2y 400(0, 112.5)(50, 75)3x 4y 450y 050 100 150 x(80, 0)P(x, y) 6x 4yP(0, 0) 6(0) 4(0) or 0P(0, 112.5) 6(0) 4(112.5) or 450P(50, 75) 6(50) 4(75) or 600P(80, 0) 6(80) 4(0) or 48050 bicycles, 75 tricycles44.4 3 8 5 4(8) 3(9) 4(5) 3(6)7 2 9 6 7(8) 2(9) 7(5) 2(6) 59 38 74 4745. 3(2x y z) 3(0)6x 3y 3z 0→3x 2y 3z 21 3x 2y 3z 219x y 213x 2y 3z 214x 5y 3z 27x 3y 233(9x y) 3(21)27x 3y 63→7x 3y 237x 3y 2320x 40x 29x y 21 2x y z 09(2) y 21 2(2) (3) z 0y 3 z 7(2, 3, 7)46. 4x 2y 7→12x 6y 21consistent and dependent47. yy x22 014 248. m 16 or 7y 2 7(x 0)y 7x 249. [f g](x) f(g(x)) f(x 6)2(x 6) 112x 23[g f ](x) g(f(x)) g(2x 11) (2x 11) 62x 550. 75 3 75 7 75 37 75 10The correct choice is B.3-2Oy x 2Families of Graphsxy 2x 7 2 y 2x 7 2 Page 142 Check for Understanding1. y (x 4) 3 72. The graph of y (x 3) 2 is a translation of y x 2three units to the left. The graph of y x 2 3 is atranslation of y x 2 three units up.3. reflections and translations4. When c 1, the graph of y f(x) is compressedhorizontally by a factor of c.When c 1, the graph of y f(x) is unchanged.When 0 c 1, the graph is expandedhorizontally by a factor of 1 c .5a. g(x) 3 x 15b. h(x) 3 x 15c. k(x) 3 x 2 16. The graph of g(x) is the graph of f(x) translatedleft 4 units.7. The graph of g(x) is the graph of f(x) compressedhorizontally by a factor of 1 3 , and then reflectedover the x-axis.8a. expanded horizontally by a factor of 58b. translated right 5 units and down 2 units8c. expanded vertically by a factor of 3, translatedup 6 unitsChapter 3 68
9a. translated up 3 units, portion of graph belowx-axis reflected over the x-axis9b. reflected over the x-axis, compressedhorizontally by a factor of 1 2 9c. translated left 1 unit, compressed vertically by afactor of 0.7510. y11.y12a.12b.O$250$200$150$100xx f(x)0 x 1 501 x 2 1002 x 3 1503 x 4 2004 x 5 250$5000 1 2 3 4 5Time (h)$250$200$150$100$500012c. $2251 2 3 4 5 6Time (h)Ox20a. reflected over the x-axis, compressedhorizontally by a factor of 0.620b. translated right 3 units, expanded vertically by afactor of 420c. compressed vertically by a factor of 1 2 , translateddown 5 units21a. expanded horizontally by a factor of 521b. expanded vertically by a factor of 7, translateddown 0.4 units21c. reflected across the x-axis, translated left 1 unit,expanded vertically by a factor of 922a. translated left 2 units and down 5 units22b. expanded horizontally by a factor of 1.25,reflected over the x-axis22c. compressed horizontally by a factor of 3 5 ,translated up 2 units23a. translated left 2 units, compressed vertically bya factor of 1 3 23b. reflected over the y-axis, translated down 7 units23c. expanded vertically by a factor of 2, translatedright 3 units and up 4 units24a. expanded horizontally by a factor of 224b. compressed horizontally by a factor of 1 6 ,translated 8 units up24c. The portion of parent graph on the left of the y-axis is replaced by a reflection of the portion onthe right of the y-axis.25a. compressed horizontally by a factor of 2 5 ,translated down 3 units25b. reflected over the y-axis, compressed verticallyby a factor of 0.7525c. The portion of the parent graph on the left of they-axis is replaced by a reflection of the portion onthe right of the y-axis. The new image is thentranslated 4 units right.26. y x 2yPages 143–145 Exercises13. The graph of g(x) is a translation of the graph off(x) up 6 units.14. The graph of g(x) is the graph of f(x) compressedvertically by a factor of 3 4 .15. The graph of g(x) is the graph of f(x) compressedhorizontally by a factor of 1 5 .16. The graph of g(x) is a translation of f(x) right 5units.17. The graph of g(x) is the graph of f(x) expandedvertically by a factor of 3.18. The graph of g(x) is the graph of f(x) reflected overthe x-axis.19. The graph of g(x) is the graph of f(x) reflected overthe x-axis, expanded horizontally by a factor of2.5, and translated up 3 units.0.2527. y x 3 428. y29.O xOyOxx69 Chapter 3
30. y31.y36a. 0OxOx32. y33.Ox4y4O 4 8x4812[7.6, 7.6] scl:1 by [5, 5] scl:136b. 0.6634.8y844 O4y f(|x|)4x[7.6, 7.6] scl:1 by [5, 5] scl:136c. 0.25y f(x)835a. 0[7.6, 7.6] scl:1 by [5, 5] scl:137a. 0[7.6, 7.6] scl:1 by [5, 5] scl:135b. 0.5[7.6, 7.6] scl:1 by [5, 5] scl:137b. 2.5[7.6, 7.6] scl:1 by [5, 5] scl:135c. 1.5[7.6, 7.6] scl:1 by [5.5] scl:1[7.6, 7.6] scl:1 by [5, 5] scl:1Chapter 3 70
37c. 0.642a. (1) y x 2 (2) y x 3(3) y x 2 (4) y x 342b. (1) y(2) y44 O44 8x4 O484 8x[7.6, 7.6] scl:1 by [5, 5] scl:138a. The graph would continually move left 2 unitsand down 3 units.38b. The graph would continually be reflected overthe x-axis and moved right 1 unit.b39. The x-intercept will be . a0.25[[x 1]] 1.50 if [[x]] x40a. y 0.25[[x]] 1.50 if [[x]] x40b.xy0 x 1 1.501 x 2 1.752 x 3 2.003 x 4 2.254 x 5 2.50Price(dollars)41a.yA 1 2 bh8 1 2 (10)(5)4 25 units 2O 4 8 12x41b. The area of the triangle is A 1 2 (10)(10) or50 units 2 . Its area is twice as large as that of theoriginal triangle. The area of the triangle formedby y c f(x) would be 25c units 2 .y84O 4 8 12x41c. The area of the triangle is A 1 2 (10)(5) 25 units 2 . Its area is the same as that of theoriginal triangle. The area of the triangle formedby y f(x c) would be 25 units 2 .y84O 4 8 12x54321001 2 3 4Fare Units5(3) y(4)4 O 4 8x4812y4 O 4 8 x442c. (1) y (x 3) 2 5 (2) y (x 3) 3 5(3) y (x 3) 2 5 (4) y (x 3) 3 543a. reflection over the x-axis, reflection over they-axis, vertical translation, horizontalcompression or expansion, and verticalexpansion or compression43b. horizontal translation44. f(x) x 17 x 15f(x) (x) 17 (x) 15 f(x) (x 17 x 15 )f(x) x 17 x 15 f(x) x 17 x 15yes; f(x) f(x)45. Let x number of preschoolers.Let y number of school-age children.x y 50x 3(10)y 5(10)y60 (0, 50) y 50x 040y 0x 30(30, 20)20(30, 0)(0, 0)x y 50O 20 40 60 xy 0I(x, y) 18x 6yI(0, 0) 18(0) 6(0) or 0I(0, 50) 18(0) 6(50) or 300I(30, 20) 18(30) 6(20) or 660I(30, 0) 18(30) 6(0) or 54030 preschoolers and 20 school-age46.0 1 5 1 2 4 3 11 0 4 3 1 51 2A(4, 5), B(3, 1), C(1, 2)47. x 2 25 9 y 12 2zx 56 z48. 5(6x 5y) 5(14) 30x 25y 70→6(5x 2y) 6(3) 30x 12y 1813y 52y 45x 2y 35x 2(4) 3x 1 (1, 4)49. The graph implies a negative linear relationship.50. 3x 4y 0 → y 3 4 xperpendicular slope: 4 3 871 Chapter 3
51. 5d 2p 500 → p 5 2 d 25025052. [f g](x) f(g(x)) f(x 2 6x 9) 2 3 (x 2 6x 9) 2 2 3 x 2 4x 4[g f](x) g(f(x)) g 2 3 x 2 2 3 x 2 2 6 2 3 x 2 9 4 9 x 2 8 3 x 4 4x 12 9 4 9 x 2 2 03x 2553. If m 1; d 1 5 01or 49.If m 10; d 10 5 010 or 5.If m 50; d 50 5 050 or 49.50If m 100; d 100 1 00or 99.550If m 1000; d 1000 1 000 or 999.95.The correct choice is A.3-3Graphs of Nonlinear InequalitiesPage 149 Check for Understanding1. Aknowledge of transformations can helpdetermine the graph of the boundary of theshaded region, y 5 x. 22. When solving a one variable inequalityalgebraically, you must consider the case wherethe quantity inside the absolute value is nonnegativeand the case where the quantity insidethe absolute value is negative.3. Sample answer: Pick a point not on the boundaryof the inequality, and test to see whether it is asolution to the inequality. If that point is asolution, shade all points in that region. If it is nota solution to the inequality, test a point on theother side of the boundary and shade accordingly.4. This inequality has no solution since the twographs do not intersectyO5. y 5x 4 7x 3 83 ? 5(1) 4 7(1) 3 83 4; yes6. y ⏐3x 4⏐ 13 ? ⏐3(0) 4⏐ 13 3; nox7. y8.9.yOOxx10. Case 1 Case 2⏐x 6⏐ 4 ⏐x 6⏐ 4(x 6) 4 x 6 4x 6 4x 2x 10x 10{x⏐x 10 or x 2}11. Case 1 Case 2⏐3x 4⏐ x⏐3x 4⏐ x(3x 4) x3x 4 x3x 4 x 2x 44x 4 x 2x 1{x⏐ 1 x 2}12a. ⏐x 12⏐ 0.00512b. Case 1 Case 2⏐x 12⏐ 0.005 ⏐x 12⏐ 0.005(x 12) 0.005 x 12 0.005x 12 0.005 x 12.005x 11.995x 11.99512.005 cm, 11.995 cmPages 150–151 Exercises13. y x 3 4x 2 2 14. y ⏐x 2⏐ 70 ? (1) 3 4(1) 2 2 8 ? ⏐3 2⏐ 70 1; no 8 8; no15. y x 11 11 ? (2) 11 11 2; yes16. y 0.2x 2 9x 763 ? 0.2(10) 2 9(10) 763 63; no17. y x2 6x9 ? (6 ) 2 6 6 9 5; yes18. y 2⏐x⏐ 3 70 ? 2⏐0⏐ 3 70 7; yesyOxChapter 3 72
19. y x 2 y x 20 ? 0 2 4 ? 1 20 2; yes 4 3; noy x 2 y x 21 ? 1 2 0 ? 1 21 3; yes 0 i 2; noy x 21 ? 1 21 3; yes(0, 0) (1, 1) and (1, 1); if these points are in theshaded region and the other points are not, thenthe graph is correct.20. y21. y30. y31.O x32. yOxyOxO x22. y23.1284O 4 8 12 x24. y25.Ox26. y27.O x28. y29.O xOyOyOyOyOxxxxx33. Case 1 Case 2⏐x 4⏐ 5 ⏐x 4⏐ 5(x 4) 5 x 4 5x 4 5 x 1x 9x 9{x⏐x 9 or x 1}34. Case 1 Case 2⏐3x 12⏐ 42 ⏐3x 12⏐ 42(3x 12) 42 3x 12 423x 12 42 3x 303x 54 x 10x 18{x⏐x 18 or x 10}35. Case 1 Case 2⏐7 2x⏐ 8 3 ⏐7 2x⏐ 8 3(7 2x) 8 3 7 2x 8 37 2x 8 3 2x 42x 18x 2x 9{x⏐2 x 9}36. Case 1 Case 2⏐5 x⏐ x⏐5 x⏐ x(5 x) x 5 x x5 x x2x 55 0; true x 2.5{x⏐x 2.5}37. Case 1 Case 2⏐5x 8⏐ 0 ⏐5x 8⏐ 0(5x 8) 0 5x 8 05x 8 0 5x 85x 8x 8 5 x 8 5 no solution38. Case 1 Case 2⏐2x 9⏐ 2x 0 ⏐2x 9⏐ 2x 0(2x 9) 2x 0 2x 9 2x 02x 9 2x 09 0; true4x 9x 9 4 all real numbers73 Chapter 3
39. Case 1 Case 2 2 3 ⏐x 5⏐ 8 2 3 ⏐x 5⏐ 8 2 3 ((x 5)) 8 2 3 (x 5) 8 2 3 (x 5) 8 2 3 x 1 0380 2 3 x 1 38 2 3 x 1 3 2 3 x 3 43x 7x 17{x⏐17 x 7}40. ⏐x 37.5⏐ 1.2Case 1 Case 2⏐x 37.5⏐ 1.2 ⏐x 37.5⏐ 1.2(x 37.5) 1.2 x 37.5 1.2x 37.5 1.2 x 38.7x 36.3x 36.336.3 x 38.741. Case 1 Case 23⏐x 7 ⏐ x 1 3⏐x 7⏐ ⏐x 1⏐3((x 7)) x 1 3(x 7) x 13(x 7) x 1 3x 21 x 13x 21 x 1 2x 204x 22 x 10x 5.5{x⏐5.5 x 10}42. 30 units 2The triangular region has vertices A(0, 10),B(3, 4), and C(8, 14). The slope of side AB is 2.The slope of side AC is 0.5, therefore AB isperpendicular to AC. The length of side AB is35. The length of side AC is 45 the area ofthe triangle is 0.5(35)(45) or 30.43. 0.10(90) 0.15(75) 0.20(76) 0.40(80) 0.15(x) 800.15x 12.55x 83 2 3 44.[1, 8] scl:1 by [1, 8] scl:144a. b 044b. none44c. b 0 or b 444d. b 444e. 0 b 445a. P(x)400300200100445b. The shaded region shows all points (x, y) wherex represents the number of cookies sold and yrepresents the possible profit made for a givenweek.46. The graph of g(x) is the graph of f(x) reflected overthe x-axis and expanded vertically by a factor of 2.147. y a14 → b a 4x-axis (b) a14b a141y-axis b (a) 4nob a14 yes1y x (a) (b ) 4 1a (bno) 4 1y x (a) ( b) 41a (bno) 4y-axis1 5 3 5 34 8 8 4 848. 21 8 3 4 549. 3 4 3 4 (8) 3 4 (7)8 74 0 3 4 (4) 3 4 (0)50.51.x f(x)2 111 80 51 82 11StateswithTeenCourts6 21 4 3 0504540353025201510500 1980 1990 2000 2010Year52. [f g](4) f(g(4)) f(0.5(4) 1) f(1) 5(1) 9 14[g f ](4) g(f(4)) g(5(4) 9) g(29) 0.5(29) 1 13.5f(x)OxO9001000 1100 1200 1300 1400 1500xChapter 3 74
53. Student A 15Student B 1 3 (15) 15 or 20Let x number of years past.20 x 2(15 x)20 x 30 2xx 10Page 151 Mid-Chapter Quiz1. x 2 y 2 9 0 → a 2 b 2 9 0x-axis a 2 (b) 2 9 0a 2 b 2 9 0 yesy-axis (a) 2 b 2 9 0a 2 b 2 9 0 yesy x (b) 2 (a) 2 9 0a 2 b 2 9 0 yesy x (b) 2 (a) 2 9 0a 2 b 2 9 0 yesx 2 y 2 9 0 → f(x) x 2 9f(x) (x) 2 9 f(x) (x 2 9)f(x) x 2 9 f(x) x 2 9yesx-axis, y-axis, y x, y x, origin2. 5x 2 6x 9 y → 5a 2 6a 9 bx-axis5a 2 6a 9 (b)5a 2 6a 9 bnoy-axis5(a) 2 6(a) 9 b5a 2 6a 9 b noy x5(b) 2 6(b) 9 (a)5b 2 6b 9 a noy x5(b) 2 6(b) 9 (a)5b 2 6b 9 ano5x 2 6x 9 y → f(x) 5x 2 6x 9f(x) 5(x) 2 6(x) 9 5x 2 6x 9f(x) (5x 2 6x 9)f(x) 5x 2 6x 9 nonone of these3. x 7 y → a 7 b 7x-axis a ( b)a 7 b noy-axis(a) 7 b a 7 b no7y x (b) (a )a 7 b yes7y x (b) ( a)a 7 b yesx 7 y → f(x) 7 x 7f(x) ( x)f(x) 7 x f(x) 7 x f(x) 7 x yesy x, y x, origin4. y ⏐x⏐ 1 → b ⏐a⏐ 1x-axis (b) ⏐a⏐ 1b ⏐a⏐ 1 noy-axis b ⏐(a)⏐ 1b ⏐a⏐ 1 yesy x (a) ⏐(b)⏐ 1a ⏐b⏐ 1 noy x (a) ⏐(b)⏐ 1a ⏐b⏐ 1 noy ⏐x⏐ 1 → f(x) ⏐x⏐ 1f(x) ⏐ x⏐ 1 f(x) (⏐x⏐ 1)f(x) ⏐x⏐ 1f(x) ⏐x⏐ 1 noy-axis5a. translated down 2 units5b. reflected over the x-axis, translated right 3 units5c. compressed vertically by a factor of 1 4 , translatedup 1 unit6a. expanded vertically by a factor of 36b. expanded horizontally by a factor of 2 andtranslated down 1 unit6c. translated left 1 unit and up 4 units7. y8.O9. Case 1 Case 2⏐2x 7⏐ 15 ⏐2x 7⏐ 15(2x 7) 15 2x 7 152x 7 15 2x 222x 8 x 11x 44 x 1110. ⏐x 64⏐ 3Case 1 Case 2⏐x 64⏐ 3 ⏐x 64⏐ 3(x 64) 3 x 64 3x 64 3 x 67x 61x 6161 x 673-4xInverse Functions and RelationsPages 155–156 Check for Understanding1. Sample answer: First, let y f(x). Theninterchange x and y. Finally, solve the resultingequation for y.2. n is odd3. Sample answer: f(x) x 24. Sample answer: If you draw a horizontal linethrough the graph of the function and it intersectsthe graph more than once, then the inverse is nota function.yOx75 Chapter 3
5. She is wrong. The inverse is f 1 (x) (x 3) 2 2,which is a function.6.7.f(x) ⏐x⏐ 1x f(x)2 31 20 11 22 3f(x)Of(x)f(x) x 3 1x f(x)2 71 00 11 22 9f(x)f(x)f 1 (x)xf 1 (x)x f 1 (x)3 22 11 02 13 2f 1 (x)x f 1 (x)7 20 11 02 19 29. f(x) 3x 2y 3x 2x 3y 2x 2 3yy 1 3 x 2 3 f 1 (x) 1 3 x 2 3 ; f 1 (x) is a function.110. f(x) x 3y x13x y13y 3 1 x y 3 1 x f 1 (x) 13 xor13x; f 1 (x) is a function.11. f(x) (x 2) 2 6y (x 2) 2 6x (y 2) 2 6x 6 (y 2) 2x 6 y 2y 2 x 6f 1 (x) 2 x; 6 f 1 (x) is not a function.12. Reflect the graph of y x 2 over the line y x.Then, translate the new graph 1 unit to the leftand up 3 units.yf 1 (x)OxOx8.f(x) (x 3) 2 1x f(x)1 32 03 14 05 3f 1 (x)f(x)f 1 (x)x f 1 (x)3 10 21 30 43 513. f(x) 1 2 x 5y 1 2 x 5x 1 2 y 5x 5 1 2 yy 2x 10f 1 (x) 2x 10[f f 1 ](x) f(2x 10) 1 2 (2x 10) 5 x[f 1 f ](x) f 1 1 2 x 5Of(x)x 2 1 2 x 5 10 xSince [f f 1 ](x) [f 1 f ](x) x, f and f 1 areinverse functions.14a. B(r) 1000(1 r) 3B 1000(1 r) 3B 10 00 (1 r) 3B 3 10 00 1 rr 1 3 B 10Chapter 3 76
14b. r 1 1 or 0.0323; 3.23%Pages 156–15815. 3 B 3 1100 10f(x) ⏐x⏐ 2x f(x)2 41 30 21 32 4Exercisesf 1 (x)x f 1 (x)4 23 12 03 14 218.f(x) x 5 10x f(x)2 421 110 101 92 22f(x)1010 O10f (x)f 1 (x)10xf 1 (x)x f 1 (x)42 211 110 09 122 216.f(x)f (x)Of(x) ⏐2x⏐x f(x)2 41 20 01 22 4f (x)f(x)f 1 (x)xf 1 (x)f 1 (x)x f 1 (x)4 22 10 02 14 219.f(x) [x]xf(x)2 x 1 21 x 0 10 x 1 01 x 2 12 x 3 2f 1 (x)x f 1 (x)2 2 x 11 1 x 00 0 x 11 1 x 22 2 x 3f(x)f 1 (x)OxOf(x)x17.f(x) x 3 2x f(x)2 101 30 21 12 6f(x)f(x) 1077 Chapter 3f 1 (x)x f 1 (x)10 23 12 01 16 220.f(x) 3x f(x)2 31 30 31 32 3f 1 (x)x f 1 (x)3 23 13 03 13 2f 1 (x)f(x)f(x)OxOxf 1 (x)
21.22.23.f(x) x 2 2x 4x f(x)3 72 41 30 41 7f(x)84f (x)f 1 (x)8 4 O 4 8x48f(x) (x 2) 2 5x f(x)4 93 62 51 60 9f 1 (x)f(x)f(x)Of(x) (x 1) 2 4x f(x)4 52 31 40 32 5f(x)Of(x)xf 1 (x)f 1 (x)x f 1 (x)7 34 23 14 07 1f 1 (x)x f 1 (x)9 46 35 26 19 0f 1 (x)x f 1 (x)5 43 24 13 05 224. f(x) x 2 4y x 2 4x y 2 4x 4 y 2y x ; 4 f 1 (x) x 4f(x) x 2 4x f(x)2 81 50 41 52 8f(x)25. f(x) 2x 7y 2x 7x 2y 7x 7 2yy x 72f 1 (x) x 72; f 1 (x) is a function.26. f(x) x 2y x 2x y 2y x 2f 1 (x) x 2; f 1 (x) is a function.27. f(x) 1 x y 1 x x 1 y y 1 x 44 O4f 1 (x) 1 x ; f 1 (x) is a function.128. f(x) x 2y x12x y12f(x)4f 1 (x)xf 1 (x) x 4x f 1 (x)8 25 14 0y 2 1 x y 1 x f 1 (x) ; 1 x f 1 (x) is not a function.29. f(x) (x 3) 2 7y (x 3) 2 7x (y 3) 2 7x 7 (y 3) 2x 7 y 3y 3 x 7f 1 (x) 3 x ; 7 f 1 (x) is not a function.Chapter 3 78
30. f(x) x 2 4x 3y x 2 4x 3x y 2 4y 3x 1 y 2 4y 4x 1 (y 2) 2x 1 y 2y 2 x 1f 1 (x) 2 x; 1 f 1 (x) is not a function.131. f(x) x 21y x 21x y y 2 1 x 2y 1 x 2f 1 (x) 1 x 2; f 1 (x) is not a function.132. f(x) (x 1) 21y (x 1) 235. Reflect the part of the graph of x 2 that lies in thefirst quadrant about y x. Then, translate 5 unitsto the left.36. Reflect the graph of x 2 about y x. Then,translate 2 units to the right and up 1 unit.f(x)Of(x)Oxx1x (y 1) 2(y 1) 2 1 x y 1 1 x y 1 f 1 (x) 1 1x1x; f 1 (x) is not a function.37. Reflect the graph of x 3 about y x to obtain thegraph of 3 x. Reflect the graph of 3 x about thex-axis. Then, translate 3 units to the left anddown 2 units.f(x)233. f(x) (x 2) 32y (x 2) 32x (y 2) 3Ox(y 2) 3 2 x y 2 3 2 x y 2 3 2 x f 1 (x) 2 3 ; 2 x f 1 (x) is not a function.34. g(x) x 2 y x 2 x y 2 y 2 2y 3 x 32x32x32yy 2 2y 1 3 x 1(y 1) 2 3 x 1y 1 3 x 1y 1 3 x 1g 1 (x) 1 3 x 138. Reflect the graph of x 5 about y x. Then,translate 4 units to the right. Finally, stretch thetranslated graph vertically by a factor of 2.Of(x)x79 Chapter 3
39. f(x) 2 3 x 1 6 y 2 3 x 1 6 x 2 3 y 1 6 x 1 6 2 3 yy 3 2 x 1 4 f 1 (x) 3 2 x 1 4 [f f 1 ](x) f 3 2 x 1 4 2 3 3 2 x 1 4 1 6 x 1 6 1 6 x[f 1 f ](x) f 1 2 3 x 1 6 3 2 2 3 x 1 6 1 4 x 1 4 1 4 xSince [f f 1 ](x) [f 1 f ](x) x, f and f 1 areinverse functions.40. f(x) (x 3) 3 4y (x 3) 3 4x (y 3) 3 4x 4 (y 3) 3 3 x 4 y 3y 3 3 x 4f 1 (x) 3 3 x 4[f f 1 ](x) f(3 3 x) 4 [(3 3 x) 4 3] 3 4 x 4 4 x[f 1 f ](x) f 1 [(x 3) 3 4] 3 3[(x 3) 4] 3 4 3 x 3 xSince [f f 1 ](x) [f 1 f ](x) x, f and f 1 areinverse functions.41a.d(x) ⏐x 4⏐d 1 (x)x d(x) x d 1 (x)6 2 2 65 1 1 54 0 0 43 1 1 32 2 2 2d 1 (x)O41b. No; the graph of d(x) fails the horizontal line test.41c. d 1 (x) gives the numbers that are 4 units from xon the number line. There are always two suchnumbers, so d 1 associates two values with eachx-value. Hence, d 1 (x) is not a function.x42a. v 2gh v 2 2ghv2h 2 gv2h 2( 32)v2h 6443a. Sample answer: y x43b. The graph of the function must be symmetricabout the line y x.43c. Yes, because the line y x is the axis ofsymmetry and the reflection line.44a.xC(x)0 x 1 $0.101 x 2 $0.202 x 3 $0.303 x 4 $0.404 x 5 $0.5044b. positive real numbers; positive multiples of 1044c.x C 1 (x)$0.10 0 x 1$0.20 1 x 2$0.30 2 x 3$0.40 3 x 4$0.50 4 x 544d. positive multiples of 10; positive real numbers44e. C 1 (x) gives the possible lengths of phone callsthat cost x.45. It must be translated up 6 units and 5 units to theleft; y (x 6) 2 5; y 6 x. 546a. KE 1 2 mv 22KE mv 2 2K Em v 2v 2K Em42b. h 6vC(x)$0.80$0.70$0.60$0.50$0.40$0.30$0.20$0.101 2 3 4 5 6 7 8x46c. There are always two velocities.47a. Yes; if the encoded message is not unique, it maynot be decoded properly.47b. The inverse of the encoding function must be afunction so that the encoded message may bedecoded.47c. C(x) 2 x 3y 2 x 3x 2 y 3x 2 y 3(x 2) 2 y 3y (x 2) 2 3C 1 (x) (x 2) 2 3O87654321O245) 24h (7 6h 87.89Yes. The pump can propelwater to a height of about88 ft.C 1 (x)20¢ x40¢ 60¢ 80¢46b. v 2K Emv 2(1 5)1v 5.477; 5.5 m/secChapter 3 80
47d. C 1 (x) (x 2) 2 3C 1 (1) (1 2) 2 3 or 6, FC 1 (2.899) (2.899 2) 2 3 or 21, UC 1 (2.123) (2.123 2) 2 3 or 14, NC 1 (0.449) (0.449 2) 2 3 or 3, CC 1 (2.796) (2.796 2) 2 3 or 20, TC 1 (1.464) (1.464 2) 2 3 or 9, IC 1 (2.243) (2.243 2) 2 3 or 15, OC 1 (2.123) (2.123 2) 2 3 or 14, NC 1 (2.690) (2.690 2) 2 3 or 19, SC 1 (0) (0 2) 2 3 or 1, AC 1 (2.583) (2.583 2) 2 3 or 18, RC 1 (0.828) (0.828 2) 2 3 or 5, EC 1 (1) (1 2) 2 3 or 6, FC 1 (2.899) (2.899 2) 2 3 or 21, UC 1 (2.123) (2.123 2) 2 3 or 14, NFUNCTIONS ARE FUN48. Case 1 Case 2⏐2x 4⏐ 6 ⏐2x 4⏐ 6(2x 4) 6 2x 4 62x 4 6 2x 22x 10 x 1x 5{x⏐5 x 1}49. both50a. a 0, b 0, 4a b 32, a 6b 5450b.a 0b a 6b 54(0, 9)O (0, 0) b 01 4 2 1 1(6, 8)4a b 32(8, 0)aG(a, b) a bG(0, 0) 0 0 or 0G(0, 9) 0 9 or 9G(6, 8) 6 8 or 14G(8, 0) 8 0 or 814 gallons51. 4x 2y 10→4x 2y 10y 6 x x y 64 2 x 101 1 y 61 2 1 2 1 2 141 1 2 1 2 4 2 x 1 2 1 2 101 4 1 1 y 1 4 6x 1 (1, 7)52. 1 2 1 9 31 2 (3)2 (9)6 6 1 1 2 (6)2 (6) 9 2 3y 3 2 37453.Oy54. 1 4 4; 1 4 4 1; neither 7 0 5555. m 2 5x y y 1 m(x x 1 ) or 1 y 7 1(x 0)y x 756. b c 180If PQ is perpendicular to QR, then m∠PQR 90.Since the angles of a triangle total 180,a d 90 180.a d 90a b c d 180 90 or 270The correct choice is C.3-5Continuity and End BehaviorPage 165 Check for Understanding1. Sample answer: The function approaches 1 as xapproaches 2 from the left, but the functionapproaches 4 as x approaches 2 from the right.This means the function fails the second conditionin the continuity test.2.a n n x → p(x) →positive even positive even positive odd positive odd a n n x → p(x) →negative even negative even negative odd negative odd 3. Infinite discontinuity; f(x) → as x → ,f(x) → as x → .4. f(x) x 2 is decreasing for x 0 and increasing forx 0, g(x) x 2 is increasing for x 0 anddecreasing for x 0. Reflecting a graph switchesthe monotonicity. In other words, if f(x) isincreasing, the reflection will be decreasing andvice versa.5. No; y is undefined when x 3.6. No; f(x) approaches 6 as x approaches 2 from theleft but f(x) approaches 6 as x approaches 2from the right.81 Chapter 3
7. a n : positive, n: oddy → as x → , y → as x → .8. a n : negative, n: eveny → as x → , y → as x → .9.f(x)OOdecreasing for x 3; increasing for x 310. yxxdecreasing for x 1 and x 1; increasing for1 x 111a. t 4 11b. when t 4 11c. 10 ampsPages 166–168 Exercises12. Yes; the function is defined when x 1; thefunction approaches 3 as x approaches 1 fromboth sides; and y 3 when x 1.13. No; the function is undefined when x 2.14. Yes; the function is defined when x 3; thefunction approaches 0 as x approaches 3 fromboth sides; and f(3) 0.15. Yes; the function is defined when x 3; thefunction approaches 1 (in fact is equal to 1) as xapproaches 3 from both sides; and y 1 whenx 3.16. No; f(x) approaches 7 as x approaches 4 fromthe left, but f(x) approaches 6 as x approaches 4from the right.17. Yes; the function is defined when x 1; f(x)approaches 3 as x approaches 1 from both sides;and f(1) 3.18. jump discontinuity19. Sample answer: x 0; g(x) is undefined whenx 0.20. a n : positive, n: oddy → as x → , y → as x → .21. a n : negative, n: eveny → as x → , y → as x → .22. a n : positive, n: eveny → as x → , y → as x → .23. a n : positive, n: eveny → as x → , y → as x → .24.y x12xy10,000 1 10 81000 1 10 6100 1 10 410 0.010 undefined10 0.01100 1 10 41000 1 10 610,000 1 10 8y → 0 as x → , y → 0 as x → .25.1f(x) x 3 2xf(x)10,000 21000 2.000000001100 2.00000110 2.0010 undefined10 1.999100 1.9999991000 1.99999999910,000 2f(x) → 2 as x → , f(x) → 2 as x → .26.[6, 6] scl:1 by [30, 30] scl:5increasing for x 3 and x 1; decreasing for3 x 127.[7.6, 7.6] scl:1 by [5, 5] scl:1decreasing for all xChapter 3 82
28.33b. Since f is odd, its graph must be symmetric withrespect to the origin. Therefore, f is increasingfor 2 x 0 and decreasing for x 2. f musthave a jump discontinuity when x 3 andf(x) → as x → .f(x)[7.6, 7.6] scl:1 by [8, 2] scl:1decreasing for x 1 and x 129.Ox34a. polynomial34b.[25, 25] scl:5 by [25, 25] scl:5increasing for x 1 and x 5; decreasing for1 x 2 and 2 x 530.[7.6, 7.6] scl:1 by [1, 9] scl:1decreasing for x 2 and 0 x 2; increasingfor 2 x 0 and x 231.[5, 80] scl:10 by [500, 12000] scl:10000.5 t 39.534c. 0 t 0.5 and t 39.535a. 1954-1956, 1960-1961, 1962-1963, 1966-1968,1973-1974, 1975-1976, 1977-1978, 1989-1991,1995-199735b. 1956-1960, 1961-1962, 1963-1966, 1968-1973,1974-1975, 1976-1977, 1978-1989, 1991-1995,1997-200436a.[7.6, 7.6] scl:1 by [1, 9] scl:1decreasing for x 3 2 and 0 x 3 2 ; increasingfor 3 2 x 0 and x 3 2 32. As the denominator, r, gets larger, the value ofU(r) gets smaller. U(r) approaches 0.33a. Since f is even, its graph must be symmetricwith respect to the y-axis. Therefore, f isdecreasing for 2 x 0 and increasing forx 2. f must have a jump discontinuity whenx 3 and f(x) → as x → .f(x)Ox[1, 8] scl:1 by [10, 1] scl:1x 436b. Answers will vary.36c. The slope is positive. In an interval where afunction is increasing, for any two points on thegraph, the x- and y-coordinates of one point willbe greater than that of the other point, ensuringthat the slope of the line through the two pointswill be positive.36d. See graph in 36a. x 436e. The slope is negative; see students’ work.37a. The function has to be monotonic. If the functionwere increasing on one interval and decreasingon another interval, the function could not passthe horizontal line test.83 Chapter 3
37b. The inverse must be monotonic. If the inversewere increasing on one interval and decreasingon another interval, the inverse would fail thehorizontal line test. That would mean thefunction fails the vertical line test, which isimpossible.38a.PercentwithSimilarComputerUsage40353025201510500 2 4 6 8 1012141638b. 0 x 1, 1 x 2, 2 x 4, 4 x 6,6 x 8, x 839. For the function to be continuous at 2, bx a andx 2 a must approach the same value as xapproaches 2 from the left and right, respectively.Plugging in x 2 to find that common value gives2b a 4 a. Solving for b gives b 2. For thefunction to be continuous at 2, b x andbx a must approach the same value as xapproaches 2 from the left and right,respectively. Plugging in x 2 gives b 2 2b a. We already know b 2, so theequation becomes 0 4 a. Hence, a 4.40. f(x) (x 5) 2y (x 5) 2x (y 5) 2x y 5y 5 xf 1 (x) 5 x41. The graph of g(x) is the graph of f(x) translatedleft 2 units and down 4 units.42. f (x, y) x 2yf(0, 0) 0 2(0) or 0f(4, 0) 4 2(0) or 4f(3, 5) 3 2(5) or 13f(0, 5) 0 2(5) or 1013, 043. 5 4 5(2) 8(4) or 428 244a. c 47.5h 3544b. c 47.5h 35c 47.5 2 1 4 35c $141.87545. f(x) 2x 2 2x 8f(2) 2(2) 2 2(2) 8 8 4 8 or 2046. The volume of the cube is x 3 .The volume of the other box isx(x 1)(x 1) x(x 2 1) or x 3 x.The difference between the volumes of the twoboxes is x 3 (x 3 x) or x.The correct choice is A.3-5BGap DiscontinuitiesPage 1701. {all real numbers x⏐x 3}[10, 10] scl:1 by [6, 50] scl:102. {all real numbers x⏐2 x 4}[9.4, 9.4] scl:1 by [6.2, 6.2] scl:13. {all real numbers x⏐x 3 or x 1}[18.8, 18.8] scl:1 by [12.4, 12.4] scl:14. {all real numbers x⏐x 3 or x 2}[4.7, 4.7] scl:1 by [25, 25] scl:105. {all real numbers x⏐x 1 or x 1}[4.7, 4.7] scl:1 by [3.1, 3.1] scl:1Chapter 3 84
6. {all real numbers x⏐x 6 or x 2}[9.4, 9.4] scl:1 by [6.2, 6.2] scl:17. {all real numbers x⏐x 3 or x 4}12. Yes; sample justification: if f(x) is a polynomialfunction, then the graph of y is like the graph of f(x), but with an infinitenumber of “interval bites” removed.13. Yes; sample justification: the equation y x 2 (x 2) (2x 4)(x 4)((x 2) or (x 4))the function described.14a.f(x)(⏐x [[x]]⏐ 0.25)is a possible equation for[15, 15] scl:2 by [10, 20] scl:214b.[9.4, 9.4] scl:1 by [3, 9.4] scl:18. {all real numbers x⏐x 2 or 1 x 1or x 2}[9.1, 9.1] scl:1 by [6, 6] scl:1[4.7, 4.7] scl:1 by [2, 8] scl:19. {all real numbers x⏐x 1 or 2 x 3 or x 4}[3, 6.4] scl:1 by [2, 8] scl:110. {all real numbers x⏐x 1 or x 2}[9.4, 9.4] scl:1 by [6.2, 6.2] scl:111. Sample answer: y x 2((x 2) or ((x 5) and (x 7)) or x 8))3-6Critical Points and ExtremaPage 176 Check for Understanding1. Check values of the function at x-values very closeto the critical point. Be sure to check values onboth sides. If the function values change fromincreasing to decreasing, the critical point is amaximum. If the function values change fromdecreasing to increasing, the critical point is aminimum. If the function values continue toincrease or to decrease, the critical point is a pointof inflection.2. rel. min.;f(0.99) 3.9997f(1) 4f(1.01) 3.9997By testing points on either side of the criticalpoint, it is evident that the graph of the functionis decreasing as x approaches 1 from the left andincreasing as x moves away from 1 to the right.Therefore, on the interval 0.99 x 1.01, (1, 4) isa relative minimum.85 Chapter 3
3. Sample answer:y(4, 6)12c. $58.80 per acre12d. Rain or other bad weather could delay harvestand/or destroy part of the crop.(3, 1)O(0, 4)4. rel. min.: (3, 2); rel. max.: (1, 6)5. rel. min.: (1, 3); rel. max.: (3, 3)6. rel. max.: (0, 0); rel. min.: (2, 16)xPages 177–179 Exercises13. abs. max.: (4, 1)14. abs. max.: (1, 3); rel. min.: (0.5; 0.5);rel. max.: (1.5, 2)15. rel. max.: (2, 7): abs. min.: (3, 3)16. rel. max.: (6, 4), rel. min.: (2, 3)17. abs. min.: (3, 8); rel. max.: (5, 2);rel. min.: (8, 5)18. no extrema19. abs. max.: (1.5, 1.75)[4, 6] scl:1 by [20, 20] scl:57. rel. min.: (2.25, 10.54)[5, 5] scl:1 by [8, 2] scl:120. rel. max.: (1.53, 1.13);rel. min.: (1.53, 13.13)[6, 4] scl:1 by [14, 6] scl:28. f(1.1) 0.907f(1) 1f(0.9) 0.913 max.9. f(2.6) 12.24f(2.5) 12.25f(2.4) 12.24min.10. f(0.1) 0.00199f(0) 0f(0.1) 0.00199 pt. of inflection11. f(0.1) 0.97f(0) 1f(0.1) 0.97 min.12a. P(x) (120 10x)(0.48 0.03x)Profit(dollars)P(x)8070605040302012b. 2 weeks10x00 2 4 6 8 1012141618Time (weeks)[5, 5] scl:1 by [16, 4] scl:221. rel. max.: (0.59, 0.07), rel. min.: (0.47, 3.51)[5, 5] scl:1 by [5, 5] scl:122. abs. min.: (1.41, 6), (1.41, 6);rel. max.: (0, 2)[5, 5] scl:1 by [8, 2] scl:1Chapter 3 86
23. rel. max.: (1, 1); rel. min.: (0.25, 3.25)35b.[5, 5] scl:1 by [5, 5] scl:124. no extrema[1, 12] scl:1 by [200, 500] scl:1002.37 cm by 2.37 cm35c. See students’ work.36a. P sd 25d s(200s 15,000) 25(200s 15,000)200s 2 20,000s 375,000[5, 5] scl:1 by [5, 5] scl:125. abs. min.: (3.18, 15.47); rel. min. (0.34, 0.80);rel. max.: (0.91, 3.04)[5, 5] scl:1 by [16, 5] scl:226. f(0.1) 0.001f(0) 0f(0.1) 0.001 pt of inflection27. f(3.9) 5.99f(4) 6f(4.1) 5.99 max.28. f(2.6) 19.48f(2.5) 19.5f(2.4) 19.48min.29. f(0.1) 6.98f(0) 7f(0.1) 6.98 max.30. f(1.9) 3.96f(2) 4.82f(2.1) 3.96 min.31. f(2.9) 0.001f(3) 0f(3.1) 0.001 pt. of inflection32. f(2.1) 4.32f(2) 4.53f(1.9) 4.32 max.33. f(0.57) 2.86f 2 3 2.85f(0.77) 2.86min.34. The point of inflection is now at x 6 and thereis now a minimum at x 3.35a. V(x) 2x(12.5 2x)(17 2x)[0, 100] scl:10 by [0, 130,000] scl:10,000abs. max.: (50, 125,000)$5036b. Sample answer: The company’s competitionmight offer a similar product at a lower cost.37a. AM 2 MB 2 AB 2AM 2 x 2 2 2AM x2 4f(x) 5000(x) 2 4 3500(10 x)37b.[10, 20] scl:2 by [0, 60,000] scl:10,000abs. min.: (1.96, 42,141.4)1.96 km from point B38. equations of the form y x n or y n x, where n isodd39.[3, 7] scl:1 by [50, 10] scl:10The particle is at rest when t 0.14 and whent 3.52. Its positions at these times ares(0.14) 8.79 and s(3.52) 47.51.87 Chapter 3
40. If a cubicle has one critical point, then it must bea point of inflection. If it were a relative maximumor minimum, then the end behavior for a cubicwould not be satisfied. If a cubic has three criticalpoints, then one must be a maximum, another aminimum, and the third a point of inflection.41. No; the function is undefined when x 5.42. yO43. Let x units of notebook paper.Let y units of newsprint.x y 200x 10y 80A(3, 4)OD(3, 1)B(2, 4)xxC(2, 1)200100Oy(10, 190)x y 200(120, 80)y 80(10, 80)x 10100200P(x, y) 400x 350yP(10, 80) 400(10) 350(80) or 32,000P(10, 190) 400(10) 350(190) or 70,500P(120, 80) 400(120) 350(80) or 76,000120 units of notebook and 80 units of newsprint44. yx47. Let x number of 1-point free throws.Let y number of 2-point field goals.Let z number of 3-point field goals.1x 2y 3z 32x y z 17y 0.50(18)1x 2y 3z 32 → 1x 2y 3z 321(x y z 17) x y z 17y 2z 15y 0.50(18) y 2z 15y 9 9 2z 15z 3x y z 17x 9 3 17x 548. y 6 4y 249. 2x 3y 15 → y 2 3 x 56x 4y 16 → y 3 2 x 4 2 3 3 2 1; perpendicular5 free throws, 9 2-point fieldgoals, 3 3-point field goals50. A relation relates members of a set called thedomain to members of a set called the range. In afunction, the relation must be such that eachmember of the domain is related to one and onlyone member of the range. You can use the verticalline test to determine whether a graph is thegraph of a function.51. The area of PTX is equal to the area of RTY.The area of STR is 25% of the area of rectanglePQRS. The correct choice is D.P XQyOx3 x 2, 1 y 445. 1 3 1(5) 2(3) or 1; yes2 546. 3A 3 4 2 5 7 3(4) 3(2)or 12 63(5) 3(7) 15 212B 23 54 32(3) 2(5) or 6 102(4) 2(3) 8 612 6 6 1015 21 8 6 12 (6) 6 1015 (8) 21 6 6 4 7 273A 2B STYRChapter 3 88
3-7Pages 185–1861.Graphs of Rational Functionsf(x)O1a. x 2, y 611b. y x 2 62. Sample graphs:Vertical AsymptoteyCheck for UnderstandingxHorizontal Asymptotey6. x 2, x 1 y (x 2x8y844 O44 8xy x 2 y y 3)( x 1)x3x 2 x 3x3 x 2 x 2x3 x 3 x 31 1 x 1 2 x xno horizontalasymptotes17. f(x) x 1 28. The parent graph is translated 4 units right. Thevertical asymptote is now at x 4. The horizontalasymptote, y 0, is unchanged.23OxOx89. The parent graph is translated 2 units left anddown 1 unit. The vertical asymptote is now at x 2 and the horizontal asymptote is now y 1.Slant AsymptoteyyOxOxx 1)3. Sample answer: f(x) x( x 14. False; sample explanation: if that x-value alsocauses the numerator to be 0, there could be ahole instead of a vertical asymptote.5. x 5xf(x) x 5xy x 5y(x 5) xxy 5y xxy x 5yx(y 1) 5y5yx y; y 1 110. 3x 5x 33x 2 24x 5 → 3x 5 x3x 2 9x5x 55x 1520y 3x 511. y12.13a.POx0 3yOxOV89 Chapter 3
13b. P 0, V 013c. The pressure approaches 0.Pages 186–188Exercises14. x 4 f(x) x215. x 6 y xxy x2x 4x 4y(x 4) 2xxy 4y 2xxy 2x 4yx(y 2) 4y4yx y; y 2y 22 6 x 2x2 x 2 6x2 x 2y no horizontalasymptote16. x 1 2 x 1, x 5 y (2x 17. x 1, x 3 y x 2 18. no vertical asymptote, y x 21)(x 5)x 1y 2x 2 9x 5y 2y ; y 02 9 x 5 xx 24x 3y 2y ; y 01 4 x 3 xy 1 1 x 6 x 2x 1 x 2 x 2 2 xx 2 9 x 5x 2 x 2 1 x 1 x 2 xx22 x 2 xx2 4 x 3x2 x 2 1 x 2 x 2x2 1 x 2x2 x 2 1x2 x 21y ; y 111 x 22219. x 1 y (xx 2 20. x 2xy 3(x 2) 4y 8x 3x 4 8x 3 24x 2 32x 16y84OyO448x4 8xy x2 y x 3x4 xx4 8 xx4 24 x 2x 4 3 2x x 4 16x4y 4 3y ; y 01 8 x 2 4x2 3 2x3 16x4121. f(x) x 3 1122. f(x) x 32 1 x 1) 2 12x 1x 2 1 x 2x2 2 x 1x2 x 2 x 2 1x2 x 21 2 x 1 x 2y ; y 111 x 223. f(x) 1 x 124. The parent graph is translated 3 units up. Thevertical asymptote, x 0, is unchanged. Thehorizontal asymptote is now y 3.25. The parent graph is translated 4 units right andexpanded vertically by a factor of 2. The verticalasymptote is now x 4. The horizontal asymptote,y 0, is unchanged.Chapter 3 90
26. The parent graph is translated 3 units left. Thetranslated graph is then expanded vertically by afactor of 2 and translated 1 unit down. Thevertical asymptote is now x 3 and thehorizontal asymptote is now y 1.4yO4 44x27. The parent graph is expanded vertically by afactor of 3, reflected about the x-axis, andtranslated 2 units up. The vertical asymptote,x 0, is unchanged. The horizontal asymptote isnow y 2.y8 4 O 4 8x488428. The parent graph is translated 3 units right. Thetranslated graph is expanded vertically by a factorof 10 and then translated 3 units up. The verticalasymptote is x 3 and the horizontal asymptoteis y 3.12y30. x 1x 4x 2 1x 3 3 → x 1 x 4x 2 4xx 3x 41 y x 131. x 3xx x 2 3 4 →x 2 x 3 4 x 3x 43x4 y x 332. x 2x 2 1xx 3 2 x 4 → x 2 x 2x 3 x 2x 2 4 2x 2 2 2 y x 233. 1 2 x 5 4 2x 3xx 2 4 1 → 1 2 x 5 4 x 2 3 2 x34. No; the degree of the numerator is 2 more thanthat of the denominator.35. y36. y(2, 4)O 5 2 x 1 5 2 x 1 54 1 41 y 1 2 x 5 4 xO2 1 1 41 2x 3x844 O44 8 12x37. y38.y812O(2, 1 4)xO(2, 0)x29. The parent graph is translated 5 units left. Thetranslated graph is expanded vertically by a factorof 22 and then translated 4 units down. Thevertical asymptote is x 5 and the horizontalasymptote is y 4.39. 40.yyy16 12884O444xO(3, 0)xOx8121691 Chapter 3
41a. C(t) 48 400 3t t0 3t0 t0 3t0 t41b. C(t) 48 410 48 4400 10t 480 3t7t 80t 11.43 L42. Sample answer: The circuit melts or one of thecomponents burns up.43. To get the proper x-intercepts, x 2 and x 3should be factors of the numerator. The verticalasymptote indicates that x 4 should be a factorof the denominator. To get point discontinuityat (5, 0), make x 5 a factor of both thenumerator and denominator with a biggerexponent in the numerator. Thus, a sampleanswer is f(x) (x 4)(x 5) .44a. V x 2 h A(x) 4x h 2x 244b. A(x)20120 x 2 h A(x) 4x 1 x2 2x 2 1 20x2 hA(x) 48 0x 2x 244c. The surface area approaches infinity.45. If the degree of the denominator is larger thanthat of the numerator, then y 0 will be ahorizontal asymptote. To make the graph intersectthe x-axis, the simplest numerator to use is x.xThus, a sample answer is f(x) x 2 .146a. A vertical asymptote at r 0 and a horizontalasymptote at F 0.46b. The force of repulsion increases without boundas the charges are moved closer and closertogether. The force of repulsion approaches 0 asthe charges are moved farther and farther apart.47a. a a47b.320300280260240220200180160140O2 9 3(x 2)(x 3)(x 5) 22 4 6 8 101214 1618 xx 2.9 2.99 3 3.01 3.1m 5.9 5.99 — 6.01 6.1The slope approaches 6.48. abs. max.: (2, 1)[1, 6] scl:1 by [5, 2] scl:149. x 2 9 yy 2 9 xy 2 x 9y x 950. f(x, y) y xf(0, 0) 0 0 or 0f(4, 0) 0 4 or 4f(3, 5) 5 3 or 2f(0, 5) 5 0 or 55; 451. 46 5 4(6) 4(5)8 4 4(8) 4(4) 24 2032 1652. Let x price of film and y price of sunscreen.8x 2y 35.103x y 14.30 8x 2y 35.10y 14.30 3x 8x 2(14.30 3x) 35.102x 28.60 35.10x 3.25y 14.30 3xy 14.30 3(3.25)y 4.55$3.25; $4.5553. x y 3 x y 30 0 ? 3 3 2 ? 30 3 no 5 3 yesx y 3 x y 34 2 ? 3 2 4 ? 32 3 no 2 3 no(3, 2)1 154. 15y x 1 → y 1x5 1 555. [f g](x) f(g(x)) f(2 x 2 ) 8(2 x 2 ) 16 8x 2[g f ](x) g(f(x)) g(8x) 2 (8x) 2 2 64x 2Chapter 3 92
56. Let x the width of each card and y the heightof each card. The rectangle has a base of 4x or 5y.The rectangle has a height of x y.A bh180 4x(x y) 4x 5y180 4xx 4 x5 y 4 x5180 36 x 25y 4( 5)525 x 2 y 45 xPerimeter 2(4x) 2(x y)P 2(4 5) 2(5 4)P 58 in.The correct choice is B.3-8Direct, Inverse, and Joint VariationPages 193–194 Check for Understanding1a. inverse1b. neither1c. direct2. Sample answer:Suppose y varies directly as x n .Then y 1 kx n 1 and y 2 kx n2y 1 kx n11 x1Division property of equality. y y k2 kx2 y 1y x 12 x2Simplify.3. The line does not go through the origin, thereforeits equation is not of the form y kx n .4a. Sample answer: The amount of money earnedvaries directly with the number of hours worked.4b. Sample answer: The distance traveled by a carvaries inversely as the amount of gas in the car.4c. Sample answer: The volume of a cylinder variesjointly as its height and the radius of its base.5. xy k xy 124(3) k 15y 1212 k y 4 5 6. y kx 2 y 2 3 x 254 k(9) 2 y 2 3 (6) 2 2 3 k y 247. y kxz 3 y 0.5xz 316 k(4)(2) 3 y (0.5)(8)(3) 30.5 k y 1088. y k xwz2xz)( 10)3 k(3 2 2y 0.4( 40.4 k y 29. y varies directly as x 4 ; 1 7 .10. A varies jointly as and w; 111. y varies inversely as x; 3.4)(20)212a. V khg 2288 k(40)(1.5) 23.2 kV 3.2hg 212b. V 3.2hg 2V 3.2(75)(2) 2V 96050 960 48,000 m 3Pages 194–196 Exercises13. y kx y 0.2x0.3 k(1.5) y 0.2(6)0.2 k y 1.214. xy k xy 5025(2) kx(40) 5050 k x 1.2515. y kxz y 15xz36 k(1.2)(2) y 15(0.4)(3)15 k y 1816. x 2 y k x 2 y 36(2) 2 (9) k 3 2 y 3636 k y 417. r kt 2 r 16t 24 k 1 2 2 r 16 1 4 216 k r 118. xy k1.21 (0.44) k0.484 kxy 0.484 or y 0.484 y 0.484 y 1.211 x19. y kx 3 z 2 1y 1x23 z 29 k(3) 3 (2) 2 1y 1(4)23 (3) 21 1 2 ky 4820. y k z 2 x y 0 z 1 6 k( 20)62.3x2( 14) y 0.35 20.3 k y 0.16821. y k xz wy 2 xz w3 k(2) (3)4y 2(4 )(2 k y 1422. y k xz239) 2y x7)42z 224) 26 k( 3 3y 2( 6 342 k y 2 723. a k b 2 ca 15 cb 2(6) 210.16 5b 245 k 12015 k 8 b24. x 2 y k yx 2 32(4) 2 (2) k 8x 2 3232 k x 293 Chapter 3
25. C varies directly as d; .26. y varies directly as x; 1 4 .27. y varies jointly as x and the square of z; 4 3 .28. V varies directly as the cube of r; 4 3 .29. y varies inversely as the square of x; 5 4 .30. y varies inversely as the square root of x; 2.31. A varies jointly as h and the quantity b 1 b 2 ; 0.5.32. y varies directly as x and inversely as the squareof z; 1 3 .33. y varies directly as x 2 and inversely as the cube ofz; 7.34. y varies jointly as the product of the cube of x andz and inversely as the square of w.35a. Joint variation; to reduce torque one must eitherreduce the distance or reduce the mass on theend of the fulcrum. Thus, torque varies directlyas the mass and the distance from the fulcrum.Since there is more than one quantity in directvariation with the torque on the seesaw, thevariation is joint.35b. T 1 km 1 d 1 and T 2 km 2 d 2T 1 T 2km 1 d 1 km 2 d 2 Substitution propertym 1 d 1 m 2 d 2 of equality35c. m 1 d 1 m 2 d 275(3.3) (125)d 21.98 d 2 ; 1.98 meters36a. tr k36b. tr k tr 36,00045(800) k t(1000) 36,00036,000 k t 36 minutes37. If y varies directly as x then there is a nonzeroconstant k such that y kx. Solving for x, we findx 1 k y 1 k is a nonzero constant, so x variesdirectly as y.38a. I dk2 76k38b. I d 2 a 2 b 2 c 2 I 5 d2k (6) 2 (25) 2 c 2 57616 6 2I 6.5 c(6 . 5) 2576 kI 13.6 lux39. a is doubleda k b2c3a a 1 4 kb 2 1 8 c 3a 2 k ck 1 2 b 2 1 2 c 3b2340b. F G m 11.99 10 20 GkL41. R r21.07 10 2 k 2 (0 .001) 21.68 10 8 k42. y8642O8642 2 4 6 8x2468consistent y and dependent46.O m 2d2(5.98 10 24 )(7.36 10 22 )(3.84 10 8 ) 26.67 10 11 G; 6.67 10 11 N k40c. F G m 1 m 2d2 (6.67 10 11 )x m2g2 3.53 10 22 N40d. 3.53 10 22 (1.99 10 20 )x178 x; about 178 times greaterR 1.6 8 10(3)(0.003) 28R 1.78 10 3 43. f(x) (x 3) 3 6y (x 3) 3 6x (y 3) 3 6x 6 (y 3) 3 3 x 6 y 3y 3 x 6 3f 1 (x) 3 x 6 3; f 1 (x) is a function.44. 1 0 1 3 1 3 1 3 1 3 0122), C(1, 4), D(3, 0)0 2 2A(1, 2), B(45. 4x 2y 7 → y 2x 7 2 12x 6y 21 → y 2x 7 2 8.6 23.2000 1995 4 .652(5.98 10 24 )(1.99 10 30 )(1.50 10 11 ) 24147. m 2y 18.6 0.92(x 2000) or 0.92y 0.92x 1858.6048. 144 4 2 9 or 12 2 112 is divisible by 3, 4, 6, and 12.4040a. F G m 1 m 2d2Chapter 3 94
The correct choice is D.Chapter 3 Study Guide and AssessmentPage 197 Understanding and Using theVocabulary1. even 2. continuous 3. point4. decreasing 5. maximum 6. rational7. inverse 8. monotonic 9. slant10. JointPages 198–200 Skills and Concepts11. f(x) 2(x) f(x) (2x)f(x) 2x f(x) 2x yes12. f(x) (x) 2 2 f(x) (x 2 2)f(x) x 2 2 f(x) x 2 2 no13. f(x) (x) 2 (x) 3f(x) x 2 x 3f(x) (x 2 x 3)f(x) x 2 x 3no14. f(x) (x) 3 6(x) 1f(x) x 3 6x 1f(x) (x 3 6x 1)f(x) x 3 6x 1no15. xy 4 → ab 4x-axis a(b) 4ab 4y-axis (a)b 4ab 4 noy x (b)(a) 4ab 4 yesy x (b)(a) 4ab 4 yesy x and y x16. x y 2 4 → a b 2 4x-axis a (b) 2 4a b 2 4noy-axis (a) b 2 4a b 2 4y x (b) (a) 2 4a 2 b 4y x (b) (a) 2 4yesnonoa 2 b 4 nox-axis17. x 2y → a 2bx-axisa 2(b)a 2b noy-axis(a) 2ba 2b noy x(b) 2(a)b 2a noy x(b) 2(a)b 2a no; none18. x 2 1 y → a 2 1 b x-axis a 2 1 ( b)a 2 1 b noy-axisby2 1 a y no; y-axis19. The graph of g(x) is a translation of the graph off(x) up 5 units.20. The graph of g(x) is a translation of the graph off(x) left 2 units.21. The graph of g(x) O is the x graph of f(x) expandedvertically by a factor of 6.O x22. The graph of g(x) is the graph of f(x) expandedhorizontally yby a factor of 4 3 and translated y down4 units.Ox(a) 2 1 b a 2 1 b yesy x (b) 2 1 (a )b 2 1 a noy x (b) 2 1 ( a)23. 24.25. 26.27. Case 1 Case 2⏐4x 5⏐ 7 ⏐4x 5⏐ 7(4x 5) 7 4x 5 74x 5 7 4x 2Ox95 Chapter 3
f(x)4x 3x121f 1 (x)x 0.5x 3x f(x)x f 1 (x){x⏐x 3 or x 0.5}2 77 228. Case 1 Case 21 44 1⏐x 3⏐ 2 11 ⏐x 3⏐ 2 110 11 0(x 3) 2 11 x 3 2 111x 5 211 2 x1 12f(x) x 6x 6 f(x){x⏐6 x 12} f 1 (x)29.O x32.f(x) (x 1) 2 4x f(x)3 02 31 40 31 0f 1 (x)f(x)Of(x)xf 1 (x)x f 1 (x)0 33 24 13 00 130.f(x) 1 4 x 5x f(x)2 5.51 5.250 51 4.752 4.5f(x)64264O 2 4 646f 1 (x)f(x)xf 1 (x)x f 1 (x)5.5 25.25 15 04.75 14.5 233. f(x) (x 2) 3 8y (x 2) 3 8x (y 2) 3 8x 8 (y 2) 3 3 x 8 y 2y 3 x 8 2f 1 (x) 3 x 8 2; yes34. f(x) 3(x 7) 4y 3(x 7) 4x 3(y 7) 4 3x (y 7) 4x 4 3 y 7xy 7 4 3f 1 (x) 7 4 3x ; no31.f(x) 2 x 3x f(x)3 2.32 21 1 1 2 10 — 1 2 71 52 43 3.7f(x)yOf 1 (x)xf 1 (x)x f 1 (x)2.3 32 21 11 1 2 7 1 2 5 14 23.7 335. Yes; the function is defined when x 2; thefunction approaches 6 as x approaches 2 from bothsides; and y 6 when x 2.36. No; the function is undefined when x 1.37. Yes; the function is defined when x 1; thefunction approaches 2 as x approaches 1 from bothsides; and y 2 when x 1.38. a n : negative, n: oddy → as x → , y → as x → .39. a n : positive, n: oddy → as x → , y → as x → .40.y x12 1xy1000 1.000001100 1.000110 1.011 210 1.01100 1.00011000 1.000001y → 1 as x → , y → 1 as x → .41. a n : positive, n: oddy → as x → , y → as x → .Chapter 3 96
42.52. x 1 y x x1y x x x x 1 x 1y ; y 11 1 x [5, 5] scl:1 by [20, 10] scl:5decreasing for x 2 and x 1;increasing for 2 x 143.[6, 6] scl:1 by [5, 20] scl:5decreasing for x 3 and 0 x 3;increasing for 3 x 0 and x 344. abs. max.: (2, 1)45. rel. max.: (0, 4), rel. min.: (2, 0)46. f(2.9) 0.029f(3) 0f(3.1) 0.031 min.47. f(0.1) 6.996f(0) 7f(0.1) 7.004 pt of inflection48. f(x) 1 x 149. f(x) 2 x 50. The parent graph is translated 2 units left andexpanded vertically by a factor of 3. The verticalasymptote is now x 3. The horizontalasymptote, f(x) 0, is unchanged.Of(x)x51. The parent graph is translated 3 units right andthen translated 2 units up. The vertical asymptoteis now x 3 and the horizontal asymptote isf(x) 2.f(x)Ox53. x 2 y x x2 1 2 x 2 1x2 x 2y y xx2 x2211 x 2 1 x 2 x 2no horizontal asymptotes54. x 3, y (x 3) 2x 2 y x2 96x 9x 2 9 x 2x2 6 x 9x2 x 2 x 2 9x 2 x 2y 1 6 x 9 x 291 x 2k140 x xk140 49 xy ; y 155. x 2xx x 2 2 1 →x 2x 2 1 x 2x2x1 yes; y x 256. y kxz y 0.625xz5 k(4)(2) y 0.625(6)(3)0.625 k y 11.2557. y y 20 10 140 k x 14x 19658. y k x2 zy 32 0x2z7.2 k(0 .3) 24y 32 0(1) 240320 k y 8Page 201 Applications and Problem Solving59. ⏐x 6.5⏐ 0.2;Case 1 Case 2⏐x 6.5⏐ 0.2 ⏐x 6.5⏐ 0.2(x 6.5) 0.2 x 6.5 0.2x 6.5 0.2 x 6.7x 6.3x 6.3 6.3 x 6.797 Chapter 3
60a.x C(x)0 x 1 0.401 x 2 0.802 x 3 1.203 x 4 1.604 x 5 2.005 x 6 2.403.603.202.802.40Cost 2.00(dollars) 1.601.200.800.4000 1 2 3 4 5 6 789Time (min)Page 201 Open-Ended Assessment1a. Sample answer: x y 2yOx1b. Sample answer: y x 2yxO60b. positive real numbers;positive multiples of $0.4060c.x C 1 (x)0.40 0 x 10.80 1 x 21.20 2 x 31.60 3 x 42.00 4 x 52.40 5 x 6Time(min)98765432100 0.40 1.20Cost (dollars)2.00 2.80 3.6060d. positive multiples of $0.40;positive real numbers60e. C 1 (x) gives the possible number of minutesspent using the scanner that cost x dollars.61a. h(t)1c. Sample answer: xy 1yxO1d. Sample answer: xy 1yO x1e. Sample answer: y x 3yO x10.5O0.5 1t2. Sample answer: 2(x 4) 2 161b. 1.08 mChapter 3 98
3a. Sample answer:(2, 1)y(0, 0)O3b. abs. min.: (2, 3); rel. max.: (0, 0); rel. min.:(2, 1)Chapter 3 SAT & ACT PreparationPage 203 SAT & ACT Practice1. Always factor or simplify algebraic expressionswhen possible. Notice that the numerator in theproblem is the difference of two squares, a 2 b 2 .Factor it.y 2 9 3 y 9 (y 3) (y 3)3y 9(2, 3)Factor the denominator. Both the numerator anddenominator contain the factor (y 3). Simplifythe fraction. (y 3) (y3y 3)9 (y 3) (y3(y 3)3) y 33 The correct answer is E.2. You need to find the statement that is not true.Compare the given information with each answerchoice. Choice A looks like x y z, except forthe numbers. Multiply both sides of the equationx y z by 2.2(x y) 2z or 2x 2y 2zSo choice A is true. For choice B, start with x yand subtract y from each side.x y y y 0So choice B is true. For choice C, start with x yand subtract z from each side.x z y zSo choice C is true. For choice D, substitute y for xand x y for z.zx 2y x y2 2 y2ySo choice D is also true. For choice E, write eachside of the equation in terms of y.z y (x y) x y2x 2yy 2ySo choice E is not true. The correct choice is E.x3. Notice that 450 miles is the distance toGrandmother’s house, not the round trip. This is amultiple-step problem. First calculate the numberof gallons of gasoline used in each direction of thetrip.miles miles gallons4.pergallon 4 5025 18 gallonsOn the trip to Grandmother’s, the cost of gasolineis 18 gallons $1.25 per gallon or $22.50.On the trip back, the gasoline cost is 18 gallons $1.50 per gallon or $27.00. The difference betweenthe costs is $4.50.Afaster way to find the cost difference is to reasonthat each gallon cost $0.25 more on the trip back.So the total amount more that was paid was18 gallons $0.25 or $4.50. The correct choice is B.OyxThe portion of the graph of f(x) which is showncrosses the x-axis 3 times.The correct choice is D.5. Notice that the denominators are all powers often. Carefully convert each fraction to a decimal.Then add the three decimals. 9 0090910 1 00 10 00 90 0.9 0.009 90.909The correct choice is C. You could also use yourcalculator on this problem.6. Combine like terms.(10x 4 x 2 2x 8) (3x 4 3x 3 2x 9) (10x 4 3x 4 ) (3x 3 ) (x 2 ) (2x 2x) (8 9) 7x 4 3x 3 x 2 0 17The correct choice is A.7. One method of solving this problem is to “plug in”a number in place of n. Choose a number thatwhen divided by 8, has a remainder of 5. Forexample, choose 21.21 2(8) 5Then use this value for n in the answer choices.Find the expression that has a remainder of 7.Choice A: n 18 21 18 2 28 2R6The remainder is 6.Choice B: n 28 21 28 2 38 2R7You could also reason that since n divided by 8 hasa remainder of 5, then (n 2) divided by 8 willhave a remainder of (5 2) or 7. The correctchoice is B.99 Chapter 3
8.Simplify the expression inside the square rootsymbol. Factor 100 from each term. Then factorthe trinomial.100x 2 0x 6000 9100(x 2 6x 9) x 3x 310xx 2 6 9x 310(x )(x 33) x 310(x ) 3 2 x 3 10 (x 3)x 3 10The correct choice is B.9. Since a b c, substitute a b for c ina c 5. So, a (a b) 5. Then b 5 orb 5. Substitute 5 for b in b c 3. So,5 c 3. Then c 8 or c 8.The correct choice is B.10. There are two equations and two variables, so thisis a system of equations. First simplify theequations. Start with the first equation. Divideboth sides by 2.4x 2y 242x y 12Now simplify the second equation. Multiply bothsides by 2x. 7 y2x 77y 7(2x)7y 14xDivide both sides by 7.y 2xYou need to find the value of x. Substitute 2x for yin the first equation.2x y 122x (2x) 124x 12x 3The answer is 3.Chapter 3 100
Chapter 4 Polynomial and Rational Functions4-1Polynomial FunctionsPages 209–210 Check for Understanding1. A zero is the value of the variable for which apolynomial function in one variable equals zero. Aroot is a solution of a polynomial equation in onevariable. When a polynomial function is therelated function to the polynomial equation, thezeros of the function are the same as the roots ofthe equation.2. The ordered pair (x, 0) represents the points onthe x-axis. Therefore, the x-intercept of a graph ofa function represents the point where f(x) 0.3. Acomplex number is any number in the forma bi, where a and b are real numbers and i isthe imaginary unit. In a pure imaginary number,a 0 and b 0. Examples: 2i, 3i;Nonexamples: 5, 1 i4. yOx5. 3; 1 6. 5; 87. no; f(x) x 3 5x 2 3x 18f(5) (5) 3 5(5) 2 3(5) 18f(5) 125 125 15 18f(5) 338. yes; f(x) x 3 5x 2 3x 18f(6) (6) 3 5(6) 2 3(6) 18f(6) 216 180 18 18f(6) 09. (x (5))(x 7) 0(x 5)(x 7) 0x 2 2x 35 0; even; 210. (x 6)(x 2i)(x (2i))(x i) (x (i)) 0(x 6)(x 2i)(x 2i)(x i)(x i) 0(x 6)(x 2 4i 2 )(x 2 i 2 ) 0(x 6)(x 2 4)(x 2 1) 0(x 3 6x 2 4x 24)(x 2 1) 0x 5 6x 4 5x 3 30x 2 4x 24 0;odd; 111. 2; x 2 14x 49 0(x 7)(x 7) 0x 7 0 x 7 0x 7 x 7f(x)6020O(0, 49)40 f(x) x 2 14x 49(7, 0)4 812x12. 3; a 3 2a 2 8a 0a(a 2 2a 8) 0a(a 4)(a 2) 0a 0 a 4 0 a 2 0a 4 a 24302 O10f(a)20 f(a) a 2 2a 2 8a10(4, 0) (0, 0) (2, 0)2 4a13. 4; t 4 1 0(t 2 1)(t 2 1) 0(t 1)(t 1)(t 2 1) 0t 1 0 t 1 0 t 2 1 0t 1 t 1 t 2 1t i(1, 0)f(t)Of (t) t 4 1(1, 0)t14a. x 2 r 2 6 2 V(x) Bhr 2 36 x 2 V(x) (36 x 2 )(2x)14b. V(x) (36 x 2 )(2x)V(x) (36 x 2 )(2x)V(x) 72x 2x 314c. V(x) 72x 2x 3V(4) 72(4) 2(4) 3V(4) 502.65 units 3Pages 210–212 Exercises15. 4; 5 16. 7; 3 17. 3; 5 18. 5; 2519. 6; 1 20. 2; 121. Yes; the coefficients are complex numbers and theexponents of the variable are nonnegativeintegers.101 Chapter 4
122. No; a 1 , which is a negative exponent.a23. yes; f(a) a 4 13a 2 12af(0) (0) 4 13(0) 2 12(0)f(0) 024. no; f(a) a 4 13a 2 12af(1) (1) 4 13(1) 2 12(1)f(1) 1 13 12f(1) 2425. yes; f(a) a 4 13a 2 12af(1) (1) 4 13(1) 2 12(1)f(1) 1 13 12f(1) 026. yes; f(a) a 4 13a 2 12af(4) (4) 4 13(4) 2 12(4)f(4) 256 208 48f(4) 027. no; f(a) a 4 13a 2 12af(3) (3) 4 13(3) 2 12(3)f(3) 81 117 36f(3) 7228. yes; f(a) a 4 13a 2 12af(3) (3) 4 13(3) 2 12(3)f(3) 81 117 36f(3) 029. f(b) b 4 3b 2 2b 4f(2) (2) 4 3(2) 2 2(2) 4f(2) 16 12 4 4f(2) 12; no30. f(x) x 4 4x 3 x 2 4xf(1) (1) 4 4(1) 3 (1) 2 4(1)f(1) 1 4 1 4f(1) 0; yes31a. 3; 1 31b. 2; 2 31c. 4; 232. (x ( 2))(x 3) 0(x 2)(x 3) 0x 2 x 6 0; even; 233. (x (1))(x 1)(x 5) 0(x 1)(x 1)(x 5) 0(x 2 1)(x 5) 0x 3 5x 2 x 5 0; odd; 334. (x (2))(x (0.5))(x 4) 0(x 2)(x 0.5)(x 4) 0(x 2 2.5x 1)(x 4) 0x 3 4x 2 2.5x 2 10x x 4 0x 3 1.5x 2 9x 4 02x 3 3x 2 18x 8 0; odd; 335. (x (3))(x (2i))(x 2i) 0(x 3)(x 2i)(x 2i) 0(x 3)(x 2 4i 2 ) 0(x 3)(x 2 4) 0x 3 3x 2 4x 12 0; odd; 136. (x (5i))(x (i))(x i)(x 5i) 0(x 5i)(x i)(x i)(x 5i) 0(x 5i)(x 5i)(x i)(x i) 0(x 2 25)(x 2 1) 0x 4 26x 2 25 0; even; 037. (x (1))(x 1)(x 4)(x (4)(x 5) 0(x 1)(x 1)(x 4)(x 4)(x 5) 0(x 2 1)(x 2 16)(x 5) 0(x 4 17x 2 16)(x 5) 0x 5 5x 4 17x 3 85x 2 16x 80 0; odd; 538. (x (1))(x 1)(x 3)(x (3)) 0(x 1)(x 1)(x 3)(x 3) 0(x 2 1)(x 2 9) 0x 4 10x 2 9 039. 1; x 8 0f(x)x 8(0, 8)(9, 0) f(a)2010 O 10 20a20406080(9, 0)f(a) a 2 81(0, 81)f(x) x 8(8, 0)40. 2; a 2 81 0(a 9)(a 9) 0a 9 0 a 9 0a 9a 941. 2; b 2 36 0b 2 36b 6i4f(t)4(2, 0) (2, 0)2 O4(0, 8)122 4 tf(t) t 3 2t 2 4t 88060f(b) b 2 3640204f(b)(0, 36)O2 O 2 4b42. 3; t 3 2t 2 4t 8 0t 2 (t 2) 4(t 2) 0(t 2)(t 2 4) 0(t 2)(t 2)(t 2) 0t 2 0 t 2 0 t 2 0t 2 t 2 t 2xChapter 4 102
43. 3; n 3 9n 0n(n 2 9) 0n(n 3)(n 3) 0n 0 n 3 0 n 3 0n 3n 3(3, 0)420102 O1020100(2.5, 0) 5042 O50100f(n)(0, 0)(3, 0)2 4nf(n) n 3 9n44. 3; 6c 3 3c 2 45c 0c(6c 2 3c 45) 0c(c 3)(6c 15) 0c 0 c 3 0 6c 15 0c 3c 1 56f(c)(0, 0)(3, 0)2 4cf(c) 6c 3 3c 2 45cc 2.547. 4; 4m 4 17m 2 4 0(4m 2 1)(m 2 4) 04m 1 0 m 2 4 0m 14m 4 m 0.5im 2if(m)f(m) 4m 4 17m 2 4Of(u)O(0, 4)48. (u 1)(u 2 1) 0(u 1)(u 1)(u 1) 0u 1 0 u 1 0 u 1 0u 1 u 1 u 1(1, 0) (1, 0)(0, 1)muf(u) (u 1)(u 2 1)49a. y49b.y45. 4; a 4 a 2 2 0(a 2 2)(a 2 1) 0(a 2 2)(a 1)(a 1) 0a 2 2 0 a 1 0 a 1 0a 2 2 a 1 a 1a 2iOxOxf(a)49c. y49d.y(1, 0)Of(a) a 4 a 2 2(1, 0)(0, 2)aOxOx46. 4; x 4 10x 2 9 0(x 2 9)(x 2 1) 0(x 3)(x 3)(x 1)(x 1) 0x 3 0 x 3 0 x 1 0 x 1 0x 3 x 3 x 1 x 1202 O10f(x)10 (0, 9)(3, 0)(1, 0) (1, 0) (3, 0)4202 4xf(x) x 4 10x 2 949e. y49f. not possibleO x103 Chapter 4
50.[5, 5] sc11 by [2, 8] sc1150a. 4 50b. 2; 1, 150c. There are 4 real roots. However, there is adouble root at 1 and a double root at 1.51a. V(x) 99,000x 3 55,000x 2 65,000x51b. r 0.15 x 1 rx 1 0.15x 1.15V(x) 99,000x 3 55,000x 2 65,000xV(1.15) 99,000(1.15) 3 55,000(1.15) 2 65,000(1.15)V(1.15) 150,566.625 72,737.5 74,750V(1.15) 298,054.125; about $298,054.1352. 1 and 3 are two of its zeros.53a. d(t) 1 2 at 2 d(t) 1 2 at 2d(30) 1 2 (16.4)(30) 2 d(60) 1 2 (16.4)(60) 2d(30) 7380 ft d(60) 29,520 ftd(t) 1 2 at 2d(120) 1 2 (16.4)(120) 2d(120) 118,080 ft53b. It quadruples; (2t) 2 4t 2 .54. Let x the width of the sidewalk.The length of the pool would be 70 2x feet.The width of the pool would be 50 2x feet.A w2400 (70 2x)(50 2x)2400 3500 240x 4x 20 4x 2 240x 11000 x 2 60x 2750 (x 55)(x 5)x 55 0 x 5 0x 55 x 5Use x 5 since 55 is an unreasonable solution.5 ft55. Let x the number of pizzas.(160 16x)(16 0.40x) 40006.4x 2 192x 2560 40006.4x 2 192x 1440 0x 2 30x 225 0(x 15)(x 15) 0x 15 0 x 15 0x 15 x 1516 0.40x 16 0.40(15) $1056. (x B)(x C) 0x 2 Cx Bx BC 0x 2 (C B)x BC 0C B B from x2 Bx C 0BC CB 1C 1 1C 2C 2Sample answer: 1; 2x 257. y x(x 2)(x 2)58a. Let x the width. The length 1 2 (52 2x) or26 x. A(x) x(26 x)58b. A(x) x(26 x)A(x) 26x x 2[5, 30] sc15 by [2, 200] sc120x 1326 x 26 13 1313 yd by 13 yd59. The graph of y 2x 3 1 is the graph of y 2x 3shifted 1 unit up.60. (6, 9)61. 15 515(3) (9)(5)9 345 45 or 0; no62. AB 2 1 3 9 23 4 57 6 2(3) (1)(5) 2(9) (1)(7)3(3) 4(5) 3(9) 4(7)2(2) (1)(6)3(2) 4(6) 1 25 1029 1 1863. x 4y 9y4y x 9y 1 4 x 9 4 64. Parallel; the lines have the same slope.Ox 4y 9xChapter 4 104
65. [f g](x) f(g(x)) f 1 2 x 6 1 2 x 6 2 4 1 4 x 2 6x 36 4 1 4 x 2 6x 32[g f ](x) g( f(x)) g(x 2 4) 1 2 (x 2 4) 6 1 2 x 2 2 6 1 2 x 2 466. The pictograph shows two more small car symbolsin the row for 1999 than it does for 2000. Thesetwo small cars represent the 270 additional carsthat were sold in 1999 compared to 2000. Sincethe two small cars represent 270 real cars, eachsmall car symbol must represent 27 0 or 135 real2cars.The correct choice is A.4-2Quadratic EquationsPages 218–219 Check for Understanding1. Add 4 to each side of the equation to get t 2 6t 4. Determine the value needed to make t 2 6ta perfect square trinomial. Add this value (9) toeach side. Take the square root of each side of theequation and solve the two resulting equations.t 3 132. Quadratic Formula; Since the leading coefficientdoes not equal 1 and the discriminant equals 185which is not a perfect square, the QuadraticFormula would be the best way to get an exactanswer. Completing the square can also be used,but errors in arithmetic are more likely. A graphwill give only approximate solutions.13 (13) p 2 5)(7) 4( 13 29 or 2(5)103a. equals 0 3b. negative number3c. positive number4. Graphingf(x) f(x)x 2 4x 54(5, 0) (1, 0)6 4 2 O 2x48Factoringx 2 4x 5 0(x 5)(x 1) 0x 5 0 x 1 0x 5 x 1Completing the Squarex 2 4x 5 0x 2 4x 5x 2 4x 4 5 4(x 2) 2 9x 2 3x 2 3x 1Quadratic Formulax 4 42 (1)(5 4)x 4 22(1)36 x 2 3x 5x 4 62x 2 3x 2 3 x 2 3x 1x 5See students’ work.5. x 2 8x 20 0x 2 8x 20x 2 8x 16 20 16(x 4) 2 36x 4 6x 4 6x 4 6x 2x 106. 2a 2 11a 21 0a 2 1 1 a2 2 21 0a 2 1 12a 2 12a 2 1 1 a2 1 12 16 2 12 1 2116a 1 41 2 2 18961 1 47a 1 4a 1 1 4 1 47a 1 1 4 1 7 4a 3 2 a 77. b 2 4ac 12 2 4(1)(36) 0; 1 realm 12 2(1)m 1 22or 68. b 2 4ac (6) 2 4(1)(13)16; 2 imaginary0(6) 16 2(1)t 6 4i2 3 2i9. p 2 6p 5 0(p 5)(p 1) 0p 5 0 p 1 0p 5 p 110. r 2 4r 10 0(4) (4)2 )(10)4(1r 2(1)2r 4 24r 4 2i6 2r 2 i6105 Chapter 4
17. t 2 3t 7 011. P 12I 0.02I 2g 2 263 28. s b b2 4 2a1600 12I 0.02I 2t 2 3t 70.02I 2 12I 1600 0I 2 t 2 3t 9 4 600I 80,000 0 7 9 4 (I 200)(I 400) 0t 3 2 2 3 74I 200 0 I 400 0t 3 2 37 2I 200 ampsI 400 ampst 3 2 37 2z 1 57 121 m 2(6)z 1 5z 1 5z 6z 4m 7 1112Pages 219–221 Exercises18. 1 2 2 1 4 12. z 2 2z 24 0z 2 2z 24z 2 2z 1 24 1(z 1) 2 2519. b 2 4ac (6) 2 4(4)(25)3642 imaginary; the discriminant is negative.20. b 2 4ac 7 2 4(6)(3) or 121; 2 real13. p 2 3p 88 0m 3p 2 2 , 1 3 3p 88p 2 3p 9 4 88 9 21. b 2 4ac (5) 2 4(1)(9) or 11; 2 imaginary4 p 3 2 2 36 1s 5 11 2(1)45 i11p 3 2 1 92p 3 2 1 9s 2222. b 2 4ac (84) 2 4(36)(49) or 0; 1 realp 11p 814. x 2 84 0 10x 21 0d 2(36)x 2 10x 21d 8 4x 2 72 or 7 6 10x 25 21 25(x 5) 2 423. b 2 4ac (2) 2 4(4)(9) or 140; 2 imaginaryx 5 2x 2 140 2(4)x 5 2x 5 2x 7 x 3x 2 2i35 815. d 2 3 4 d 1 1 i35 8 x 04d 2 3 4 d 1 24. 3p 2 4p 88 3p 2 4p 8 0d 2 3 4 9d 6 4 1 8 9 6b 2 4ac 4 2 4(3)(8) or 112; 2 real4d 3 8 2 4 112 p 6142(3)d 3 8 1 4 478 p 6d 3 8 1 8 d 3 8 1 8 2 27p 3d 1 2 d 1 4 25. 2k 2 5k 916. 3g 2 2k 2 5k 9 0 12g 4b 2 4ac 5 2 4(2)(9) or 97; 2 realg 2 4g 4 3 5 97k 2(2)g 2 4g 4 4 3 45 97 k 4(g 2) 2 8 3 26. 7 i5 27. 5 2ig 2 263s 5 (5)2 )(9)4(32(3)s 5 83 6 6s 5 i83Chapter 4 106
29. x 2 3x 28 0(x 7)(x 4) 0x 7 0 x 4 0x 7x 430. 4w 2 19w 5 0(4w 1)(w 5) 04w 1 0 w 5 04w 1w 5w 1 4 31. 4r 2 r 54r 2 r 5 0(4r 5)(r 1) 08 2 4(1)(c) 064 4c 04c 64c 1636a. A bhA 12(16)A 192(12 2x)(16 2x) 1 2 (192)(12 2x)(16 2x) 9636b. (12 2x)(16 2x) 96192 56x 4x 2 964x 2 56x 96 0x 2 14x 24 04r 5 04r 5r 5 4 32. p b b2 4 2a35. b 2 4ac 0f(x)20r 1 0r 1p 2 22 410p p 33. x 2(1)2 28 22 2i7 2p 1 i7x x 26 (26) 2 1)(2 4( )2(1)26 32226 42 2x 6 2234a. P 0.01A 2 0.05A 107P 0.01(25) 2 0.05(25) 107P 6.25 1.25 107P 114.5 mm Hg34b. P 0.01A 2 0.05A 107125 0.01A 2 0.05A 1070 0.01A 2 0.05A 18A b b2 ac 4A A O2a0.05 0.052 .01)(4(018)2(0.01)0.05 0.72250.020.05 0.722 50.02A or A A 40A 4540 years old34c. P150125100 P 75 0.01A 2 0.05A 107502525 50 75 100A0.05 0.722 50.02As a woman gets older, the normal systolicpressure increases.O102010 2036c. roots: 2, 1212 2x 12 2(2) or 816 2x 16 2(2) or 128 ft by 12 ft12 2x 12 2(12) or 1216 2x 16 2(12) or 8∅37a. d(t) v 0 t 1 2 gt 2d(t)d(t) 5t 1 2 (32)t 2d(t) 5t 16t 2xf(x) x 2 14x 2437b. 0 and about 0.337c. The x-intercepts indicate when the woman is atthe same height as the beginning of the jump.37d. d(t) 5t 16t 250 5t 16t 237e. 50 5t 16t 216t 2 5t 50 0t b b2 ac 42a6)(54(10)2(16)325 3225 32t 5 (5)2t 5 3225t t O5 3225 32t 1.93 s t 1.62about 1.93 std(t) 5t 16t 2107 Chapter 4
38. ax 2 bx c 0x 2 b xca a 0x 2 ab x ac x 2 b x a b2a 2 c a b2a 2bx 2 a 2 c b2 a 4 a2bx 2 a 2 4a c b24a2bx 2 a b2 ac 4 2abx 2x y2 01 10 21 12 0x a b2 ac 4 2ab bac2 4 2a39. 2; 18a 2 3a 1 0(3a 1)(6a 1) 03a 1 0 6a 1 03a 1 6a 1a 1 3 a 1 6 40.41. f(x) (x 9) 2y (x 9) 2x (y 9) 2x y 9y x 9f 1 (x) x 942. 3x 4y 3752(5x 2y) 2(345)→3x 4y 3753(45) 4y 375y 60 (45, 60)43. m 6 2.82m 4019– 595– 2.4. 4f(a)f(a) 18a 2 3a 1O3x 4y 37510x 4y 6907x 315x 4560 619 – x 2.8 – 3.2619 x 24m 60 x $64344. 3y 8x 123y 8x 12y 3x 4; 8 3 45. x 2 x 20 (x 5)(x 4)The correct choice is A.ayOy |x| 2x4-3The Remainder and FactorTheoremsPage 226 Check for Understanding1. The Remainder Theorem states that if apolynomial P(x) is divided by x r, the remainderis P(r). If a division problem has a remainder of 0,then the divisor is a factor of the dividend. Thisleads to the Factor Theorem which states that thebinomial x r is a factor if and only if P(r) 0.2. (x 3 4x 2 7x 8) (x 5); x 2 x 2; 23. The degree of a polynomial is one more than thedegree of its depressed polynomial.4. Isabel; if f(3) 0, then (x (3)) or (x 3) is afactor.5. 2⏐ 1 1 4 6. 5⏐ 1 1 17 152 25 20 151 1 61 4 3 0x 1, R6x 2 4x 37. f(x) x 2 2x 15f(3) (3) 2 2(3) 15 9 6 15 or 0; yes8. f(x) x 4 x 2 2f(3) (3) 4 (3) 2 2 81 9 2 or 92; no9. f(x) x 3 5x 2 x 5f(1) (1) 3 5(1) 2 1 5 1 5 1 5 or 01⏐ 1 5 1 51 4 51 4 5 0x 1 is a factor x 2 4x 5 (x 5)(x 1)(x 5), (x 1), (x 1)10. f(x) x 3 6x 2 11x 6f(1) (1) 3 6(1) 2 11(1) 6 1 6 11 6 or 01⏐ 1 6 11 61 5 61 5 6 ⏐ 0x 1 is a factor x 2 5x 6 (x 2)(x 3)(x 1), (x 2), (x 3)11. 1⏐ 1 0171k66 k 2k 41 1 6 6 k12a. 12 12b. 12 12c. 1112d. f(x) x 7 x 9 x 12 2x 2 x 12 x 9 x 7 2x 2 x(x 11 x 8 x 6 2x) x 2 (x 10 x 7 x 5 2)x, x 2 , x 11 x 8 x 6 2x, or x 10 x 7 x 5 213. h r 4 V r 2 hV r 2 (r 4)5 r 2 (r 4)5 r 3 4r 20 r 3 4r 2 50 (r 3 4r 2 5)1⏐ 1 4 0 51 5 51 5 5 0r 1 0 h r 4r 1 in.h 1 4 or 5 in.Chapter 4 108
Pages 226–228 Exercises14. 7⏐ 1 20 91 15.7 911 13 0x 13x 2 6x 9, R 116. 2⏐ 1 12060121241 3 6 12 23x 3 3x 2 6x 12, R2317.2⏐ 1 0 8 0 162 4 8 161 2 4 8 0x 3 2x 2 4x 818.1⏐ 3 2 5 4 23 5 10 143 5 10 14 123x 3 5x 2 10x 14, R1219.2 2 0 32x 2 2x, R 320. f(x) x 2 2 21. f(x) x 5 32f(1) (1) 2 2 f(2) (2) 5 32 1 2 or 1; no 32 32 or 0;yes22. f(x) x 4 6x 2 8f(2) (2) 4 6(2) 2 8 4 12 8 or 0; yes23. f(x) x 3 x 6f(2) (2) 3 2 6 8 2 6 or 12; no24. f(x) 4x 3 4x 2 2x 3f(1) 4(1) 3 4(1) 2 2(1) 3 4 4 2 3 or 13; no25. f(x) 2x 3 3x 2 xf(1) 2(1) 3 3(1) 2 1 2 3 1 or 0; yes26a-d.1⏐ 2 0 2 32 2 027. (6) 4 36 36 36 or 028. 1⏐ 1 7116771 6 7 0x 2 6x 7 (x 1)(x 7)(x 1)(x 1)(x 7)29. 2⏐ 1 1246441 3 2 0x 2 3x 2 (x 1)(x 2)(x 2), (x 1), (x 2)30. 1⏐ 1 1149049491 0 49 0x 2 49 (x 7)(x 7)(x 1), (x 7), (x 7)3⏐ 1 9 27 283 18 271 6 9 1r 1 3 2 81 1 4 2 61 1 2 4 42 1 5 8 82 1 1 4 0d31. 4⏐ 1 5424881 1 2 0x 2 x 2 (x 2)(x 1)(x 4), (x 2), (x 1)32. 2⏐ 1 2240881 0 4 0x 2 4 (x 2)(x 2)(x 2), (x 2), (x 2)33. 1⏐ 1 4115441 5 4 0x 2 5x 4 (x 1)(x 4)(x 1), (x 1), (x 4)34. 1⏐ 1 3132111 2 1 0x 2 2x 1 (x 1)(x 1)(x 1), (x 1), (x 1)35. 2⏐ 1 029401024200816161 2 5 10 4 8 0x 5 2x 4 5x 3 10x 2 4x 82⏐ 1 2 5 10 4 82 8 6 8 81 4 3 4 4 0x 4 4x 3 3x 2 4x 42⏐ 1 4 3 4 42 12 30 521 6 15 26 562 times36. 1⏐ 1 2111221 1 2 0x 2 x 2 (x 2)(x 1)1 time; 2, 137. f(x) 2x 3 x 2 x kf(1) 2(1) 3 (1) 2 1 k0 2 1 1 k2 k38. f(x) x 3 kx 2 2x 4f(2) (2) 3 k(2) 2 2(2) 40 8 4k 4 40 4k 82 k39. f(x) x 3 18x 2 kx 4f(2) (2) 3 18(2) 2 k(2) 40 8 72 2k 40 2k 6834 k40. f(x) x 3 4x 2 kx 1f(1) (1) 3 4(1) 2 k(1) 10 1 4 k 10 k 44 k109 Chapter 4
41. d(t) v 0 t 1 2 at 225 4t 1 2 (0.4)t 20 0.2t 2 4t 255⏐ 0.2 4 251 250.2 5 ⏐ 0t 5 0t 5 s42. 1⏐ 1 1 7 a b1 2 5 5 a1 2 5 5 a ⏐5 a b2⏐ 1 2 5 5 a 5 a b2 0 10 10 2a1 0 5 5 a ⏐15 a b5 a b 0 5 a b 015 a b 0 5 a 10 020 2b 0 a 5 02b 20a 5b 1043a. V(x) (3 x)(4 x)(5 x)V(x) (12 7x x 2 )(5 x)V(x) x 3 12x 2 47x 6043b. V(x)OV(x) x 3 12x 2 47x 6043c. V w hV 3 4 5 or 60 3 5 V 3 5 (60) 36V(x) x 3 12x 2 47x 6036 x 3 12x 2 47x 6043d. 36 x 3 12x 2 47x 600 x 3 12x 2 47x 24x44b.25O20O15O10O50OV(x)V(x) 2x 3 38x 2 180x4 8 1244c. V(x) 2x 3 38x 2 180x224 2x 3 38x 2 180x44d. 224 2x 3 38x 2 180x0 2x 3 38x 2 180x 2240 x 3 19x 2 90x 1122⏐ 1 19 90 1122 34 1121 17 56⏐ 0 2 in.45. P(3 4i) 0 and P(3 4i) 0 implies that theseare both roots of ax 2 bx c. Since thispolynomial is of degree 2 it has only these tworoots.x 3 4ix 3 4i(x 3) 2 16x 2 6x 9 16x 2 6x 25 0a 1, b 6, c 2546. r 2 5r 8 0r 2 5r 8r 2 5r 2 54 8 2 54r 5 2 2 5 74r 5 2 57 2r 5 2 57 247a. f(x) x 4 4x 3 x 2 4xf(2) (2) 4 4(2) 3 (2) 2 4(2)f(2) 16 32 4 8 or 12; no47b. f(0) (0) 4 4(0) 3 (0) 2 4(0)f(0) 0 0 0 0 or 0; yes47c. f(2) (2) 4 4(2) 3 (2) 2 4(2)f(2) 16 32 4 8 or 36; no47d. f(4) (4) 4 4(4) 3 (4) 2 4(4)f(4) 256 256 16 16 or 0; yes48. f(x) x 5 32x[3, 10] sc11 by [200, 100] sc125about 0.60 ft44a. 1 2 (20 2x) or 10 xw 18 2xh xV(x) (10 x)(18 2x)(x)V(x) (180 38x 2x 2 )(x)V(x) 2x 3 38x 2 180x(0, 32); point of inflection[4, 4] sc11 by [50, 10] sc11049. wider than parent graph and moved 1 unit leftChapter 4 110
6. p: 1, 3q: 1, 2 P q : 1, 3, 1 2 , 3 2 r 2 3 8 31 2 5 3 01 2 1 9 12 1 2 2 4 6 0 1 2 2 2 6 1.53 2 9 19 603 2 3 1 0rational roots: 3, 1 2 , 17. 2 or 0; f(x) 8(x) 3 6(x) 2 23(x) 6f(x) 8x 3 6x 2 23x 6; 1 p q : 1, 2, 3, 6, 1 2 , 3 2 , 1 4 , 3 4 , 1 8 , 3 8 r 8 6 23 61 8 2 21 152 8 10 3 08x 2 10x 3 0(4x 1)(2x 3) 04x 1 0 2x 3 04x 12x 3x 1 4 x 3 2 or 1 1 2 1 1 2 , 1 4 , 28. 1; f(x) (x) 3 7(x) 2 7(x) 15f(x) x 3 7x 2 7x 15; 2 or 0 p q : 1, 3, 5, 15r 1 7 7 151 1 8 15 01 1 6 1 163 1 4 5 05 1 2 3 05, 3, 19. r 2 15 2 x 2 V 1 3 r 2 h1152 1 3 (15 2 x 2 )(15 x)3456 (15 2 x 2 )(15 x)3456 3375 225x 15x 2 x 3x 3 15x 2 225x 81 0Possible rational roots: 1, 9, 81f(x) x 3 15x 2 225x 81 0f(1) 128 f(1) 320f(9) 0 f(9) 2592f(81) 611,712 f(81) 414,720x represents 9 cm.Pages 234–235p10. q :1, 2, 3, 6Exercisesr 1 2 5 61 1 3 2 82 1 4 3 0x 2 4x 3 0(x 3)(x 1) 0x 3, x 1rational roots: 3, 1, 211. p q : 1, 2, 3, 6, 9, 18r 1 2 1 181 1 3 4 142 1 4 9 0x 2 4x 9 0does not factorrational root: 212. p q : 1, 2r 1 5 9 7 21 1 4 5 2 0x 3 4x 2 5x 2r 1 4 5 22 1 2 1 0x 2 2x 1 0(x 1)(x 1) 0x 1, x 1rational roots: 1, 213. p q : 1, 2, 4, 5, 10, 20r 1 5 4 201 1 4 8 121 1 6 2 182 1 3 10 0x 2 3x 10 0(x 5)(x 2) 0x 5, x 2rational roots: 2, 2, 514. p: 1, 3q: 1, 2 p q : 1, 3, 1 2 , 3 2 r 2 1 0 6 3 1 2 2 0 0 6 02x 3 6 0x 3 3x 3 3rational root: 1 2 Chapter 4 112
15. p: 1q: 1, 2, 3, 6 p q : 1, 1 2 , 1 3 , 1 6 r 6 35 1 7 1 1 2 6 38 18 2 06x 3 38x 2 18x 2r 6 38 18 2 1 3 6 36 6 06x 2 36x 6 0x 2 6x 1 0does not factorrational roots: 1 3 , 1 2 16. 4; 3 or 1;f(x) (x) 4 2(x) 3 7(x) 4(x) 15f(x) x 4 2x 3 7x 4x 15; 1 negative1 positive17. f(x) x 3 7x 60 or 2 negativer 1 0 7 61 1 1 6 121 1 1 6 0x 2 x 6 0(x 3)(x 2) 0x 3, x 2rational zeros: 2, 1, 318. 1 positivef(x) x 3 2x 2 81 negativef(x) x 3 2x 2 8x0 x(x 2 2x 8) x(x 4)(x 2)x 0, x 4, x 2rational zeros: 2, 0, 419. 1 positivef(x) x 3 3x 2 10x 242 or 0 negativer 1 3 10 243 1 6 8 14x 2 6x 8 0(x 4)(x 2) 0x 4, x 2rational zeros: 4, 2, 320. 2 or 0 positivef(x) 10x 3 17x 2 7x 21 negativer 10 17 7 2 1 2 10 22 4 010x 2 22x 4 05x 2 11x 2 0(5x 1)(x 2) 0x 1 5 , x 2rational zeros: 1 2 , 1 5 , 221. 2 or 0 positivef(x) x 4 2x 3 9x 2 2x 82 or 0 negativer 1 2 9 2 81 1 3 6 8 0x 3 3x 2 6x 8r 1 3 6 81 1 2 8 0x 2 2x 8 0(x 4)(x 2) 0x 4, x 2rational zeros: 4, 1, 1, 222. 2 or 0 positivef(x) x 4 5x 2 42 or 0 negativer 1 0 5 0 41 1 1 4 4 0x 3 x 2 4x 4r 1 1 4 41 1 0 4 0x 2 4 0(x 2)(x 2) 0x 2, x 2rational zeros: 2, 1, 1, 223a. f(x) (x 2)(x 2)(x 1) 20 (x 2)(x 2)(x 1) 2x 2 0 x 2 0 (x 1) 2 0x 2 x 2 x 1 0x 123b. f(x) (x 2)(x 2)(x 1) 2f(x) (x 2 4)(x 2 2x 1)f(x) x 4 2x 3 3x 2 8x 423c. 1 positivef(x) x 4 2x 3 3x 2 8x 43 or 1 negative23d. There are 2 negative zeros, but according toDescartes’ Rule of Signs, there should be 3 or 1.This is because 1 is actually a zero twice.24a. Let the length.w 4h 2 1V() w hV() ( 4)(2 1)V() ( 2 4)(2 1)V() 2 3 9 2 4113 Chapter 4
24b. V() 2 3 9 2 42208 2 3 9 2 424c. 2208 2 3 9 2 40 2 3 9 2 4 2208r 2 9 4 220812 2 15 184 0 12 w 4 h 2 1w 12 4 or 8 h 2(12) 1 or 2312 in. 8 in. 23 in.25a. Sample answer: x 4 x 3 x 2 x 3 025b. Sample answer: x 3 x 2 2 025c. Sample answer: x 3 x 026a. Let the length.h 9V() 1 3 BhV() 1 3 ( 2 )( 9)V() 1 3 3 3 226b. V() 1 3 3 3 26300 1 3 3 3 226c. 6300 1 3 3 3 20 1 3 3 3 2 63000 3 9 2 18,900r 1 9 0 18,90030 1 21 630 0 30 h 9h 30 9 or 21base: 30 in. by 30 in., height: 21 in.27. d 0.0000008x 2 (200 x)0.8 0.0000008x 2 (200 x)0 0.00016x 2 0.0000008x 3 0.80 8x 3 1600x 2 8,000,0000 x 3 200x 2 1,000,000r 1 200 0 1,000,000100 1 100 10,000 0x 100 ft28. The graphs are reflections of each other over thex-axis. The zeros are the same.29. 7⏐ 1 1 567 561 8 0x 830. b 2 4ac 6 2 4(4)(25)364; 2 imaginary31. (x 1)(x (1))(x 2)(x (2)) 0(x 1)(x 1)(x 2)(x 2) 0(x 2 1)(x 2 4) 0x 4 5x 2 4 032. y 4.3x 8424.3y 4.3(2008) 8424.3y $210.1033. 2x 3x 3 x22(2x 3) x(3 x)4x 6 3x x 2x 2 x 6 0(x 3)(x 2) 0x 3 0 x 2 0x 3 x 2The correct choice is A.Page 235 Mid-Chapter Quiz1. (x 1)(x (1))(x 2i)(x (2i)) 0(x 1)(x 1)(x 2i)(x 2i) 0(x 2 1)(x 2 4) 0x 4 3x 2 4 02. 3; x 3 11x 2 30x 0x(x 2 11x 30) 0x(x 6)(x 5) 0x 0 x 6 0 x 5 0x 6 x 53. x 2 5x 150x 2 5x 2 54 150 2 54x 5 2 2 62 54x 5 2 2 5225x 5 225x 5 2x 5 2x 10x 154. b 2 4ac (39) 2 4(6)(45) 441; 2 real rootsb b 39 139 441 2(6) 212 21225b 39 1 or b 39 1b 5b 3 2 5. 2⏐ 1 3222881 1 4 0x 2 x 46. 4⏐ 1 4420681 0 2 22; no7. 1⏐ 1 2151661 1 6 0x 2 x 6 (x 3)(x 2)(x 3)(x 1)(x 2)8. p q : 1, 3r 1 6 10 33 1 3 1 0x 2 3x 1 0does not factorrational root: 3 212Chapter 4 114
9. 1 positiveF(x) x 4 4x 3 3x 2 4x 43 or 1 negativex 3 5x 2 8x 4 0x 2 4x 4 0(x 2)(x 2) 0x 2, x 2rational zeros: 2, 1, 110. Let r radius.h r 6V 1 3 r 2 h27 1 3 r 2 (r 6)0 1 3 r 3 2r 2 270 r 3 6r 2 81r 3 h r 6h 3 6 or 9r 3 cm, h 9 cm4-5r 1 4 3 4 41 1 5 8 4 0r 1 5 8 41 1 4 4 0r 1 6 0 813 1 9 27 0Locating Zeros of a PolynomialFunctionPages 239–240 Check for Understanding1. If the function is negative for one value andpositive for another value, the function must crossthe x-axis in at least one point between the twovalues.f(x)f(b)Of(a)a(a, f(a))(b, f(b))bx2. Use synthetic division to find the values of thepolynomial function for consecutive integers.When the values of the function change frompositive to negative or from negative to positive,there is a zero between the integers.3. Use synthetic division to find the values of thepolynomial function for consecutive integers. Aninteger that produces no sign change in thequotient and the remainder is an upper bound.To find a lower bound of a function, find an upperbound for the function of x. The lower bound isthe negative of the upper bound for the functionof x.4. Nikki; the sign changes between 2 and 1.5. r211462101 1 5 30 1 4 21 1 3 52 1 2 63 1 1 54 1 0 25 1 1 34 and 5, 1 and 06. r 1 3 2 42 1 5 8 121 1 4 2 20 1 3 2 41 1 2 4 0 ←2 1 1 4 43 1 0 2 24 1 1 2 122 and 1, at 1, 3 and 47. r0224400331 2 2 2 52 2 0 0 33 2 2 6 15approximate zero: 2.38. r21131201 1 2 0zeros: 2, 19. Sample answer:r 1 0 0 8 21 1 1 1 7 52 1 2 4 0 2upper bound: 2f(x) x 4 8x 2r 1 0 0 8 20 1 0 0 8 2lower bound: 010. Sample answer:r 1 0 1 0 31 1 1 2 2 12 1 2 5 10 17upper bound: 2f(x) x 4 x 2 3r 1 0 1 0 31 1 1 2 2 12 1 2 5 10 17lower bound: 211a. Let x amount of increase.V(x) (25 x)(30 x)(5 x)V(x) (750 55x x 2 )(5 x)V(x) x 3 60x 2 1025x 3750115 Chapter 4
11b. V w h 1.5V 1.5(3750)V 25(30)(5) 1.5V 5625V 3750V(x) x 3 60x 2 1025x 37505625 x 3 60x 2 1025x 375011c. 5625 x 3 60x 2 1025x 37500 x 3 60x 2 1025x 1875r 1 60 1025 18751 1 61 1086 7892 1 62 1149 423x 1.725 x 25 1.7 30 x 30 1.7 26.7 31.75 x 5 1.7 6.7about 26.7 cm by 31.7 cm by 6.7 cmPages 240–242 Exercises12. r1110101230 1 0 0 21 1 1 1 12 1 2 4 63 1 3 9 251 and 213. r12257180 2 5 11 2 3 22 2 1 13 2 1 40 and 1, 2 and 314. r21124081152281 1 3 3 2 00 1 2 0 1 21 1 1 1 0 12 1 0 0 1 0at 1, at 215. r 1 0 8 0 103 1 3 1 3 192 1 2 4 8 6 1 1 1 7 7 30 1 0 8 0 101 1 1 7 7 32 1 2 4 8 63 1 3 1 3 193 and 2, 2 and 1, 1 and 2, 2 and 316. r 1 0 3 12 1 2 1 11 1 1 2 30 1 0 3 11 1 1 2 12 1 2 1 32 and 1, 0 and 1, 1 and 217. r3220611936031832 2 4 9 21 451 2 2 3 6 90 2 0 1 3 31 2 2 3 0 32 2 4 9 16 35no real zeros18. r6662412541831115 6 6 24 117yes; f(6) 111, f(5) 11719–25. Use the TABLE feature of a graphingcalculator.19. 0.7, 0.7 20. 2.6, 0.421. 2.5 22. 0.4, 3.423. 1, 1 24. 1.3, 0.9, 7.425. 1.2426. Sample answers:r 3 2 5 11 3 1 6 5upper bound: 1f(x) 3x 3 2x 2 5x 1r 3 2 5 10 3 2 5 1lower bound: 027. Sample answers:r 1 1 11 1 0 12 1 1 1upper bound: 2f(x) x 2 x 1r 1 1 11 1 2 1lower bound: 128. Sample answers:r 1 6 2 6 131 1 5 3 3 102 1 4 6 6 253 1 3 7 15 584 1 2 6 18 855 1 1 3 9 586 1 0 2 18 95upper bound: 6f(x) x 4 6x 3 2x 2 6x 13r 1 6 2 6 131 1 7 9 3 102 1 8 18 30 47lower bound: 2Chapter 4 116
29. Sample answers:r 1 5 3 201 1 6 3 172 1 7 11 2upper bound: 2f(x) x 3 5x 2 3x 20r 1 5 3 201 1 4 7 132 1 3 9 23 1 2 9 74 1 1 7 85 1 0 3 56 1 1 3 38lower bound: 630. Sample answers:r 1 3 2 3 51 1 2 4 1 62 1 1 4 5 153 1 0 2 3 144 1 1 2 11 39upper bound: 4f(x) x 4 3x 3 2x 2 3x 5r 1 3 2 3 51 1 4 2 1 62 1 5 8 13 21lower bound: 231. Sample answers:r 1 5 3 20 0 151 1 6 3 23 23 8upper bound: 1f(x) x 5 5x 4 3x 3 20x 2 15r 1 5 3 20 0 151 1 4 7 27 27 82 1 3 9 38 76 1373 1 2 9 47 141 4084 1 1 7 48 192 7535 1 0 3 35 175 8606 1 1 3 2 12 577 1 2 11 57 399 2808lower bound: 732a. 4 32b. 1, 532c. 3 or 1; f(x) x 4 3x 3 2x 2 3x 51 negative real zero32d. r 1 3 2 3 55 1 2 8 43 2104 1 1 2 11 393 1 0 2 3 142 1 1 4 5 151 1 2 4 1 60 1 3 2 3 51 1 4 2 1 62 1 5 8 13 213 1 6 16 45 1302 and 1, 3 and 432e. Sample answers:upper bound: 4 (See table in 32d.)f(x) x 4 3x 3 2x 2 3x 5r 1 3 2 3 51 1 4 2 1 62 1 5 8 13 21lower bound: 232f. 1.4, 3.4 (Use TABLE feature of a graphingcalculator.)33a. 1890: P(0) 0.78(0) 4 133(0) 3 7500(0) 2 147,500(0) 1,440,000 1,440,0001910: P(20) 0.78(20) 4 133(20) 3 7500(20) 2 147,500(20) 1,440,000 2,329,2001930: P(40) 0.78(40) 4 133(40) 3 7500(40) 2 147,500(40) 1,440,000 1,855,2001950: P(60) 0.78(60) 4 133(60) 3 7500(60) 2 147,500(60) 1,440,000 1,909,2001970: P(80) 0.78(80) 4 133(80) 3 7500(80) 2 147,500(80) 1,440,000 1,387,200The model is fairly close, although it is lessaccurate at for 1950 and 1970.33b. 1980 1890 90P(90) 0.78(90) 4 133(90) 3 7500(90) 2 147,500(90) 1,440,000P(90) 253,80033c. The population becomes 0.33d. No; there are still many people living inManhattan.34. Sample answer:f(x) (x 2)(x 2)(x 1)f(x) (x 2 2)(x 1)f(x) x 3 x 2 2x 2; 2, 1f(x)f(x) x 3 x 2 2x 2O35a. 37.44 60x 3 60x 2 60x35b. f(x) 60x 3 60x 2 60x 37.4435c. f(x)f(x) 4O60x 3 60x 2 60x 37.442Oabout 1 2 35d. 0.4 (Use TABLE feature of a graphingcalculator.)36. Sample answer: f(x) x 2 1x2 1 O 1 2x2040117 Chapter 4
37a. f (x)100,00080,00060,00040,0002O,000Of(x) 0.125x 5 3.125x 4 40004 8 12 16 20 24x42. 7 9 7(6) 3(9) or 153 643. (x, y) 3 82, 2 42 (2.5, 1)44. x 2y 4 0y 1 2 x 2; 1 2 ; 245.A37b. f(0) 0.125(0) 5 3.125(0) 4 4000f(0) 4000 deer37c. 1920 1905 15f(15) 0.125(15) 5 3.125(15) 4 4000f(15) 67,281.25about 67,281 deer37d. in 193038a. 81.58 6x 4 18x 3 24x 2 18x38b. 81.58 6x 4 18x 3 24x 2 18x0 6x 4 18x 3 24x 2 18x 81.58about 1.1 (Use TABLE feature of a graphingcalculator.)38c. x rate 11.1 rate 10.1 rateabout 10%39. 2 or 0; f(x) 2x 3 5x 2 28x 151 negative zeror 2 5 28 153 2 11 5 02x 2 11x 5 0(2x 1)(x 5) 0x 0.5 x 5rational zeros: 3, 0.5, 540. d(t) v 0 t 1 2 gt 241.yO1750 4t 1 2 (9.8)t 24.9t 2 4t 1750 0t b b2 ac 4 2at 4 (4)2 .9)(14(4750)2(4.9)69.84 34,31 6 9.8t 4 34,31t 4xy x 1xt 4 34,31 6 9.8t 19.3 t 18.5about 19.3 sC By x cannot be true.The correct choice is B.4-6xyRational Equations and PartialFractionsPage 247 Check for Understanding1. Multiply by the LCD, 6(b 2). Then, solve theresulting equation.2. If a possible solution causes a denominator toequal 0, it is not a solution of the equation.3. Decomposing a fraction means to find twofractions whose sum or difference equals theoriginal fraction.224. x x 2 2 x 222x x 2 (x 2) 2 x 2 (x 2)x(x 2) 2 2(x 2) 2x 2 2x 2 2x 4 2x 2 4x 4 0(x 2)(x 2) 0x 2 0 x 2 0x 2 x 2If you solve the equation, you will get x 2.However, if x 2, the denominators will equal 0.5. b 5 b 4b 5 b b (4)bb 2 5 4bb 2 4b 5 0(b 5)(b 1) 0b 5 0 b 1 0b 5b 1D9 36. b 5 b 3 9b 5 (b 5)(b 3) 3b 39(b 3) 3(b 5)9b 27 3b 156b 42 0b 7 (b 5)(b 3)Chapter 4 118
7. t 4 3 16t t 4 t 2 4t t 4 3t t 4 (t)(t 4) 16t 2 4t(t 4)(t 4) 3(t) 16t 2 16 3t 16t 2 3t 0t(t 3) 0t 0 t 3 0t 3But t 0, so t 3.8. 3 p 1 3p 1p 2 1 (p 1) (p 1) 3 p 1 A Bp 2 1 p 1 p 1 (t)(t 4)3p 1 A(p 1) B(p 1)Let p 1.3(1) 1 A(1 1) B(1 1)2 2B1 BLet p 1.3(1) 1 A(1 1) B(1 1)4 2A2 A 3 p 1 2 1p 2 1 p 1 p 19. 5 1 x 1 ;x6exclude: 05x 1 165x 15x 31 16Test x 1: 5 ( 1) ( 1)4 16 trueTest x 1: 5 1 1 1 616 16 falseTest x 4: 5 1 4 1 645 1 4 4 true<strong>Solution</strong>: x 0, x 3510. 1 a 1 7 6 ; exclude: 16(a 1) 30 7(a 1)6a 6 30 7a 731 a5Test a 1: 1 1 1 7 6 3 2 7 6 true5Test a 2: 1 2 7 6 6 7 6 false5Test a 36: 1 36 1 7 6 1 1 7 7 6 <strong>Solution</strong>: a 1, a 3111a. 3 60 203 x11b. 3 60 203 x 57.141 57.14 4 842 4 942 true3 60 20 57.14(3 x)200 171.42 57.14x0.50 x; 0.50 hPages 247–250Exercises12. 1 t2 t 8 0 1 t2 t 8t (0)t12 t 2 8t 0t 2 8t 12 0(t 6)(t 2) 0t 6 0 t 2 0t 6 t 2 34m 2113. m m 21 2m 2 m 22m2 mm 2 34m 2 34m 2m 20 m 2 34m0 m(m 34)m 0 m 34 0m 34But m 0, so m 34.214. y 2 3 y y y 2 2y 2 3 y (y)(y 2) yy 2 (y)(y 2)2y 3(y 2) y 25y 6 y 2y 2 5y 6 0( y 3)(y 2) 0y 3 0 y 2 0y 3y 2But y 2, so y 3.15. n 2 0n1 2 1 2 n5n110 1 2 n n 51 2 n 5n 1n51 (n 1)(n 1) 2 n (n 1)(n 1)10 (2n 5)(n 1) (2n 5)(n 1)2n 2 3n 5 2n 2 3n 56n 10n 5 3 1 1 316. b 2 b 2 b 1 1 1b 2 b 2 (b 2)(b 1) 3b 1b 1 b 1 3(b 2)2b 2 3b 64 b (b 2)(b 1)7a5 3a17. 3a 3 4a 4 2a 27a5 3a 3(a 1) 4(a 1) 2(a 1) 7a53(a 1) 4(a 1) (12)(a 1)(a 1) 3a2(a 1) (12)(a 1)(a 1)4(a 1)7a 3(a 1)5 6(a 1)3a28a 2 28a 15a 15 18a 2 18a10a 2 25a 15 02a 2 5a 3 0(2a 1)(a 3) 02a 1 0 a 3 0a 1 2 a 3119 Chapter 4
18. 1 1 1aa a 11 aa a 11(1 a)(a 1) 1 a 1 a 2 a a 1 a(1 a)a 2 2a 1 a 2 2a 10 0all reals except 12q2q19. 2q 3 2q 3 2q2q2q 3 2q 3q 1 (1 a)(a 1) (2q 3)(2q 3) 1(2q 3)(2q 3)2q(2q 3) 2q(2q 3) (2q 3)(2q 3)4q 2 6q 4q 2 6q 4q 2 90 4q 2 12q 912 144 9) 4(4)(2 412 288812 122 83 32 2120. 3 m 6m 9 3m 33m 4m1 9 3 3m 6m (12m) 3m 3m4m(12m)4 4(6m 9) 3(3m 3)4 24m 36 9m 915m 2347 321. x 1 2 x x 14 x 1m 2 135 (x 1)(2 x)(x 1) 2 73 x x 1(x 1)(2 x)(x 1)4(2 x)(x 1) 7(x 1)(x 1) 3(x 1)(2 x)4(x 2 x 2) 7(x 2 1) 3(x 2 3x 2)4x 2 4x 8 4x 2 9x 135 13x 153 x22a. (n 1)(n 2) 22b. 1, 2 6 4 n 22c. 1 n n 1 21 n 64(n 1)(n 2) n 1n (n 1)(n 2)2(n 1)(n 2) (n 2)(n 6) 4(n 1)n 2 n 2 n 2 4n 12 4n 42n 2 n 18 0n 1 1 (2)(1 48)2 2 1 145 4x 6 x 623. x 2 2xx( x 2)x 6x 2 A 2xx Bx 2x 6 A(x 2) B(x)Let x 2.2 6 A(2 2) B(2)4 2B2 BLet x 0.0 6 A(0 2) B(0)6 2A3 Ax 6x 2 3 2xx 2x 224. 5 m 4 5m 4m 2 4 (m 2)( m 2) 5 m 4 A Bm 2 4 m 2 m 25m 4 A(m 2) B(m 2)Let m 2.5(2) 4 A(2 2) B(2 2)6 4B1.5 BLet m 2.5(2) 4 A(2 2) B(2 2)14 4A3.5 A 5 m 4 3.51.5m 2 4 m 2 m 24y4y 4y 1 (3y 1)(y 1)4yA B 3y 2 4y 1 3y 1 y 125. 3y 2 26. 9x 2 4y A(y 1) B(3y 1)Let y 1.4(1) A(1 1) B(3(1) 1)4 2B2 BLet y 1 3 .4 1 3 A 1 3 4 3 2 A 32 A 1 B3 1 3 14y2 2 3y 2 4y 1 3y 1 y 19x9 9x 9 (x 3)(x 3) 9 9xA Bx 2 9 (x 3) (x 3)Let x 3.9 9(3) A(3 3) B(3 3)18 6B3 BLet x 3.9 9(3) A(3 3) B(3 3)36 6A6 A 9 9x63x 2 9 x 3 x 363 x, 3 x 327a. a(a 6)Chapter 4 120
27b. a 2 a 4a a 6 a 2(a)(a 6) a 4aa 6x 2 1630. 0x 2 4x 5(a)(a 6)x 2 16 (x 0; exclude 5, 15)(x 1)(a 2)(a 6) (a 4)(a)x 2 16 0a 2 8a 12 a 2 4ax 2 1612 4ax 43 a(5)Test x 5: 2 16 0(5 5)(5 1)27c. 0, 69 27d. Test a 1: 1 2 1 1 44 0 true0 (2)1 6Test x 2: 2 16 03 5 7 false(2 5)(2 1) 12 Test a 1: 1 2 1 47 0 false1 1 60 16Test x 0: 1 3 5 true0 2 0 4(0) 5 16 0 trueTest a 4: 4 2 4 454.54 4 6Test x 4.5: 2 16 0 1 2 0 false(4.5 5)(4.5 1)4.25 Test a 7: 7 2 7 4 0 false2.757 7 66Test x 6: 2 16 0 5 7 3 true6 2 24 5 2 0 0 true7<strong>Solution</strong>: 0 a 3, 6 a<strong>Solution</strong>: x 4, 1 x 4, x 5228. w 3 29 ; exclude: 0w131. 4 a 5 8 a 1 2 ; exclude: 02 3w 291w 94 a 5 8 a 1 2 229Test w 1: 3 1 2 5 4a11 29 true 7Test w 1: 2 4 a1 3 2 91 51Test a 1: 4( 1) 8( 1) 1 2 5 29 false2Test w 10: (1 3 2 90) 10 7 8 1 2 false 3 210 2 91 5 trueTest a 1: 104( 1) 8( 1) 1 2 <strong>Solution</strong>: w 0, w 9 7 8 1 2 true( x 3)( x 4)29. 0; exclude 5, 6( x 5)( x 6)21 5Test a 2: 4( 2) 8( 2) 1 2 (x 3)(x 4) 07 x 3 0 x 4 01 6 1 2 falsex 3 x 4<strong>Solution</strong>: 0 a 7( 0 3)( 0 4)4 Test x 0: 0( 0 5)( 0 6)212 0 true180(3.5 3)(3.5 4)Test x 3.5: 0(3.5 5)(3.5 6) 20.25 0 false9.375(4.5 3)(4.5 4)(4.5 5)(4.5 6) 20.75 1.125(5.5 3)(5.5 4)(5.5 5)(5.5 6) 23.75 0 . 1256.5 3)(6.5 4)(6.5 5)(6.5 6) 2Test x 4.5: 0 0Test x 5.5: 0 0Test x 6.5:( 08.75 0 . 375 0<strong>Solution</strong>: x 3, 4 x 5truefalsefalse121 Chapter 4
1 1 832. 2b 1 b 1 1 51 1 8 2b 1 b 1 1 5; exclude: 1 2 , 115(b 1) 15(2b 1) 8(2b 1)(b 1)45b 30 16b 2 24b 80 16b 2 21b 220 (16b 11)(b 2)16b 11 0 b 2 011 b 1 6 b 211 8Test b 2: 2(2 ) 1 2 1 1 5Test b 0.8: 2(0.18) 1Test b 0.6: 2(0.16) 1Test b 0: 2(0)1 1Test b 3: 2(3)1 1 4 3 8 1 5false1 8 (0.8 ) 1 1 5 1 0 83 1 51 8 (0.6 ) 1 1 5 5 2 8 1 51 8 0 1 1 582 1 51 8 3 1 1 511 8 2 8 1 5truefalse11 <strong>Solution</strong>: 1 b , 1 6 1 2 b 2733. y 1 7; exclude 17 7(y 1)1 y 10 y7Test y 2: 2 1 77 7 false7Test y 0.5: 0.5 1 714 7 true7Test y 1: 1 71 7 2 7 false<strong>Solution</strong>: 1 y 034. Let x the number.4 1 x x 10 2 5 20 5x 2 52x5x 2 52x 20 0(5x 2)(x 10) 05x 2 0 x 10 0x 2 5 x 10truefalse35. x 2x 5 x 2x 5 0.30 0.30; exclude 5x 2 0.30(x 5)x 2 0.30x 1.50.7x 3.5x 5Test x 6: 6 26 5 0.300.36 0.30 trueTest x 0: 0 20 5 0.300.4 0.30 falseTest x 6: 6 26 5 0.308 0.30 true<strong>Solution</strong>: x 5 or x 536a. 1 8 1 1 d 336b. 1 8 d1i2 1 8 (32d i ) d1 i 3 21 1 i 3 24d i 32 d i3d i 32d i 10 2 3 cm37. Sample answer: x x (32di )13 x 238. Let x capacity of larger truck. 5 2 x x 35(x 3) 2x5x 15 2x3x 15x 5 tons1 139a. 1 0 2 r39b. 11 110 1 r 2110 2 r1r0 1 r 21 (20r) 2 1 r 1 2 0 (20r)2r 10 20 rr 302r 2(30) or 60; 60 ohms, 30 ohms40. Let x the number of quiz questions to beanswered.11 x 2 0 x 0.70011 x 0.70(20 x)11 x 14 0.70x0.3x 3x 10 questions41. Let x the speed of the wind.1062738 20 0 x 20 0 x1062(200 x) 738(200 x)212,400 1062x 147,600 738x64,800 1800x36 x; 36 mphChapter 4 122
42. a1 1 b 1 c a1 1 b (a)(b)(c) 1 c (a)(b)(c)bc ac abbc ab acbc a(b c)bc b a c43a. 1 x 1 2 1 y 1 z 1 x 1 2 31 10 4 5 1 10 4 5 1 x 1 1 6 0 9 01 10 9 043b. 1 x 1 2 3 1 x (360x) 6 (360x)360 6x 4x360 10x36 x44. Let x number of gallons of gasoline. 20 mg 15,0 00mx g583 1 3 y 15,000y 25.7; about 25.7 mpg45. T d s 10 2 3 s26 26 5 s 52626 5 s 5 (3)(s 5)(s 5)10 2 3 (3)(s 5)(s 5) s32(s 5)(s 5) 26(3)(s 5) 26(3)(s 5)32s 2 800 78s 390 78s 39032s 2 156s 800 08s 2 39s 200 0(8s 25)(s 8) 08s 25 0 s 8 05s 2 8s 88 mph46. 3x 5 5yy 3 x5 y 5 y5y47.20x 15,000x 750 gallons750 $1.20 $900x $1.20 $900 $200x 583 1 3 gallonsLet y number of miles per gallon. y 15,000 mgm583 1 3 g 11 1 10r 1 2 3 53 1 1 0 52 1 0 3 11 1 1 4 10 1 2 3 51 1 3 0 52 1 4 5 548. 5⏐ 1 0 30 05 25 251 5 5 ⏐25no49. 2; 12x 2 8x 15 0(6x 5)(2x 3) 06x 5 0 2x 3 0x 5 6 x 3 2 50. ⏐3x⏐ 12 0⏐3x⏐ 123x 12 3x 12x 4x 451. y 2x 3x3 ? 2(6) 363 5 2 falseno52. y 2 121x 2 → b 2 121a 252a. (b) 2 121a 2 52b. b 2 121(a) 2b 2 121a 2 yes b 2 121a 2 yes52c. (a) 2 121(b) 2 52d. (a) 2 121(b) 2a 2 121b 2 no a 2 121b 2 no53a. Let x short answer questions and y essay questions.x y 2020y2x 12y 60x 0y 01612x y 208 2x 12y 60(0, 5)4(18, 2)x 0xO 4 8 12 16 20(0, 0) y 0 (20, 0)S(x, y) 5x 15yS(0, 0) 5(0) 15(0) or 0S(0, 5) 5(0) 15(5) or 75S(18, 2) 5(18) 15(2) or 120S(20, 0) 5(20) 15(0) or 10018 short answer and 2 essay for a score of120 points3 and 2, 2 and 1, 1 and 2123 Chapter 4
53b. Let x short answer questions and y essay questions.yx y 20202x 12y 12016x 0x y 20y 012 (0, 10)2x 12y 1208(12, 8)x 04xO 4 8(0, 0) y 012 16 20(20, 0)S(x, y) 5x 15yS(0, 0) 5(0) 15(0) or 0S(0, 10) 5(0) 15(10) or 150S(12, 8) 5(12) 15(8) or 180S(20, 0) 5(20) 15(0) or 10012 short answer and 8 essay for a score of 180points54.1 1 3 51 135 x 1(3) 1(5) 1(5)1(3) 1(3) 1(5) 1(5) x0 00 0 x55. y y 1 m(x x 1 )y 1 2(x (3))y 1 2x 62x y 7 056a. m 300 0 500020 6056b. $2000; $50m 50y 3000 50(x 20)y 50x 2000C(x) 50x 200056c.C(x)Cost$4000$3000$2000$100000Televisions Produced57. A of JKL 1 2 (9)(7) or 31.52C(x) 50x 2000A of small triangle 1 2 (5)(3) or 7.5A of shaded region 31.5 7.5 or 24The answer is 24.468x4-7Radical Equations andInequalitiesPages 254–255 Check for Understanding1. To solve the equation, you need to get rid of theradical by squaring both sides of the equation. Ifthe radical is not isolated first, a radical willremain in the equation.2. The process of raising to a power sometimescreates a new equation with more solutions thanthe original equation. These extra or extraneoussolutions do not solve the original equation.3. When solving an equation with one radical, youisolate the radical on one side and then squareeach side. When there is more than one radicalexpression in an equation, you isolate one of theradicals and then square each side. Then youisolate the other radical and square each side. Inboth cases, once you have eliminated all radicalsigns, you solve for the variable.4. 1 t 4 21 4t 44t 3t 3 4 1 4 3 4 1 3 22 2 5. 3 x 4 12 3 Check: 3 x 4 12 3 3 x 4 9 3 733 4 12 3x 4 729 3 729 12 3x 733 9 12 33 3 6. 5 x 4 2 Check: 5 x 4 2x 4 35 13 4 2x 4 9 5 9 2x 13 5 3 2no real solution7. 6x 4 2x 0 16x 4 2x 104x 14x 3.5Check: 6x 4 2x 0 1 2 7 2 106 7 2 4Check: 1 t 4 221 4 7 0 117 17 8. a 4 a 3 7a 4 7 a 3a 4 49 14a 3 a 342 14a 31764 196(a 3)9 a 312 aCheck: a 4 a 3 712 4 12 3 716 9 74 3 7 Chapter 4 124
9. 5x 4 8 5x 4 05x 4 645x 45x 60x 0.8x 12Test x 1: 5(1) 4 81 8 meaninglessTest x 0: 5(0) 4 84 8 trueTest x 13: 5(13) 4 869 8 false<strong>Solution</strong>: 0.8 x 1210. 3 4a 5 10 4a 5 04a 5 7 4a 54a 5 49 a 1.254a 54a 13.5Test a 0: 3 4(0) 5 103 5 10 meaninglessTest a 2: 3 4(2) 5 104 3 10 trueTest a 14: 3 4(14) 5 103 51 10 false<strong>Solution</strong>: 1.25 a 13.511a. v v 64h 2 0 90 10 2 64h90 100 64h11b. 90 100 64hCheck: 90 100 64h8100 100 64h 90 100 5) 64(128000 64h 90 8100 1125 h; 125 ft 90 90 Pages 255–257 Exercises12. x 8 5 Check: x 8 5x 8 2517 8 5x 17 25 55 5 13. 3 y 7 4 Check: 3 y 7 4y 7 64 3 71 7 4y 71 3 64 44 4 14. 8n 5 1 28n 5 3Check: 8n 5 1 2 87 4 5 1 28n 5 914 5 1 28n 14 3 1 2n 7 4 2 2 15. x 16 x 4x 16 x 8x 160 8x0 x0 xCheck: x 16 x 40 6 1 0 416 44 4 16. 43m 2 15 43m 2 15 13m 2 15 13m 2 16m 2 1 63m 4 3 3Check: 43m 2 15 44 3 4 3 32 15 441 44 4 Check: 43m 2 15 44 3 4 3 32 15 441 44 4 17. 9u 4 7u 20 9u 4 7u 202u 16u 8Check: 9u 4 7u 20 9(8) 4 7(8) 2076 76 no real solution18. 3 6u 5 2 3 3 6u 5 56u 5 1256u 120u 20Check: 3 6u 5 2 3 3 6(20 ) 5 2 3 3 125 2 35 2 33 3 19. 4m 2 3m 2 2m 5 04m 2 3m 2 2m 54m 2 3m 2 4m 2 20m 2523 23m1 mCheck: 4m 2 3m 2 2m 5 04(1) 1) 2 3( 2 2(1) 5 09 2 5 03 3 00 0 20. k 9 k 3k 9 3 kk 9 3 23k k6 23k36 4(3k)36 12k3 kCheck: k 9 k 33 9 3 312 3 323 3 33 3 125 Chapter 4
21. a1 2 1 a2 1a 21 2a1 2 1 a 122a1 2 104(a 21) 100a 21 25a 4Check: a1 2 1 a2 14 1 2 1 4 2 125 1 165 1 44 4 22. 3x 4 2x 7 33x 4 3 2x 73x 4 9 62x 7 2x 7x 2 62x 7x 2 4x 4 36(2x 7)x 2 68x 256 0(x 4)(x 64) 0x 4 0 x 64 0x 4 x 64Check: 3x 4 2x 7 33(4) 4 2(4) 7 316 1 34 1 33 3 Check: 3x 4 2x 7 33(64) 4 2(64) 7 3196 121 314 11 33 3 23. 2 3 7b 1 4 02 3 7b 1 48(7b 1) 647b 1 87b 9b 9 7 Check:2 3 7b 1 4 02 3 7 9 7 1 4 02 3 8 4 04 4 00 0 24. 4 3t 2 0 Check: 4 3t 2 0 4 3t 2 4 3 1 63 2 03t 16 4 16 2 0t 1 632 2 00 0 25. x 2 7 x 9x 2 14x 2 49 x 914x 2 42196(x 2) 1764x 2 9x 7Check: x 2 7 x 97 2 7 7 93 7 44 4no real solution26. 2x 1 2x 6 52x 1 5 2x 62x 1 25 102x 6 2x 630 102x 63 2x 69 2x 63 2x 3 2 xCheck: 2x 1 2x 6 52 3 2 1 2 3 2 6 54 9 52 3 55 5 27. 3x 0 1 x 11 13x 10 x 11 2x 11 12x 2 2x 11x 1 x 11x 2 2x 1 x 11x 2 3x 10 0(x 5)(x 2) 0x 5 0 x 2 0x 5x 2Check: 3x 0 1 x 11 13(5) 10 5 1 1 125 16 15 4 15 3Check: 3x 0 1 x 11 13(2) 10 2 11 14 9 12 3 12 2 <strong>Solution</strong>: x 228a. 3t 4 1 t 63t 4 1 6 t3t 14 36 12t t 20 t 2 15t 500 (t 5)(t 10)t 5 0 t 10 0t 5 t 10Check: 3t 4 1 t 63(5) 14 5 61 5 61 5 66 6 Check: 3t 4 1 t 63(10) 14 10 616 10 64 10 614 61028b. 5Chapter 4 126
29. 2x 7 52x 7 252x 32x 162x 7 02x 7Test x 0: 2(0) 7 5 34. m 27 5meaninglessTest x 4: 2(4) 7 58 7 51 5x 7 2 false 3m 4 0Test x 17: 2(17) 7 53m 427 5m 4 3true<strong>Solution</strong>: x 1630. b 4 6 Test b 5: 5 4 6b 4 361 6b 32meaninglessb 4 0Test b 0: 0 4 6b 44 62 6trueTest b 33: 33 4 637 6false<strong>Solution</strong>: 4 b 3231. a 5 4 Test a 0: 0 5 4a 5 165 4a 21meaninglessa 5 0Test a 6: 6 5 4a 51 41 4trueTest a 22: 22 5 417 4false<strong>Solution</strong>: 5 a 2132. 2x 5 6 Test x 0: 2(0) 5 62x 5 365 62x 41meaninglessx 20.5 Test x 5: 2(5) 5 62x 5 05 62x 5truex 2.5 Test x 22: 2(22) 5 639 6false<strong>Solution</strong>: 2.5 x 20.533. 4 5y 9 2 Test y 0: 4 5(0) 9 25y 9 16 4 9 25y 25meaninglessy 5 Test y 2: 4 5(0) 9 25y 9 0 4 1 25y 9truey 1.8 Test y 6: 4 5(6) 9 2 4 21 2false<strong>Solution</strong>: 1.8 y 5 3m 4m 2 3m 42m 2m 1m 2 0m 2Test m 3: 3 2 3(3) 41 5 meaninglessTest m 1.6: 1.6 2 3(1. 6) 40.4 0.8 meaninglessTest m 1.2: 1.2 2 3(1. 2) 40.8 0.4 falseTest m 0: 0 2 3(0) 42 4 true<strong>Solution</strong>: m 135. 2c 5 7 Test c 0: 2(0) 5 72c 5 495 72c 54meaninglessc 27 Test c 5: 2(5) 5 72c 5 05 72c 5falsec 2.5 Test c 28: 2(28) 5 751 7true<strong>Solution</strong>: c 2736a. t 2 sg3 2(7 .2)g3 14 .4g 36b. 3 14 .4g9 14 .4g9g 14.4g 1.6 m/s 237. x 5 3 x 3x 5 3 (x ) 32x 5 3 xx 2 6 9(x 5) 3 x 2 6x 9x 3 15x 2 75x 125 x 2 6x 9x 3 16x 2 81x 134 0Use a graphing calculator to find the zero.[2, 10] sc11 by [10, 10] sc11about 7.8838a. s 30fd s 30(0.6 )(25) s 450 s 21.2 mph38b. s 30fd 35 30(0.6 )d1225 18d68.06 d; about 68 ft127 Chapter 4
38c. No; it is not a linear function. t (200)22 50 108 t 2002 t 20 0 22 28 20 10 14250048. 4(a b c) 4(6)→ t 416 t2 400t 40,000 4 00 2544. p q : 6, 3, 2, 139a. T 2 g39b. t 2 gr 1 5 5 5 6T 2 19.8 T 2 18.9 1 1 6 11 6 0T 2.01 sT 2.11 sx 3 6x 2 11x 6 039c. Let x the new length of the pendulum.r 1 6 11 622 g 2 x g1 1 5 6 04 g 2 x x 5x 6 0g(x 3)(x 2) 02 g x x 3 0 x 2 0gx 3x 2 x4 g g 3, 2, 1, 14 x45a. point discontinuityIt must be multiplied by 4.45b. jump discontinuity40. T aT r 3 a b rb45c. infinite discontinuitypv 2 25687 67,20 0 ,000 346a. p 10Or bw50,625 4 71,969 3.03 1023p 10 56 w10Orb 50,625r 3 b 1.43 10 29Ot 2 832t 173,056 t 400t 50,000 4 44(a b c) 4(6)t 2 832t 173,056 t 2 → 400t 50,000r 3 b 2.83 10 24r 141,433,433.8; about 141,433,434 mi41. 2x 9 a b10O2x 9 a b46b. x- and y-axes 46c. It increases.2x 9 0, so a b 046d. It is halved.no real solution when a b 042. T t c2 t c2 p 247. 0 34 1 62 24 0 2 108 t ( 5 1200)2a 3 2 a 1 2 y 4 7 x 3 7 3 2 perpendicular slope: 7 4 123,056 1232t12b99.88 t; about 99.88 psi43.a 2 2 a 1 a 3 3 4a ;2exclude: 1 2a 7b 21 a b c 62 2a 7(1) 21 7 1 c 6 a 22a 1 (6)(2a 1) a 3 3 2(2a 1) (6)(2a 1)a 7(7, 1, 2)6(a 2) a(2)(2a 1) 3(3)6a 12 4a 2 49. y 3.54x 7125.4 2a 90 4a 2 y 3.54(2010) 7125.4 4a 3y 10 students0 (2a 3)(2a 1)2a 3 0 2a 1 050. 7y 4x 3 0Chapter 4 1281056p w4(0) (1)(2) 6(5)4(0) 0(2) 2(5)4(3)(1)(2) 6(1)4(3) 0(2) 2(1) 10Ow4a 4b 4c 242a 3b 4c 3 2a 3b 4c 32a 7b 214a 4b 4c 244a 8b 4c 12 4a 8b 4c 1212b 1y 5 7 4 (x 2)y 7 4 x 3 2 51. A r 2 A r 2 1 2 2 (1) 2 1 4 1 4 1 5 4 The correct choice is C. 1c 2
4-8Modeling Real-World Data withPolynomial FunctionsPages 261-262 Check for Understanding1a. Sample answer:y1b. Sample answer:1c. Sample answer:2. You need to recognize the general shape so thatyou can tell the graphing calculator which type ofpolynomial function to use as a model.3. Sample answer: If companies use less packagingmaterials, consumers keep items longer, and oldbuildings are restored instead of demolished, theamount of waste will decrease more rapidly. Ifconsumers buy more products, companies packageitems in larger containers, and many old buildingsare destroyed, the amount of waste will increaseinstead of decrease.4. quartic5. Sample answer:f(x) 1.98x 4 2.95x 3 5.91x 2 0.22x 4.896. Sample answer: f(x) 3.007x 2 0.001x 7.8967a. Sample answer: f(x) 0.48x 58.07b. Sample answer: 2010 1950 60f(x) 0.48x 58.0 0.48(60) 58.0 86.8%7c. Sample answer: f(x) 0.48x 58.089 0.48x 58.064 x1950 64 2014OyOyOxxx12. f(x) 1.25x 513. f(x) 8x 2 3x 914. Sample answer:f(x) 1.03x 4 5.16x 3 6.08x 2 0.23x 0.9415. Sample answer:f(x) 0.09x 3 2.70x 2 24.63x 65.2116. Sample answer:f(x) 4.05x 4 0.09x 3 6.69x 2 222.03x 2697.7417. Sample answer:f(x) 0.02x 3 8.79x 2 3.35x 27.4318a. Sample answer: f(x) 1.99x 2 1.74x 2.7618b. Sample answer:f(x) 0.96x 3 0.56x 2 0.36x 4.0518c. Sample answer: Cubic; the value of r 2 for thecubic function is closer to 1.19a. Sample answer: f(x) 0.126x 22.73219b. Sample answer:2010 1900 110f(x) 0.126x 22.732f(110) 0.126(110) 22.732f(110) 36.5923719c. Sample answer:2025 1900 125f(x) 0.126x 22.732f(125) 0.126(125) 22.732f(125) 38.4823820. Sample answer:x 1 2 3 4 5 6 7 8 9f(x) 1 3 6 3 –13 –49 –112–209 –34721a. Sample answer:f(x) 0.008x 4 0.138x 3 0.621x 2 0.097x 18.96121b. Sample answer: 1994 1992 2f(x) 0.008x 4 0.138x 3 0.621x 2 0.097x 18.961f(2) 0.008(2) 4 0.138(2) 3 0.621(2) 2 0.097(2) 18.961f(2) 20.663about 21%22. Asixth-degree polynomial; there are 5 changes indirection.23a. Sample answer:f(x) 0.109x 2 0.001x 48.696Pages 262-264 Exercises8. cubic 9. quadratic10. linear 11. quadratic129 Chapter 4
23b. Sample answer:f(x) 0.109x 2 0.001x 48.696100 0.109x 2 0.001x 48.6960 0.109x 2 0.001x 51.30427.r 2 1 0 1 21 2 3 3 2 00 2 1 0 1 21 2 1 1 2 01, 128a. Let x number of weeks.P (120 10x)(0.48 0.03x)P 57.6 1.2x 0.3x 2[5, 25] sc11 by [50, 10] sc15root: (21.7, 0)1985 22 200723c. Sample answer: 1998 1985 13f(x) 0.109x 2 0.001x 48.696f(13) 0.109(13) 2 0.001(13) 48.696f(13) 67.104No; according to the model, there should havebeen an attendance of only about 67 million.Since the actual attendance was much higherthan the projected number, it is likely that therace to break the homerun record increased theattendance.24a. Sample answer:f(x) 0.033x 3 1.471x 2 1.368x 5.56324b. Sample answer: 1996 1990 6f(x) 0.033x 3 1.471x 2 1.368x 5.563f(6) 0.033(6) 3 1.471(6) 2 1.368(6) 5.563f(6) 43.183about 43.18 million24c. Sample answer:f(x) 0.033x 3 1.471x 2 1.368x 5.563200 0.033x 3 1.471x 2 1.368x 5.5630 0.033x 3 1.471x 2 1.368x 194.437[5, 25] sc15 by [300, 50] sc150root: (14.8, 0)14 1990 2004about 200425. 5 b 2 0 Check: 5 b 2 05 b 2 5 23 2 025 b 2 5 25 023 b 5 5 00 0 6 p26. p 3 p 3 6 pp 3 p 3 1 (p 3)(p 3) 1(p 3)(p 3)6(p 3) p(p 3) (p 3)(p 3)6p 18 p 2 3p p 2 99p 9p 1[20, 20] sc12 by [40, 60] sc15maximum: (2, 58.8)2 weeks28b. $58.80 per tree29. x 0.10x 0.90x0.90x 0.10(0.90x) 0.90x 0.09x 0.99xThe correct choice is B.4-8BFitting a Polynomial Function to aSet of PointsPage 2661. y 7x 3 4x 2 17x 152. y 7x 3 4x 2 17x 15; yes3. Sample answer: y 5x 6 2x 5 40x 4 2x 3 x 2 8x 44. Infinitely many; suppose that you are given a setof n points in a coordinate plane, no two of whichare on the same vertical line. You can pick aninfinite number of other points with different x-coordinates. You could find polynomial functionsthat went through the original n points and anynumber of the other points.5. There is no problem with using L1 0 with list L 1for the example. However, if you are using adifferent list which happens to have 0 as one of itselements, using L1 0 will result in an errormessage, since 0 0 is undefined.Chapter 4 Study Guide and AssessmentPage 267 Understanding and Using theVocabulary1. Quadratic Formula 2. Integral Root Theorem3. zero 4. Factor Theorem5. polynomial function 6. lower boundChapter 4 130
7. Extraneous 8. complex roots9. complex numbers 10. quadratic equation20. b 2 4ac 4 2 4(1)(4) 0; 1 real4 0a 2(1)65 10 m 318. b 2 4ac (1) 2 4(1)(6)23; 2 imaginaryx 1 23 2(1)1 i23 x 219. b 2 4ac 3 2 4(2)(8) 73; 2 real3 73y 2(2)y 3 73 4a 4 2 Pages 268-270 Skills and Concepts11. no; f(a) a 3 3a 2 3a 4f(0) (0) 3 3(0) 2 3(0)f(0) 412. yes; f(a) a 3 3a 2 3a 4f(4) (4) 3 3(4) 2 3(4)f(4) 013. no; f(a) a 3 3a 2 3a 4f(2) (2) 3 3(2) 2 3(2)f(2) 1814. f(t) t 4 2t 2 3t 1f(3) (3) 4 2(3) 2 3(3)f(3) 73no15. 3; x 3 2x 2 3x 0x(x 2 2x 3) 0x(x 3)(x 1) 0x 0 x 3 0 x x 3f(x)f(x) x 3 2x 2 3xO x16. b 2 4ac (7) 2 4(2)(4) 81; 2 real7 81 x 2(2)a 221. b 2 4acr 1 1 r 22. f(x) f(2) 823. f(x) 2xf(5) 2(5) 25024. f(x) f 1 2 25. f(x) x 4f(3) (3) 8126. p q : 1, 2x 2 (x 2)(xx 2 x rational27. p q : 1(1)199;x 2 45x 2 5(5)1254 1 2 10x 210(3)90 roots: 1,r 4(5)(10)2 imaginary10x(2)207x35719or1,110(2)orornoyes11x0 8yes161; no21 011121201 4 2199 42(5)i199 10x 4 8(2) 3 2 8 0; 1 1 2 7(5) 1 14x 3 7x 13 1 2 148 7 2 or 4;1 0 x 1 2 9 9 0;r1x 2 0 1) 00 12x2x 7 941 1 1 0 1 21 1 1 0 1 0x 7 94x 7 9rational root: 1428. p: 1, 2, 4x 4x 1 r 2 2 2 42 q: 1, 21 2 0 2 617. b 2 4ac (10) 2 4(3)(5) p q : 1, 2, 4; 1 2 2 2 2 2 0 40; 2 real2x 2 2x 2 010 40m x 2 x 1 02(3)10 2 10m does not factorrational root: 2131 Chapter 4
29. p: 1, 3q: 1, 2 p q : 1, 3, 1 2 , 3 2 r 2 3 6 11 31 2 5 1 12 151 2 1 7 4 13 2 9 21 52 1533 2 3 3 20 57 1 2 2 4 4 13 1 2 1 2 2 2 7 1 5 3 24 3 2 2 6 3 3 12 10 54 3 2 2 0 6 2 0rational root: 3 2 30. p q : 1, 2, 4r 1 0 7 1 12 41 1 1 6 5 7 32 1 2 3 5 2 0x 4 2x 3 3x 2 5x 2 0r 1 2 3 5 22 1 0 3 1 0x 3 3x 1 0r 1 0 3 12 1 2 1 11 1 1 2 34 1 4 13 534 1 4 13 51rational roots: 2, 231. p: 1, 2, 4, 8q: 1, 3 p q : 1, 2, 4, 8, 1 3 , 2 3 , 4 3 , 8 3 r 3 7 2 81 3 10 8 03x 2 10x 8 0(3x 4)(x 2) 03x 4 0 x 2 0x 4 3 x 2rational roots: 2, 4 3 , 132. p: 1, 2q: 1, 2, 4 p q : 1, 2, 1 4 , 1 2 r 4 1 8 2 1 4 4 0 8 0933. p q : 1, 5r 1 0 4 0 51 1 1 5 5 0x 3 x 2 5x 5 0r 1 1 5 51 1 0 5 0x 2 5 0does not factorrational roots: 1, 134. 1 positivef(x) x 3 x 2 34x 562 or 0 negativer 1 1 34 567 1 6 8 0x 2 6x 8 0(x 4)(x 2) 0x 4 0 x 2 0x 4x 2rational zeros: 4, 2, 735. 2 or 0 positivef(x) 2x 3 11x 2 12x 91 negativer 2 11 12 9 1 2 2 12 18 02x 2 12x 18 0x 2 6x 9 0(x 3)(x 3) 0x 3 0 x 3 0x 3 x 3rational zeros: 1 2 , 336. 2 or 0 positivef(x) x 4 13x 2 362 or 0 negativer 1 0 13 0 362 1 2 9 18 0x 3 2x 2 9x 18 0r 1 2 9 183 1 5 6 0x 2 5x 6 0(x 3)(x 2) 0x 3 0 x 2 0x 3 x 2rational zeros: 3, 2, 2, 34x 2 8 0x 2 2 0does not factorrational root: 1 4 Chapter 4 132
37.r 3 0 0 12 3 6 12 231 3 3 3 20 3 0 0 11 and 038.r 1 4 20 1 4 21 1 3 12 1 2 23 1 1 14 1 0 20 and 1, 3 and 439.r 1 3 31 1 4 10 1 3 31 1 2 52 1 1 53 1 0 34 1 1 11 and 0, 3 and 440.r 1 1 0 12 1 3 6 111 1 2 2 10 1 1 0 11 1 0 0 11 and 041.r 4 2 11 32 4 7 3 31 4 3 8 110 4 1 11 31 4 5 6 32 4 9 7 172 and 1, 0 and 1, 1 and 242.r 9 25 24 61 9 34 58 640 9 25 24 61 9 16 8 22 9 7 10 140 and 143. Use the TABLE feature of a graphing calculator.4.9, 1.8, 2.2644. n n 5 06n n 5(n) 0(n)n 2 5n 6 0(n 6)(n 1) 0n 6 0 n 1 0n 6 n 145. 1 x x 2 1 x (2x 2 ) x 2 3x 2 3x 2 (2x 2 )2x x 3x 346. 5 6 2m 2m 5 6 6(m 1)(m 1) 2(m1 2 3m 32m1 1) 3(m 1) 6(m 1)(m 1)5(m 1)(m 1) (2m)(3)(m 1) 2(m 1)5m 2 5 6m 2 6m 2m 20 m 2 8m 3m 8 (8)2 )(3)4(12(1)8 52 2m m 4 1347. 3 y 2 5 y ; exclude: 0 3 y 2y 5 y y3 2y 5y 135Test y 2: 2 2 2 7 2 5 2 true35Test y 0.5: 0 .5 2 0 .58 10falseTest y 1: 3 1 2 5 1 1 5 true<strong>Solution</strong>: y 1, y 01 1 x ;1exclude 1, 11 (x 1)(x 1) 1 x (x 1)(x 1)248. x 1 2x 12(x 1) (x 1)(x 1) (x 1)2x 2 x 2 x 20 x 2 3x0 x(x 3)x 0 x 3 0x 321Test x 2: 2 1 1 2 12 4 3 true21Test x 0.5: 0.5 1 1 0.5 14 1 false21Test x 0.5: 0.5 1 1 0.5 1 4 3 3 true21Test x 2: 2 1 2 2 3 0 false21Test x 4: 4 1 4 1111 2 5 2 3 true<strong>Solution</strong>: x 1, 0 x 1, x 349. 5 x 2 0 Check: 5 x 2 05 x 2 5 23 2 025 x 2 5 25 023 x 5 5 00 0 1133 Chapter 4
50. 3 4a 1 8 5 Check: 3 4a 1 8 5 3 4a 1 3 3 4(6. 5) 1 8 54a 1 27 3 27 8 54a 26 3 8 5a 6.55 5 51. 3 x 8 x 359 6x 8 x 8 x 356x 8 18x 8 3x 8 9x 1Check: 3 x 8 x 353 1 8 1 5 33 9 363 3 66 6 52. x 5 7 x 5 0x 5 49 x 5x 54Test x 0: 0 5 75 7 meaninglessTest x 10: 10 5 75 7 trueTest x 60: 60 5 755 7 false<strong>Solution</strong>: 5 x 5453. 4 2a 7 6 2a 7 02a 7 22a 72a 7 4a 3.52a 3a 1.5Test a 5: 4 2(5) 7 64 3 6 meaninglessTest a 2: 4 2(2) 7 64 3 6 falseTest a 0: 4 2(0) 7 64 7 6 true<strong>Solution</strong>: a 1.554. cubic55. f(x) 2x 2 x 3Page 271 Applications and Problem Solving56. Let x width of window.Let x 6 height of window.A w315 x(x 6)315 x 2 6x0 x 2 6x 3150 (x 21)(x 15)x 21 0 x 15 0x 21 x 15Since distance cannot be negative, x 15 andx 6 21. the window should be 15 in. by 21 in.57. Let x width.Let x 6 length.(x 12)(x 6) (x 6)(x) 288x 2 18x 72 x 2 6x 28812x 72 288x 18x 6 2418 ft by 24 ft58a.58b. g(x) 0.006x 4 0.140x 3 0.053x 2 1.79x x(0.006x 3 0.140x 2 0.053x 1.79) x(x 3 23.3x 2 8.83x 298.3)r 1 23.333 8.833 298.3331 1 22.333 13.503 311.8365 1 18.333 82.835 712.50823.5 1 0.167 12.758 0rational zeros: 0, about 23.559. T 2 g1.6 2 90.25 9.8 .8 0.06 9.8 0.64 ; about 0.64 mPage 271 Open-Ended Assessmentx 21. Sample answer: x 2x xx 3200100g(x)g(x) 0.006x 4 0.140x 3 0.053x 2 1.79xO 10203 12 1 (x 3)(2x 1) 2x (x 3)(2x 1)x(2x 1) 2(x 3)2x 2 x 2x 62x 2 x 6 0(2x 3)(x 2) 02x 3 0 x 2 0x 3 2 x 22a. Sample answer: x 4 x 22b. Sample answer: x 4 x 2(x 4) 2 x 2x 2 8x 16 x 2x 2 9x 18 0(x 6)(x 3) 0x 6 0 x 3 0x 6 x 3Check: x 4 x 2 x 4 x 26 4 6 2 3 4 3 22 4 1 12 2 1 1The solution is 6. Since 1 1, 3 is anextraneous root.3a. Sample answer:x 3 2 1 0.5f(x) 12 0 2 1.125x 0 0.5 1 2f(x) 0 0.625 0 83b. Sample answer: f(x) x 3 x 2 2x3c. Sample answer: 2, 0, 1xChapter 4 134
Chapter 4 SAT & ACT PreparationPage 273 SAT and ACT Practice1. There are two ways to solve this problem. You canuse the distance formula or you can sketch agraph.d (x x 2 1 ) 2 (y 2 y 1 ) 2 (1 2) ( 2 (1)) (3 2 3 2 4 2 9 6 1 25 or 5yO(2, 1) 3(1, 3)4xWhen you sketch the points and draw a righttriangle as shown above, you can see that this is a3-4-5 right triangle. Using the PythagoreanTheorem, you can calculate that the length of thehypotenuse is 5.5 2 4 2 3 2 The correct choice is C.2. Points on the graph of f(x) are of the form (x, f(x)).To move the entire graph of f(x) up 2 units,2 must be added to each of the second coordinates.Points on the translated graph are of the form(x, f(x) 2). The function which represents thetranslation of the graph up 2 units is f(x) 2.The correct choice is E.3. You need to find both the x- and y-coordinates ofpoint C. Use the properties of a parallelogram.First find the y-coordinate. Since opposite sides ofa parallelogram are parallel and side AD is on thex-axis, point C must have the same y-coordinateas point B. So the y-coordinate is b. This meansyou can eliminate answer choices A and BNow find the x-coordinate. Since opposite sides ofa parallelogram have equal length and side ADhas length d, side BC must also have length d.Point B is a units from the y-axis, so point C mustbe a d units from the y-axis. The x-coordinate ofpoint C is a d. So point C has coordinates(a d, b). The correct choice is E.4.You may want to draw a diagram.ss 6Use the formula for the perimeter of a rectangle,where represents the length and w representsthe width.2 2w PReplace w with s. Replace with s 6.2(s 6) 2s 60The correct choice is E.5. First find the slope of the given line. Write theequation in the form y mx b.3x 6y 126y 3x 12y 1 2 x 2 The slope is 1 2 .So the slope of the line perpendicular to this lineis the negative reciprocal of this slope. The slopeof the perpendicular line is 2. The correct choiceis A.6. Be sure to notice the small piece of giveninformation: x is an integer. You need to find thenumber, written in scientific notation, that couldbe x 3 . This means that the cube root of thenumber is an integer.Take the cube root of each of the answer choicesand see which one is an integer. You can use yourcalculator or do the calculations by hand. Noticethat 2.7 is one-tenth of 27, which is 3 3 .2.7 10 13 27 10 12 327 10 12 3 10 4 or 30,000. 30,000 is aninteger.When you try the same calculation with each ofthe other answer choices, the resulting power of10 has a fractional exponent. So the numbercannot be an integer. The correct choice is C.7. This is a system of equations, but you do not needto solve for x or y. You need to find the value of6x 6y.Notice that the first equation contains 5y and thesecond contains 1y. If you subtract the secondfrom the first, you have 6y. Similarly, subtractionof the x values gives a result of 6x. Use the samestrategy that you would for solving a system.Subtract the second equation from the first.10x 5y 144x 5y 26x 6y 12The correct choice is C.135 Chapter 4
8. You can solve this problem using the midpointformula or by sketching a graph.The midpoint formula:x 2 y 2 (4) 3 1, 8 2 1 2 ,4 x 1 , y 1 2 2 3 2y( )(4, 3)1 , 42 (3, 5)O, 5 2x 2The correct choice is B.9. The expression x 2 2ax a 2 is a perfect squaretrinomial and can be factored as (x a) 2 . Thesquare of a real quantity is never negative.The correct choice is A.(x a 2 x 2 2ax a 2 (x 2 a 2 ) 2axSo the quantity in Column A equals the quantityin Column B plus the sum of the squares of x anda. Since neither x nor a equal 0, their squaresmust be greater than 0. So the quantity inColumn A is always greater than the quantity inColumn B. The correct choice is A.10. Since the problem does not include a figure, drawone. Label the four points.3 2 8E F G HOne method of solving this problem is to “plug-in”numbers for the segment lengths. SinceEG 5 3 EF, let EF 3. Then EG 5. This meansthat FG must equal 2, since EF FG EG.HF 5FG 5(2) 10HG HF FG 10 2 8EF H G 3 8 The answer is .375 or 3/8.Chapter 4 136
Chapter 5 The Trigonometric Functions5-1Angles and Degree MeasurePages 280–281 Check for Understanding1. If an angle has a positive measure, the rotation isin a counterclockwise direction. If an angle has anegative measure, the rotation is in a clockwisedirection.2. Add 29, 4 560 26, and 3 600 .3. 270° 360k° where k is an integer4.y1260°5. 34.95° 34° (0.95 60) 34° 5734° 576. 72.775° (72° (0.775 60))(72° 46.5)(72° 46 (0.5 60))(72° 46 30)72° 46 307. 128° 30 45 128° 30 1°60 45 136128.513°8. 29° 6 6 29° 6 1°60 6 1°3600 29.102°9. 2 (360°) 720°10. 4.5 360° 1620°11. 22° 360k°; Sample answers:22° 360k° 22° 360(1)° or 382°22° 360k° 22° 360(1)° or 338°12. 170° 360k°; Sample answers:170° 360k° 170° 360(1)° or 190°170° 360k° 170° 360(1)° or 530°53 98 1.26 2.2213. 4 360 360(1)° 453° 360 453° 93°; II15. 180° 227° 180° 47°16. 360° 210° 150°180° 180° 150° 30°117. 2(360°)4 15°1 6 0 1 2 41 6 0 1 6 0 1 2 4Ox14. 7 360°00 2.22 2 0.220.22 360° 78°360° 78° 282°; IV 360° 0.25°, or 0.25(60) 15 360° 0.0042°,or 0.0042(60)(60) 15Pages 281–283 Exercises18. 16.75° (16° (0.75 60))(16° 45)16° 4519. 168.35° 168° (0.35 60) 168° 21168° 2120. 183.47° (183° (0.47 60))(183° 28.2)(183° 28 (0.2 60)(183° 28 12)183° 28 1221. 286.88° 286° (0.88 60) 286° 52.8 286° 52 (0.8 60) 286° 52 48286° 52 4822. 27.465° 27° (0.465 60) 27° 27.9 27° 27 (0.9 60) 27° 27 5427° 27 5423. 246.876° 246° (0.876 60) 246° 52.56 246° 52 (0.56 60) 246° 52 33.6246° 52 33.61°24. 23° 14 30 23° 14 60 30 36 23.242°1°1°0025. 14° 5 20 14° 5 60 20 3614.089°1°26. 233° 25 15 233° 25 60 15 36 233.421°1°27. 173° 24 35 173° 24 60 35 36 173.410°1°1°00 1°001°0028. 405° 16 18 405° 16 60 18 36405.272°1°00 29. 1002° 30 30 1002° 30 60 30 3600 1002.508°30. 3 360° 1080° 31. 2 360° 720°32. 1.5 360° 540° 33. 7.5 (360°) 2700°34. 2.25 360° 810° 35. 5.75 (360°) 2070°36. 4 360° 1440°37. 30° 360k°; Sample answers:30° 360k° 30° 360(1)° or 390°30° 360k° 30° 360(1)° or 330°38. 45° 360k°; Sample answers:45° 360k° 45° 360(1)° or 315°45° 360k° 45° 360(1)° or 405°39. 113° 360k°; Sample answers:113° 360k° 113° 360(1)° or 473°113° 360k° 113° 360(1)° or 247°1°1° 137 Chapter 5
40. 217° 360k; Sample answers:217° 360k° 217° 360(1)° or 577°217° 360k° 217° 360(1)° or 143°41. 199° 360k°; Sample answers:199° 360k° 199° 360(1)° or 161°199° 360k° 199° 360(1)° or 559°42. 305° 360k°; Sample answers:305° 360k° 305° 360(1)° or 55°305° 360k° 305° 360(1)° or 665°43. 310° 360k° 310° 360(0)° or 310°44. 60° 360k° 60° 360(2)° or 780°60° 360k° 60° 360(3) or 1020°45. 4 00°360° 1.11 46. 280°360° 0.78 360(1)° 400° 360(1)° 280° 360° 400° 360° 280° 40°; I 80°; I47. 9 40°360° 2.61 48. 1 059°360° 2.94 360(2)° 940° 360(2)° 1059° 720° 940° 720° 1059° 220°; III 339°; IV49. 624°360° 1.731.73 1 0.730.73 360° 264°360° 264° 96°; II50. 989°360° 2.752.75 2 0.750.75 360° 269°360° 269° 91°; II51. 1 275°360° 3.54 360(3)° 1275° 1080° 1275° 195°; III52. 360° 360° 327° or 33°53. 180° 180° 148° or 32°54. 563° 360° 203° 180° 203° 180° or 23°55. 420° 360° 60° 56. 360° 197° 163°360° 60° 300° 180° 163° 17°360° 300° 60°57. 1 045°360° 2.90 360(2)° 1045° 720° 1045° 325°360° 360° 325° or 35°58. 20°180° 180° 20° or 160°180° 180° 20° or 200°360° 360° 20° or 340°evolutions360° rev 32,400°/second59. 90 r secondolution 90 r evolseco utionsnd 6 0 seconds360°1 minute rev olution1,944,000°/minute60. 90k, where k is an integerevolutions360° rev 34,200°/minute61. 95 r minuteolution30 seconds 1 2 minute34,200° 1 2 17,100°30,000 revolutions360°62. minute rev olution10,800,000 or 1.08 10 7 100,000 revolutions 360° minute rev olution36,000,000 or 3.6 10 71.08 10 7 to 3.6 10 7 degreesrotations360° ro 22,320° second63. 62 secondtation 62 rotations360°second ro tation 60 secondsminute1,339,200°/minute 62 rotations360°second ro tation 60 secmin ondsute 60 m inutesho ur80,352,000°/hour 62 rotations360°second ro tation 60 secmin ondsute 60 m inutesho ur24 h oursday 1,928,448,000°/day64. 25° 120k°, where k is an integer65a. 44.4499° 44° (0.4499 60) 44° 26.994 44° 26 (0.994 60) 44° 26 59.6444° 26 59.6468.2616° 68° (0.2616 60) 68° 15.696 68° 15 (0.696 60) 68° 15 41.7668° 15 41.7665b. 24° 33 32 24° 33 1°60 32 1°3600 24.559°81° 45 34.4 81° 45 1°60 34.4 136 81.760°66a. Sydney:1 rotation 7 0 minutes 60 m inho utesur 24 hod a ursy 20.6 r otationsd ayevolutionSan Antonio: 1r 1 ho ur 24 d ay24 20.6 3.4 revolutionsabout 3.4 revolutions66b. Sydney:otationsdays360° ro 51,840° 20.6 r da y 7 w eektationolutions°00ho urs 24 rev olutionsd aydays360° revSan Antonio: 24 rev da y 7 w eekolution60,480°60,480° 51,840° 8640°67a. Use graphing calculator to find cubic regression.Sample answer: f(x) 0.00055x 3 0.0797x 2 3.7242x 76.214767b. 2010 1950 60f(x) 0.00055x 3 0.0797x 2 3.7242x 76.2147f(60) 0.00055(60) 3 0.0797(60) 2 3.7242(60) 76.2147 20.8827Sample answer: about 20.9%Chapter 5 138
68. 3 6n 5 15 10 3 6n 5 56n 5 1256n 120n 20Check: 3 6n 5 15 10 3 6(20) 5 15 10 3 125 15 105 15 1010 1069. x 3x 2 x 3x 2(x 2)(x 3) x 3x 2f(x)Of(x) |(x 1) 2 2|x 2 x 2 53 2 (x 2)3x 6(x 3) (x 2)(x 3)(2) (x 2)(x 3) (x 2)(x 3)(x 3) (x 2)(x 3)(2) 3x 2 6x 9 2x 2 10x 12 30 x 2 4x0 x(x 4)x 0 or x 4 0x 470. 2 1 1 0 8 12 4 241 2 12 252571. (x (5))(x (6))(x 10) 0(x 5)(x 6)(x 10) 0(x 2 11x 30)(x 10) 0x 3 x 2 80x 300 072. r 1 t 1 r 2 t 218(3) r 2 (11) 18 (3)11 r 24.91 r 2about 4.9173. x 1 0x 1Point discontinuity74.yO3(x 3) y x2 1x 1decreasing for x 1, increasing for x 175. expanded vertically by a factor of 3, translateddown 2 unitsx76.(1, 5)yO5248(3, 5)(0, 3)77. [f g](x) f(g(x)) f(x 0.3x) (x 0.3x) 0.2(x 0.3x) x 0.3x 0.2x 0.06x 0.56x78. m∠EOD 180° m∠EOA m∠BOD 180° 85° 15° 80°m∠OED m∠EDOm∠OED 1 2 (180° m∠EOD) 1 2 (180° 80°) 50°m∠ECA 180° m∠EOC m∠OED 180° (80° 15°) 50° 35°The correct choice is D.5-2xTrigonometric Ratios in RightTrianglesPage 284 Graphing Calculator Exploration1. Sample answers:2052. R 1 1or about 0.38463592369R 1 1 3or about 0.3846R 2 1 1or about 0.9231R 2 3 3or about 0.92315R 3 1or about 0.41672562369R 3 1 3or about 0.41673. R 1 1 1or about 0.9231R 1 3 3or about 0.92315R 2 1or about 0.38463592R 2 1 3or about 0.3846R 3 1 5or 2.4R 3 3 615or 2.44. Each ratio has the same value for all 22.6° angles.5. yes 6. Yes; the triangles are similar.262410139 Chapter 5
Pages 287–288 Check for Understanding11. (AC) 2 (CB) 2 (AB) 21. The side opposite the acute angle of a right8 2 5 2 (AB) 2triangle is the side that is not part of either side of89 (AB) 2the angle. The side adjacent to the acute angle is89 ABthe side of the triangle that is part of the side ofsin A s ideoppositehypotenusecos A s ideadjacenthypotenusethe angle, but is not the hypotenuse.5 5898sin A or cos A or 889 2. cosecant; secant; cotangent89 8989 893. sin A a c , cos A b c , tan A a sideoppositeb ,tan A s ideadjacentccsc A ,casec A ,bbcot A tan A 5a8 4. sin A cos B, csc A sec B, tan A cot B12. (AC) 2 (BC) 2 (AB) 25. (TV) 2 (VU) 2 (TU ) 2(AC) 2 12 2 40 217 2 15 2 (TU ) 2(AC) 2 1456514 (TU ) 2AC 1456 or 491514 TUsin A s ideoppositehypotenusecos A s ideadjacenthypotenusesin T s ideoppositehypotenusecos T s ideadjacenthypotenusesin A 1 240 491 91 or 130cos A or 40 1015 1551417sin T or cos T or 17514 sideopposite514 514514 514tan A s ideadjacentsideopposite12 391 tan T s ideadjacenttan A or 491 91tan T 1 513. tangent1711116. csc v si n v7. tan v co t v14. cot v ta n v15. csc v si n v1csc v or 5 2 1tan v 1 .5 11or about 0.6667 cot v or 3 csc v or 7 3 2 5 8. (PS) 2 (QS) 2 (QP) 2(PS) 2 6 2 20 2(PS) 2 364PS 364 or 291sin P s ideoppositehypotenusecos P s ideadjacenthypotenuse6 3291sin P 2or 1cos P or 91 tan P ss0 0ideoppositeideadjacent6 391 291 91ypotenuseideadjacentcsc P shtan P or csc P 2 6hsec P scot P s s 20 10ypotenuseideopposite0 or 1 03ideadjacentideopposite20 1091sec P or 291cot P or 91 291 I9. cos v t Io 91 63 1 3 1116. cos v se17. sin v csc v 1 5 9 3 7 c vcos v or 9 5 1sin v 2 .5 or 0.41118. tan v co t v19. sec v co s v1tan v 0. 75 or about 1.3333 sec v 0.1125or 820. (RT ) 2 (TS) 2 (RS) 214 2 (TS) 2 48 2(TS) 2 2108TS 2108 or 2527 sin R s hideoppositeypotenuse2527 527 48 24cos R s hsin R or cos R 1 4sideoppositehtan R scsc R sideadjacentideadjacentypotenuse48 7or 2 4 ypotenuseideopposite 527 48cos 45° tan R orcsc R or24527 Io 14 72527 527Ithypotenuseside adjacentsec R cot R 2 2 2 4 It I o0.5I 0 I tIt Io sec R 4 12527side adjacent84 or 2 47side opposite142527 527 cot R or7527 Pages 288–290 Exercises10. (AC) 2 (CB) 2 (AB) 280 2 60 2 (AB) 210,000 (AB) 2100 ABsin A s ideoppositehypotenusecos A s idehy60sin A 1 00or 3 5 80cos A 1 00sideoppositetan A stan A 6 8ideadjacent00or 3 4 adjacentpotenuse or 4 5 Chapter 5 140
21. (ST) 2 (TR) 2 (SR) 238 2 (TR) 2 40 2(TR) 2 156TR 156 or 239sin R s ideoppositehypotenusecos R s hsin R 3 84sidtan R s0 or 1 92e oppositeideadjacent38 1939 239 39ypotenuseideadjacent40 2039 239 39ideadjacentypotenuse0 239 cos R or 39 40 20hypotenusecsc R stan R or csc R 4 3sec R shcot R s sv 72° 74° 76° 78° 80°sin 0.951 0.961 0.970 0.978 0.985cos 0.309 0.276 0.242 0.208 0.174v 82° 84° 86° 88°sin 0.990 0.995 0.998 0.999cos 0.139 0.105 0.070 0.035ideopposite08 or 2 019 ideadjacentideopposite239 38sec R or cot R or22. (ST ) 2 (TR) 2 (SR) 2(7) 2 9 2 (SR) 288 (SR) 288 SR; 88 or 222sin R s hideoppositeypotenuse7 154 222 44ideoppositeideadjacentcos R s h39 19ideadjacentypotenuse9 222 44ypotenuseideoppositesin R or cos R or 922tan R sscsc R sh7222tan R csc R or 2154 sec R sh 9ypotenuseideadjacent222 9cot R s ssec R cot R or23. cot (90° v) tan v 24a. 0.7963540136cot (90° v) 1.324b. 0.186524036 24c. 35.3401510624d. 1.3763819225.25a. 1 25b. 026.v 18° 16° 14° 12° 10°sin 0.309 0.276 0.242 0.208 0.174cos 0.951 0.961 0.970 0.978 0.985tan 0.325 0.287 0.249 0.213 0.176v 8° 6° 4° 2°sin 0.139 0.105 0.070 0.035cos 0.990 0.995 0.998 0.999tan 0.141 0.105 0.070 0.03526a. 0 26b. 126c. 0sin27. v is invrsin 45° sin 27°55 n n1.5103 n 7ideadjacentideopposite9 97 7 7728. sin R s hideoppositeypotenusesin R 3 7 a 2 b 2 c 23 2 b 2 7 2b 2 40b 40 or 210cos R s ideadjacenthypotenusetan R ssideoppositeideadjacentcos R tan R 7210 orhypotenusehypotenuscsc R ssec R sideoppositeeideadjacentcsc R 7 3 sec R orcot R s scot R 210 ideadjacentideopposite210 329a. tan v gv 2 r 29b. tan v gv 2 r 215.5)25.5)vvtan 11° 9.8(tan 13° 9.8( 129.53 v 2 35.07 v 25.4 v 5.9 vabout 5.4 m/sabout 5.9 m/s29c.vtan v 2 g r29d. increasev2tan 15° 9.8( 15.5)40.70 v 26.4 vabout 6.4 m/s30. sin v s ideopposite cos v s ideadjhypotenusehypote s ideoppositesinv hypotenuse c osv —— s ideadjacenthypotenusesinv c osv s ideoppositehypote nuse s ideadjacenth ypotenusesinv c osv s ideoppositehypotenuse hypotenuses ideadjacentsinv sideopposite c osvs ideadjacentsinv c osv tan v31a. ∠ 90° L 23.5° cos (N 3∠ 90° 26° 23.5° cos (172 36∠ 90° 26° 23.5° (0.99997)∠ 87.5°∠ 90° L 23.5° cos (N 10365∠ 90° 26° 23.5° cos (355 36∠ 90° 26° 23.5° 1∠ 40.5°31b. ∠ 90° L 23.5° cos (N 37 210 acentnuse10)3606510)3605 )360 10)3605 10)3606510)3605 ∠ 90° 64° 23.5° cos (172 36∠ 90° 64° 23.5° 0.99997∠ 49.5°∠ 90° L 23.5° cos (N 10)360365∠ 90° 64° 23.5° cos[ (355 10)360]365∠ 90° 64° 23.5° 1∠ 2.5°3310 20710 20141 Chapter 5
31c. 87.5° 40.5° 47°49.5° 25° 47°neither32. x t sin c( B A)osA 60° 41°)os41°x 10 sin( cx 10(0.4314)x 4.31; about 4.31 cm33. 88.37° 88° (0.37 60) 88° 22.2 88° 22 (0.2 60) 88° 22 1288° 22 1234. positive: 1f(x) x 4 2x 3 6x 1negative: 3 or 135. 35a. 23 employees35b. $1076[10, 50] scl:10 by [10, 1200] scl:1007 3 50 1 4 14 0 12 0 8 08 2 0 7(2) (3)(8) 5(8) 7836. 7 (3) 5 4 0 37. m 3 65 22 or 1 2 m 4y y 1 m(x x 1 )y 3 1 2 (x 6)y 1 2 x 638. A 1 2 bh 2x 2(2) or 412 1 2 (2x)(3x) 3x 3(2) or 612 3x 2 a 2 b 2 c 24 x 2 4 2 6 2 c 22 x 52 c 2The correct choice is C.5-352 c; 52 or 213Trigonometric Functions on theUnit CirclePage 296 Check for Understanding1. Terminal side of a 180° angle in standard positionis the negative x-axis which intersects the unitcircle at (1, 0). Since csc v 1 y , csc 180° 1 0 which is undefined.822.As v goes from 0° to 90°, the y-coordinateincreases. As v goes from 90° to 180°, they-coordinate decreases.3. cot v x y c osvsinv4.15. (1, 0); tan 180° y x 0or ;106. (0, 1); sec(90°) 1 x or 1 0 ; undefined3 17. , 2yO 2sin 30° ysin 30° 1 2 tan 30° y x tan 30° tan 30° tan 30° sec 30° 1 x sec 30° 1 2 32133 31 32x1 y O 1 x1FunctionQuadrantI II III IVsin or cos cos or sec tan or cot sec 30° 323sec 30° 23cos 30° xcos 30° csc 30° 1 y csc 30° csc 30° 2cot 30° x y cot 30° 3 21 1 2 32 1 2 cot 30° 3Chapter 5 142
8. terminal side — Quadrant III12. cos v x rreference angle: 225° 180° 45° r 2 x 2 y 22 2cos v , 2 2 1 2 2 2 (1) 2 y 2x 1, r 2 3 y 2sin 225° ysin 225° tan 225° y x tan 225° tan 225° 1sec 225° 1 x sec 225° 2 2 22 22sec 225° 29. r x 2 y 2 r 3 2 4 2 t 25 or 5cos 225° xcos 225° 22csc 225° 22csc 225° 2cot 225° x y sin v y r cos v x r tan v y x sin v 4 5 cos v 3 5 tan v 4 3 csc v yr sec v xr cot v x y csc v 5 4 sec v 5 3 cot v 3 4 10. r x 2 y r (6) 2 6 2r 72 or 62sin v y r cos v x r tan v y x 6 626 626sin v cos v tan v or 1sin v csc v yr 2 262 6cos v 2 2sec v xr 62 6csc v or 2 sec v or 2cot v x y cot v 66or 111. tan v y x r 2 x 2 y 2tan v 1 r 2 1 2 (1) 2x 1, y 1 r 2 2r 2sin v y r cos v x r sin v 1 2cos v 1 22sin v 2cos v 22csc v yr 2 1sec v xr 2 1csc v or 2 sec v or 2cot v x y 1 221cot v 1or 1csc 225° 1 y csc 225° cot 225° sec 225° 2 2 cot 225° 11 22 22 2263 yQuadrant II, so y 3sin v y r tan v y x csc v yr 33sin v tan v or 3 csc v 2 21323csc v 3rsec v x cot v x y 2sec v 1or 2cot v 1 3cot v 13. C 2r cos L C 2r cos LC 2(3960) cos 0° C 2(3960) cos 90°C 24,881.41 C 0The circumference goes from about 24,881 miles to0 miles.Pages 296–298 Exercises14. (0, 1); sin 90° y or 115. (1, 0); tan 360° y x or 0 1 ; 016. (1, 0); cot(180°) x y or 10; undefined17. (0, 1); csc 270° 1 y 1or ;1118. (0, 1); cos(270°) x or 019. (1, 0); sec 180° 1 x 1or ;1120. Sample answers: 0°, 180° 21. undefined2 222. , 2 2sin 45° ysin 45° 2 2tan 45° y x 22tan 45° or 1 223 3cos 45° xcos 45° 22csc 45° 1 y 2csc 45° 2csc 45° 2sec 45° 1 x cot 45° x y sec 45° 1 222 2sec 45° sec 45° 2csc 45° 1 22 22cot 45° or 1 22143 Chapter 5
23. terminal side — Quadrant IIreference angle: 180° 150° 30°3 , 1 2 sin 150° ycos 150° xsin 150° 1 2 3cos 150° 2tan 150° y x csc 150° 1 y tan 150° 1 2 321tan 150° 3csc 150° 2tan 150° 3 3csc150° 1 1 2 25. terminal side — Quadrant IIIreference angle: 210° 180° 30°3 , 1 2 2 sin 210° ycos 210° xsin 210° 1 2 3cos 210° 2tan 210° y x csc 210° 1 y csc 210° 1tan 210° 132 2tan 210° 13csc 210° 2tan 210° 3 3 1 2 2sec 210° 1 x cot 210° x y sec 150° 1 x sec 150° sec 315° 1 x 1 22cot 150° x y csc 315° 2 2csc 315° 2cot 315° x y 22sec 315° cot 315° or 1sec 315° 1 32sec 150° 2 3sec 150° 2 2sec 315° 2cot 150° 24. terminal side — Quadrant IVreference angle: 360° 315° 45°2 2 , 2 2sin 315° ysin 315° tan 315° y x 23 32 2 22 32 1 2 cot 150° 3cos 315° xcos 315° csc 315° 1 y tan 315° or 1 csc 315° 1 222 2 22 22sec 210° sec 210° 1 322 323 3cot 210° cot 210° 3sec 210° 26. terminal side — Quadrant IVreference angle: 360° 330° 30°3 , 1 2 2sin 330° ysin 330° 1 2 tan 330° y x tan 330° 1 2 321tan 330° 3tan 330° 33sec 330° 1 x sec 330° 1 32cos 330° xcos 330° 32csc 330° 1 y csc 330° 32 1 2 1 1 2 csc 330° 2cot 330° x y cot 330° 32 1 2 sec 330° 2 3 cot 330° 3sec 330° 23 3Chapter 5 144
27. terminal side — Quadrant Ireference angle: 420° 360° 60° 1 2 3, 2 sin 420° ycos 420° xsin 420° cos 420° 1 2 3 2tan 420° y x csc 420° 1 y 2sin v 2cos v 22 3r r2csc 420° 1 csc v y sec v x cot v x ytan 420° 6262csc v sec v cot v 6 1 3 2 26csc 420° 2 66tan 420° 33or 2 or 2 or 1csc 420° 2332. r x 2 y sec 420° 1 3x sec 420° 1 cot 420° x r 2 2 0 y 1 2 1 r 4 or 22 sec 420° 2cot 420° sin v y r cos v x r tan v y x 32sin v 0 2 or 0 cos v 2 2 or 1 tan v 0 2 or 0cot 420° 13rcsc v yrsec v x cot v x y 3cot 420° 3csc v 2 0 sec v 2 2 or 1 cot v 2 0 28. terminal side — Quadrant IVundefinedundefinedreference angle: 45°33. r x 2 y 2 2 , 2 2 r 18) 2 ( 2cot (45°) x y r 652sin v y r cos v x r tan v y x cot (45°) or 1 812sin v cos v tan v 8 65 65 1or 82865 sin v cos v 65 29. terminal side — Quadrant 16565r rreference angle: 390° 360° 30°csc v y sec v x3 , 1 2 2 2csc 390° 1 y 1csc 390° or 2 1 2 30. r x 2 y 2 r (4) 2 ) (32r 25 or 5sin v y r cos v x r tan v y x sin v 3 5 cos v 4 5 tan v rcsc v yrsec v x cot v x y csc v 5 3 sec v 5 4 cot v 34or 3 4 43or 4 3 31. r x 2 y 2 r (6) 2 6 2r 72 or 62sin v y r cos v x r tan v y x 6626626sin v cos v tan v or 165 865 865 1csc v or sec v or 65cot v x y 1cot v 8or 1 8 34. r x 2 y 2 r 53) 2 ( 2r 34sin v y r cos v x r tan v y x 3sin v cos v tan v 34 34 5334 sin v cos v 534 csc v yr 345 34sec v xr 34csc v 34or sec v 34 cot v x y 3 35cot v 3or 5 3 35. r x 2 y 2 r (8) 2 15 2r 289 or 17sin v y r cos v x r tan v y x 5sin v 1 17cos v 178 8 or 157 15tan v 8r rcsc v y sec v x cot v x y csc v 1 71715sec v or 1 7 8 cot v 815863 or 3 5 or 1 588 or 1 5145 Chapter 5
36. r x 2 y r x 2 y r 56) 2 ( 2 r (5) 2 6 241. tan v y x r 2 x 2 y 2tan v 2 rr 61r 61 1 2 2 2sin v y r sin v y y 2, x 1 r 2 5r r 56sin v sin v 6 sin v y61 61 r cos v x r rcsc v y21661 661 sin v cos v csc v 5sin v sin v 552616125The sine of one angle is the negative of the sine ofsin v cos v 555sec v 5612cos v 1I t I othe other angle.rsec v x cot v x y 37. If sin v 0, y must be negative, so the terminal5side is located in Quadrant III or IVsec v or 5 cot v 1 2 138. cos v x r r 2 x 2 y 2r42. sec v x r 2 x 2 y 2cos v 1 21313 2 (12) 2 y 2sec v 3 (3) 2 1 2 y 2x 12, r 13 25 y 2r 3, x 1 2 y 25 y2 yQuadrant III, so y 5Quadrant IV, so y 2sin v y r tan v y x sin v y r cos v x r tan v y x sin v 2125 555sin v cos v tan v 13or 1 3tan v 12or 331or 21 26r rcsc v y sec v sin v cos v 333x13csc v 5or 1 3135sec v 12or 1 3rcsc v y cot v x y 123rcot v csc v cot v 1 22ycot v 125625 csc v cot v 222r39. csc v y r 2 x 2 y 243. cot v x y r 2 x 2 y 2csc v 2 2 2 x 2 1 2cot v 1 (Quadrant III) r 2 (1) 2 (1) 2r 2, y 1 3 x 2x 1, y 1 r 2 23 xr 2Quadrant II, so x 3 sin v sin v y r cos v x r tan v y y r cos v x r tan v y x x 11sin v cos v tan v 1 sin v 1 2 3 3cos v or tan v 1 221or 12223sin v cos v 2322tan v r r 3csc v rsec v x cot v x y sec v xy 22csc v or 2 sec v 2311or 2sec v cot v 31or 3rsec v 2344. csc v y r 2 x 2 y 2340. sin v y csc v 2 2 2 x 2 (1) 2r r 2 x 2 y 2r 2, y 1 3 x 2sin v 1 5 5 2 x 2 (1) 23 xQuadrant III, so x 3y 1, r 5 24 x 2tan v y x 26 xQuadrant IV, so x 26 tan v 13cos v x r tan v y x r 3csc v ytan v 32615cos v tan v csc v 526 1or 5 45. g sin v cos v 0tan v 6sin v 0 or cos v 012v 0° v 90°rsec v x cot v x y 46a. k is an even integer. 46b. k is an odd integer.sec v 526 cot v 261or 26 47. cos v tIocos v 1v 0°Chapter 5 146
48. Let x 1. y 3(1)55. f(x) x 2 16f(x)y 3y x 2 6 16r 2 x 2 y 2x y 2 f(x) x 162 16 4r 2 (1) 2 (3) 2x 16 y 22r 2 10x 16 yr 1015 10 5 O 5 xsin v y r cos v x r tan v y 2x 4313sin v cos v tan v 1or 310 10 6310 sin v cos v 10 101056. 2 1r r 2(2) (3)1csc v y sec v x3 210 10 7csc v sec v 31or 10 1 7 2 1cot v 3 2xy 57. 3(8m 3n 4p) 3(6) 24m 9n 12p 184m 9n 2p 4 →4m 9n 2p 428m 14p 14cot v 13or 1 3 49a. 4 2(36) 76 ft49b. ABC is equilateral.m∠BCA 60°m∠ACD m∠BCA 90°ADsin v 5 7 51. 8403603660˚ 2.33360° 120° 240°; III52. 5 b 2 05 b 225 b 223 b53. x b b2 ac 42a)(5)4(42(4)x 9 (9)2B36C4300˚m∠ACD 60° 90°m∠ACD 30°Since AC 36, AD 18.18 4 22 ft49c. Refer to 49b for diagram and reasoning.Since AC 30, AD 15.15 4 19 ft49d. 1 2 r 4150. sin v cs c vsin v 1 7 5 x 9 1 81 360(2)° 840° 720° 840° 120°x 9 8or x 9 8x 1 08or 1.25 x 8 8 or 154. k y x y kx9k 15 y (0.6)(21)k 0.6y 12.614(8m 3n 4p) 4(6) →32m 12n 16p 246m 12n 5p 1 6m 12n 5p 138m 11p 2311(28m 14p) 11(14)14(38m 11p) 14(23)↓308m 154p 154532m 154p 322224m 168m 3 4 38m 11p 2338 3 4 11p 23p 1 2 4m 9n 2p 44 3 4 9n 2 1 2 4n 2 3 3 4 , 2 3 , 1 2 58. 2x 4y 7 2x 4y 72(9) 4(3) ? 7 2(1) 4(2) ? 76 7; yes 10 7; yes2x 4y 72(2) 4(2) ? 712 7; no59. absolute value; f(x) 2 1 2 x 60. A of square A of circle As 2 r 2 A2 2 (1) 2 A0.86 AThe correct choice is C.5-4Applying Trigonometric FunctionsPages 301–302 Check for Understanding1a. cos or sec 1b. tan or cot1c. sin or csc2. Sample answer: Find a. A147 Chapter 510C38˚aB
3. ∠DCB; ∠ABC; the measures are equal; if parallellines are cut by a transversal, the alternateinterior angles are congruent.4. Sample answer: If you know the angle of elevationof the sun at noon on a particular day, you canmeasure the length of the shadow of the buildingat noon on that day. The height of the buildingequals the length of the shadow times the tangentof the angle of elevation of the sun.5. tan A a b 6. sin B b c atan 76° 1 13 sin 26° 1 8 c13 tan 76° a c sin 26° 181852.1 a c sin 26°c 41.17. cos B a c cos 16° 45 — 1a3 —13 cos 16° 45 a12.4 a8b. Let x 1 2 of the base.xcos 55° 30 1 010 cos 55° 30 x5.66 xbase 2xbase 2(5.66)base 11.3 cm9. tan 13° 15 — 17 5—xx tan 13° 15 175175x tanPages 302–30413°15x 743.2 ftExercises8a. Let x altitude.sin 55° 30 1x0 10 sin 55° 30 x8.2 xabout 8.2 cm10. tan A — a b — 11. cos B — a c —tan 37° a 6 acos 67° — 1 6 —6 tan 37° a 16 cos 67° a4.5 a 6.3 a12. sin B — b c — 13. sin A — a c —bsin 62° —— 2 4sin 29° — 4 .6 —c24 sin 62° b c sin 29° 4.64. 621.2 b c — sinc 9.514. cos B — a c — 15.btan B — — a—29°cos 77° — 17 .3—ctan 61° — 33 .2—ac cos 77° 17.3 a tan 61° 33.217.333.2c —— coa — tas 77°8c. A — 1 2 — bhA — 1 2 — (11.3)(8.2)A 46.7 cm 2—n 61°c 76.9 a 18.416. tan B — ab —tan 49° 13 — 1 0 —aa tan 49° 13 101a — tan 4a 8.617. sin A — a c —asin 16° 55 ——13 .713.7 sin 16° 55 a4.0 a18. cos B — a c —cos 47° 18 — 22 .3—cc cos 47° 18 22.322.c — cos0—9°133—47°18c 32.919.hsin 30° — 1 2 — ncos 30° — 1 2 —12 sin 30° h 12 cos 30° n6 h 10.4 n6tan 45° m6sin 45° — — pm tan 45° 6 p sin 45° 66m tan 45°6p sinm 6 p 8.520a. cos 36° — 10 .8—xx cos 36° 10.810.8x — co—s 36°x 13.3 cm20b. tan 36° 10.8 tan 36° 1 2 s2 10.8 tan 36° s15.7 sabout 15.7 cm20c. P 5sP 5(15.7)P 78.5 cm21a. cos 42° 30 1 2 s 10.8 1 2 (14.6) xx cos 42° 30 — 1 2 — (14.6) 1 2 (14.6)cos 42° 30x x 9.9 m21b. tan 42° 30 — 1 2 — (14.6) tan 42° 30 x6.7 xabout 6.7 mx 1 2 (14.6)45°Chapter 5 148
21c. A — 1 2 — bh22a. r — 1 2 — (6.4) or 3.2A — 1 2 — a(14.6)(6.7)cos 30° — 3 .2 —A 48.8 m 23.2 cos 30° a2.771281292 aabout 2.8 cm22b. Let x side of hexagon. 22c. P 6s 1 2 P 6(3.2)xsin 30° P 19.2 cm 3.232 sin 30° — 1 2 — x2 3.2 sin 30° x3.2 x; 32 cm22d. A — 1 2 — paA — 1 2 — (19.2)(2.771281292)A 26.6 cm 25.823. sin 10° 21 36 — 19 —xx sin 10° 21 36 195.8195.8x — —sin 1 0°21 36x 1088.8 ft24. height: V — 1 3 — area of base height25a.xtan V — 1 3 — (s 2 ) —1 2 — s tan — 1 2 — s tan x60˚8 ft 1 2 sV — 1 6 — s 3 tan 84 ft25b. 84 8 7625c. sin 60° — 7 —x6tan 60° — 7 —x6x sin 60° 7676x tan 60° 76x ——sin 60°76x ——tan 60°x 87.8 ftx 43.9 ft26a. tan 6° — 39 —x00x tan 6° 39003900x ——t an6°x 37,106.0 ft26b. sin 6° — 39 —x00x sin 6° 39003900x ——s in6°x 37,310.4 ft27. Yacht: Barge:tan 20° — 20 8—xtan 12° 30 — 20 8—xx tan 20° 208 x tan 12° 30 208208208x —— tax ——tann 20°x 571.5 x 938.2938.2 571.5 366.8 ft; no12°3028. Let M represent the point of intersection of thealtitude and EF. Since GEF is isosceles, thealtitude bisects EF. EMG is a right triangle.Therefore, sin v — a ạs — or s sin v a and tan v — 0 5b —or 0.5b tan v a.29. Latasha: Markisha:xxsin 35° —— 25 0sin 42° ——22 5250 sin 35° x 225 sin 42° x143.4 x 150.6 x1506 143.4 7.2Markisha’s; about 7.2 ft30. Let x the height of the building.Let y the distance between the buildings.tan 47° 30 — x y —tan 54° 54 40 xyy tan 47° 30 x y tan 54° 54 40 xx40 xy —— tan 4 7° 30y ——tan 54°54x40 x tan 4 7° 30 ——tan 54°54tan 54° 54(x) tan 47° 30(40 x)x tan 54° 54 40 tan 47° 30 x tan 47° 30x(tan 54° 54 tan 47° 30) 40 tan 47° 30x x 131.7 ft31. terminal side — Quadrant IIreference angle: 180° 120° 60°— 1 2 — ,3 240 tan 47° 30———tan 54° 54 tan 47° 30sin 120° ycos 120° x3sin 120° cos 120° — 1 2 — 2tan 120° y x 32 1 2 csc 120° 1 y csc 120° 1tan 120° 23tan 120° 3 csc 120° 3sec 120° — 1 x — cot 120° — x y —sec 120° 1 1 2 1 cot 120° 2 32sec 120° 2cot 120° 332. (PR) 2 (RQ) 2 (PQ) 27 2 2 2 (PQ) 253 (PQ) 253 PQsin P — P r — cos P — q r —2sin P cos P 753 53 253 753sin P cos P 53tan P — P q — 53tan P — 2 7 —33. 43° 15 35 43° 15° 1°60 35 136 43.260° 323°00149 Chapter 5
34.yOy |x 2|x35. Let x the cost of notebooks and y the cost ofpencils.3x 2y 5.894x y 6.203x 2y 5.80 → 3x 2y 5.804. r x 2 y 2 r 25) 2 ( 2r 29sin v — y r — cos v x r tan v y x sin v cos v tan v — 25— or 5 2 sec G 61111sin v cos v rcsc v — — yrsec v xcsc v 29 29 29 or sec v 5 52cot v — x y —2(4x y) 2(6.20)8x 2y 12.402cot v —— 5or 2 5 5x 6.60 5. tan 27.8° — 55 0—x4x y 6.20 x $1.32x tan 27.8° 5504(1.32) y 6.20550y $0.92x tan 27.8°36. — m milesh hoursxh — milesx 1043.2 ftThe correct choice is E.5-5Page 304 Mid-Chapter QuizSolving Right Triangles1. 34.605 ° 34° (0.605 60) 34° 36.3 34° 36 (0.3 60)Pages 308–3091a. linearCheck for Understanding1b. angle 34° 36 182. They are complementary.34° 36 182. — 3. Sample answer:400°—360° 1.111.11 1 0.110.11 360° 40°360° 40° 320°; IV60˚30˚3. (GI ) 2 (IH ) 2 (GH ) 24. Marta; they need to find the inverse of the cosine,(GI ) 2 10 2 12 21(GI ) 2 not — 44c os — .5. 60°, 300° 6. 150°, 330°GI 44 or 211sin G — s ideoppositehypotenuseideadjacent7. sin sin 1 3 2 h —ypotenuseLet A sin 1 33 . Then sin A 22.sin G — 1 0—12or 5 6 211 11cos G or 12 6sin sin 1 3 3 sideoppositehypotenuse2 2tan G — — s ideadjacentcsc G — —s ideopposite8. Let A cos 1 — 310tan G csc G 1 2 10or — 6 5 — . Then cos A — 3 5 —5 —211r 2 x 2 y 25115tan G 3 2 y 2 1116 y 2hypotenusesec G — — s ideadjacentcot G — s ideadjacent—sideopposite4 y12211sec G cot G or 11 tan A — 4 3 — ; tan cos 1 — 3 5 — — 4 3 —21110 5529 529 29229 229 29Chapter 5 150
9. tan R — r s —7tan R ——1 0R tan 1 7——1 0S cos 1 — 1 2R 35.0°S 53.1°11. A 78° 90°A 12°Find b. Find c.btan B — — a cos B — a c —btan 78° —— 4 1cos 78° — 4 —c141 tan 78° b c cos 78° 4141192.9 b c ——cos 78°c 197.2A 12°, b 192.9, c 197.212. a 2 b 2 c 2 Find B.11 2 21 2 c 2 btan B — — a562 c tan B — 2 123.7 c B tan 1 — 2 1B 62.4°Find A.A 62.35402464 90A 27.64597536c 23.7, A 27.6°, B 62.4°13. 3.2° B 90°B 58°Find a. Find b.sin A — a c — cos A — b c —asin 32° — 1 3 — cos 32° ——1b313 sin 32° a 13 cos 32° b6.9 a 11.0 bB 58°, a 6.9, b 11.014a. tan x — 1 28—02100x tan 1 — 1 28—02100x 31.4°14b. tan 38° — 12 —x80x tan 38° 12801280x ——ta n 38°x 1638.3 ft14c. tan 65° — 12 —x80x tan 65° 12801280x — ta—n 65°x 596.9 ft10. cos S — r t —cos S — 1 220 —Pages 309–312 Exercises15. 90° 16. 120°, 300°17. 30°, 330° 18. 90°, 270°19. 225°, 315° 20. 135°, 315°21. Sample answers: 30°, 150°, 390°, 510°11 —11 —20 —22. Let A arccos — 4 5 — . Then cos A — 4 5 — .cos arccos — 4 5 — — 4 5 —23. Let A tan 1 — 2 3 — . Then tan A — 2 3 — .tan tan 1 — 2 3 — — 2 3 —24. Let A cos 1 — 2 5 — . Then cos A — 2 5 — .1sec A — co sA —sec A sec A — 5 2 —sec cos 1 — 2 5 — — 5 2 —25. Let A arcsin 1. Then sin A 1.1csc A ——sin Acsc A — 1 1 — or 1csc (arcsin 1) 126. Let A cos 1 55— 1—.3Then cos A ——1 3r 2 x 2 y 213 2 5 2 y 2144 y 212 ytan A — 1 25—; tancos 1 5—— 1 — 1 2—527. Let A sin 1 — 2 5 — . Then sin A — 2 5 —r 2 x 2 y 25 2 x 2 2 221 x 221 xcos A 1 2 5 21 528. tan N — mn —tan N — 1 5—9N tan 1 — 1 5—9N 59.0°30. cos M — n p —cos M — 2 230 —M cos 1 — 2 230 —M 42.8°32. cos N — m p —7.2cos N ——13; cos sin 1 — 2 5 — 21 529. sin M — m p —8sin M ——1 4M sin 1 8— 1 4 —M 34.8°31. tan N — mn —tan N — 1 18.8—4.3N tan 1 — 1 1N 52.7°33. sin M — m p —2.5—4.7sin M — 3 7.15N cos 1 7.2——1M sin7.11 — 3 5N 65.1°M 36.5°34. tan A — 1 8—24tan B 2 418A tan 1 — 1 824 — B tan 1 — 2 418 —A 36.9° B 53.1°8.8—4.32.5—4.7151 Chapter 5
35. — 1 2 — (14) 7base angles:vertex angle 2m∠Btan A — 8 7 — tan B — 7 8 —A tan 1 — 8 7 — B tan 1 — 7 8 —A 48.8° B 41.2°2m∠B 82.4°about 48.8°, 48.8°, and 82.4°36. a 2 b 2 c 221 2 b 2 30 2b 459 b 21.444.427004 B 90B 45.6°37. 35° B 90°B 55°sin A a c sin A — 2 130 —A sin 1 — 2 3A 44.4°tan A — a b — cos A — b c —tan 35° — a 8 — cos 35° — 8 c —8 tan 35° a c cos 35° 885.6 a c ——cos 35°c 9.838. A 47° 90°A 43°btan B — — a sin B — b c —tan 47° — 12 .5—asin 47° — 12 .5—ca tan 47° 12.5 c sin 47° 12.512.512.5a —— ta n 47°c ——si m 47°a 11.7 c 17.139. a 2 b 2 c 2 tan A — a b —3.8 2 4.2 2 c 2 tan A — 3 —4.232.08 c A tan1 — 3 .4 —8. 25.7 c A 42.1°42.13759477 B 90B 47.9°40. a 2 b 2 c 2sin B — ba 2 3.7 2 9.5 2c —sin B — 3 .—7a 76.56 9.5a 8.7B sin 1 — 3 .—79.5A 22.92175446 90 B 22.9°A 67.1°41. 51.5° B 90°B 38.5°tan A — a b — sin A — a c —tan 51.5° — 13 .3—bsin 51.5° — 13 .3—cb tan 51.5° 13.3 c sin 51.5° 13.313.313.b —— tanc — sin51.5°. 810 —3—51.5°b 10.6 c 17.042. A 33° 90°A 57°sin B — b c — cos B — a c —bsin 33° —— 15 .2acos 33° — 1515.2 sin 33° b 15.2 cos 33° a8.3 b 12.7 a43. 14° B 90°B 76°sin A — a c — cos A — b c —asin 14° — 9 .8 — bcos 14° — 9 .8 —9.8 sin 14° a 9.8 cos 14° b2.4 a 9.5 b64744a. sin v ——1 020v sin 1 647——1 020v 39.4°44b. tan 39.4° — 64 7—xx tan 39.4° 647647x ——tan 39.4°x 788.5 ft45a. Since the sine function is the side oppositedivided by the hypotenuse, the sine cannot begreater than 1.45b. Since the secant function is the hypotenusedivided by the side opposite, the secant cannotbe between 1 and 1.45c. Since cosine function is the side adjacent dividedby the hypotenuse, the cosine cannot be lessthan 1.46. 10 6 44tan v ——1 5v tan 1 4——1 5v 14.9°847a. tan v — 1 00 — 547b. tan v — 1 00 —v tan 1 8— 1 00 — v tan 1 5— 1 00 —v 4.6° v 2.9°4548. tan v — 2200 —v tan 1 4— 2 2v 1.2°500 —49. — 65 mihou— lesr — 52 80 feet1 hourm —ile— 360 0 secondstan v — gv 2 r —tan v — 3295.32—( 1200)5.32—1200)v tan 1 9— 32 (v 13.3°— 95.3 — sefeet—cond—.2Chapter 5 152
sin50. — v is invr— s in 60°sinvr— si n 60°2.42— n55. x-axis y 3 x 2 2411sin F 157cos F 15The correct choice is A.(y) 3 x 2 2— 2.42y 3 x 2 2; no— sin v y-axis y 3 x 2 2ry 3 (x) 2 20.3579 sin v rsin 1 y 3 x 2 2;0.3579 v ry x y 3 x 2 221.0° v r(x) 3 (y) 2 2yes51. Draw the altitude from Y to XZ. Call the point ofx 3 y 2 2; nointersection W.y x y 3 x 2 2m∠X m∠XYW 90°(x) 3 (y) 2 230° m∠XYW 90°x 3 y 2 2; nom∠XYW 60°y-axisIn XYW:cos 30° — X W—16sin 30° — W Y56. 1 0 5 5 3 1—0 1 34 6 3 21616 cos 30° XW 16 sin 30° WY 1(5) 0(3) 1(5) 0(4)13.9 XW 8 WY0(5) 1(3) 0(5) 1(4)In ZYW:1(3) 0(6) 1(1) 0(3)880(3) 1(6) 0(1) 1(3)sin Z —— 2 4tan 19.5° ——W ZZ sin 1 81(2) 0(2)—— 2 4WZ tan 19.5° 80(2) 1(2) 8Z 19.5° WZ ——tan 19.5°5 5 3 1 234 6 3 2WZ 22.68(5, 3), (5, 4), (3, 6), (1, 3), (2, 2)cos m∠WYZ ——2 44 3 2 2 2 2m∠WYZ cos 1 8——257.48 2 0 5 1 1 m∠WYZ 70.5°9 6 3 7 2 2Y m∠XYW m∠WYZ y XW WZ4 (2) 3 2 2 (2)Y 60° 70.5° y 13.9 22.6 8 (5) 2 1 0 1Y 130.5° y 36.59 (7) 6 2 3 (2)52. baseball stadium: football stadium:2 1 0tan 63° — 10 00 —xtan 18° — 10 00 3 1 1— y2 8 5x tan 63° 1000 y tan 18° 100022.2 42.51000100058. m x —— ta n 63°y ——1 950 1880ta n 18°x 509.5 y 3077.7m — 2 0.3—70or 0.29distance x yy 22.2 0.29(x 1950)distance 509.5 3077.7y 0.29x 587.7distance 3587.2 ft59. 2x 5y 10 053. (FD) 2 (DE) 2 (FE) 25y 2x 107 2 (DE) 2 15 2(DE) 2 y — 2 5 — x 2 176— 2 DE 176 or 4115 — ; 2sin F — s ideoppositehypotenuseideadjacenth —acypotenuse60. —— a c b —— a c d 10 — ab cd—a c 10sideoppositehypotenusetan F — — scsc F — —sideadjacent411ideopposite15 411ideadjacent—ideopposite7411tan F 7csc F orhypotenusesec F — — scot F — s sideadjacent5sec F 1 7cot F or54. Use TABLE feature of a graphing calculator.0.3, 1.4, 4.31511 44711 445-6The Law of SinesPage 316 Check for Understandingx3— 2x1. — sinx30°x 1 2 sin 60°x3 32 sin 90°2x 12x 2x 2x60˚x2xx 330˚153 Chapter 5
2. Sample answer:Cb35˚A103. Area of WXYZ Area of triangle ZWY Area oftriangle XYW.m∠X m∠Ztriangle ZWY:triangle XYW:1K —— 2a bsin Z k — 1 2 — ab sin X1K —— 2a bsin XK — 1 2 — ab sin X 1 2 ab sin XK ab sin X4. Both; if the measures of two angles and a nonincludedside are known or if the measures of twoangles and the included side are known, thetriangle is unqiue.5. C 180° (40° 59°) or 81°a cb c—— sin A —— —— sin Csin B ——sin Ca 14b 14—— sin 40° —— —— sin 81°sin 59° ——sin 81°a — 14 sinsin— 40°81°b — 14 sin59°si —n 81°a 9.111200533 b 12.14992798C 81°, a 9.1, b 12.16. C 180° (27.3° 55.9°) or 96.8°b ac a—— sin B —— —— sin Asin C ——sin Ab8.6c8.6—— sin 5 5.9° —— —— sin 27.3°sin 9 6.8° ——sin 27.3°b — 8.6 sin 5sin 27— 5.9°. 3°c — 8.6 sin 96.8°sin —27.3°b 15.52671055 c 18.61879792C 96.8°, b 15.5, c 18.67. A 180° (17° 55 98° 15) or 63° 50c a—— sin C ——sin Ac17—— sin 98 °15 ——sin 6 3°50c — 17 sin98°15—sin63°50c 18.78. K — 1 2 — bc sin AK — 1 2 — (14)(12) sin 78°K 82.2 units 29. C 180° (22° 105°) or 53°K — 1 2 — A sin Cb 2 sin sinB22°sin53°—in105°K — 1 2 — (14) 2 — sin sK 30.4 units 210. Let d the distance from the fan to the pitcher’smound.v 180° (24° 12 5° 42) or 150° 6d60.5—— sin 1 50° 6 ——sin 5°42d — 60.5 sin 150°6—sin 5° 42c40˚d 303.7 ftBPages 316–318 Exercises11. B 180° (40° 70°) or 70°b ac a—— sin B —— —— sin Asin C ——sin Ab 20c 20—— sin 70° —— —— sin 40°sin 70° ——sin 40°b — 20 sinsin— 70°40°c — 20 sin70°si —n 40°b 29.238044 c 29.238044B 70°, b 29.2, c 29.212. A 180° (100° 50°) or 30°a cb c—— sin A —— —— sin Csin B ——sin Ca 30b 30—— sin 30° —— —— sin 50°sin 100° ——sin 50a — 30 sinsin— 30°50°b — 30 sin100°s —in 50°a 19.58110934 b 38.56725658A 30°, a 19.6, b 38.613. C 180° (25° 35°) or 120°a bc b—— sin A —— —— sin Bsin C ——sin Ba 12c 12—— sin 25° —— —— sin 35°sin 120° ——sin 35°a — 12 sinsin— 25°35°c — 12 sin120°s —in 35°a 8.84174945 c 18.11843058C 120°, a 8.8, c 18.114. C 180° (65° 50°) or 65°a cb c—— sin A —— —— sin Csin B ——sin Ca 12b 12—— sin 65° —— —— sin 65°sin 50° ——sin 65°a — 12 sinsin— 65°65°b — 12 sin50°si —n 65°a 12 b 10.14283828C 65°, a 12, b 10.115. A 180° (24.8° 61.3°) or 93.9°b ac a—— sin B —— —— sin Asin C ——sin Ab8.2c8.2—— sin 2 4.8° —— —— sin 93.9°sin 6 1.3° ——sin 93.9°b — 8.2 sin 2sin 93— 4.8°. 9°c — 8.2 sin 61.3°sin —93.9°b 3.447490503 c 7.209293255A 93.9°, b 3.4, c 7.216. B 180° (39° 15 64° 45) or 76°a cb c—— sin A —— —— sin Csin B ——sin Ca19.3b 19.3—— sin 3 9 51 —— —— sin 64°45 sin 76° ——sin 64°45a — 19ṡ 3 sin39in 64°— ° 1545b — 1 9.3sin76°s —in 64°45a 13.50118124 b 20.7049599B 76°, a 13.5, b 20.717. C 180° (37° 20 51° 30) or 91° 10a c—— sin B ——sin Cb125—— sin 5 1° 30 ——sin 91°10b — 12 5 sin51°31—sin91°10b 97.818. A 180° (29° 34 23° 48) or 126° 38a b—— sin A ——sin B1— — sin 2— sin 12a6° 381—9°34a — 11 sin126°38—sin29°34a 17.9Chapter 5 154
19. K — 1 2 — bc sin A28c. P 112.7 72.7 80P 265.4 ftr s—— sin R — sin S— r si n Ss7°sin84°— m s in N—n — r si n Sssin N sin S, so — m n — — r s —b 2 — sin sinBMNP RST.15°sin52°—30. 360° 5 72°in113°soccer fields.x4—— sin 6 2° 30 ——sin 9 7° 15x — 4 sin62°30—sin97°15B sin C—in A1sin 60.8° sin 65.4°— ———2 sin 53.8°K — 1 2 — (14)(9) sin 28°K 29.6 units 220. A 180° (37° 84°) or 59°K — 1 2 — a 2 — sin B sin C—sinAK — 1 2 — (5) 2 — sin 3 —sin59°K 8.7 units 221. C 180° (15° 113°) or 52°K — 1 2 — A sin C—K — 1 2 — (7) 2 — sin sK 5.4 units 222. K — 1 2 — bc sin AK — 1 2 — (146.2)(209.3) sin 62.2°K 13,533.9 units 223. K — 1 2 — ac sin BK — 1 2 — (12.7)(5.8) sin 42.8°K 25.0 units 224. B 180° (53.8° 65.4°) or 60.8°K — 1 2 — a 2 — sin sK (19.2) 2K 181.3 units 225. K ab sin X (formula from Exercise 3)K (14)(20) sin 57°K 234.8 cm 226. Area of pentagon 5 Area of triangle360° 5 72°9 72˚ 9K — 1 2 — (9)(9) sin 72° 5K 5(38.51778891)K 38.51778891 5K 192.6 in 227. Area of octagon 8 Area of triangle360° 8 45°5545˚K — 1 2 — (5)(5) sin 45° 8K 8(8.838834765)K 8.838834765 8K 70.7 ft 228a. 180° (95° 40°) 45°x 80y 8028b. —— sin 95° —— —— sin 45°sin 40° ——sin 45°x — 80 sinsin— 95°45°y — 80 sin40°si —n 45°x 112.7065642 y 72.72311643about 112.7 ft and 72.7 ft29. Applying the Law of Sines, — sinmn—M — sin Nn N— and—. Thus sin M — n si —nand sin R —. Since ∠M ∠R, sin M sin R and—. However, ∠N ∠S andand — m r — — n s — . Similarproportions can be derived for p and t. Therefore,triangle: K — 1 2 — (300)(300)sin 72°K 42,797.54323pentagon: 5K 5(42,797.54323)5K 213,987.7 ft31a. v 180° (20° 15 62° 30) or 97° 15Let x the distance from the balloon to thex 3.6 mi31b. v 180° (20° 15 62° 30) or 97° 15Let y the distance from the balloon to thefootball field.— sin 240° 154— ——sin 9 7° 15y — 4 sin20°15—sin97°15y 1.4 mi32. 180° 30° 150°v 180° (26.8° 150°) or 3.2°Let x the length of the track.x 100—— sin 2 6.8° ——sin 3.2°x — 100 sin26.8°—sin3. 2°x 807.7 ft33a. Let x the distance of the second part of theflight.v 180° (13° 160°) or 7°x 80—— sin 13° ——si n 7°x — 80 sin13°—sin7°x 147.6670329distance of flight 80 147.7 or about 227.7 mi33b. Let y the distance of a direct flight.y 80—— sin 160° ——si n 7°y — 80 sin160°—sin7°y 224.5 mi155 Chapter 5
34. x90° 63° 27°55˚ 63˚ 180° (55° 63°) 62°y62˚Let x the verticaldistance.Let y the length of the13 ftoverhang.27˚63˚x 13y 66—— sin 27° —— —— sin 63°sin 63° ——sin 62°x — 13 sinsin— 27°63°y — 6.6 sin 63°sin —62°x 6.623830843 y 6.684288563about 6.7 fta b35a. —— sin A ——sin B— a b — sinA ——s inBa c35b. —— sin A ——sin C— a c — sinA ——s inC— a c — sinA 1 —— s inC 1— a c — — c c — sin — s in— AC — s ins —CinC— a c—c— sin A sin C—sin C35c. From Exercise 34b, — a c—c— sin A sin C—sin Cor — sin A sa — in Cc — sin C—.ca c—— sin A ——sin C— a c — sinA ——s inC— a c — sinA 1 —— s inC 1— a c — — c c — sin — s in— AC — s ins —CinC— a c—c— sin A sin C—sin C— sin C—c— sin A sin C—a cTherefore, — sin A sa — in Cc— sin A sin Ca — cor — a c—a c— s inA sinCsinA —.sinCa b35d. —— sin A ——sin B— a b — sinA ——s inB— a b — sinA 1 —— s inB 1— a b — — b b — sinA —— s inB — s in—BsinB— a b—b— sin A sin B—sin Bb sin B—— a b— —sin A sin B36. tan v — 4 520 —v tan 1 — 4 2v 66.0°50 —37. sin v — y r — r 2 x 2 y 2sin v — 1 6 — 6 2 x 2 (1) 2y 1, r 6 35 x 235 xQuadrant IV, so x 35cos v — x r — tan v — y x — csc v — yr —35 61 35 35 356cos v tan v csc v —— or 6ytan v sec v — xr — cot v — x y —6 35 635 3535 1sec v cot v or 35sec v 38. 83° 360k°39. Let x standard carts and let y deluxe carts.2 x 84 y 11x y 151612y x 2 x 8x y 15(4, 11) y 118(8, 7)(2, 11)y 4(2, 4) (8, 4)xO 4 12 16M(x, y) 100x 250yM(2, 4) 100(2) 250(4) or 1200M(2, 11) 100(2) 250(11) or 2950M(4, 11) 100(4) 250(11) or 3150M(8, 7) 100(8) 250(7) or 2550M(8, 4) 100(8) 250(4) or 18004 standard carts, 11 deluxe carts40. 4x y 2z 03x 4y 2z 207x 5y 2z 203(3x 4y 2z) 3(20)2(2x 5y 3z) 2(14)↓9x 12y 6z 604x 10y 6z 285x 22y 885(7x 5y) 5(20) → 35x 25y 1007(5x 22y) 7(88) 35x 154y 616129y 516y 47x 5y 20 4x y 2z 07x 5(4) 20 4(0) 4 2z 0x 0z 2(0, 4, 2)41. 6 3x y 3x y 12y 3x 6 y 3x 1216 3x y 12OxChapter 5 156
42. Area of one face of the small cube 1 2 or 1 in 2 .Surface area of the small cube 6 1 or 6 in 2 .Area of one face of large cube 2 2 or 4 in 2 .Surface area of large cube 6 4 or 24 in 2 .Surface area of all small cubes 8 6 or 48 in 2 .The difference in surface areas is 48 in 2 24 in 2or 24 in 2 .The correct choice is A.Page 319 History of Mathematics1. See students’ work; the sum is greater than 180°.In spherical geometry, the sum of the angles of atriangle can exceed 180°.2. See students’ work. Sample answer: Postulate 4states that all right angles are equal to oneanother.3. See students’ work.5-7The Ambiguous Case for the Lawof SinesPage 323 Graphing Calculator Exploration1. B 44.1°, C 23.9°, c 1.82. B 52.7°, C 76.3°, b 41.0; B 25.3°,C 103.7°, b 22.03. The answers are slightly different.4. Answers will vary if rounded numbers are used tofind some values.Page 324 Check for Understanding1. A triangle cannot exist if m∠A 90° anda b sin A or if m∠A 90° and a b.2.BA30˚106 10 sin 30° si n Bn30°56.4˚693.6˚ Csin B 10 si 6B sin 1 10 si n30°6 AB6 123.6˚30˚ 26.4˚10C180° 180° 56.4° 123.6°C 180° (30° 123.6°) 26.4°B 56.44269024C 180° (30° 56.4°) 93.6°3. Step 1: Determine that there is one solution forthe triangle.Step 2: Use the Law of Sines to solve for B.Step 3: Subtract the sum of 120 and B from 180 tofind C.Step 4: Use the Law of Sines to solve for c.4. Since 113° 90°, consider Case II.15 8; 1 solution5. Since 44° 90°, consider Case I.a sin B 23 sin 44°a sin B 23 (0.6947)a sin B 15.9771425212 16.0; 0 solutions6. Since 17° 90°, consider Case I.a sin C 10 sin 17° 2.9237170472.9 10 11; 1 solutionc a sin C sin A11 10 sin 17° si n Asin A 10 s in 17°11in 17° A sin 1 10 s 11A 15.41404614B 180° (15.4° 17°) or about 147.6°c b sin C sin B11b sin 17° sin 1 47.6°b 11 s in147.6°sin17°b 20.16738057A 15.4°, B 147.6°, b 20.27. Since 140° 90°, consider Case II.3 10; no solutions8. Since 38° 90°, consider Case I.b sin A 10 sin 38°b sin A 6.1566147536.2 8 10; 2 solutionsa b sin A sin B8 10 sin 38° si n Bn38°sin B 10 si 8B sin 1 10 si n38°8 B 50.31590502180° 180° 50.3° or 129.7°<strong>Solution</strong> 1C 180° (50.3° 38°) or 91.7°a c sin A sin C8c sin 38° sin 9 1.7°c 8s in91.7°sin38°c 12.98843472<strong>Solution</strong> 2C 180° (129.7° 38°) or 12.3°a c sin A sin C8c sin 38° sin 1 2.3°c 8s in12.3°sin38°c 2.768149638B 50.3°, C 91.7°, c 13.0; B 129.7°,C 12.3°, c 2.8157 Chapter 5
9. Since 130° 90°, consider Case II.17 5; 1 solutionc b sin C sin B175 sin 130° sin B130°sin B 5sin 17B sin 1 5sin 130°17 B 13.02094264A 180° (13.0 130°) or 37.0°c a sin C sin A17a sin 130° sin 3 7.0°a 17 sin37.0°sin130°a 13.35543321A 37.0°, B 13.0°, a 13.410a.45 ft10b. 90° 10° 80°180° 80° 100°45 70100˚80˚45 70100˚80˚70 ft10˚x˚39.3˚y10˚7045 sin 100° si n xsin x 45 si n 100°70n 100° x sin 1 45 si 70x 39.3°10c. v 180° (100° 39.3°)or about 40.7°10˚y70 sin 4 0.7° sin 100°y 70 sin40.7°sin100°y 46.4 ft13. Since 61° 90°, consider Case I.a sin B 12 sin 61°a sin B 10.495436498 10.5; 0 solutions14. two angles are given; 1 solution15. Since 100° 90°, consider Case II.15 18; 0 solutions16. Since 37° 90°, consider Case I.a sin B 32 sin 37°a sin B 19.2580807427 19.3; 2 solutions17. Since 65° 90°, consider Case I.b sin A 57 sin 65°b sin A 51.6595438655 51.7; 2 solutions18. Since 150° 90°, consider Case II.6 8; no solution19. Since 58° 90°, consider Case I.b sin A 29 sin 58°b sin A 24.5933947926 24.6; 2 solutionsa b sin A sin B26 29 sin 58° si n Bsin B 29 s in 58°26in 58° B sin 1 29 s 26B 71.06720496180° 180° 71.1° or 108.9°<strong>Solution</strong> 1C 180° (58° 71.1°) or 50.9°a c sin A sin C26 c sin 58° sin 5 0.9°c 26 sin50.9°sin58°c 23.80359004<strong>Solution</strong> 2C 180° (58° 108.9)° or 13.1°a c sin A sin C26 c sin 58° sin 1 3.1°c 26 sin13.1°sin58°c 6.931727606B 71.1°, C 50.9°, c 23.8; B 108.9°,C 13.1°, c 6.9Pages 324–326 Exercises11. Since 57° 90°, consider Case I.b sin A 19 sin 57°b sin A 15.9347407911 15.9; 0 solutions12. Since 30° 90°, consider Case I.c sin A 26 sin 30°c sin A 1313 13; 1 solution.Chapter 5 158
20. Since 30° 90°, consider Case I.b sin A 8 sin 30°b sin A 44 4; 1 solutiona b sin A sin B4 8 sin 30° sin B30°sin B 8sin 4B sin 1 8sin 30°4 B 90C 180° (30° 90°) or 60°a c sin A sin C4 c sin 30° sin 60°C 4 sin60°sin30°C 6.92820323B 90°; C 60°, c 6.921. Since 70° 90°, consider Case I.a sin C 25 sin 70°a sin C 23.4923155224 23.5; 2 solutionsc a sin C sin A24 25 sin 70° si n Asin A 25 s in 70°24in 70° A sin 1 25 s 24A 78.1941432180° 180° 79.2° or 101.8°<strong>Solution</strong> 1B 180° (70° 79.2°) or 31.8°c b sin C sin B24 b sin 70° sin 3 1.8°b 24 sin31.8°sin70°b 13.46081025<strong>Solution</strong> 2B 180° (70° 101.8°) or 8.2°c b sin C sin B24 b sin 70° sin 8.2°b 24 sin8.2°sin70°b 3.640196918A 78.2°, B 31.8°, b 13.5; A 101.8°,B 8.2°, b 3.622. C 180° (40° 60°) or 80°c ac b sin C sin Asin C sin B20 a20°b sin 80° sin 40°sin 80° sin 60°a 20 sin40°sin80°b 20 sin60°sin80°a 13.05407289 b 17.58770483C 80°, a 13.1, b 17.623. Since 90° 90°; consider Case II.12 14; no solution24. Since 36° 90°, consider Case I.c sin B 30 sin 36°c sin B 17.6335575719 17.6; 2 solutionsb c sin B sin C19 30 sin 36° si n Csin C 30 s in 36°19in 36° C sin 1 30 s 19C 68.1377773180° 180° 68.1° or 111.9°<strong>Solution</strong> 1A 180° (36° 68.1°) or 75.9°b a sin B sin A19 a sin 36° sin 7 5.9°a 19 sin75.9°sin36°a 31.34565276<strong>Solution</strong> 2A 180° (36° 111.9°) or 32.1°b a sin B sin A19 a sin 36° sin 3 2.1°a 19 sin32.1°sin36a 17.1953669A 75.9°, C 68.1°, a 31.3; A 32.1°,C 111.9°, a 17.225. Since 107.2° 90°, consider Case II.17.2 12.2; 1 solutiona c sin A sin C17.212.2 sin 107.2° s inCsin C 12.2 s in107.2°17.2C sin 1 12.2 s in107.2°17.2C 42.65491459B 180° (107.2° 42.7°) or 30.1°a b sin A sin B17.2b sin 107.2° sin 3 0.1°b 17 . 2 sin30.1°sin107.2°b 9.042067456B 30.1°, C 42.7°, b 9.026. Since 76° 90°, consider Case I.b sin A 20 sin 76°b sin A 19.405914535 19.4; no solution159 Chapter 5
27. Since 47° 90°, consider Case I.16 10; 1 solutionc a sin C sin A16 10 sin 47° si n Asin A 10 s in 47°16in 47° A sin 1 10 s 16A 27.19987995B 180° (47° 27.2) or 105.8°c b sin C sin B16b sin 47° sin 1 05.8°b 16 s in105.8°sin47°b 21,0506609A 27.2°, B 105.8°, b 21.128. Since 40° 90°, consider Case I.c sin B 60 sin 40°c sin B 38.5672565842 38.6; 2 solutionsb c sin B sin C42 60 sin 40° si n Csin C 60 s in 40°42in 40° C sin 1 60 s 42C 66.67417652180° 180° 66.7° or 113.3°<strong>Solution</strong> 1A 180° (40° 66.7°) or 73.3°b a sin B sin A42 a sin 40° sin 7 3.3°a 42 sin73.3°sin40°a 62.58450564<strong>Solution</strong> 2A 180° (40° 113.3°) or 26.7°b a sin B sin A42 a sin 40° sin 2 6.7°a 42 sin26.7°sin40°a 29.33237132A 73.3°, C 66.7°, a 62.6; A 26.7°,C 113.3°, a 29.329. Since 125.3° 90°, consider Case II.32 40; no solution21.7 19.330. sin 57.4° s inxsin x 19.3 sin57.4°21.7x sin 1 19.3 sin57.4°21.7x 48.52786934v 180 (57.4° 48.5°) or 74.1°21.7 y sin 57.4° sin 7 4.1°y 21ṡ 7 sin74.1°in 57.4°y 24.7692241719.3 cm 74.1˚ 21.7 cm57.4˚ 48.5˚24.8 cma b31. sin A sin B15 20 sin 29° si n Bsin B 20 s in 29°15in 29° B sin 1 20 s 15B 40.27168721180° 180° 40.3° or 139.7°<strong>Solution</strong> 1C 180° (29° 40.3°) or 110.7°a c sin A sin C15c sin 29° sin 1 10.7°c 15 sin110.7°sin29°c 28.93721187Perimeter a b c 15 20 28.9 or about 63.9 units<strong>Solution</strong> 2C 180° (29° 139.7°) or 11.3°a c sin A sin C15 c sin 29° sin 1 1.3°c 15 sin11.3°sin29°c 6.047576406Perimeter a b c 15 20 6.0 or about 41.0 unitsb a32. sin B sin A13 15 sin 55° si n Asin A 15 s in 55°13in 55° A sin 1 15 s 13A 70.93970395180° 180° 70.9° or 109.1°<strong>Solution</strong> 1C 180° (70.9° 55°) or 54.1°b c sin B sin C13 c sin 55° sin 5 4.1°c 13 sin54.1°sin55°c 12.8489656Perimeter a b c 15 13 12.8 or about 40.8<strong>Solution</strong> 2c 180° (109.1° 55°) or 15.9°b c sin B sin C13 c sin 55° sin 1 5.9°c 13 sin15.9°sin55°c 4.35832749Perimeter a b c 15 13 4.4 or about 32.4A 70.9°, B 55°, C 54.1°Chapter 5 160
33. side opposite 37° 15 18 or 33side between v and 37° 15 22 or 37Let x the measure of the third angle.33 37 sin 37° si n xsin x 37 s in 37°33in 37° x sin 1 37 s 33x 42.43569405v 180 (37° 42.4°) or about 100.6°34a. a b sin A 34b. a b sin Aa 14 sin 30° a 14 sin 30°a 7 a 7 or a 1434c. a b sin Aa 14 sin 30°a 7 and a 147 a 14184.5 14035. s in 59° s in xsin x 140 sin59°184.5x sin 1 140 sin59°184.5x 40.57365664v 180° (59° 40.6°) or about 80.4°90° 80.4° 9.6°36a. 12° 90° and 316 450 sin 12°; 2 solutions316450 si n 12° s invsin v 450 sin12°316v sin 1 450 sin12°316v 17.22211674180° 180° 17.2° or 162.8°turn angle 180° 162.8° or 17.2°about 17.2° east of north36b. v 180° (162.8° 12°) or about 5.2°x 316 sin 5.2° si n 12°x 316 sin 5.2°sin 12°x 138.3094714d rt138.3 23t6.013455278 t; about 6 hr36c. 180° 20° 160°180° (160° 12°) 8°200c si n 8° sin 160°c 200 sin160°sin8°c 491.5032301Since 491.5 450, the shipwill not reach port.20˚200 mi.Portc160˚12˚37. Distance from satellite to center of earth is 3960 1240 or 5200 miles.angle across from 5200 mi side 45° 90° or 135°5200 si n 135° 3 960sinxsin x 3960 sin135°5200x sin 1 3960 sin135°5200x 32.58083835v 180° (135° 32.6°) or about 21.4° 2 1.4°360° (2 hours) 0.0689953425 hours or about 4.1minutes38. P turns 20(360°) or 7200° every second whichequals 72° every 0.01 second.PQOP si n O si n Q15 5 sin 72° sin Qsin Q 5si n 72°15n 72° Q sin 1 5si 15Q 18.48273235m∠P 180° (72° 18.5°) or about 89.5°QOPQ s in P si n OQO15 sin 89.5° sin 72°QO 15 sin89.5°sin72°QO 15.77133282QO 5 15.8 5 or about 10.8 cm39a. b c sin B12 17 sin B2 sin B 1 1727sin 1 1 1 B44.90087216 BB 44.9°39b. b c sin B12 17 sin B 1 21727 sin Bsin 1 1 1 B44.90087216 BB 44.9°39c. b c sin B12 17 sin B 1 21727 sin Bsin 1 1 1 B44.90087216 BB 44.9°40. Area of rhombus 2(Area of triangle)triangle:K 1 2 bc sin AK 1 2 (24)(24) sin 32°K 152.6167481rhombus:A 2(152.6) or about 305.2 in 2161 Chapter 5
41. tan 22° 7 4. Sample answers:x575x AA 180° (90° 53°) or 37°tan 22°sin B b cx 185.6 m42. 3; 1 2 ⏐ 4 4 13 6sin 53° 1 010cc102 1 6c C53˚ Bsin 53°4 2 12 ⏐ 0a4x 2 c 12.52135658 2x 12 0bx 2 tan B (2)2 )(12)4(4a2(4)tan 53° 1 0 ax 2 188 108a tan 53°x 2 2i47 8a 7.5355405011 i47 x A 37°, a 7.5, c 12.54C 180° (5.5° 45°) or 80°43. noAb a3x3x sin B 4 x 1sin Ax 1 x 1x 1 1x 1 x 1b 10 4x 1 sin 45° 9x9x9x xsin 55°3 3xb55˚ c x 1 x 1x 1b 10 sin45°sin55°44. 5x 2y 9 y 3x 1C45˚B105x 2(3x 1) 9 y 3(7) 1b 8.6321799c a5x 6x 2 9 y 22 sin C sin Ax 7c 10 sin 80° (7, 22)sin 55°c 10 sin80°45. 2x 5y 7sin55°y 2 c 12.02228285 x 7 5 C 80°, b 8.6, c 12.0perpendicular slope: 5 2 c 2 a 2 b 2 2ab cos Cy 4 5 c 2 10 2 8 2 2(10)(8) cos 50°2 (x (6))c 2 61.15398245y 4 5 2 x 15c 7.8201011792y 8 5x 30a c Asin A sin C5x 2y 22107. 8 si n A 46. Perimeter of XYZ 4 8 9 or 21sin 50°8 csin A 10 s in50°length of AB 1 3 of perimeter7.8C50˚B A sin 1 1010 s in50° 1 3 (21) or 77.8 The answer is 7.A 78.4024367B 180° (78.4° 50°) or 51.6°A 78.4°, B 51.6°, c 7.85. a 2 b 2 c 2 2bc cos A5-8 The Law of Cosines32 2 38 2 46 2 2(38)(46) cos A 322 382 46 22(38)(46) cos APages 330–331 Check for Understandingcos 1 322 38 2 4622(38)(46) A1. The Law of Cosines is needed to solve a triangle ifthe measures of all three sides or the measures of43.49782861 Aa btwo sides and the included angle are given. sin A sin B2. Sample answer: 1 in., 2 in., 4 in.3238 sin 43.5° si n B3. If the included angle measures 90°, the equationsin B 38 si n 43.5°becomes c 2 a 2 b 2 2ab cos C. Since32cos 90° 0, c 2 a 2 b 2 2ab(0) or c 2 a 2 b 2 .B sin 1 38 si n 43.5°32B 54.8C 180° (43.5° 54.8°) or 81.7°A 43.5°, B 54.8°, C 81.7°Chapter 5 162
6. c 2 a 2 b 2 2ab cos Cc 2 25 2 30 2 2(25)(30) cos 160°c 2 2934.538931c 54.1713848c a sin C sin A54.2 25 sin 160° si n Asin A 25 s in160°54.2A sin 1 25 s in160°54.2A 9.1B 180° (9.1° 160°) or 10.9°A 9.1°, B 10.9°, c 54.27. The angle with greatest measure is across fromthe longest side.21 2 18 2 14 2 2(18)(14) cos v 212 182 14 22(18)(14) cos vcos 1 182 14 212 22(18)(14) v81.0 vabout 81.0°8. s 1 2 (2 7 8) 8.5K 8.5(8. 5 )(8.5 27)(8.5 8) 6.4 units 29. s 1 2 (25 13 17) 27.5K 27.5(2 7.5 7.5 25)(27.5 13)(217) 102.3 units 210. a 2 b 2 c 265 ft65 2 65 2 c 28450 c 2c91.92388155 ccos 1 65 ft65 ftc65 2 65 2 91.9 22(65)(91.9)2(65)(91.9) v45.01488334 v65 2 65 2 91.9250 ft45.0˚a 2 b 2 c 2 2bc cos v65 2 65 2 91.9 2 2(65)(91.9) cos v cos vc 2 a 2 b 2 2ab cos Cc 2 65 2 50 2 2(65)(50) cos 45.0°c 2 2128.805922c 46.1 ftPages 331–332 Exercises11. a 2 b 2 c 2 2bc cos Aa 2 7 2 10 2 2(7)(10) cos 51°a 2 60.89514525a 7.803534151a b sin A sin B7. 8 7 sin 51° sin Bsin B 7si n 51°7.8n 51° B sin 1 7si 7.8B 44.22186872C 180° (51° 44.2°) or 84.8°B 44.2°, C 84.8°, a 7.812. c 2 a 2 b 2 2ab cos C7 2 5 2 6 2 2(5)(6) cos C 72 52 6 22(5)(6) 52 6 22( 5)(6) cos Ccos 1 72 C78.46304097 Ca c sin A sin C5 7 sin A sin 7 8.5°78.5°sin A 5sin 7A sin 1 5sin 78.5°7 A 44.42268919B 180° (44.4° 78.5°) or 57.1°A 44.4°, B 57.1°, C 78.5°13. c 2 a 2 b 2 2ab cos C7 2 4 2 5 2 2(4)(5) cos C 72 42 5 22(4)(5) 42 5 22( 4)(5) cos Ccos 1 72 C101.536959 Ca c sin A sin C47 sin A sin 1 01.5°101.5°sin A 4sin 7A sin 1 4 sin 101.5°7 A 34.05282227B 180° (34.1° 101.5°) or 44.4°A 34.1°, B 44.4°, C 101.5°14. b 2 a 2 c 2 2ac cos Bb 2 16 2 12 2 2(16)(12) cos 63°b 2 225.6676481b 15.02223845a b sin A sin B1615.0 si n A si n 63°sin63°5.0A sin 1 16 sin63°15.0sin A 16 1A 71.62084388C 180° (71.6° 63°) or 45.4°A 71.6°, C 45.4°, b 15.0163 Chapter 5
15. b 2 a 2 c 2 2ac cos B13.7 2 11.4 2 12.2 2 2(11.4)(12.2) cos Bcos 1 cos B B70.8801474 Ba b sin A sin B11.413.7 s in A sin 70.9°sin70.9°3.7A sin 1 11.4 sin70.9°13.7sin A 11.4 113.7 2 11.4 2 12.222(11.4)(12.2)A 51.84180107C 180° (51.8° 70.9°) or 57.3°A 51.8°, B 70.9°, C 57.3°16. c 2 a 2 b 2 2ab cos Cc 2 21.5 2 13 2 2(21.5)(13) cos 79.3°c 2 527.462362c 22.96654876a c sin A sin C21.523.0 s inA sin 79.3°sin79.3°3.0A sin 1 21.5 sin79.3°23.0sin A 21.5 213.7 2 11.4 2 12.2 22(11.4)(12.2)A 66.90667662B 180° (66.9° 79.3°) or 33.8°A 66.9°, B 33.8°, c 23.017. 14.9 2 23.8 2 36.9 2 2(23.8)(36.9) cos v14.9 2 23.8 2 36.9 22(23.8)(36.9) cos vcos 1 14.92 23.8 2 36.922(23.8)(36.9) v13.75878964 vabout 13.8°18. d 2 1 40 2 60 2 2(40)(60) cos 132°d 2 1 8411.826911d 1 91.71601229180° 132° 48°d 2 2 40 2 60 2 2(40)(60) cos 48°d 2 2 1988.173089d 2 44.58893461about 91.7 cm and 44.6 cm19. s 1 2 (4 6 8) 9K 9(9 4)(9 6)(9 8) 11.6 units 220. s 1 2 (17 13 19) 24.5K 24.5(2 4.5 4.5 17)(24.5 13)(219) 107.8 units 221. s 1 2 (20 30 40) 45K 45(45 )(45 205 30)(40) 290.5 units 222. s 1 2 (33 51 42) 63K 63(63 )(63 333 51)(62) 4 690.1 units 223. s 1 2 (174 138 188) 250K 250 76 12 12 6 11,486.3 units 224. s 1 2 (11.5 13.7 12.2) 18.7K 187(1 8.7 (18.7 11.5) 7)(18.13.2.2) 1 66.1 units 225a. d 2 30 2 48 2 2(30)(48) cos 120°d 2 4644d 68.1 in.25b. Area of parallelogram 2(Area of triangle)K 1 2 (30)(48) sin 120°K 623.53829072K 2(623.5382907) or about 1247.1 in 226a. s 1 2 (15 15 24.6) 27.3K 27.3(2 7.3 7.3 15)(27.3 15)(224.6) 105.6Area of rhombus 2(105.6) 211.2 cm 226b. 24.6 2 15 2 15 2 2(15)(15) cos v24.6 2 15 2 1522(15)(15) cos vcos 1 24.62 15 2 1522(15)(15) v110.1695875 v180° 110.2° 69.8°about 110.2°, 69.8°, 110.2°, 69.8°27. The angle opposite the missing side 45°.x 2 400 2 90 2 2(400)(90) cos 45°x 2 117,188.3118x 342.3 ft28.74 ft88 ft38 2 74 2 88 22(74)(88) 742 88 2cos 1 382 2(74)(88) v25.28734695 vsin v s ideopposithypotenusex38 ft38 2 74 2 88 2 2(74)(88) cos v cos vesin 25.3° 7x4 31.60970664 xabout 31.6 ft29a. x 2 100 2 220 2 2(100)(220) cos 10°x 2 15,068.45887x 122.7536511about 122.8 mi29b. (100 122.7536511) 220 2.7536511about 2.8 miChapter 5 164
30.202 ft82.5˚I201.5 ftxyII125 ft180.25 ftIII75˚158 ftI: K 1 2 (201.5)(202) sin 82.5°K 20,177.3901II: x 2 201.5 2 202 2 2(201.5)(202) cos 82.5°x 2 70,780.6348x 266.046302y 2 158 2 180.25 2 2(158)(180.25) cos 75°y 2 42.711.98851y 206.6687894206.7 2 266.0 2 125 2 2(266.0)(125) cos v206.7 2 266.0 2 12522(266.0)(125) cos vcos 1 206.72 266.0 2 12522(266.0)(125) v48.93361962 vK 1 2 (266.0)(125) sin 48.9°K 12,536.58384III: K 1 2 (180.25)(158) sin 75°K 13,754.54228Area of pentagon I II III 20,177.4 12,536.6 13,754.5 46,468.5 ft31. I: 24 2 35 2 40 2 2(35)(40) cos v 242 352 40 22(35)(40) 352 40 22( 35)(40) cos vcos 1 242 v36.56185036 vII:24 2 30 2 20 2 2(30)(20) cos v 242 302 20 22(30)(20) 302 20 22( 30)(20) cos vcos 1 242 v52.89099505 vthe player 30 ft and 20 ft from the posts32a. sin 6° 20, 000x32b. sin 3° 15, y0,000x 2 sin6°x 191,335.4 ft0005,000y 1 sin3°y 286,609.8 ft32c. 6° 3° 3°d 2 191,335.4 2 286,609.8 2 2(191,335.4)(286,609.8) cos 3°d 2 9,227,519,077d 96,060.0 ft33. Since 63.2° 90°, consider Case I.b sin A 18 sin 63.2°b sin A 16.0665447317 16.1; 2 solutions34. tan v ssytan v 5 7C51˚O A(0, 0) 50ideoppositeideadjacent70007000v tan 1 5 7v 39.2°35. 775° 2(360°) 55°reference angle ⏐55°⏐ or 55°36. 3⏐ 1 7 k 63 12 36 3k1 4 12 k ⏐30 3k30 3k 0k 1037. m 55m 4 3t tt 2tttor 4 3 3 x6338. 2 x 2 3y 2 y 8 x6y3The correct choice is A.5-8BGraphing Calculator Exploration:Solving TrianglesPage 3341. AB 12.1, B 25.5°, C 119.5°2.40xB(50, 0)Find y using AC.y (tan 51°)xFind y using BC.50 si n Cin 51°0 sin4051°sin C 50 s 4in 51°0 C sin 1 50 s 4C 76.27180414B 80 51 76.27180414 52.72819586yC4051˚xO A(0, 0) 50 B(50, 0)ytan(180 52.72819586) x 50(x 50) tan(127.2718041) ySet the two values of y equal to each other.(tan 51°)x (x 50), tan 127.2718041°(tan 51°)x x(tan 127.2718041° 50(tan 127.2718041°)x 50(tan 127.2718041°)tan 51° tan 127.271804°x 25.77612538y (tan 51°) (25.77612538) 31.83086394165 Chapter 5
C could also equal 180 76.27180414 or103.728195°B 180 51 103.7281959° 25.2718041ytan (180 25.2718041) x 50(x 50) tan 154.7281959° ySet the two values of y equal to each other.(tan 51°) x (x 50)tan 154.7281959°(tan 51°) x x(tan 154.7281959°) 50(tan154.7281959°)x 50(tan 154.7281959°)tan 51° tan 154.7281959°x 13.82829048y (tan 51°) (13.82828048) 17.07651659B 52.7°, C 76.3°, b 40.9; B 25.3°,C 103.7°, b 2203. Law of Cosines4. Sample answer: put vertex A at the origin andvertex C at (3, 0).Chapter 5 Study Guide and AssessmentPage 335 Understanding and Using theVocabulary1. false; depression 2. false; arcosine3. true 4. false; adjacent to5. true 6. false; coterminal7. true 8. false; Law of Cosines9. false; terminal side 10. truePages 336–338 Skills and Concepts11. 57.15° 57° (0.15 60) 57° 957° 912. 17.125° (17° (0.125 60))(17° 7.5)(17° 7 (0.5 60))(17° 7 30)17° 7 3060°146° 2.39 3.1813. 8 360° 360(2)° 860° 720° 860° 140°; II14. 1 360° 360(3)° 1146° 1080° 1146° 66°; I15. 156°360°16. 9 3 0.43 360(1)° 156° 360° 156° 204°; III98°60° 2.77 360(2)° 998° 720° 998° 278°; IV17. 300°360° 0.83 360(1)° 300° 360° 300° 60°; I18. 1 072°360°19. 6 3 2.98 360(2)° 1072° 720° 1072° 352°; IV54°60° 1.82 360(1)° 654° 360° 654° 294°; IV20. 832°360° 2.31 360(2)° 832° 720° 832° 112°360° 112° 248°; III21. 284° has terminal side in first quadrant.360° 284° 76°22. 5 92°360° 1.64 360(1)° 592° 360° 592° 232°terminal side in third quadrant232° 180° 52°23. (BC) 2 (AC) 2 (AB) 215 2 9 2 (AB) 2306 (AB) 2306 AB334 ABsin A o ppositesidehypotenusecos A a djacentsihypotenusdeesin A or cos A 334 334 orsideoppositetan A stan A 1 915ideadjacent5 or 5 3 534 349334 34Chapter 5 166
24. (PM) 2 (PN) 2 (MN) 28 2 12 2 (MN) 2208 (MN) 2208 MN413 MNsin M o ppositesidehypotenusesin M or cos M orsideoppositehtan M scsc M stan M 1 28hsec M ssec M or cot M 1825. (MP) 2 (PN) 2 (MN) 2(MP) 2 10 2 12 2(MP) 2 44sin M o hsin M 1 1stan M stan M or csc M 1 1hsec M ssec M or cot M or126. sec v co s v1cos v se cvcos v 27. r x 2 y 2 r 3 2 3 2 r 18 or 32sin v y r cos v x r tan v y x sin v cos v tan v 3 3 or 1sin v csc v yr sec v xr csc v or 2 sec v or 2cot v x y 32 3cot v 3 3 or 128. r x 2 y 2 r (5) 2 12 2r 169 or 13sin v y r cos v x r tan v y x 2sin v 1 13cos v 135 5 or 1 3 tan v 12r rcsc v y sec v x cot v x y csc v 1 31312sec v or 1 35cot v 512532 35or 1 255 or 1 229. r x 2 y 2 r 82) 2 ( 2r 68 or 217sin v y x cos v x y tan v y x cos M a djacentside28 hypotenuse217 217 12 313 8 213 17 417 1313sin v cos v 413413 1717ypotenusercsc v ideadjacentideoppositey or 3 217 2 413 13 csc M or 212 3ypotenuseideadjacentcot M s ideadjacentcot v x y sideopposite8413 13 2or 2 cot v 3 2or 48 230. r x 2 y 2 r (2) 2 0 2MP 44 or 211ppositesideypotenusecos M a djacentsidesin v 0 2 or 0 cos v 22hypotenuse02or 5 6 211 11cos M or 12 6csc v 2ideoppositehypotenuse0 2sec v 2ideadjacentcsc M s ideoppositeundefined10 5112 0or 6 5 211 1131. r x 2 y 2 ypotenuseideadjacentcot M s ideadjacentr 4 2 5 2 sideopposite12r 41611211 11 211 1110 554 41 41 541 441 sin v cos v 14141 or 5 7 7 5 41 41 5432. r x 2 y 2 r (5) 2 ) (92r 106 33 32 3295 22106 106 cos v 2 29106 5106 sin v cos v 106106sin v cos v tan v 8sec v xr 217 8csc v or 17 sec v or 17 r 4 or 2sin v y r cos v x r tan v y x 2 or 1 4 0 or 1 tan v or 0csc v yr sec v xr cot v x y or 1 cot v 0undefinedsin v y r cos v x r tan v y x sin v cos v tan v 5 4 csc v yr sec v xr cot v x y csc v sec v cot v 4 5 sin v y r cos v x r tan v y x sin v cos v tan v csc v yr sec v xr csc v or sec v or cot v x y cot v 9or 5 9 33. r x 2 y 2 r (4) 2 4 2r 32 or 425sin v y r cos v x r tan v y x 42295or 9 5 4sin v cos v tan v or 1sin v 106 94 422 2cos v csc v yr sec v xr cot v x y 42106 94 422 242106 5csc v sec v cot v 444or 1csc v 2 sec v 244106 5167 Chapter 5
34. r x 2 y 2 r 5 2 0 2 43. a 2 b 2 c 2cos A — ba 2 15 2 20 2c —r 25 or 5cos A — 1 5sin v y r cos v x r tan v y a 175 20 —x a 13.2A cos 1 — 1 5sin v 0 5 or 0 cos v 5 5 or 1 tan v 0 20 —5 or 0A 41.4°r rcsc v y sec v x cot v x 41.40962211° B 90°y B 48.6°csc v 5 0 sec v 5 5 or 1 cot v 5 0 a 13.2, A 41.4°, B 48.6°undefinedundefined44. 64° B 90°35. cos v x B 26°r r 2 x 2 y 2cos v 3 sin A — a8 8 2 (3) 2 y 2c — cos A — b c —asin 64° x 3, r 8 55 y 2— 2 8 — cos 64° ——2b855 y28 sin 64° a 28 cos 64° bQuadrant 11, so y 5525.2 a 12.3 bsin v y r tan v y B 26°, a 25.2, b 12.3x 45. A 180° (70° 58°) or 52°55 55 55sin v tan v or 8 3rcsc v yrsec v x cot v x y 8csc v 55 sec v 83or 8 3 3cot v 5 5csc v 8 55 55cot v 3 5536. tan v y x r 2 x 2 y 255 tan v 3; Quadrant III r 2 (1) 2 (3) 2y 3, x 1 r 2 10r 10sin v — y r — cos v — x r — rcsc v — — x3sin v 1 1cos v 01 csc v 1 0 03or 3sin v 3 10 cos v 10 1010rsec v — — x cot v — x y —sec v 10 1or 10 cot v or 1 1 3 3 37. sin B — b c —bsin 42° ——1 515 sin 42° b10.0 b39. tan B — ab — 338. sin A — a c —sin 38° — 2 c4 —10 c sin 38° 2424c ——sin 38°c 39.040. 30°, 210°tan 67° — 2 4 —aa tan 67° 2424a tan 67°a 10.241. 180°42. A 49° 90°A 41°btan B — — a cos B — a c —btan 49° —— 1 6cos 49° — 1 —c616 tan 49° b c cos 49° 161618.4 b c ——cos 49°c 24.4A 41°, b 18.4, c 24.4b ac a—— sin B —— —— sin Asin C ——sin Ab 84c 84—— sin 70° —— —— sin 52°sin 58° ——sin 52°b — 84 sinsin— 70°52°c — 84 sin58°si —n 52°b 100.1689124 c 90.39983243A 52°, b 100.2, c 90.446. A 180° (57° 49°) or 74°a cb c—— sin A —— —— sin Csin B ——sin Ca 8b 8—— sin 74° —— —— sin 49°sin 57° ——sin 49°a — 8 sinsin— 74°49°b — 8 sin57°s —in 49°a 10.1891739 b 8.889995197A 74°, a 10.2, b 8.947. B 180° (20° 64°) or 96°K — 1 2 — a 2 — sin B sin C—sinA6°sin64°K — 1 2 — (19) 2 — sin 9 —sin20°K 471.7 units 248. C 180° (56° 78°) or 46°K — 1 2 — b 2 — sin A sin C—sinBK — 1 2 — (24) 2 — sin 5 sinK 175.6 units 249. K — 1 2 — bc sin A6°sin46°—78°K — 1 2 — (65.5)(89.4) sin 58.2°K 2488.4 units 250. K — 1 2 — ac sin BK — 1 2 — (18.4)(6.7) sin 22.6°K 23.7 units 2Chapter 5 168
51. Since 38.7° 90°, consider Case I.c sin A 203 sin 38.7°c sin A 126.9242592172 126.9; 2 solutionsa c—— sin A ——sin C172203 sin 38.7° ——s in Csin C — 203 s in38.7°—172C sin 1 —203 s in38.7°—172C 47.55552829180° 180° 47.6° or 132.4°<strong>Solution</strong> 1B 180° (38.7° 47.6°) or 93.7°b a—— sin B ——sin Ab 172—— sin 937° ——sin 38.7°b — 17 2 sin93.7°—sin38.7°b 274.5059341<strong>Solution</strong> 2B (180° (38.7° 132.4°) or 8.9°b a sin B sin Ab 172—— sin 8.9° ——sin 38.7°b — 17 2 sin8.9°—sin38.7°b 42.34881128B 93.7°, C 47.6°, b 274.5; B 8.9°,C 132.4°, b 42.352. Since 57° 90°, consider Case I.b sin A 19 sin 57°b sin A 15.9347407412 15.9; no solution53. Since 29° 90°, consider Case I.c sin A 15 sin 29°c sin A 7.27214430412 7.3; 2 solutionsa c—— sin A ——sin C12 15 sin 29° si n Csin C — 15 s in 29°—12in 29°— C sin 1 —15 s 12C 37.30170167180 180° 37.3° or 142.7°<strong>Solution</strong> 1B 180° (29° 37.3°) or 113.7°a b—— sin A ——sin B12b sin 29° sin 1 13.7°b — 12 sin113.7°—sin29°b 22.6647614<strong>Solution</strong> 2B 180° (29° 142.7°) or 8.3°a b—— sin A ——sin B12 b—— sin 29° ——sin 8.3°b — 12 sin8.3°—sin29°b 3.573829815B 113.7°, C 37.3°, b 22.7;B 8.3°, C 142.7°, b 3.654. Since 45° 90°, consider Case I.83 79; 1 solutiona b—— sin A ——sin B83 79—— sin 45° ——si n Bsin B — 79 s in 45°—83in 45°— B sin 1 —79 s 83B 42.30130394C 180° (45° 42.3°) or 92.7°a c—— sin A ——sin C83 c—— sin 45° ——sin 9 2.7°c — 83 sin92.7°—sin45°c 117.2495453B 42.3°, C 92.7°, c 117.255. a 2 b 2 c 2 2bc cos Aa 2 40 2 45 2 2(40)(45) cos 51°a 2 1359.446592a 36.87067388a b—— sin A ——sin B36.940—— si n 51° ——si n Bsin B — 40 sin51°—36.9sin51°— B sin 1 —40 36.9B 57.39811237C 180° (51° 57.4) or 71.6°a 36.9, B 57.4°, C 71.6°56. b 2 a 2 c 2 2ac cos Bb 2 51 2 61 2 2(51)(61) cos 19°b 2 438.9834226b 20.95193124b a—— sin B ——sin A21.051—— si n 19° ——si n Asin A — 51 sin19°—21.0sin19°— A sin 1 —51 21.0A 52.4178316C 180° (52.4° 19°) or 108.6°b 21.0, A 52.4°, C 108.6°57. c 2 a 2 b 2 2ab cos C20 2 11 2 13 2 2(11)(13) cos C— 202 112 13 22(11)(13) 112 13 22( 11)(13)— cos Ccos 1 —202 — C112.6198649 Ca c—— sin A ——sin C1120—— si n A ——sin 1 12.6°112.6°—sin A — 11 sin 20A sin 1 —11 sin 112.6°20A 30.51023741B 180° (30.5° 112.6°) or 36.9°A 30.5, B 36.9°, C 112.6°— 169 Chapter 5
58. b 2 a 2 c 2 2ac cos By 2 1b 2 42 2 6.5 2 2(42)(6.5) cos 24°b 2 1307.45418b 36.15873588b c—— sin B ——sin C36.26.5—— si n 24° ——si n Csin C — 6.5 sin24°—36.2C sin 1 —6.5 sin24°—36.2C 4.192989407A 180° (24° 4.2°) or 151.8°b 36.2, A 151.8°, C 4.2°Page 339 Applications and Problem Solving859a. sin v ——1 2v sin 1 8——1 2v 41.8°59b. cos v — 1x2 —cos 41.8° — 1x2 —12 cos 41.8° x8.94427191 xabout 8.9 ft60a. x 2 4.5 2 8.2 2 2(4.5)(8.2) cos 32°x 2 24.9040505x 5.0 mi8.25. 060b. —— s in v ——sin 32°sin v — 8.2 s in32°—5.0v sin 1 —8.2 s in32°—5.0 4.v 60.54476292180 v 180 60.5 or about 119.5°D 45˚Page 339 Open-Ended AssessmentC1. K — 1 2 — ab sin C125 — 1 2 — ab sin 35°435.86 abSample answer: about 40 cm and 10.9 cm2a. Sample answer: a 10, b 24, A 30°;10 24, 10 24 sin 30°2b. Sample answer: b 18; 10 18, 10 18 sin 30°Chapter 5 SAT & ACT PreparationPage 341 SAT and ACT Practice1. There are several ways to solve this problem. Usethe Pythagorean Theorem on the large triangle.(2y 3y) 2 4 2 3 2(5y) 2 16 925y 2 25— 2 5y 2—25 — 2 525 —y 1Since y is a length, use only the positive root.Another method is to use the Triangle InequalityTheorem. The hypotenuse must be shorter thanthe sum of the lengths of the other two sides.5y 3 45y 7Which of the answer choices make this inequalitytrue?5(1) 5 75(2) 10 7The correct choice is A.2. If you recall the general form of the equation of acircle, you can immediately see that this equationrepresents a circle with its center at the origin.(x h) 2 (y k) 2 r 2If you don’t recall the equation, you can try toeliminate some of the answer choices. Since theequation contains squared variables, it cannotrepresent a straight line. Eliminate choice D.Similarly, eliminate choice E. Since both the x andy variables are squared, it cannot represent aparabola. Eliminate choice C. The choicesremaining are circle and ellipse. This is a goodtime to make an educated guess, since you have a50% chance of guessing correctly. It represents acircle. The correct choice is A.3. Use factoring and the associative property.999 111 3 3 n 2(9 111) 111 3 3 n 2 ,3 3 (111) 2 3 3 n 2 .So n must equal 111. The correct choice is C.45˚7A77BSince ABC is an equilateral triangle and one sideis 7 units long, each side is 7 units long. so AC 7.AD is the hypotenuse of right triangle ACD. Oneleg is 7 units long. One angle is 45°, so the otherangle must also be 45°. A 45°45°90° triangle isa special right triangle. Its hypotenuse is 2times the length of a leg. (The SAT includes thistriangle in the Reference Information at thebeginning of the mathematics sections.) Thehypotenuse is 72. The correct choice is B.Chapter 5 170
5. You need to find the fraction’s range of values,from the minimum to the maximum. Theminimum value of the fraction occurs when a is assmall as possible and b is as large as possible.Since the smallest value of a must be slightlygreater than 4, and the largest value of b must beslightly less than 9, this minimum value of thefraction must be larger than — 4 9 — . The maximumvalue of the fraction occurs when a is as large aspossible and b is as small as possible. Thismaximum must be smaller than — 7 7 — or 1. Thecorrect choice is A.6. Start by making a sketch of the situation.36,00090009000900090001:00 P.M.10:00 A.M.By 1:00 P.M. the pool is three-fourths full. Threefourths of 36,000 gallons is 27,000 gallons. Thepool contained 9,000 gallons at the start. So27,000 9,000 or 18,000 gallons were added in3 hours. The constant rate of flow is 18,000gallons 3 hours or 6,000 gallons per hour. Tofill the remaining 9,000 gallons at this same ratewill take 9,000 gallons 6,000 gallons per houror 1.5 hours. One and a half hours from 1:00 P.M.,is 2:30 P.M. The correct choice is C.7. There are two right triangles in the figure. Youneed to find the length of one leg of the largertriangle, but you don’t know the length of theother leg. Use the Pythagorean Theorem twice—once for each triangle. Let y represent the lengthof side AC.In the smaller right triangle,y 2 4 2 6 2y 2 16 36x 2 52You do not need to solve for y.In the larger triangle,10 2 x 2 y 2100 x 2 52x 2 48x 48x 43The correct choice is B.8. Factor the polynomial in the numerator of thefraction. Simplify the fraction. Solve for x.— x2 7x 12x 4— (x 3)(x 4)x 4— 5— 5x 3 5x 2The correct choice is B.9.T is on the circle, so OT is a radius of the circle.The length of OT is 3. Since TA is tangent to thecircle, OTA is a right angle, and OTA is a righttriangle. In particular, OTA is a 30°-60°-90° righttriangle. In a 30°-60°-90° right triangle, the lengthof the hypotenuse is 2 times the length of theshorter leg.OA 2(OT )OC CA 2(OT )3 CA 2(3) or 6CA 3The correct choice is B.10. Draw a diagram from the information given in theproblem. Drawing a valid diagram is the mostdifficult part of solving this problem. Yourdiagram could be different from the one below andstill be valid.CA54OTX3C5430 ABDSince two segments bisect each other, you knowthe length of each half of the segment. Notice thatBD is a side of a right triangle. It is a 3-4-5 righttriangle. So BD 3. The answer is 3.171 Chapter 5
Chapter 6 Graphs of Trigonometric Functions6-1Pages 347–348Angles and Radian MeasureCheck for Understanding1. y2. 90°; — 4 —3. Divide 10 by 8.4. Let R 2r. For the circle with radius R, s Rv or2rv which is 2(rv). Thus, s 2s. For the circlewith radius R, A 1 2 R 2 v or 1 2 (2r) 2 v which is 1 2 (4r 2 )v or 4 1 2 r 2 v. Thus, A 4A.5. 240° 240° 186. 570° 570° 180° 4 3 19 67. 3 2 3 2 18 0° 270°8. 1.75 1.75 18 0° 100.3°9. reference angle: 3 4 or 4 ; Quadrant 2sin 3 4 2210. reference angle: 11 6 or 5 ;6 Quadrant 3tan 11 6 3311. s rv 12. 77° 77° 18s 15 5 6s 39.3 in.13. A 1 2 r 2 v 7 17 A 1 2 (1.4 2 ) 2 3A 2.1 units 215. 30° 30° 18 0° 6 Os rvs 1.4 6 s 0.7 m341x80s rvs 15 7 7180s 20.2 in.0°14. 54° 54° 18 — 3 10 —A 1 2 r 2 v0°0°A 1 2 (6 2 ) 3 10 A 17.0 units 2Pages 348–35116. 135° 135° 18Exercises0°17. 210° 210° 180° 3 4 7 618. 300° 300° 18 0°19. 450° 450° 18 5 3 5 220. 75° 75° 1821. 1250° 1250° 1822. 7 12 7 10° 5 12 12 5182 18 0°23. 11 3 11 3 18 0° 105° 660°0°24. 17 17 18 25. 3.5 3.5 18 974.0°200.5°26. 6 .2 6 .2 18 0°27. 17.5 17.5 18 0°29.0° 1002.7°28. reference angle: 2 5 3 or 3 ; Quadrant 4sin 5 3 230°0°29. reference angle: 7 6 or 6 ; Quadrant 3tan 7 6 3330. reference angle: 5 4 or 4 ; Quadrant 3cos 5 4 2231. reference angle: 7 6 or 6 ; Quadrant 3sin 7 6 1 2 32. 14 3is coterminal with 2 3reference angle: 2 3 or 3 ; Quadrant 2tan 14 3333. 19 6is coterminal with 5 6reference angle: 5 6 or 6 ; Quadrant 2cos 19 6 3234. s rv 35. s rvs 14 2 3 s 14 5 12 s 29.3 cms 18.3 cm36. 150° 150° 1837. 282° 282° 180° 5 6 4 730s rvs rvs 14 5 6 s 14 4 730s 36.7 cms 68.9 cm38. s rv 39. 320° 320° 18s 14 3 11 16 9s 12.0 cms rvs 14 16 9 s 78.2 cm0°0°0°Chapter 6 172
40. r 1 2 d 78° 78° 183r 1 2 (22) 1 30r 11s rvs 11 1 330s 15.0 in.41. s rv d 2r70.7 r 5 4 d 2(18.0)18.00360716 r d 36.0 m42. 60° 60° 18 0°s rv 3 43. A 1 2 r 2 vA 1 2 (10 2 ) —5 12 — A 65.4 units 245. A 1 2 r 2 vA 1 2 (7 2 ) — 8 — A 9.6 units 214.2 r 3 0°13.56 r; about 13.6 cm44. 90° 90° 18 0°46. A 1 2 r 2 vA 1 2 (12.5 2 ) —4 7A 140.2 units 247. 225° 225° —— 1848. 82° 82° ——18A — 1 2 — r 2 v0° — 5 —4 — 4 1—90A — 1 2 — r 2 v —A 1 2 (6 2 ) 5 40A 70.7 units 2 A 38.1 units 249a. s rv6 r(1.2)5 r; 5 ft50a. 135° 135° 18 A 1 2 (7.3 2 ) 4 190° 2 A 1 2 r 2 vA 1 2 (22 2 ) 2 A 380.1 units 249b. A 1 2 r 2 v0°A 1 2 (5 2 )(1.2)A 15 ft 250b. A 1 2 r 2 v — 3 —4A 1 2 (48.4 2 ) 3 4s rv A 2757.8 mm 2114 r 3 4 48.38 r; about 48.4 mm51a. A 1 2 r 2 v15 1 2 r 2 (0.2)150 r 212.247 rabout 12.2 in.52a. A 1 2 r 2 v15.3 1 2 (3 2 )v3.4 v; 3.4 radians52c. s rvs 3(3.4)s 10.2 m51b. s rvs 12.2(0.2)s 2.4 in. 52b. 3.4 3.4 18 0° 194.8°53a. 225° 225° 18s rvs 2 5 4 s 7.9 ft 5 40°53b. s rv 2.5 2.5 18 0°5 2v 143.2°2.5 v54. 330° 330° 18 11 60°s rvs rvs 2 11 6 11.5 8vs 11.5 in.1.44 v; about 1.4 radians55. s rv 0.5 0.5 18 0°10.5 22.9v 26.3°0.46 v; about 0.556a. 45° 34° 11° 56b. 45° 31° 14°11° 11° 1814° 14° 180° 1 1180 — 7—90s rvs rvs 3960 1 1180 s 3960 7 9s 760.3 mis 967.6 mi56c. 34° 31° 3°3° 3° 18 0° 6 0 s rvs 3960 — —60s 207.3 mi57. 84.5° 84.5° 1880° 80° 180° 1 6936 0 4 9s rvs rvs 0.70 1 6936 s 0.67 04 9s 1.03 mis 0.94 mi1.03 1.46 0.94 1.8 5.23 mi58a. r 1 2 d 1.5 rotations 1.5 2 radiansr 1 2 2 1 2 3 radiansr 1.25s rvs 1.25(3)s 11.8 ft58b. s rv 3.6 3.6 18 0°4 1 2 1.25v 206.3°3.6 v59a. v 2 2 or 3 2A 1 2 r 2 vA 1 2 (15 2 ) —3 2A 530.1 ft 2 —59b. A 1 2 r 2 v 750 1 2 r 2 3 2318.3098862 r 217.84124116 r; about 17.8 ft0°0 0°173 Chapter 6
60. 3.5 km 350,000 cms rv350,000 32v10,937.5 v; 10,937.5 radians61. Area of segment Area of sector Area of triangleA 1 2 r 2 a — 1 2 — r r sin aA 1 2 r 2 (a sin a)62. s 1 2 (6 8 12) 13K s(s a)(s b)(s c)K 13(13 13 6)( 8)(13 12)K 455 K 21.3 in 263. Since 152° 90°, consider Case II.10.2 12, so there is no solution.64. C 180° 38° 27° 115°560a sin 115° sin 27°a 280.52xsin 38° 280 .52x 172.7 yd65. I, III66a. Find a quadratic regression line using agraphing calculator. Sample answer: y 102x 2 505x 18,43066b. 2020 1970 50y 102x 2 505x 18,430y 102(50) 2 505(50) 18,430y 248,180Sample answer: about 248,18067.r 1 3 2 6 101 1 2 4 2 122 1 1 4 2 63 1 0 2 0 104 1 1 2 14 66f(x) (x) 4 3(x 3 2(x) 2 6(x) 10f(x) x 4 3x 3 2x 2 6x 10r 1 3 2 6 101 1 4 2 4 62 1 5 8 10 30Sample answers: 4; 268. 2 1 6 12 122 8 81 4 4 4No; there is a remainder of 4.69. x 2 y 2 16 → a 2 b 2 16x-axis a 2 b 2 16a 2 (b) 2 16a 2 b 2 16; yesy-axis a 2 b 2 16(a) 2 b 2 16a 2 b 2 16; yesy x a 2 b 2 16(b) 2 (a) 2 16a 2 b 2 16; yesy x a 2 b 2 16(b) 2 (a) 2 16a 2 b 2 16; yesall70. 4x 2y 3z 65x 4y 3z 759x 6y 812(4x 2y 3z) 2(6) 8x 4y 6z 123(3x 3y 2z) 3(2)→9x 9y 6z 617x 5y 65(9x 6y) 5(81)45x 30y 4056(17x 5y) 6(6)→102x 30y 36147x 441x 39x 6y 814x 2y 3z 69(3) 6y 81 4(3) 2(9) 3z 6y 9 z 8(3, 9, 8)71. b72. Since q 0, q 0. Given that p 0,p q p ⏐q⏐ and p ⏐q⏐ 0. So theexpression p q is nonnegative.The correct choice is B.6-2Page 3551.Linear and Angular VelocityCheck for Understanding5 rev2. 1 min 2 rad1 r iansev 1 min6 0 s3. Linear velocity is the movement along the arcwith respect to time while angular velocity is thechange in the angle with respect to time.4. Both individuals would have the same change inangle during the same amount of time. However,an individual on the outside of the carousel wouldtravel farther than an individual on the insideduring the same amount of time.5. Since angular velocity is — v t — , the radius has noeffect on the angular velocity. Let R 2r. Fora circle with radius R, v R v t or (2r) v t which is2r v t . Thus v 2v.6. 5.8 2 11.6 or about 36.4 radians7. 710 2 1420 or about 4461.1 radians8. 3.2 2 6.4 9. 700 2 1400q v t q v t q 6. 4 7q 14 0015q 2.9 radians/sq 293.2 radians/minChapter 6 174
10. v rq 11. v rqv 12(36)v 7(5)v 432 in./sv 110.0 m/min12a. r 3960 22,300 or 26,260 mis rvs 26,260(2)s 164,996.4 mi12b. v r v t v 26,260 2 24 v 6874.9 mphPages 355–358 Exercises13. 3 2 6 or about 18.8 radians14. 2.7 2 5.4 or about 17.0 radians15. 13.2 2 26.4 or about 82.9 radians16. 15.4 2 30.8 or about 96.8 radians17. 60.7 2 121.4 or about 381.4 radians18. 3900 2 7800 or about 24,504.4 radians19. 1.8 2 3.6 20. 3.5 2 7q v t q v t q 3. 6 9q 7 3q 1.3 radians/sq 7.3 radians/min21. 17.2 2 34.4 22. 28.4 2 56.8q v t q v t . 4. 8q 34 12q 56 19q 9.0 radians/sq 9.4 radians/s23. 100 2 200 24. 122.6 2 245.2q v t q v t 05.2q 20 1 6q 24 2 7q 39.3 radians/min q 28.5 radians/minrevolution2 radians 1 0.1 radian/s25. 1 50secondsrevolution26. — 500 revolutions1 minute2 radians—1 minute —— 6 0 seconds — 1 revolution27. 8 5 rad1 sec iansond 6 0 seco1 min ndsute 1 revolution2 radians— 52.4radians/s 811.7 rpm28. v rq 29. v rqv 8(16.6) v 4(27.4)v 132.8 cm/sv 109.6 ft/s30. v rq 31. v rqv 1.8(6.1)v 17(75.3)v 34.5 m/minv 4021.6 in./s32. v rq 33. v rqv 39(805.6)v 88.9(64.5)v 31,418.4 in./min v 18,014.0 mm/min120°34a. 1s econd 6 0 seco1 min ndsute 1re volution3 60°34b. 120° 120° 18 0° 2 3q v t q 2 31q 2 3 20 rpmv rqv 5 2 3 v 10.5 in./s135a. In 1 second, the second hand moves 6(360°)0or 6°. 6° 6° 18 0°or 3 0 v rvv 30 30 v 3.1 mm/s35b. In 1 second, the minute hand moves1 6 0 1 6 0 (360°) or 0.1°.0.1° 0.1° 18 0°or 0 . 1180v rvv 27 0 . 1180v 0.05 mm/s35c. In 1 second, the hour hand moves1121 6 0 1 6 0 (360°) or about 0.008°.0.008° 0.008° 18 0°or 0. 008180v rvv 18 0. 008180v 0.003 mm/s36a. r 1 2 d v r v t r 1 2 (80) or 40 v 40 2 45v 5.6 ft/s36b. v r v t 8 40 2 t t 31 s37a. 3 2 6 radians 1 minute 60 secondsv r v t v 22— 1 2 — 6 60 v 7.1 ft/s37b. v r v t 37c. 7.1 3.1 4 ft/s3.1 r 6 60 9.87 r; about 9.9 ft38a. 35° 35° 18 7 36 0°lighter child: q v t q q 1.2 radians/sheavier child: q v t q 736 1 2 736 1 2 q 1.2 radians/s175 Chapter 6
38b. lighter child: v rqv 9(1.2)v 11.0 ft/sheavier child: v rqv 6(1.2)v 7.3 ft/s39a. 3 miles 190,080 inchesr 1 2 ds rvr 1 2 (30)190,080 15vr 1512,672 v12,672 1rev olution2 2017 revolutions39b. 2.75 revoseco lutionsnd 6 0 seco1 min ndsute 60 minutes2 radians1 hour 1 revolution 19,800 radians/hourv rqv 15(19,800)v 933,053.0181933,053.0181 inches 14.7 mph40a. Mercury: Venus:v r v t v r v t 2v 2440 1407.6 v 6052 583v 10.9 km/h v 6.5 km/hEarth:Mars:v r v t v r v t 222.5 2 v 6356 23.935 v 3375 24.623v 1668.5 km/h v 861.2 km/h40b. The linear velocity of Earth is about twice thatof Mars.41a. v v m cos qtv 4 cos t41b. v 4 cost1mile0 4 cos t0 cos tt 2 or t 3 2t 1 2 or 0.5 s t 3 2 or 1.5 s42a. 3960 200 4160 milesC 2rt C speedC 2(4160) t 26,138.05088 17,000C 26138.05088 t 1.537532405q v t 2q 1 .54q 4.1 radians/h42b. q v t t C speed4 2 t 2 2r 17,000— 2 — t — 2 — (17,000) 2r 2 (17,000) 2 r4250 r4250 3960 290; about 290 mi42c. 3960 500 4460; C 2(4460) or28023.00647t 28,023.00647 17,000 or 1.648412145q v t 2q 1 .65q 3.8Its angular velocity is between 3.8 radians/h and4.1 radians/h.43a. B clockwise; C counterclockwise43b. v A r A v t Av A 3.0 12 01 v A 360The linear velocity of each of the three rollers isthe same.v B r B v t Bv C r C v t C360 2.0 vB 1360 4.8 vC 1180 v B 75 v C180 rpm 75 rpm44. 105° 105° 18 7 12 A 1 2 r 2 vA 1 2 (7.2 2 ) 7 12 A 47.5 cm 245. r 1 2 dr yx0°v 360° 10 or 36°sin v x r r 1 2 (7.3)r 3.65cos v y r xsin 36° 3. 65 ycos 36° 3. 65 x 2.145416171 y 2.952912029A 1 2 bhA 1 2 (2.15)(2.95)A 3.16761261Area of pentagon 10(3.17) or about 31.68 cm 21°1° 46. 35°2055 35° 20 60 55 3600 35.349°47. 10 k 5 8 Check: 10 k 5 8k 5 2 10 9 5 8k 5 4 10 4 8k 9 10 2 812 8no real solution48. (x (4))(x 3i)(x (3i)) 0(x 4)(x 3i)(x 3i) 0(x 4)(x 2 9) 0x 3 4x 2 9x 36 0Chapter 6 176
49.yy x 3 1Oxmaximum value of the cosine function occurswhen x n, where n is an even integer, and itsminimum value occurs when x n, where n isan odd integer.5. yes; 4 6. 0 7. 18. 3 2 2n, where n is an integer9.y50. m 0 56 85 514or 1 4m y y 1 m(x x 1 )5y 0 1(x4 (6))5y 1x4 1 5751. P 2a 2bP 2 3 4 b 2bP 3 2 b 2b10.1O15y sin x6y cos x7y1xP 7 2 bP b The correct choice is D. 2 7432O1x6-3Graphing Sine andCosine FunctionsPage 363 Check for Understanding1. Sample answer:yOperiod: 62. Sample answers: 3 , 2 2 , 5 23. cos x cos(x 2)4. yy sin x1O12y cos xBoth functions are periodic functions with theperiod of 2. The domain of both functions is theset of real numbers, and the range of bothfunctions is the set of real numbers between 1and 1, inclusive. The x-intercepts of the sinefunction are located at n, but the x-intercepts ofthe cosine function are located at 2 n, where nis an integer. The y-intercept of the sine functionis 0, but the y-intercept of the cosine function is 1.The maximum value of the sine function occurswhen x 2 2n and its minimum value occurswhen x 3 2 2n, where n is an integer. Thex 3 4 5x11. Neither; the period is not 2.12. April (month 4):y 49 28 sin 6 (t 4)y 49 28 sin 6 (4 4)y 49October (month 10):y 49 28 sin 6 (t 4)y 49 28 sin 6 (10 4)y 49The average temperatures are the same.Pages 363–366 Exercises13. yes; 6 14. no 15. yes; 20 16. no17. no 18. no 19. 1 20. 021. 0 22. 1 23. 1 24. 125. sin cos 0 (1)126. sin 2 cos 2 0 1127. 2n, where n is an integer28. 2 2n, where n is an integer29. 2 n, where n is an integer30. v 2n, where n is an integer31.y5y sin x431O1x177 Chapter 6
32. y1y cos xO 8 9133.y cos x5 4310 xy1O x143a. csc v 1 si n v43b. csc v 1 si n v11 —— si n v1 1 si n vsin v 1sin v 1 2 2n, where n 3 2 2n, where nis an integeris an integer43c. csc v is undefined when sin v 0.n, where n is an integer44a. sec v 1 co s v44b. sec v 1 co s v1 1 co s v1 1 co s vcos v 1cos v 12n, where n 2n, where nis an integeris an integer44c. sec v is undefined when cos v 0. 2 n, where n is an integer34.y45.1y sin xO56x1[0, 2] sc1— 2 — by [2, 2] sc1135.yx 3 , 7 4 4y cos x146.32Ox136.y1y sin x[0, 2] sc1— 2 — by [2, 2] sc110 x 4 , 5 4 x 2O4 5 x47.137. y cos x; the maximum value of 1 occurs whenx 4, the minimum value of 1 occurs whenx 5, and the x-intercepts are 7 , 9 ,2 2 and 11 2.38. Neither; the graph does not cross the x-axis.39. y sin x; the maximum value of 1 occurs whenx 11 2, the minimum value of 1 occurswhen x 13 2, and the x-intercepts are 7,6, and 5.40. Sample answer: a shift of 2 to the left41. x 2 n, where n is an integer42. x n, where n is an integer[0, 2] sc1 2 by [2, 2] sc11none48.[0, 2] sc1 2 by [2, 2] sc11x 0, 2 x , 3 2 x 2Chapter 6 178
49.50.[0, 2] sc1 2 by [2, 2] sc11x 0, 2 , 2[0, 2] sc1 2 by [2, 2] sc11x 4 , 5 451a. July (month 7):y 43 31 sin 6 (t 4)y 43 31 sin 6 (7 4)y 74January (month 1):y 43 31 sin 6 (t 4)y 43 31 sin 6 (1 4)y 1274 12 62; it is twice the coefficient.51b. Using answers from 51a., 74 12 86; it istwice the constant term.52a. n, where n is an integer52b. 2 52c. 2 52d. 252e.y252f. It expands the graph vertically.53a. P 100 20 sin 2tP 100 20 sin 2(0) or 100P 100 20 sin 2(0.25) or 120P 100 20 sin 2(0.5) or 100P 100 20 sin 2(0.75) or 80P 100 20 sin 2(1) or 10053b. 0.25 s 53c. 0.75 s21O12y 2 sin x2 x54a. v 3.5 cos t — k —mv 3.5 cos 0.9 —1 1v 3.3 cmv 3.5 cos t mk v 3.5 cos 1.7 —1 1v 2.0 cm—9.6.99—9.6.9954b. v 3.5 cos t — k —m0 3.5 cos t —1 9.61.9— 90 cos t —1 9.61.9— 9cos 1 0 t —1 9.61.9— 91.570796327 t —1 9.61.9— 90.5005164776 t; about 0.5 s54c. v 3.5 cos t — k —m3.5 3.5 cos t —1 9.61.9— 91 cos t —1 9.61.9— 9cos 1 1 t —1 9.61.9— 92 t —1 9.61.9— 92.00206591 t; about 2.0 s55a. 4 n2, where n is an integer55b. 1 55c. 1 55d. 55e.y21O1y cos 2x2 x56a. P 500 200 sin [0.4(t 2)]P 500 200 sin [0.4(0 2)] or about 357pumasD 1500 400 sin (0.4t)D 1500 400 sin (0.4(0)) or 1500 deer56b. P 500 200 sin [0.4(t 2)]P 500 200 sin [0.4(10 2)] or about 488pumasD 1500 400 sin (0.4t)D 1500 400 sin (0.4(10)) or about 1197 deer56c. P 500 200 sin [0.4(t 2)]P 500 200 sin [0.4(25 2)] or about 545pumasD 1500 400 sin (0.4t)D 1500 400 sin (0.4(25)) or about 1282 deer57. 500 revolutions1 minute 1 minute2 rad6 0 seconds ians1 revolution 52.4 radiansper second58. 1.5 1.5 18 0° 85.9°59. 45°, 135°179 Chapter 6
2 x60. x 2 2 x 2 4x x2 41(x 2)(x 2) 2x 2 1(x 2)(x 2) 2 xx 2 42 4 (1)(x 2)(x 2) x x1(x 2)(2) (x 2)(x) (1)(x 2 4)2x 4 x 2 2x x 2 4x 2But, x 2, so there is no solution.61. 1 positive real zerof(x) 2x 3 3x 2 11x 62 or 0 negative real zeros2⏐ 2 3 11 64 14 62 7 3 ⏐ 02x 2 7x 3 0(2x 1)(x 3) 02x 1 0 or x 3 0x 1 2 x 33, 1 2 , 262. 1⏐ 1 2 9 181 3 61 3 6 ⏐ 1212; no63. g(x)vertical:x 2 x 0g(x) x2x 2 xx(x 1) 0O xx 0 or x 1 0x 1horizontal:y 164. reflected over the x-axis, expanded vertically by afactor of 32 4 165. 1 1 0 3 4 51 0 1 0 1 14 5 3 5 3 42(5) 4(5) 1(1)1166.1 0 3 2 10 1 24 62 4 (1) 1(3) 0(2) 1(2) 0(4)0(3) 1(2) 0(2) 1(4)3 2 1 2 4 6A(3, 2), B(2, 4), C(1, 6)67. x 3 2 yy 2 3 xyART1(1) 0(6)0(1) 1(6) 3x y 268. Perimeter of square RSVW RS SV VW WR 5 5 5 5 or 20Perimeter of rectangle RTUW RT TU UW WR (5 2) 5 (5 2) 5 2424 20 4The correct choice is B.Page 3671.n n 2 n 31 22 123 364 805 1506 2527 3928 5769 81010 110011 145212 187213 236614 294015 360016 435217 520218 615619 722020 840021 970222 11,13223 12,69624 14,40025 16,25026 18,25227 20,41228 22,73629 25,23030 27,900History of Mathematics[0, 30] sc15 by [0, 30,000]sc15000The graph is not a straight line. It curves upward,increasing more rapidly as the value of nincreases.2. See students’ work.OxChapter 6 180
6-4Page 369Amplitude and Period of Sine andCosine FunctionsGraphing Calculator Exploration7. 2 4 2 y1y sin 4 1.O23218. ⏐10⏐ 10; 2 2 2. The graph is shrunk horizontally.3. The graph of f(x) sin kx for k 0 is the graph off(x) sin ⏐k⏐x reflected over the y-axis.y84y 10 sin 2Pages 372–373 Check for Understanding1. Sample answer: y 5 sin 2v2. The graphs are a reflection of each other over thev-axis.3. A: period 2 2 or B: period 2 52C: period or 4 1 2 D: period 2C has the greatest period.4. Period and frequency are reciprocals of each other.5. yy 3 cos y cos 32O22y cos All three graphs are periodic and curve above andbelow the x-axis. The amplitude of y 3 cos v is 3,while the amplitude of y cos v and y cos 3v is1. The period of y cos 3v is 2 ,3 while the periodof y cos v and y 3 cos v is 2.6. ⏐2.5⏐ 2.5y2O2y 2.5 cos 2 3 43O489. ⏐3⏐ 3; 2 2 y321O12310. ⏐0.5⏐ 0.5; 12y11O11. ⏐ 1 5 ⏐ 1 5 ; 8y0.2O0.22y 3 cos 22332 1 6 y 0.5 sin 62 4 6 8 102 1 4 241y 5cos64812. ⏐A⏐ 0.8 2 k A 0.8y 0.8 sin 2vk 2 or 2181 Chapter 6
13. ⏐A⏐ 7 2 k 3 2A 7 k or 6y 7 sin 6v14. ⏐A⏐ 1.5 2 k 5A 1.5 k 2 5 or 2 5 y 1.5 cos 2 5 v15. ⏐A⏐ 3 4 2 k 6A 3 4 k 2 6 or 3 y 3 4 cos 3 v16. ⏐A⏐ 0.25 2 1k 2 94 A 0.25k 588y 0.25 sin (588 t)Pages 373–37717. ⏐2⏐ 2y2Oy 2 sin Exercises 2 3 3 20. 2 2 221. 8 1 4 y1O1y1O122. 2 6 3 y1y cos 2 2 3 y cos 42 4 6 8y sin 62O218. 3 4 3 4 1y0.5O3y 4cos 2 3 23. ⏐5⏐ 5; 2 1 2y4y 5 cos 0.5O23419. ⏐1.5⏐ 1.5y21y 1.5 sin 4224. ⏐2⏐ 2; 0 4y. 5O12 2 3 2Oy 2 cos 0.5 2 342Chapter 6 182
25. 2 5 2 5 ; 2 2930. ⏐3⏐ 3; 0 . 5 4227. ⏐3⏐ 3; 4 32. 12 3 1 3 2; 6 1 3 yy4y 3 sin 121 2y 3 sin 3 O 3 2 5 O 2 3222224128. 2 3 2 3 2; 1 423 3 33. ⏐4⏐ 4; 4 172 yy2y 3 cos 0.5y 5 sin 90.42O42O 2 30.42226. ⏐8⏐ 8; 0 . 5 431. 1 3 1 3 ; 2 3yy8 y 8 sin 0.5114y 3 cos 3O 2 3 4 5 O 2 34814445 5 5y0.82 3y 3 cos 7 y4y 4 sin 20.4O0.423225232O223450.8429. ⏐3⏐ 3; 2 2 x2y 3 sin 22 1 5 34. ⏐2.5⏐ 2.5; 10yy 2.5 cos 52O22345O2234535. ⏐0.5⏐ 0.5; 6298 1 3 49 183 Chapter 6
50. sine curve A 0.5 36. ⏐A⏐ 0.4 2 k 10 2 k vy 2 cos 2A 0.4k 1 5 k 2v y 0.5 sin 2vy 0.4 sin 537. ⏐A⏐ 35.7 2 k 51. cosine curve A 3 2 4 k 2k 1A 35.7 k 8y 3 cos vy 35.7 sin 8v38. ⏐A⏐ 1 4 2 k 52. sine curve A 1.5 2 3 k 4A 1 k 14 2 k 6vy 1 y 1.5 sin 24 sin 6v39. ⏐A⏐ 0.34 2 53. ⏐A⏐ 3.8 2 1 k 1 20 k 0.75A 3.8k 240A 0.34k 8 3 y 3.8 sin(240 t)y 0.34 sin 8 3 v54. ⏐A⏐ 15 2 1 k 3 6 40. ⏐A⏐ 4.5 2 k 5 A 15k 724y 15 cos (72 t)A 4.5k 8 5 55.y 4.5 sin 8 5 v41. ⏐A⏐ 16 2 k 30A 16k 1 5 y 16 sin 1 5 v42. ⏐A⏐ 5 2 k 2All the graphs have the same shape, but haveA 5 k 1been translated vertically.y 5 cos v43. ⏐A⏐ 5 8 2 k 56a. ⏐A⏐ A (A)7 2 2 k 8A 5 k 8 4 k 14y 5 ⏐A⏐ 38 2 cos 14v44. ⏐A⏐ 7.5 2 A 1.5; down first, so A 1.5 k 6y 1.5 sin A 7.5k 1 4 t3 56b. y 1.5 sin v4 t 56c. y 1.5 sin 4 ty 7.5 cos 3 y 1.5 sin 45. ⏐A⏐ 0.5 2 4 (3)y 1.5 sin 4 (12)k 0.3y 1.1 fty 0 ftA 0.5 k 2 057a. Maximum value of sin v 1.3y 0.5 cos 2 0Maximum value of 2 sin v 2 1 or 33v57b. Minimum value of sin v 146. ⏐A⏐ 2 5 2 k 3 5 Minimum value of 2 sin v 2 (1) or 1k 1 03A 2 5 257c. 2 1 y 2 5 cos 1 03v37d. y47. ⏐A⏐ 17.9 2 k 16y 2 sin A 17.9k 8 2y 17.9 cos 8 v48. ⏐A⏐ 1.5 2 k 2 A 1.5 k 4O 2 3 4 5 y 1.5 sin 4v, y 1.5 cos 4v49. cosine curve A 2 2 k 458a. ⏐A⏐ 0.2 2 1k 2 62 k 1 2 A 0.2k 524y 0.2 sin (524 t)Chapter 6 184
58b. ⏐A⏐ 1 2 (0.2) 2 1k 5 24 A 0.1k 1048y 0.1 sin (1048 t)58c. ⏐A⏐ 2(0.2) 2 1k 1 31 A 0.4k 262y 0.4 sin (262 t)59a. y A cos t g y 1.5 cos t 9 659b. y 1.5 cos t 9 6.8 .8 .8 y 1.5 cos 4 9 6y 0.6about 0.6 m to the right59c. y 1.5 cos t 9 .8 6.8 y 1.5 cos 7.9 9 6y 1.2about 1.2 m to the left60a. 2 n, where n is an integer60b. 160c. 2 1 260d. yy cos( )1O161a. y 1.5 cos t k m8.5 2 k 6.8y 1.5 cos t 1 0.4k 0.9 s/cycley 1.5 cos 6.8t1frequency: 0 .9 1.1 hertz61b. y 1.5 cos t k m8.5 2 k 5.6y 1.5 cos t 1 0.6k 1.1 s/cycley 1.5 cos 5.6t1frequency: 1 .1 0.9 hertz61c. y 1.5 cos t k m8.5 2 k 4.8y 1.5 cos t 1 0.8k 1.3 s/cycle1y 1.5 cos 4.8t frequency: 1 .3 0.8 hertz61d. It increases. 61e. It decreases.62. 063. 84 2 168 radiansq v t 234564. 73° 73° 18 7 31800°s rv s 9 7 1380s 11.5 in.65. a 2 b 2 c 2 tan A a b 5.19.515.1 2 19.5 2 c 2 tan A 1 124.66292764 c A tan 1 1 5.119.5A 37.75273111B 180° (90° 37.8°) or 52.2°c 24.7, A 37.8°, B 52.2°66. T 2 g4.1 2 90.6525352667 9.8 0.458022743 9 .8 4.17 ; about 4.17 m67. b 2 4ac 5 2 4(3)(10)952 imaginary roots68a. Let x the number of Model 28 cards and lety the number of Model 74 cards.30x 15y 240 y4020x 30y 36012x 10y 480x 0y 03224x 0P(x, y) 100x 60yP(0, 0) 100(0) 60(0) or 0P(0, 12) 100(0) 60(12) or 720P(3, 10) 100(3) 60(10) or 900P(0, 8) 100(0) 60(8) or 4803 of Model 28, 10 of Model 7468b. $90069.1 0 2 1 3 30 1 11 4 2 1(2) 0(1) 1(1) 0(1)0(2) (1)(1) 0(1) (1)(1)1(3) 0(4) 1(3) 0(2)0(3) (1)(4) 0(3) (1)(2) 2 1 3 31 1 4 2(2, 1), (1, 1), (3, 4), (3, 2)70.g(x).8 12x 10y 48016 30x 15y 240(0, 12) (3, 10)8 20x 30y 360(0, 0) (8, 0)y 0 xO 8 16 24 32 40q 16 8 6q 88.0 radians/sOx185 Chapter 6
71. y 14.7x 140.1y 14.7(20) 140.1y $434.1072.x x 2 y4 (4) 2 163 (3) 2 92 (2) 2 4{(4, 16), (3, 9), (2, 4)}; yes73. A s 2 radius 1 2 (10) or 5100 s 2 A r 210 s A (5) 2 or 254(25) 100The correct choice is C.6-5Page 3781.Translations of Sineand Cosine FunctionsGraphing Calculator Exploration2. The graph shifts farther to the left.3. The graph shifts farther to the right.Page 3771. 5 6 5 6 18 0°Mid-Chapter Quiz 150°2. r 1 2 d s rvr 1 2 (0.5) or 0.25 s 0.25 5 3s 1.3 m3. A 1 2 r 2 v A 1 2 (8 2 ) 2 5A 40.2 ft 24. 7.8 2 15.6 or about 49.0 radians5. 8.6 2 17.2q v t q 17 .2 7q 7.7 radians/s7. 18. yy cos x16. v rqv 3(8)v 75.4 meters/sPage 383 Check for Understanding1. Both graphs are the sine curve. The graph of y sin x 1 has a vertical shift of 1 unit upward,while the graph of y sin (x 1) has a horizontalshift of 1 unit to the left.2. sine function3a. increase ⏐A⏐3b. decrease h3c. increase ⏐k⏐3d. increase c4. Graph y sin x and y cos x, and find the sum oftheir ordinates.c5. Jamal; k — or 36. 2 y2O2 2 6 y 3 cos( 2 )23 4 5O 7812 1 3 9. ⏐7⏐ 7; 69x7. 3; y 3A 1; 2 22y4y sin 2 3y8y 7 cos 3O23 4 54O4246 8 10 12 810. ⏐A⏐ 5 2 k 3 A 5 k 6y 5 sin 6vChapter 6 186
8. ⏐2⏐ 2; 2 2 ; 2 ; 5 1 1 2 2 yy13c.y120O 2 3 4 5 802 y 2 sin (2 ) 54046 6 .2 2 4 4 y 3 1cos ( 12 2 ) 2O1O 2 3 4 5 y1O1c 2. 136.2 13. 2 9. ⏐ 1 2 ⏐ 1 2 ; 4; 2 ; 310. ⏐A⏐ 20 2 k 1 c 0A 20 k 2 h 100y 20 sin 2v 10011. ⏐A⏐ 0.6 2 k 12.4A 0.6k 6 .2 c k 2.13 h 7c 2.13y 0.6 cos 6.2 v 2. 6 712.x sin x cos x y0 sin 0 cos 0 1 2 sin 2 cos 2 1 sin cos 1 3 2 sin 3 2 cos 3 2 12 sin 2 cos 2 1y1y sin x cos xPages 383–386 Exercisesc14. k 2 1 or 2; A 1; 2 1 2c15. k 2 ; A 1; 2 2 2 — 1 4 c16. 2k or 2; A 2; 8y21O12OP17. 1 2 ; y 1 2 2A 1; 4 1 2 1P 30 sin 2t 100P 30 sin 2t 1002y sin( 2)23 4 5 63y sin(2 )2344 1 4 55 2 3 4 5 6 7 y 2 cos 4( )2tO12xy21y sin 1 2 213a. 130 702 100; P 10013b. ⏐A⏐ 130 702⏐A⏐ 30 2 k 1k 2A 30 P 30 sin 2t 100O123 4 5187 Chapter 6
5; 18. 4; y 424. ⏐5⏐A 5; 2 1 2y 2 ;3 3 ; 2021. ⏐3⏐ 3; 2 2 y1 2; 2 y 10 sin ( 4 4) 51; 05y30y 20 5 cos (3 )O 2 3 4 5 22041068O 2 3 4 5 y 5 cos 425. ⏐ 1 4 ⏐ 1 2 04 ; 4; 0; 319. 7; y 7 1 1A 1; 2 2 2 2 yy8y 7 cos 2O 2 3 4 5 621 y 4 cos 4 3424O 2 3 4 5 2 426. ⏐10⏐ 10; 8; 16; 5c20. k 4 1 1 2 or 2; 34 4 y2y 3 cos( 2 )O51023 4 5O223 4 52 1 223. ⏐1⏐ 1; 6; 4 ; 2 1 3 22. ⏐6⏐ 6; 2 3 1 2; 1 3 ; 2y8y 6 sin( 3 ) 2642O2 2 3 4 5 4yO242y 2 sin 3 1 3 3 4 5( )121527. ⏐A⏐ 2 (6)24; 2 to the left or 2 ;⏐A⏐ 4down 2, or 2A 4; 428. ⏐A⏐ 7 2 c 3 h 7k 2 3 A 7 k 2 3c 2 3y 7 sin 2 3 v 2 3 729. ⏐A⏐ 50 2 k 3 4 c 2 h 25 8 3 A 50 k 8 3 c 4 3y 50 sin 8 3 v 4 3 2530. ⏐A⏐ 3 4 2 k 5 c 1 0 A 3 4 k 10 c 10h 1 4 y 3 4 sin (10v 10) 1 4 31. ⏐A⏐ 3.5 2 k 2 c 4 4 h 7A 3.5 k 4 c y 3.5 cos (4v ) 732. ⏐A⏐ 4 5 2 k 6 c 1 2 3 h 7 5 A 4 5 k 12 c 4y 4 5 cos (12v 4) 7 5 Chapter 6 188
33. ⏐A⏐ 100 2 c 45 0 h 110kA 100 k 2 45 c 0y 100 cos 2 45 v 11034. ⏐A⏐ 1 (3)2 2 k 4 h 1⏐A⏐ 2 k 1 2 A 2; 2y 2 cos v2 135. ⏐A⏐ 3.5 (2.5)236.x sin x sin x x0 0 0 2 1 2 1 2.57 2 42 c k 2 0 h 3⏐A⏐ 0.5 k 2 c 0A 0.5; 0.5y 0.5 sin 2v 338.x sin x 2x sin 2x sin x sin 2x0 0 0 0 0 4 0.71 2 1 1.71 2 1 0 1 0 2 0 0 3 2 1 3 0 12 0 4 0 0y21O12y sin x sin 2x2x5 189 Chapter 637. 0 3.14 1 3.71 3 2 1 3 22 0 2 6.28y8642Oy x sin x23x cos x sin x cos x sin x0 1 0 1 2 0 1 1 1 0 1 0 1 1 3 22 1 0 1y21y cos x sin xx39.40.41a. 2000 1000 30002000 1000 100041b. 10,000 5000 15,00010,000 5000 500041c.y4224y42O2Oy 2 sin x 3 cos xy 3 cos xy cos 2x cos 3x2y cos 3x2y 2 sin x3232y cos 2xxxO12x2[0, 24] sc11 by [0, 16,000] sc1100041d. months number 3 and 1541e. months number 0, 12, 2441f. When the sheep population is at a maximum,the wolf population is on the increase because ofthe maximum availability of food. The upswingin wolf population leads to a maximum later.
42. y349. f(x) y xx 133y x 123x y 1xx(y 1) 3O 2 3 4y 1 3 x y 3 x 1y cos x2y x cos x50.1 1 3 5 X31 1 3 5 1(3) 1(5) 1(5)1(3) 1(3) 1(5) 1(5) X43a. 46 42 4 ft0 043b. r 1 X2 d t 21 40 0r 1 51. 7(3x 5y) 7(4) 21x 35y 282 (42) t 2514x 35y 21→14x 35y 21r 2135x 4943c. 3 revolutionsx 1.460sec onds 1 revolutionx seconds3x 5y 4x 20 s3(1.4) 5y 443d. ⏐A⏐ 21 2 y 0.04 (1.4, 0.04)k 20 h 2552.A 21; 21 k 1 0 yth 25 21 sin 1 0x x 4 y43e. h 25 21 sin t106 ⏐6 4⏐ 246 25 21 sin t4 ⏐4 4⏐ 0102 ⏐2 4⏐ 21 sin t100 ⏐0 4⏐ 4Oy |x 4| xsin 1 sin t1053. 3x y 7 0 y y 1 m(x x 1 )5 t; 5 sy 3x 7 y (2) 3(x 3)43f. h 25 21 sin t10slope: 3 y 2 3x 93x y 11 0h 25 21 sin 101054. 4 inches 1 3 footh 25 ftc 75 42 44. 1 3 1050 cubic feetk 0 2 or 01050 7.48 7854 gal c 2 The correct answer is 7854.k or 4 2There is a 4 phase difference.45a. y sin x45b. y co s xx45c. y cos x 2 45d. y sin x46. 2 1k 2 94 k 58847. v rqv 7(19.2)v 134.4 cm/s48. asymptote: x 2y x 3x 2y(x 2) x 3yx 2y x 32y 3 x yx2y 3 x(1 y) 2y 31 y xasymptote: y 1y 0.25 sin 588tyOy x 3x 2x6-6Modeling Real-World Datawith Sinusoidal FunctionsPages 390–391 Check for Understanding1. any function that can be written as a sine functionor a cosine function2. Both data that can be modeled with a polynomialfunction and data that can be modeled with asinusoidal function have fluctuations. However,data that can be modeled with a sinusoidalfunction repeat themselves periodically, and datathat can be modeled with a polynomial functiondo not.3. Sample answers: the amount of daylight, theaverage monthly temperatures, the height of aseat on a Ferris wheelChapter 6 190
4a. y 5 cos 6 ty 5 cos 6 0y 55 units below equilibrium4c. y 5 cos 6 ty 5 cos 6 74b. 5 units aboveequilibriumy 4.33about 4.33 units above equilibrium5. A 140 802h 140 802A 30 h 110 2 k 1k 2 P 30 sin 2t 11041°6a. A 66° 26b. h 66° 2A 12.5° h 53.5°6c. 12 months6d. A 12.5 2 k 12 h 53.5k 6 41°y 12.5 cos 6 t c 53.541 12.5 cos 6 1 c 53.512.5 12.5 cos 6 c1 cos 6 ccos 1 1 6 ccos 1 1 6 c0.5 cSample answer: y 12.5 cos 6 t 0.5 53.56e. y 12.5 cos 6 t 0.5 53.5y 12.5 cos 6 (2) 0.5 53.5y 42.82517529Sample answer: About 42.8°; it is somewhatclose to the actual average.6f. y 12.5 cos 6 t 0.5 53.5y 12.5 cos 6 (10) 0.5 53.5y 53.20504268Sample answer: About 53.2°; it is close to theactual average.8d. h 3 cos 5 3 t 3.5h 3 cos 5 3 (25) 3.5h 2 units9a. R 1200 300 sin 2 tR 1200 300 sin 2 0R 12009b. H 250 25 sin 2 t 4 H 250 25 sin 2 0 4 H 2329c. R: 1200 300 1500H: 250 25 275 no9d. R 1200 300 sin 2 t1500 1200 300 sin 2 t300 300 sin 2 t1 sin 2 tsin 1 1 2 tsin 1 1 2 t1 tJanuary 1, 19719e. 250 25 225H 250 25 sin 2 t 4 225 250 25 sin 2 t 4 25 25 sin 2 t 4 1 sin 2 t 4 sin 1 1 2 t 4 2 sin 1 1 4 t0.5 tJuly 1, 1969; 2 2k 2 k 4next minimum: July 1, 19739f. See students’ work.10. A 4 2 2 k 10A 2k 5 y 2 cos 5 t11. h 4.25; A 3.55; 2 12.40; 4.68k 6 .2 Pages 391–394 Exercisesk 6 .2 c 2. 347a. 0.53.1y 3.55 sin 6.2 t 2. 347b. 2 k 6603.1 4.24112a. h 47.5; A 23.5; 2 k 3 30 k 12; 417c. 1 330 hertzc 2 3 30 k 6 38a. 3.5 ⏐3⏐ 6.5 units8b. 3.5 3 0.5 units28c. 2 k 5 3or 6 5 y 23.5 sin 6 t 2 3 47.5c 6 c191 Chapter 6
12b. y 23.5 sin 6 t 2 3 47.5y 23.5 sin 6 3 2 3 47.5y 35.75about 35.8°12c. y 23.5 sin 6 t 2 3 47.5y 23.5 sin 6 8 2 3 47.5y 67.9°13a. A 81° 73°213b. h 81° 73°2A 4° h 77°13c. 12 months13d. A 4 2 k 12 h 77k 6 y 4 cos 6 t c 7773 4 cos 6 1 c 77 c1 cos 6 c4 4 cos 6 cos 1 1 6 ccos 1 1 6 c0.5235987756 cSample answer: y 4 cos 6 t 0.5 7713e. y 4 cos 6 t 0.5 77y 4 cos 6 8 0.5 77y 80.41594391Sample answer: About 80.4°; it is very close tothe actual average.13f. y 4 cos 6 t 0.5 77y 4 cos 6 5 0.5 77y 79.08118409Sample answer: About 79.1°; it is close to theactual average.c14. k 1or increase shift by 2 ; 2 3 2c k 3 2c 1 3 2c 3 2 5Sample answer: y 3 cos x 3 215a. A 13.25 1.88215b. h 13.25 1.882A 5.685 fth 7.565 ft15c. 4:53 P.M. 4:30 A.M. 12:23 or about 12.4 h15d. A 5.685 2 k 12.4 h 7.565k 6 .2 4:30 A.M. 4.5 hrsh 5.685 sin 6.2 t c 7.56513.25 5.685 sin 6.2 4.5 c5.685 5.685 sin 4 6.5.2.5.2 c c1 sin 4 6sin 1 1 4 .56.2 csin 1 1 4 .56.2 c0.7093918895 cSample answer: h 5.685 sin 6.2 t 0.71 7.56515e. 7:30 P.M. 19.5 hrsh 5.685 sin 6.2 t 0.71 7.565h 5.685 sin 6.2 19.5 0.71 7.565h 8.993306129Sample answer: about 8.99 ft16a. Table at bottom of page.MonthSunrise A.M. Time Sunset P.M. Time Daylight HoursA.M. in Decimals P.M. in Decimals (P.M.-A.M.)January 7:19 7.317 4:47 16.783 9.47 hFebruary 6:56 6.933 5:24 17.4 10.47 hMarch 6:16 6.267 5:57 17.95 11.68 hApril 5:25 5.416 6:29 18.483 13.07 hMay 4:44 4.733 7:01 19.017 14.28 hJune 4:24 4.4 7:26 19.433 15.03 hJuly 4:33 4.55 7:28 19.467 14.92 hAugust 5:01 5.017 7:01 19.017 14 hSeptember 5:31 5.517 6:14 18.233 12.72 hOctober 6:01 6.017 5:24 17.4 11.38 hNovember 6:36 6.6 4:43 16.717 10.12 hDecember 7:08 7.133 4:28 16.467 9.33 hChapter 6 192
16b. A 15.03 9.33216c. h 15.03 9.332A 2.85 hh 12.18 h16d. 12 months16e. A 2.85 2 k 12 h 12.18k 6 y 2.85 cos 6 t c 12.189.47 2.85 cos 6 1 c 12.182.71 2.85 cos 6 c0.950877193 cos 6 ccos 1 0.950877193 6 ccos 1 0.950877193 6 c0.2088597251 cSample answer: y 2.85 cos 6 t 0.21 12.1817. 70.5 19.5 51y 70.5 19.5 sin 6 t c51 70.5 19.5 sin 6 1 c19.5 19.5 sin 6 c1 sin 6 sin 1 1 6 c csin 1 1 6 c2.094395102 cSample answer: about 2.0918a. 14 revolutions1 minute2 radians1 minute 6 0 seconds 1 revolution 7 1y 3.5 cos 7 15 t18b. y 3.5 cos 7 15 ty 3.5 cos 7 1y 3.197409102about (4, 3.20)19. A 120 2A 120(120)V R 120 sin 30 t5 420. See students’ work.21. ⏐3⏐ 3; 2 2 ; 2 ; 5y8642O 2 k 60k 3 0 y 3 cos(2 ) 523 4 55 rad/s24. 40 2 32 2 20 2 2(32)(20) cos vcos v 402 322 2022(32)(20) 322 20 2v cos 1 402 2(32)(20)v 97.9°180° 97.9° 82.1°about 97.9°, 82.1°, 97.9°, 82.1°25. 2 m 162m 16m 2 16 (m 4) (m 4)2m 16 A B (m 4) (m 4) m 4 m 42m 16 A(m 4) B(m 4)Let m 4.2(4) 16 A(4 4) B(4 4)8 8B1 BLet m 4.2(4) 16 A(4 4) B(4 4)24 8A3 AA B 3 1 m 4 m 4 m 4 m 426. 2⏐ 2 k 1 64 2k 8 4k 142 k 4 2k 7 ⏐ 4k 204k 20 0k 527. f(x)increasing: x 1; decreasing: x 128. The correct choice is E.6-7f(x) 2x 1 5OGraphing Other TrigonometricFunctionsPage 400 Check for Understanding1. Sample answers: , , 22. The asymptotes of y tan v and y sec v are thesame. The period of y tan v is and the periodof y sec v is 2.3. Sample answers: 2 , 3 24. 0 5. 16. n, where n is an odd integer7. 4 n, where n is an integerx22. 2n where n is an integer23. 800° 18 0° 40 9193 Chapter 6
8. 1 4 24. ; 4 4 n, where n is an integer125. 4 n, where n is an integery26. 3 y tan( 4 n, where n is an integer44 )27. 2 n, where n is an odd integer228. n, where n is an integer2 O 2 29. 1 2 2 ; 1 2 4y9. 2 2 ; 2 ; h 142yy sec(2 ) 184y cot( 2 )2 O4 282 O 2242 1 3 30. 6; no phase shift; no vertical shift10. k: 2 k 3 c: 3 h: h 4y csc 2 3 v 2 9 4c 2 911. k: ck 2 c: 4 h: h 0 1 2 k 1 2 c 8 y cot 1 2 v 8 12a. f ma 12b. F f sec vf 73(9.8)F 715.4 sec vf 715.4 N12c. 2 1 2; no phase shift, no vertical shifty12d. 715.4 N12e. The tension becomes greater.Pages 400–403 Exercises13. 0 14. 015. undefined 16. 117. 1 18. undefined19. undefined 20. 021. n, where n is an integer22. n, where n is an even integer 2n, where n is an integer23. 3 2k 2 3 2800400Oc 2 3 F 715.4 sec 231. 2 1 2; no phase shift; h 5 4 32. 2; 2 ; h 1 1 2 2 O2 24 1 2 y42y8642y sec 32 O 2y42y csc 5( )2 O2 24y tan 2 4 1Chapter 6 194
33. 2 2 ; 2 ; h 334. 6; 2 ; h 235.2 1 3 22 O 22 6 1 3 468122, , 0, , 2yy csc(2 ) 3y2 y sec( 3 6 ) 2O2 2 2468y42O24y sec c 1 2 y cos 36. k: k 2 c: 0 h: h 6k — 1 2 —y tan 2v 6c 02 37. k: k 2 c: 2c 8 h: h 7k 2 c — 4 —y cot 2v 4 738. k: 2 k cc: 2 4 h: h 10k 2c 2 y sec 2v 2 1039. k: 2 ck 3 c: h: h 1k 2 3 y csc 2 3 v 2 3 1c 2 340. k: k 5 c: h: h 12k 1 5 y cot v 5 5 12 2 3 c 1 5 c 5 41. k: 2 k 3 cc: 6 2 h: h 5k 6c 3y csc (6v 3) 542. k: 2 k 3 c: h: h 8c 2 3 k 2 3 k 2 4 0 y sec 2 3 v 2 3 8c 2 343. k: k 2 c: 2c 4 h: h 7y tan 2v 2 744a. 40no phase shiftno vertical shiftc 2 44b. d 10 tan 4 0 td 10 tan 4 0 (3)d 2.4 ft from the center44c. d 10 tan 4 0 td 10 tan 4 0 (15)d 24.1 ft from the center45.y42d4020The graph of y csc v has no range valuesbetween 1 and 1, while the graphs of y 3 csc vand y 3 csc v have no range values between3 and 3. The graphs of y 3 csc v and y 3 csc v are reflections of each other.2020 10 O 10 20402 O2 2446a. f m 9.8f 7 9.8f 68.6 N246c. 4 1 2 F4020OF 34.3 sec 2d 10 tan 40 t46b. F 1 2 f sec 2v F 1 2 (68.6) sec 2v F 34.3 sec 2v t195 Chapter 6
46d. 34.3 N52. C 180° (62°31 75°18) or 42°1146e. The tension becomes greater.a b sin A sin B47a. 220 A57.3b 2147b. 6 0 3 0ssin 62°31 sin 7 5°18 b 57ṡ 3 sin75°186 in 62°31147c. b 62.4750578360 360 47d. I 220 sin 60t 6 I 220 sin 60 60 6 I 110 A48. y 1 tan v 2 0.550.5549a. A 3.99 249b. h 3.99 2A 1.72 fth 2.27 ft49c. 12:19 P.M. 12:03 A.M. 12:16 or about 12.3 hr49d. A 1.72 2 k 12.3 h 2.272k 1 2.312:03 0.05 hr since midnighth 1.72 sin 212.3t c 2.273.99 1.72 sin 212.3 0.05 c 2.271.72 1.72 sin 0 . 112.3 c. 11 sin 0 12.3sin 1 1 0 . 112.3 csin 1 1 0 . 112.3 c1.545254923 cSample answer: h 1.72 sin 212t.3 1.55 2.2749e. noon 12 hrs since midnighth 1.72 sin 212t.3 1.55 2.27h 1.72 sin 212.3 12 1.55 2.27h 3.964014939Sample answer: 3.96 ft250. ⏐2⏐ 2; 4 1 2 51. s rvs 18 3 s 6 cmy21 c2 O 212y 2 cos 253a.a c sin A sin C57.3c sin 62°31 sin 4 2°11c 57 .3 sin42°11sin62°31c 43.37198044C 42°11, b 62.5, c 43.473˚4 m53b. tan 73° 4x x 4 tan 73°x 13.1 m53c. cos 73° 4 y 54. a 2 b 2 c 2 sin A a c 7 2 4 2 c 2 sin A 74y cos 73°y 13.7 m6565 65 c sin A 7 65cos A b c tan A a b 4cos A tan A 7 4 6565 cos A 4 65x55. — 2 4—x 02 3x 10— ( x 2)( x 2)— 0 zeros: 2, 2( x 5)( x 2)excluded values: 5, 2(3) 2 4Test 3: 0Test 0: 0 2 Test 3: 3 2 Test 6: 6 2 (3) 2 3(3) 1002 43(0) 104 1032 43(3) 105 0 0 062 43(6) 10 3 28 5 5 0 falsefalse10 0 true 0 0false2 x 5656. k 0 .5 t krk 1210 12rr 0.83Chapter 6 196
57. 3x 2y 8y 3 2 x 4y 2x 12y x 4y 1 2 x 2Oyx6-8Trigonometric Inverses and TheirGraphsPage 410 Check for Understanding1. y sin 1 x is the inverse relation of y sin x, y (sin x) 1 1is the function y sin,xand y sin(x 1 )is the function y sin 1 x .58. y 17.98x 35.47; 0.8859. A: impossible to tellB: 6(150) 10(90)900 900; trueC: impossible to tellD: 150 30 2(90)150 210; falseE: 3(90) 30 2(150)270 330; falseThe correct choice is B.6-7BPage 4041.Sound Beats2. For every y value there are more than one x value.The graph of y cos 1 x fails the vertical linetest.3. The domain of y Sin x is the set of real numbersbetween 2 and 2 , inclusive, while the domain ofy sin x is the set of all real numbers. The rangeof both functions is the set of all real numbersbetween 1 and 1, inclusive.4. Restricted domains are denoted with a capitalletter.5. Akikta; there are 2 range values for each domainvalue between 0 and 2. The principal values arebetween 0 and , inclusive.6. y Arcsin xx Arcsin ySin x y or y Sin x2y Arcsin xy1y Sin xy1 O 1x 2O x2 21the third graph2.Sample answer: The graph seems to stay abovethe x-axis for an interval of x values, and thenstay below the x-axis for another interval of xvalues.3. 0.386235834. no5. 1.78043; yes; the value for f(x) is negative andcorresponds to a point not graphed by thecalculator.6. Sample answer: As you move 1 pixel to the left orright of any pixel on the screen, the x-value for theadjacent pixel decreases or increases by almost 7.Thus, the “find” behavior of the function cannot beobserved from the graph unless you change theinterval of numbers for the x-axis.7. See students’ work.8. Yes; no, they only provide plausible visualevidence.7. y Cos x 2 x Cos y 2 Cos 1 x y 2 2y Cos 1 x 2 y1y CosO x 1 O 121( x 2 )y2y Cos 1 x 2 28. Let v Arctan 1. 9. Let v Tan 1 1.Tan v 1 Tan v 1v 4 v 4 cos (Tan 1 1) cos v cos 4 22x197 Chapter 6
10. Let v Cos 1 22 .Cos v 22v — 4 —cos Cos 1 22 2 cos v 2 cos 4 2 cos — 4 — 2211. true12. false; sample answer: x 1; when x 1,Cos 1 (1) , Cos 1 (1) 013a. C 2r 13b. C 40,212 cos vC 2(6400)C 40,212 km13c. C 40,212 cos v3593 40,212 cos v3593 4 cos v0,21235930,212cos 1 41.48 v; about 1.48 radians13d. C 40,212 cos vC 40,212 cos 0C 40,212 km vPages 410–412 Exercises14. y arccos xx arccos ycos x y or y cos xy16. y arctan xx arctan ytan x y or y tan x17. y Arccos 2xx Arccos 2yCos x 2yy2 O 1 2 Cos x y or y 1 2 Cos xy Arccos 2xy arctan x18. y 2 Arcsin x2x 2 Arcsin yx — 2 — Arcsin ySin x 2 y or y Sin x 2 x1 O 1 xyyy tan x1y1y1O22yO1y Cos x2y Sin ( x )2xx2y1y cos x2y Arcsin x21 O 1 xO12 xy arccos xO2x19. y tan 2x 1 O 1 x115. y Sin xx Sin ySin 1 x y or y Sin 1 x1y2yyx tan 2tan 1 yx 22 tan 1 x y; y 2 tan 1 xy2 y tan x 2yy 2 tan 1 xy Sin xy Sin 1 xOx2O2x 2O x 1 O 12x21 2Chapter 6 198
20. y Tan x 2 x Tan y 2 21.Tan 1 x y 2 Tan 1 x 2 yNo; the inverse is y Tan 1 x 2 .y42O24y Cot x222. Let v Sin 1 0.Sin v 0v 024. Let v Tan 1 3.3Tan v 33v 6 26. If y Cos 1 22 x, then y — 4 — .sin 2 Cos 1 22 sin (2y) sin 2 4 sin 2 127. If y Tan 1 3, then y 3 .cos (Tan 1 3) cos y cos 3 1 2 28. Let a Tan 1 1 and Sin 1 1.Tan a 1 Sin 1a 4 y2y Arccot x4 2 O 2 4 x23. Let v Arccos 0.Cos v 0v — 2 —25. If y tan 4 , theny 1.Sin 1 tan 4 Sin 1 y Sin 1 1 2 cos (Tan 1 1 Sin 1 1) cos (a ) cos 4 2 cos 4 2229. Let a Cos 1 0 and Sin 1 1 2 .Cos a 0Sin 1 2 a 2 6 cos Cos 1 0 Sin 1 1 2 cos (a ) cos 2 6 cos 2 3 1 2 — 2 —30. Let a Sin 1 1 and Cos 1 1 2 .Sin a 1a 2 Cos 1 2 3 sin Sin 1 1 Cos 1 1 2 sin (a ) sin 2 3 sin 6 1 2 31. No; there is no angle with the sine of 2.32. false; sample answer: x 2; when x 2,Cos 1 (cos 2) Cos 1 1, or 0, not 2.33. true34. false; sample answer: x 1; when x 1,Arccos (1) and Arccos ( (1)) 0.35. true 36. true37. false; sample answer: x 2 ; when x 2 , cos 1 2 is undefined.38.yy tan(Tan 1 x)O39. y 54.5 23.5 sin 6 t 2 354.5 54.5 23.5 sin 6 t 2 30 23.5 sin 6 t 2 30 sin 6 t 2 3sin 1 0 6 t 2 3 0 6 t 2 3 or 6 t 2 3 2 3 6 t 5 3 6 t4 t 10 tApril and October40. P VI Cos v7.3 122(0.62) Cos v0.0965097832 Cos vCos 1 0.0965097832 v1.47 v; about 1.47 radians41. 4 n, where n is an integer42. I I 0 cos 2 v1 8 cos 2 v 1 8 cos 2 v 1 8 cos vcos 1 1 8 v1.21 v; about 1.21 radians43a. 6:18 12:24 18:42 or 6:42 P.M.43b. 12.4 h43c. A 7.05 (0.30)2A 3.675 ftx199 Chapter 6
43d. A 3.675 2 k 12.4 h 7.05 2k 6 .2 h 3.3756:18 6.3 hy 3.675 sin 6.2 t c 3.3757.05 3.675 sin 6.2 6.3 c 3.3753.675 3.675 sin 6 .36.2 c1 sin 6 .36.2 csin 1 1 6 .36.2 csin 1 1 6 .36.2 c1.621467176 cSample answer:y 3.375 3.675 sin 6.2 t 1.6243e. y 3.375 3.675sin 6.2 t 1.6244.(0.30)6 3.375 3.675sin 6.2 t 1.622.625 3.675 sin 6.2 t 1.62 2 . 6253.675 sin 6.2 t 1.62sin 1 2 . 6253.675 — 6 .2 — t 1.62sin 1 2 . 6253.675 1.62 6 .2 t.2 sin1 2 6253.. 675 1.62 t4.767243867 t0.767243867 60 46.03463204;Sample answer: about 4:46 A.M. 6 y sin (Tan 1 x)y148.11y cos x10y19 O49. v 25°30 30xa 180° (25° 25°)25˚or 130°30 3030 x sin 25° sin 130°x 30 sin130°sin25°x 54.4 units30 y sin 65° sin 50°y 30 sin50°sin65°y 25.4 units50. 210° 180° 30°51. p: 1, 2, 3, 6q: 1, 252.30 3030 30— p q — : 1, 2, 3, 6, 1 2 , 3 2 g(x)y1g(x) x 2 31xv 2(25°) or 50°a 1 2 (180° 50°) or 65°2 O 2xOx45a. v cos 1 D 21 dc410)v cos 1 6 2(v 1.47 radians45b. L D (d D)v 2C sin vL (6) (4 6)1.47 2(10) sin 1.47L 35.81 in.46. n, where n is an integer47. ⏐A⏐ 5 2 c 3 k h 8A 5 k — 2 3 —y 5 sin 2 3 v 2 3 8 2 3 c 2 3decreasing for x 2 and x 253. [f g](x) f(g(x)) f(3x) (3x) 3 1 27x 3 1[g f ](x) g(f(x)) g(x 3 1) 3(x 3 1) 3x 3 354. D 4, F 6, G 7, H 8value: (4 6 7 8)4 (25)4 or 100The correct choice is D.Chapter 6 200
Chapter 6 Study Guide and Assessment34. ⏐0.5⏐ 0.5; 2 4 2 y1y 0.5 sin 4O 2 3135. ⏐ 1 3 ⏐ 1 3 2; 4 1 2 0°y10°14. 5 6 5 1 y cos3 26 4 3 150°O16. 2.4 2.4 18 0°2 4 6315° 137.5°1 0° 5 12 s rvs 15 5 12 y 4 sin (4v 8) 1s 19.6 cm37. ⏐A⏐ 0.5 2 k0°20. s rv 5 6s 15 5 y 0.5 sin 2v 2 3 3s 9.4 cm 38. ⏐A⏐ — 3 4 — 2 ky 3 4 cos 8v 539. A 120 802 2 ky 20 sin 2t 100q v t 40. A 130 1002— 2 ky 15 sin 2t 115q 11.3 radians/min41. period: 2 1yq v t 1. 85q 10 10y csc 312q 26.2 radians/minO 2 3 4Page 413 Understanding and Using theVocabulary1. radian 2. angular3. the same 4. amplitude5. angle 6. phase7. radian 8. frequency9. sunusoidal 10. domainPages 414–416 Skills and Concepts11. 60° 60° 18 0°12. 75° 75° 18 3 5 12 13. 240° 240° 18 18 0°15. 7 4 7 4 18 0°17. s rv 18. 75° 75° 18s 15 3 4s 35.3 cm19. 150° 150° 18s rvs 15 5 6s 39.3 cm21. 5 2 10 or about 31.4 radians22. 3.8 2 7.6 or about 23.9 radians23. 50.4 2 100.8 or about 316.7 radians24. 350 2 700 or about 2199.1 radians25. 1.8 2 3.6 26. 3.6 2 7.2q v t q 3. 6 5q 7. 2 2q 2.3 radians/s27. 15.4 2 30.8 28. 50 2 100q v t q 30 1q 6.5 radians/s29. 1 30. 0 31. 1 32. 033. ⏐4⏐ 4; 2 2y42y 4 cos 236. ⏐A⏐ 4 2 k 2 c 4 2 h 1A 4 k 4 c 8 2c 3 h 3A 0.5 k 2 c 2 3 4 8c 0 h 5A 3 4 k 8 c 0 1 h 120 2A 20 k 2 h 100 — 1 h 130 2A 15 k 2 h 1151 80100 or 2, no phase shift, no vertical shiftO2234201 Chapter 6
42. 3 2 Page 417 Applications and Problem Solving; 6 3; no vertical shift50a. A 11.5 2 k c 12 3 h 64 y6 k 6 8c 2 y 2 tan 3 4( 2 )y 11.5 sin 6 t 2 6450b. April: month 4O2 y 11.5 sin4 6 t 2 64843. vertical shift: 4y642O244. vertical shift: 2y2y sec 423y 11.5 sin 6 4 2 64y 69.75; about 69.8°50c. July: month 7y 11.5 sin 6 t 2 64y 11.5 sin 6 7 2 64y 74.0°F51. B IL s0.04 5.0(100.04(5.0(1) sin v) 0.20sin v 0.04sin v 1v 2 in v.2) sin v.2(5.0)(1)O2623y tan 246. Let v Sin 1 1.45. Let v Arctan 1.v 4 v 2 Tan v 1Sin v 147. If y tan 4 , then y 1.Cos 1 tan 4 Cos 1 y Cos 1 1Let v Cos 1 1.Cos v 1v 048. If y Sin 1 3, then y — 23 — .sinSin 1 32 sin y sin — 3 — 3249. Let a Arctan 3 and Arcsin 1 2 .Tan a 3a 3 Sin 1 2 6 cos (Arctan 3 Arcsin 1 2 cos (a ) cos 3 6 cos 2 0Page 4171. A 1 2 r 2 vOpen-Ended Assessment26.2 1 2 r 2 vSample answer: r 5 in., v 2 32a. Sample answer: If the graph does not cross they-axis at 1, the graph has been translated. Thefirst graph has not been translated and thesecond graph has been translated.y1O1y1O12b. Sample answer: If the equation does not havethe form y Acos kv, the graph has beentranslated. The graph of y 2 cos 2v has notbeen translated. The graph of y 2 cos (2v ) 3 has been translated vertically andhorizontally.22xxChapter 6 202
Chapter 6 SAT & ACT PreparationPage 419 SAT and ACT Practice1. Since there is no diagram, draw one. Sketch aright triangle and mark the information given.Notice that this is one of the “special” righttriangles. Its sides are 3-4-5. So the hypotenuse is5. The sine is opposite over hypotenuse (SOH).sin v 4 5 The correct choice is B.2. Let x be the smaller integer. The numbers are twoconsecutive odd integers. So, the larger integer is 2more than the first integer. Represent the largerinteger by x 2. Write an equation that says thatthe sum of these two integers is 56. Then solve for x.x (x 2) 562x 2 562x 54x 27Be sure to read the question carefully. It asks forthe value of the larger integer. The smaller integeris 27 and the larger integer is 29.The correct choice is C.3. Factor the numerator.a 2 b 2 (a b)(a b)(sin v cos v)(sin v cos v)sin v cos v3 sin v cos vThe correct choice is B.4. First find the coordinates of point B. Notice thatthere are two right triangles. One has ahypotenuse of length 15 and a side of length 12.This is a 3-4-5 right triangle. The coordinates ofpoint B are (9, 12).Since point A has coordinates (0, 0), each point onside AB must have coordinates in the ratio of 9 to12 or 3 to 4.The only point among the answer choices that hasthis ratio of coordinates is (6, 8).Aslightly different way of solving this problem isto write the equation of the line containing pointsA and B.y 1 29xThen test each point to see whether it makes theequation a true statement.You could also plot each point on the figure andsee which point seems to lie on the line segment.The correct choice is E.5. Factor the polynomial on the left side of theequation.x 2 2x 8 0(x 4)(x 2) 04If either of the two factors equals 0, then thestatement is true. Set each factor equal to 0 andsolve for x.x 4 0 or x 2 0x 4x 2The solutions of the equation are 4 and 2. Tofind the sum of the solutions, add 4 2 2. Thecorrect choice is D.6. You may want to label the triangle with opposite,adjacent, and hypotenuse.Cadjacent5hypotenuseTo find cos v, you need to know the length of theadjacent side. Notice that the hypotenuse is 5 andone side is 3, so this is a 3-4-5 right triangle. Theadjacent side is 4 units.Use the ratio for cos v.adjacentcos v hy potenuse 4 5 The correct choice is C.7. Look at the powers of the variables in theequation. There is an x 2 term, an x term, and a yterm, but no y 2 term. It cannot represent a line,because of the x 2 term. It cannot represent a circleor an ellipse or a hyperbola because there is no y 2term. So, it must represent a parabola.The general form of the equation of a parabola isy a(x h) 2 k. The correct choice is A.8. Factor each of the numerators and determine ifthe resulting expression could be an integer, thatis, the numerator is a multiple of the denominator.n 16 (n 1)I 16; an integer 16 n 1 16 n 1II 16n 16 16n 16( n 1 6 1)n n 1nIII 16n 2 n16n n(16 n 1)16n 16n 116opposite3; not an integer; not an integerOnly expression I is an integer.The correct choice is A.9. Since x 1, 1 x 0. So x 1 x 1 x x .1Since x 1, x x 1 1 1. So x x 1 1.The correct choice is D.AB203 Chapter 6
10. Notice that the triangles are not necessarilyisosceles. In ADC, the sum of the angles is 180°,so m∠CAD m∠ACD 80. Since segment ADbisects ∠BAC, m∠BAD m∠CAD. Similarly,m∠BAC m∠ACD. So, m∠BAD m∠BCD 80.Add the two equations. m∠CAD m∠BAD m∠ACD m∠BCD 160, so two of the angles inABC have the combined measure of 160°.Therefore, the third angle in this triangle, ∠B,must measure 20°. The correct answer is 20.Chapter 6 204
Chapter 7 Trigonometric Identities and Equations7-1 Basic Trigonometric Identities12. 7 3 2 3 cos 2 3 Page 427 Check for Understanding1. Sample answer: x 45°2. Pythagorean identities are derived by applyingthe Pythagorean Theorem to a right triangle. Theopposite angle identities are so named because Ais the opposite of A.113. tan v co t,vcot v ta n, c osvv sinv cot v,1 cot 2 v csc 2 vsin( A)4. tan(A) cos( A) sinAcosAsinAc osAtan A5. Rosalinda is correct; there may be other values forwhich the equation is not true.6. Sample answer: v 0°sin v cos v tan vsin 0° cos 0° tan 0°0 1 01 07. Sample answer: x 45°sec 2 x csc 2 x 1sec 2 45° csc 2 45° 1(2) 2 (2) 2 12 2 14 1118. sec v co s v9. tan v co t v1sec v tan v 1 2 3 sec v 3 2 10. sin 2 v cos 2 v 1 1 5 2 cos 2 v 11 2 5 cos 2 v 1cos 2 v 2 425 cos v 2 65Quadrant III, so 2 6511. tan 2 v 1 sec 2 v 4 7 2 1 sec 2 v 1 64 6 54 65 79 1 sec 2 v9 sec 2 v sec vQuadrant IV, so 765 52tan v 2 5tan v 2 55cos 7 3 cos 3 13. 330° 360° 30°1csc (330°) sin (330°)1sin (360° 30°)1 sin 30° csc 30°114. c scv sicot vn —v c osvs in v1 sin si n v vc osv1 co s v sec v17. B F c sc vIBI F csc vBIF c sc vF BI 1csc v F BIsin vPages 427–4302 2 2 21 sinx c 15. cos x csc x tan x cos x sin x osx 116. cos x cot x sin x cos x c osxsinx sin x c os2xsinx cos2 x sin 2 xsin x1 sin x csc xExercises18. Sample answer: 45°sin v cos v cot vsin 45° cos 45° cot 45° 1 1 2 119. Sample answer: 45°sec v t sin vanvsec45° t an45° 2 21 2 sin 45°2 22 sin x205 Chapter 7
20. Sample answer: 30°sec 2 x 1 c osxcscxsec 2 30° 1 c c23 2 1 3os30°sc 30°32 2 1 29 1 34 1 3 3 421. Sample answer: 30°sin x cos x 1sin 30° cos 30° 1 1 2 32 1 1 32 122. Sample answer: 0°sin y tan y cos ysin 0° tan 0° cos 0°0 0 10 123. Sample answer: 45°tan 2 A cot 2 A 1tan 2 45° cot 2 45° 11 1 12 124. Sample answer: 0cos v 2 cos v cos 2 cos 0 2 cos 0 cos 2 cos 2 cos 0 cos 2 0 1 00 11125. csc v si n v26. cot v ta1csc v cot v 1 2 5 29. 1 cot 2 v csc 2 v1 cot 2 v 11 31 cot 2 v 1 91 2cot 2 v 2 9 cot v 23Quadrant II, so 2330. tan 2 v 1 sec 2 vtan 2 v 1 5 4 2tan 2 v 1 2 516 tan 2 9v 1 6tan v 3 4 Quadrant II, so 3 4 31. sin 2 v cos 2 v 1 1 3 2 cos 2 v 1 1 9 cos2 v 1cos 2 v 8 9 cos v 2 23Quadrant III, so cos v 2 23sintan v vc osv 1 3 tan v — 2 23 or 22432. tan 2 v 1 sec 2 v1tan v 2n v 4 9 34 13343cot v 13cos v se c vcos v 1 1333cos v 1351633. cos v se c v4cos v 115 7 5 4cos v 5 7 3csc v 5 2 cot v 4 327. sin 2 v cos 2 v 1 1 4 2 cos 2 v 11 1 6 cos 2 v 1Quadrant I, socos 2 v 1 1cos v 1528. sin 2 v cos 2 v 1sin 2 v 2 3 2 1sin 2 v 4 9 1Quadrant II, so 5sin 2 v 5 9 sin v 53 2 3 2 1 sec 2 v 1 sec2 v 1 3 sec92 v sec vQuadrant III, so sec v 133313or 13sin 2 v cos 2 v 1Quadrant III, so 2 67sin 2 v 5 7 2 1sin 2 v 2 549 1sin 2 v 2 449 sin v 2 67Chapter 7 206
134. sec v co s v1sec v 40. 19 5tan 2 v 1 64 1 8 sec v 8Quadrant IV, so 3735. 1 cot 2 v csc 2 1v sin v cs c v1 4 3 2 1 csc 2 v sin v 5 3 1 1 6 csc92 v sin v 3 5 2 5 csc92 v 5 3 csc vQuadrant IV, so 5 3 36. 1 cot 2 v csc 2 v1 (8) 2 csc 2 v1 64 csc 2 v65 csc 2 v65 csc vQuadrant IV, so 65137. sec v co s vsec v 1 344sec v or 4 33 3Quadrant II, so 13 4sintan v vc osv134tan v — 34tan v 1 3 or 33sec 2 A tan 2 A2sin 2 A 2cos 2 A39 tan 2 v 1 sec 2 vtan 2 v 1 8 2 2 3 2 4 3 39 3 2 134 2 2 34 28 3 99 2 1 316 3 2 16 4 9 9 9 3 216 1 2 38. 390° 360° 30°sin 390° sin (360° 30°) sin 30°39. 27 3 3 88cos 27 cos8 3 3 8cos 3 8tan 2 v 63tan v 37sin 2 v cos 2 v 1sin 2 v 34 2 1sin 2 3v 1 16sin 2 v 1 316sin v 134 2(2) 5 tan 19 5 sin 19 5—cos 19 5sin 2(2) 5 ——cos 2(2) 5 sin 5 —cos 5 tan 5 41. — 10 —3 3 3 csc — 10 1—3sin 10 31sin 3 3 1sin 3 csc 3 42. 1290° 7(180°) 30°1sec (1290°) cos (1 co s30°sec 30°43. 660° 2(360°) 60°cot (660°) c os( 660°)sin( 660°)1sec x co s x44. —— t anx—sinxc osxcot v45. —— c osv1 sin x csc x c osvsinvcos v1 si n v csc vcos (2(360°) 60°) sin (2(360°) 60°) c os60°sin60° cot 60°sin( v )sinv46. —— c os( v ) —— cosv1290°)1cos (7(180°) 30°) tan v47. (sin x cos x) 2 (sin x cos x) 2 sin 2 x 2sin x cos x cos 2 x sin 2 x 2sin x cos x cos 2 x 2sin 2 x 2 cos 2 x 2(sin 2 x cos 2 x) 2207 Chapter 7
48. sin x cos x sec x cot x sin x cos x — cos x cos x49. cos x tan x sin x cot x cos x — sinxcosx sin x cos x50. (1 cos v)(csc v cot v) (1 cos v) — si1— — c osx—sinx — sin x —c s (1 cos v) —1 sin — 1 cos 2 v—sinv s in2vsinv sin v51. 1 cot 2 v cos 2 v cos 2 v cot 2 v 1 cot 2 v cos 2 v(1 cot 2 v) csc 2 v cos 2 v(csc 2 v) csc 2 v(1 cos 2 v) csc 2 v(sin 2 v)1 —sin—(sin 2 v)2 v 1sin x sin x52. 1 cos x 1 cos xsin x sin x cos x1 cos 2 x2 sinx 1 cos2 x 2 sinxsin2xsin x sin x cos x1 cos 2 xosx—in x 1—n v — c osv—sinv cos v—v 2 sin x 2csc x53. cos 4 a 2cos 2 a sin 2 a sin 4 a (cos 2 a sin 2 a) 2 1 2 or 154. I I 0 cos 2 v0 I 0 cos 2 v0 cos 2 v0 cos vcos 1 0 v90° v55. Let (x, y) be the point where the terminal side ofA intersects the unit circle when A is in standardposition. When A is reflected about the x-axis toobtain A, the y-coordinate is multiplied by 1,but the x-coordinate is unchanged. So,sin (A) y sin A andcos (A) x cos A.yOAA(x, y)(x, y)x56a. e W Asec vseAs W sec veAs s ecv WW eAs cos v56b. W eAs cos vW 0.80(0.75)(1000) cos 40°W 459.6266659459.63 W57. F N mg cos v 0F N mg cos vmg sin v m k F N 0mg sin v m k (mg cos v) 0m k (mg cos v) mg sin vmgsinvm k m g cosvsinm k vc58.59.v 3 260°n 18 0°nhosvm k tan vThe area of the isosceles triangle is 1 2 (a) a 2 a 24a a 2 a, tan v , so h 2ta n v a 2 cot v.hcot 18 0°n 0° cot 18 n . There are n such triangles, soA 1 4 na 2 cot 18 0°n .sin v EF and cos v OF since the circle is a unitCDcircle. tan v O D C 1D CD.COsec v O C 1O CO. EOF OBA, soD O F BAEF O A B A1os v O FEFF O B1 BA. Then cot v c sinvAlso by similar triangles, E OEF O B 1O,Aor E1 1Then csc v si n v E F O B1 OB.60. Cos 1 2 2 135°yOAEBCxF D BA..Chapter 7 208
61.y1O162. 2(3° 30) 7° 7° 7° 18s rvs 20 7180s 2.44 cm63. B 180° (90° 20°) or 70°sin A a c 7 1 800°cos A b c asin 20° 3 5 cos 20° 3b535 sin 20° a 35 cos 20° b11.97070502 a 32.88924173 ba 12.0, B 70°, b 32.964. 2 ⏐ 2 1 8 4 4 1 0 4 2 5 2 ⏐ 02x 2 5x 2 0(2x 1)(x 2) 02x 1 0 or x 2 0x 1 2 x 22, 1 2 , 265. 2x 2 7x 4 0x 2 7 2 x 2 0( )y cos x 6 2 7 5 136 3 6 3 6x 2 7 2 x 2x 2 7 2 x 4 916 2 4 1x 7 4 2 8 116x 7 4 9 4 96 x 7 4 9 4 x 0.5 or 466. continuous67. 4(x y 2z) 4(3) 4x 4y 8z 124x y z 0 → 4x y z 03y 9z 12x y 2z 3x 5y 4z 114y 2z 144(3y 9z) 4(12) → 12y 36z 483(4y 2z) 3(14) 12y 6z 4230z 90z 33y 9z 12 x y 2z 33y 9(3) 12 x (5) 2(3) 3y 5 x 2(2, 5, 3)x68. m 4 24 5 y y 1 m(x x 1 )2 9or 2 9 69. m∠BCD 40°40 1 2 m(BC)-80 m(BC)-m∠BAC 1 2 m -BCm∠BAC 1 2 (80)m∠BAC 40°The correct choice is C.7-2y 4 2 9 (x (4))y 2 9 x 2 8 9Verifying Trigonometric IdentitiesPage 433 Graphing Calculator Exploration1. yes 2. no 3. no4. No; it is impossible to look at every window sincethere are an infinite number. The only way anidentity can be proven is by showing algebraicallythat the general case is true.5.[2, 2] sc1 2 by [2, 2] sc11sin xPages 433–434 Check for Understanding1. Answers will vary.2. Sample answer: Squaring each side can turn twounequal quantities into equal quantities. Forexample, 1 1, but (1) 2 1 2 .3. Sample answer: They are the trigonometricfunctions with which most people are mostfamiliar.4. Answers will vary.co5. cos x t xccos x scx c osxsinx—1sin xs xcos x co 1cos x cos x209 Chapter 7
1cosx6. tan x sec x sin x 11cosxsinx 1 sin x 1c osx co s x1cosxsinx 1 sin x 1cosxcosx cosx sin x 1 sin x 1cot v1csc v cot v csc v cot v c scv cotvcscv cotvcscv cotcsc v cot v c sc2v cot vv 2csc v cot vcsc v cot v (1 cot 2 v) cot 2 vcot v7. csc v cot v csc v 1csc v cot v csc v 1csc v cot v csc v cot v8. sin v tan v sec v cos v1sin v tan v co s v cos v1 os2vsin v tan v cosin v tan v 1 co Pages 434–436 Exercisess v c cosvcos 2 vs vin2vosvinvosvsin v tan v s cssin v tan v sin v csin v tan v sin v tan v9. (sin A cos A) 2 1 2 sin 2 A cot Asin 2 A 2 sin A cos A cos 2 A 1 2 sin 2 A cot A1 2 sin A cos A 1 2 sin 2 A cot A1 2 sin A cos A s inAsinA 1 2 sin 2 A cot A1 2 sin 2 A c osAsinA 1 2 sin 2 A cot A1 2 sin 2 A cot A 1 2 sin 2 A cot A10. Sample answer: sin x 1 4 tan x 1 4 sec x t anxsec x 1 4 sinxc osx— 11 4 co s xsin x 1 4 11. Sample answer: cos x 1cot x sin x cos x cot x c osxsinx sin x cos x c scos x sin 2 x cos 2 xcos 2 x sin 2 x cos x1 cos xcos x 112. I c oR s v2 I2cotv R cscv I c os vR2I c osvsinv—R 2 1si n v I c os v I c osvsinvR2 s in— vsinvR 2 1si n v I c os vR 2 I c os vR 2osxin x13. tan A s ctan A coecAsc A1s A—1sin AinAosAstan A ctan A tan A14. cos v sin v cot vcos v sin v c osvsinvcos v cos vsin x15. sec x tan x 1 cos x1 sinsec x tan x co s x xc osxsec x tan x sec x tan x tanx sec x116. si n x cosxsin1 xc osx c osx cosxsin x cos x sec xsin1 xc osx sec xsin x cos x sec xcos x sin xcos x(sin x cos x)1 co s x sec xsec x sec x17. sec x csc x tan x cot xsinx osxsec x csc x csec x csc x cssec x csc x cossisec x csc x sin c1sec x csc x cos x sin x1 1sec x csc x co s x sin xsec x csc x sec x csc x2 sin 18. sin v cos v 2 s vosx c sinxin xosx s inxsinx c ossin xx c osc xosxn 2 cox sixn x sin xs 2 xcos x2 x cos2 xos x sinxin v1cosvsin v cos v 2 sin2 v (sin 2 v cos 2 v)sin v cos v s ssin v cos v sin v cos vin 2 v cos2vin v cosv(sin v cos v)(sin v cos v)sin v cos vsin v cos v sin v cos v19. (sin A cos A) 2 2 sec A csc Asec A csc A(sin A cos A) 2 2 sec A csc A s ecA cscAsecA cscA(sin A cos A) 2 1 2 se c 1A cs c A 1(sin A cos A) 2 2 cos A sin A 1(sin A cos A) 2 2 cos A sin A sin 2 A cos 2 A(sin A cos A) 2 (sin A cos A) 2Chapter 7 210
20. (sin v 1)(tan v sec v) cos vsin v tan v tan v sin v sec v sec v cos vsinv sinv1 1sin v c c sin v co co cos vosvosvcos y21. 1 sin y 1 sin ycos ycos y 1 sin y 1 siny1 siny 1 sin ycos y cos y(11 s siny)in2y 1 sin ycos y cos y (1 sin y)cos2 y 1 sin ycos y 1 sicos n yy 1 sin ycos ys vs vsin 2 v sin v sin v 1cos v sin 2 v 1cosv22. cos v cos (v) sin v sin (v) 1cos v cos v sin v(sin v) 1cos 2 v sin 2 v 11 1cox23. csc x 1 cscsc x 1 c ccsc x 1 csc x 1 csc x 124. cos B cot B csc B sin B1cos B cot B sin B sin B1 in2Bcos B cot B sincos B cot B 1 sicos B cot B c st 2c x 1sc2x 1scx 1(csc x 1)(csc x 1)csc x 1B s sinBsin2 Bn Bos2BinBosBin Bcos B cot B cos B c scos B cot B cos B cot B25. sin v cos v tan v cos 2 v 1sinvsin v cos v c cos 2 v 1osvsin 2 v cos 2 v 11 126. (csc x cot x) 2 1 1 cos v cos v cos 2 vcosv cosx cosx cosx cosx1 1 sin 2 x 2 sin x c ossin xx c os2xs in2x 1 cosx1 cosx1 2 cos x cos 2 x 1 cosx sin 2 x1 cosx (1 co1 c s )os xx22 1 cosx1 cosx(1 cos x) 1 cosx2(1 cos x)(1 cos x) 1 cosx 1 cosx1 cosx 1 cosx1 cosxcsc 2 x 2 csc x cot x cot 2 x 1 1 cos vcos v cos v27. sin x cos x 1 cos x sin xtan x 1 cot xcos xsin x cos x sin xsin1 c osx1 xs c osxin xcos xsin x cos x c osx sin x cosxsin1 c osx1 xs c osxin xcoxsin2xsin x cos x cossinsin x cos x sincsin x cos x s ssin x cos x s 2x sin x x cos xos 2 xsin2xx cos xsin x cos xin 2 x cos2xin x cosx(sin x cos x)(sin x cos x)sin x cos x s inxsinxsin x cos x sin x cos x28. sin v cos v tan v sin v sec v cos v tan vsinvsin v cos v csin v sec v cos v tan vosvin2vsin v cos v s cosv sec v cos v tan vsin v c os2cos vv s in2vc osv sec v cos v tan vsin v cos2 v sin 2 vcos v sec v cos v tan v1sin v co s v sec v cos v tan vsin v sec v sec v cos v tan vsin v c osvcosv sec v sec v cos v tan vsinvcos v c osv sec v sec v cos v tan vcos v tan v sec v sec v cos v tan vsec v cos v tan v sec v cos v tan v29. Sample answer: sec x 2 c scxcot x1si n v— c osvs in v 2 21 co s x 2sec x 230. Sample answer: tan x 2 1 tanx1 cot xinxosxs1 c——1 c osxs in x 2 cos x sin xcos x—— sin x cos xsi n xsinx c osx 2 2 2tan x 231. Sample answer: cos x 01 co t x s ecxcscx1co s x—1sin xin xosxtan x cos x cos xstan x c cos xtan x tan x cos x0 cos x211 Chapter 7
32. Sample answer: sin x 1 2 1 cos x sin xsin x 1 cos x1 2 cos x cos 2 x sin 2 xsin x(1 cos x) sin x( 1 cos x)1 2 cos x cos 2 x sin 2 xsin x(1 cos x)2 2 cosx sin x(1 cosx)2( 1 cosx) sin x(1 cosx) 4 4 4 4 42 sin x 42 4 sin x 1 2 sin x33. Sample answer: sin x 1cos 2 x 2 sin x 2 01 sin 2 x 2 sin x 2 00 sin 2 x 2 sin x 10 (sin x 1) 20 sin x 1sin x 134. Sample answer: cot x 1csc x sin x tan x cos xsinxcsc x sin x c osx cos xcsc x s in2cos xx c os2xc osx1csc x co s x1 1 sin x co s x c osxsinx 1cot x 135. t an3v 1tanv 1(tan v 1)(tan 2 v tan v 1)tan v 11cot v ta n vcot v 1 1 cot v 136. no37. yes sec 2 v 1 0 (tan 2 v 1) 1 0tan 2 v tan v 1 tan 2 v 1 1 0tan v 1 0tan v 1[2 , 2 ] sc1 2 by [2, 8] sc11[2 , 2 ] sc1 2 by [4, 4] sc1138. yes39. no[2, 2] sc1 2 by [4, 4] sc11[2, 2] sc1 2 by [4, 4] sc1140a. P I 2 0 R sin 2 2 ftP I 2 0 R(1 cos 2 2pft)40b. P I 2 0 R sin 2 2ftI RP cs41. f(x) 0 2c 2 2ftx1 x 4 2 1 2 tan vf(x) 141 2 tanv2 1 2 tan vf(x) 1 n ta 2 v 1 2 tan vf(x) sec 2 v 1 2 tan vf(x) sec vf(x) f(x) 1 2 sin vina42. sin a sin a sin c ⇒ sin a s sinccosbcos b s in a⇒ cos b sin a cos bcosccos c cos a cos b ⇒ cos b cThen cos b sin a cos b43. y 2v0 1 2 sin vc osv s ina coscsinc c osasina c osa c oscsinc tan a cot cgv22 cos 2 x sinvv cosvgv21co s vosay 2v 2 sec 2 v x tan v0gx2y 2(1 tan 2 v) x tan vv 20Chapter 7 212
44. We find the area of ABTP by subtracting the areaof OAP from the area of OBT. 1 2 OB BT 1 2 OA AP 1 2 1 tan v 1 2 cos v sin v 1 2 sinvcosv cos v sin v 1 2 sin v 1cos v cos v 1 2 sin v 1cos v c os2vcosv 1 2 sin v 1 cos2 vcosv 1 2 sin v s in2vcosv 1 2 sinv csin 2 vb45. By the Law of Sines, sinThenA 1 2 ab sin A 1 2 a a sinbsina sin A a2 sin sin 2 sinaaA 2 sin b sin 2 sin (180° (b ))A a 2 sinb sin2 sin( b ) 1 2 osvtan v sin 2 v,aso b a sinbsin.aab sinsinxsin c osx cos x sin x xtan x cos x sin x tan xc osx46. sec x tan x1 sin co s x xc osxsin x cos2 x sin 2 xcos x 1 sin xcos x sin x 1 cos xcosx 1 sin x 147. ⏐A⏐ 2 36 0°ck 180° 2 45°A 2 k 2 c 90°y 2sin (2x 90°)48. 1 51 6 1 51 6 18 0° 168.75°168.75° 168° 0.75° 6 01° 168° 45168° 4549. 3 3y 1 2 0 Check: 3 3y 1 2 0 3 3y 1 2 3 3(3) 1 2 03y 1 8 3 8 2 0y 32 2 0 50. x 1 0x 13xf(x) x 13xy x 1y(x 1) 3xyx y 3xy 3x yxy x(3 y)y 3 xy3 y 0y 351. Let x the number of shirts and y the numberof pants.yx 1.5y 100 1002.5x 2y 1802.5x 2y 18080 (0, 65)1.5x 3y 195x 1.5y 100x 060(40, 40)y 01.5x 3y 19540x 0(72, 0)20(0, 0)xO 20 40 60y 080 100P(x, y) 5x 4.5yP(0, 0) 5(0) 4.5(0) or 0P(0, 65) 5(0) 4.5(65) or 292.50P(40, 40) 5(40) 4.5(40) or 380P(72, 0) 5(72) 4.5(0) or 36040 shirts, 40 pants52. {16}, {4, 4}; no, 16 is paired with two elements ofthe range53. a ba b b ab a a ba b b ab a a b a ba b 1 (a b)1The correct choice is D.7-3Sum and Difference IdentitiesPages 441–442 Check for Understanding1. Find a counterexample, such as x 30° andy 60°.2. Find the cosine, sine, or tangent, respectively, ofthe sum or difference, then take the reciprocal.3. The opposite side for 90° A is the adjacent sidefor A, so the right-triangle ratio for sin (90° A) isthe same as that for cos A.90˚ A14. cot (a b) tan (a b)1 Atana tanb1 tana tanb 1 tana tanbtana tanb11 co t 1a co t b1 co t 1a co t b c ota cotb 1cota cot b c ota cotbcota cotb213 Chapter 7
5. cos 165° cos (45° 120°)10. sin (90° A) cos A cos 45° cos 120° sin 45° sin 120°sin 90° cos A cos 90° sin A cos A2 1 2 2 31 cos A 0 sin A cos A2 2 2cos A cos A2 611. tan v 2 4 cot v6. tan 1 2 tan 3 4 sin v 2 tan 3 tan —— cot v4 cos v ——2 1 tan 3 tan 4 sin v cos 2 cos v sin 2 3 1——— cot v1 3 1cos v cos 2 sin v sin 2 cot v4 23 2 2 37. 795° 2(360°) 75°sec 795° sec 75°cos 75° cos (30° 45°) cos 30° cos 45° sin 30° sin 45°3 2 1 2 2 2 26 2 44sec 795° 6 2 6 28. cos x 1 n si2 x 2cos y 1 n si2 y 21 1 4 21 4 9 6 581 65or 1 5 91 6orsin (x y) sin x cos y cos x sin y19. csc x sin x 5 3 sin1x 4 9 15 65 1 4 4 415 65 36cos x 1 n si2 x 2 1 3 5sin x 3 5 1 62 5or 4 5 sintan x xcosx 3 5 4 5 or 3 4 sin y 1 s co2 y5 2 1 1 3 1 4416 9or 1 213tanx tan (x y) 1tany tanx tany 3 4 1 25——1 3 4 1 5 6 320 — 4 5 6 316 2 9tan y csinyosy 1 213—51 3 or 1 515 42(sin v) 0 (cos v) 1(cos v) 0 (sin v) 1 c osvsinv12. sin (x y) 1 cot vcot v cot vcotx tan ycscx secyosx sinin x yc osy1 1 sin x co s yosx sinin x yc osy s inx cosy1 1 sinx cosy sin x co s ysin x cos y cos x sin y1Exercises2 2 2 2 21 c ssin (x y) ——1 c ssin (x y) ——sin (x y) sin (x y) sin (x y)13. sin (nq 0 t 90°) sin nq 0 t cos 90° cos nq 0 t sin 90° sin nq 0 t 0 cos nq 0 t 1cos nq 0 tPages 442–44514. cos 105° cos (45° 60°) cos 45° cos 60° sin 45° sin 60° 1 2 32 6 415. sin 165° sin (120° 45°) sin 120° cos 45° cos 120° sin 45°3 2 1 2 2 2 26 2 42 2 2 216. cos 7 12 cos 4 3 2 cos 4 cos 3 sin 4 sin 3 1 2 32 6 417. sin 12 sin 3 4 2 sin 3 cos 4 cos 3 sin 4 3 2 1 2 2 2 26 2 4 2Chapter 7 214
18. tan 195° tan (45° 150°)tan 45° tan 150°1 tan 45° tan 150° 1 33 1 1 33 3 33—3 3312 63 619. cos 12 cos 4 3 or 2 3 cos 4 cos 3 sin 4 sin p 3 2 1 2 2 3 22 6 4 220. tan 165° tan (45° 120°)3 2tan 45° tan 120° 1 tan 45° tan 120°1 (3)1 1 (3)1 31 3 4 232 or 2 3 21. tan 2 12 tan 4 5 3tan 4 tan 5 3——1 tan 4 tan 5 31 (3)1 1 (3) 4 232 or 2 322. 735° 2(360°) 15°sin 735° sin 15°sin 15° sin (45° 30°) sin 45° cos 30° cos 45° sin 30° 22 32 22 1 2 6 2423. 1275° 3(360°) 195°sec 1275° sec 195°cos 195° cos (150° 45°) cos 150° cos 45° sin 150° sin 45° 32 22 1 2 22 6 244sec 1275° 6 224. sin 5 1 2 62 sin 6 4 sin 6 cos 4 cos 6 sin 4 1 2 22 32 22 2 64csc 5 42 2 6 6 225. 11 312cot 11 312tan 1 7127272 4(2) 1 1 cot 1 1 tan 7 6 4 tan 6 tan 4 ——1 tan p 6 tan 4 33 1——1 33 13 33 3 2cot 11 3112 3 2 2 326. sin x 1 s co2 xcos y 1 n si2 y 8 7 21 1 2 3 7 2 1 1 2 25289or 1 517 1 2213 569or 3 3sin (x y) sin x cos y cos x sin y 1 517 3 5 8 237 6 2162927. sin x 1 s co2 x1 3 5 1 27 17 1 357 sin y 1 s co2 y 21 4 5 265 or 4 5 925cos (x y) cos x cos y sin x sin y 3 5 4 5 4 5 3 5 2 425 28. cos x 1 n si2 x or 3 5 sin y 1 s co2 y 8 7 21 3 5 21 1 2 25289or 1 517 1 625 or 4 5 sintan x xsin ctan y ycosx81 7— 1 517—3 33osy 4 5 — 3 5 8 1 5 4 3 tanx tanytan (x y) 1 tanx tany815 4 3 81 1 4 3 5 1 215— 7 7467 3 75 215 Chapter 7
29. sec x tan 2 x 1 5 3 2 cos y 1 n si2 y 1 1 1 3 2 3 49or 3433cos x 34 or 3 3434sintan x xc osx 5 3 —sin x3 3434sin x 5 3434cos (x y) cos x cos y sin x sin y 3 3434 2 23 5 3434 1 3 6 610 82 5 3410 2 1217 534 102 8 9 or 2 31130. tan x co t xcos y se1 — —1 5 6 6 5 tan (x y) 1ta 3 2 2 3 c y2sin y 1 s co2 y 21 2 3 5 9 or 53sintan y yc osy 53— 2 3 n x tany tanx tany 5 6 525 ——1 5 6 2 10 6512—— 12 5512 1 0 6512 55 270 122519 or 25131. sin x cs c x1—sec y tan 2 y 1 1 1 52 2 5 3 3 5 1 6925or 1 35cos x 1 n si 2 1xcos y se c y1 3 5 21— 1 35 1 625or 4 5 153sin y 1 s co2 y 5 3 21 1 1 4416 9or 1 1cos(x y) cos x cos y sin x sin y5 4 5 13 3 5 1 1 5 665 1sec (x y) cos (x y)1 — 5 655 6 556 32. cos a 1 n si2 a23 23sin b 1 s co2 b 21 2 7 21 1 5 2 425or 2 65 4 549 or 3 57sin (a b) sin a cos b cos a sin b 1 5 2 7 2 65 3 57 30 2 6 3533. sin x 1 s co2 xsin y 1 s co2 y 21 3 4 21 1 3 8 9 or 2 23 716or 74cos (x y) cos x cos y sin x sin y 1 3 3 4 2 23 74 14 3 2 1234. cos 2 x sin xcos 2 cos x sin 2 sin x sin x0 cos x 1 sin x sin xsin x sin x35. cos (60° A) sin (30° A)cos 60° cos A sin 60° sin A sin 30° cos A cos 30° sin A 1 2 cos A 32sin A 1 2 cos A 32sin A36. sin (A ) sin Asin A cos cos A sin sin A(sin A)(1) (cos A)(0) sin Asin A sin AChapter 7 216
37. cos (180° x) cos xcos 180° cos x sin 180° sin x cos x1 cos x 0 sin x cos xcos x cos x38. tan (x 45°) 1 tanx1 tanxtan x tan 45° 1 tanx 1 tan x tan 45° 1 tanxtan x 1 1 (tanx) (1) 1 tanx1 tanx 1 tanx1 tanx 1 tanx1 tanxn A tanBec A secBsinA sin c osA+ Bc osBsin (A B) ——1 co s 1A co s BsinA sin c osA+ Bc osBsin (A B) —— c osA cosB1cosA cosB co s 1A co s B39. sin (A B) ta ssin A cos B cos A sin B1sin (A B) sin (A B) sin (A B)40. cos (A B) 1 tan A tan BsecA sec BsinA sin1 c osA Bc osBcos (A B) ——s1 ccos (A B) ——cos (A B) cos (A B) cos (A B)secA sec B41. sec (A B) 1 1 co s 1A co s BinAsinosA Bc osB c osA cosB1cosA cosB co s 1A co s Bcos A cos B sin A sin B1tan A tan B1 co s 1A co s Bsec (A B) ——sinA sin1 c BcosA osB1 co s 1A co s Bsec (A B) ——sinA sin1 c Bcsec (A B) osAosB1cos A cos B sin A sin B c osA cosBcosA cosB1sec (A B) cos (A B)sec (A B) sec (A B)42. sin (x y) sin (x y) sin 2 x sin 2 y(sin x cos y cos x sin y)(sin x cos y cos x sin y) sin 2 x sin 2 y(sin x cos y) 2 (cos x sin y) 2 sin 2 x sin 2 ysin 2 x cos 2 y cos 2 x sin 2 y sin 2 x sin 2 ysin 2 x cos 2 y sin 2 x sin 2 y sin 2 x sin 2 y cos 2 x sin 2 y sin 2 x sin 2 ysin 2 x(cos 2 y sin 2 y) sin 2 y(sin 2 x cos 2 x) sin 2 x sin 2 y(sin 2 x)(1) (sin 2 y)(1) sin 2 x sin 2 ysin 2 x sin 2 y sin 2 x sin 2 y43. V L I 0 qL cos qt 2 V L I 0 qLcos qt cos 2 sin qt sin 2 V L I 0 qL(cos qt 0 sin qt 1)V L I 0 qL(sin qt)V L I 0 qL sin qt44. n n n sin 1 2 (a 60°)sin 6 0° 2sin a 2 30° sin 30°sin a 2 cos 30° cos a 2 sin 30° 1 2 n 2sin a 2 32 cos a 2 1 2 n 3 sin a 2 cos a 2 45. The given expression is the expanded form of thesine of the difference of 3 A and 3 A. We havesin 3 A 3 A sin (2A)sin 2A46a. f(x h ) f(x)h46b.n sin 1 2 (a b)sin b 2 y10.5O0.5146c. cos xs47. tan (a b) c12tan (a b) sin (x h) sin x hsin x cos h cos x sin h sin xhsin x cos 0.1 cos x sin 0.1 sin xy 0.134 5 6 7 8xin ( a b)os( a b)sin a cos b cos a sin bcos a cos b sin a sin bsina cosb cosa sinb c osa cosb c osa cosb c osa cosb sina sinbcosa cosb c osa cosbn a tanb tana tanbtan (a b) ———tatan (a b) 1Replace b with b to find tan(a b).tan a tan (b)tan (a (b)) tatan (a b) 148a. Answers will vary.1 tan a tan (b)n a tanb tana tanb217 Chapter 7
48b. tan A tan B tan C tan A tan B tan Ctan A tan B tan (180° (A B)) tan A tan B tan(180° (A B))tan A tan B tan A tan B tan 180° tan (A B)1 tan 180° tan (A B)tan 180° tan (A B)1 tan 180° tan (A B)0 tan (A B)1 0 tan (A B)0 tan (A B)1 0 tan (A B) tan A tan B tan A tan Btan A tan B tan (A B) tan A tan B tan (A B)(tan A tan B) 1 tanA tanB1 tanA tanB tan (A B) tan A tan B (A B)tan (A B)(1 tan A tan B) tan (A B) tan A tan B (A B)(1 tan A tan B 1) tan (A B) tan A tan B (A B)tan A tan B tan (A B) tan A tan B (A B) cos2x49. sec 2 x 1 1 sin2x csc 2 x cot 2 xsec 2 x 1 cos2 xcos2x 1 cot 2 x cot 2 xsec 2 1x cos 2 x c os2xcos2x 1sec 2 x sec 2 x 1 1sec 2 x sec 2 x50. sin 2 v cos 2 sinv 1 tan v vc 1 8 2 cos 2 v 1cos 2 v 6 364 cos v 3 78Quadrant III, so 3 7851. Arctan 3 3 sin (Arctan 3) sin 3 3252. k, where k is an integer53. A 86 502 2 4 2 h 86 2A 18 68 68y 18 sin 2 t c 6850 18 sin 2 1 c 68 c c18 18 sin 2 1 sin 2 sin 1 ( 1) 2 c 3 2 2 c cy 18sin 2 t 6854. ⏐8⏐ 8; 36 01 360; 3 0° 1 30°55. sin (540°) sin (360° 180°) 0osv 1 8 — 3 781 3 7 7215056. s rv A 1 2 r 2 v18 r(2.9) A 1 2 (6.2) 2 (2.9)6.2 r; 6.2 ft A 55.7 ft 257. c 2 70 2 130 2 2(70)(130) cos 130°c 2 33498.7345c 183 miles58. 120° 90°, consider Case 2.4 12, 0 solutions59.35 ft37˚12′ x6˚40′v 37° 12 6° 40 or 30° 32a 90° 6° 40 or 96° 40b 180° (30° 32 96° 40) or 52° 4835x sin 3 0°32 sin 5 2° 48x 35 sin52°48sin30°32x 54.87 ft60. 4x 3 3x 2 x 0x(4x 2 3x 1) 0x(4x 1)(x 1) 0x 0 or 4x 1 0 or x 1 0x 1 4 x 161. Case 1 Case 2⏐x 1⏐ 4 ⏐x 1⏐ 4(x 1) 4 x 1 4x 1 4 x 3x 5x 5 {x⏐x 5 or x 3}1 262. 1(6) 3(2)3 6 6 6 or 1263. fg(4) f(g(4)) f(5(4) 1) f(21) 3(21) 2 4 1319gf(4) g(f(4)) g(3(4) 2 4) g(44) 5(44) 1 22164. (8) 62 8 62 ( 8)6286 2 88The correct choice is A. 62 (1) 62 1Chapter 7 218
Page 445 Mid-Chapter Quiz11. csc v si n v1 — 2 7 7 2 Quadrant 1, so 3 522. tan 2 v 1 sec 2 1v cos v se 4 3 2 1 sec 2 v 1 69 1 sec 2 v 2 59 sec 2 v 5 3 sec vQuadrant II, so 5 3 3. 19 4 5 4 cos 5 4 cos 19 4cos 4 114. 1 t an 2 x 1 c ot 2 x 1 1cos 2 x sin 2 x 11 11 1 sec 2 x csc 2 x5. csc2 v sec 2 vsec 2 vcsc2v s ec2v s ec2vsec2v1sin 2 v—1cos 2 v c os2vsin2v csc 2 v csc 2 v 1 csc 2 v 1 csc 2 vc v 3 5 cot 2 v 1 csc 2 vcsc 2 v csc 2 v6. cot x sec x sin x 2 tan x cos x csc xosx 1sinx1 co sin x 2 c cos x sin c sinxs x1 2 11 1 cota tanbot a tanbosx7. tan (a b) 1 c11 ta n a tan btan (a b) —— tan b ta1n a1 cot 2 v csc 2 v1 cot 2 v 7 2 21 cot 2 v 4 94cot 2 v 4 54cot v 3 521 5 3 x11 ta n a tan btan (a b) —— t ana1 tanata n a tan btana tanbtan (a b) 1 tana tanbtan (a b) tan(a b)8. cos 75° cos(30° 45°) cos 30° cos 45° sin 30° sin 45° 32 22 1 2 22 6 249. cos x 1 n si2 xcos y 1 n si2 y 21 3 4 21 2 3 5 9 or 53 716 or 74cos (x y) cos x cos y sin x sin y 53 74 2 3 3 4 3 5 61210. tan x 5 4 tan y sec 2 y 1tan (x y) 1ta7-3Bn x tany tanx tany 5 4 31 5 4 35 4 344 5 345 43 4 53 80 413 59or 80 41359 Reduction Identities 22 1 3Page 4471. sin, cos, sin 2. cot, tan, cot3. tan, cot, tan 4. csc, sec, csc5. sec, csc, sec6a. (1) cos, sin, cos(2) sin, cos, sin(3) cot, tan, cot(4) tan, cot, tan(5) csc, sec, csc(6) sec, csc, sec6b. Sample answer: If a row for sin a were placedabove Exercises 1-5, the entries for Exercise 6acould be obtained by interchanging the first andthird columns and leaving the middle columnalone.7a. (1) cos, sin, cos(2) sin, cos, sin(3) cot, tan, cot(4) tan, cot, tan(5) csc, sec, csc(6) sec, csc, sec7b. Sample answer: The entries in the rows for cos aand sec a are unchanged. All other entries aremultiplied by 1.8a. Sample answer: They can be used to reducetrigonometric functions of large positive ornegative angles to those of angles in the firstquadrant.8b. Sample answer: sum or difference identities219 Chapter 7
7-4Double-Angle and Half-AngleIdentities8. sin 2 v cos 2 v 1 tan v cs 2 5 2 cos 2 v 1Page 453 Check for Understanding 2151. If you are only given the value of cos v, thencos 2v 2 cos 2 cos 2 v 2 125 2 or 2 21v 1 is the best identity to use.21 21If you are only given the value of sin v, thencos v 21cos 2v 1 2 sin 2 5(Quadrant I)v is the best identity to use. Ifsin 2v 2 sin v cos vyou are given the values of both cos v and sin v,then cos 2v cos 2 v sin 2 v is just as good as the 2 2 5 21 5 other two.2. cos 2 1 2 sin 2 4 2125vcos 2v 1 2 sin 2 vcos 2v cos 2 v sin 2 v cos 2v 12 sin 2 v 21 25 2 5 2 1 c os 2v2 sin 2 v 1 725 1 c o s2v2 tanv2 sin vtan 2v 1 tanLetting v a 2 yields sin a 2 2 v1 cos 2 2 2 ,2 21 21 2 ——or sin a 2 1 c os 2 .1 2 21221 3a. III or IV 3b. I or II 3c. I, II, III or IV4 2121—4. sin 2v 2 sin v or 4 21sin 2 2 2 sin 1 7 172 2 1 sin 2 sin 9. tan 2 v 1 sec 2 v sin 2 v cos 2 v 12 4 3 2 1 sec 2 v sin 2 v 3 5 2 10 2(1)0 2 2 5Sample answer: v 9 sec 2 v sin 2 v 1 6252 5 3 sec v (Quadrant III) sin v 4 5 5. Both answers are correct. She obtained two1cos v different representations of the same number. Onese c v(Quadrant III)way to verify this is to evaluate each expression1 or 3with a calculator. To verify it algebraically, square 5 5 3each answer and then simplify. The same result is sin 2v 2 sin v cos vobtained in each case. Since each of the original 2 4answers is positive, and they have the same5 3 5 square, the original answers are the same 2 425 number.cos 2v cos 2 v sin 2 v6. sin 8 4 3 sin 5 2 4 5 22721 cos 54 2 tanv (Quadrant I)tan 2v 2 11 22 2 2 2 27. tan 165° tan 33 0°2 1 c os330° 1 cos330°1 33—1 33(2 3) 3 2(Quadrant II) tan2 v2 4 3 —1 4 3 2 8 3 — 7 9 or 2 7210. tan 2v cot v tan v2tan 2v cot v tan 2v 2tan 2v 1tan 2v tan 2v4tan v t anvtanv2 tan vcot v tan v tan 2 vtanv tan2 vinvosv 2 5 —Chapter 7 220
11. 1 1 2 sin 2A sec A se sin Ac A co1s A sin A—— co1s A 16. tan 5 12 tan1 1 2 sin 2A 1 cos 1 co1 1 s A sin A 5 6——1 cos 5 62 ——sin 2A c os1 A cosAco s A 1 1 2 sin 2A 1 sin A cos A1 32 1 1 2 sin 2A 1 1 1 32 2 2 sin A cos A1 1 2 sin 2A 1 1 2 sin 2Ax x12. sin 2 cos 2 sin x 2 32—2 322x x2 sin 2 cos 2—— sin x2 2sin 2 x2 sin —x(2 3) 2 4 322 2 3 sin x2 sin x213. cos 2v 2 cos 2 17. sin 3 3 48 sin 2v 1cos 2v 1 2 cos 2 v1 cos 1 2 cos 2v 1 2 cos 2 vP I 2 0 3 4——2R sin 2 qtP I 2 0 R (1 cos 2 qt)P I 2 0 1 22— 2R1 1 2 cos 2qt 1 2 P I 2 0R 1 2 1 2 cos 2qt 2 22P 1 2 I 2 0 R 1 2 I 2 0 R cos 2qt18. cos 7 7 12 6 cos 21 cos 7 π6Pages 454–455 Exercises——14. cos 15° cos 3 0° 22 1 c os 30°2 (Quadrant I) 2 1 32— 2 3 2219. tan 22.5° tan 4 5° 2 2 32 1 c os45°15. sin 75° sin 15 1 cos45°0°2 1 cos150 °2(Quadrant I) 1 32 1 22—1 22 22 —2 2 3 2 5 6 2 (2 3)(2 3)(2 3)(2 3) 1 32 2 22 2(2 2)(2 2)(2 2)(2 2)(2 2)2 4 2 2 2 2 22 22 2 1 (Quadrant I)(Quadrant I)(Quadrant II) (Quadrant I)221 Chapter 7
20. tan 2v 1 1 c osv cos v1 1 4 1 1 4 3 4 — 5 4 3 5 or 51521. sin 2 v cos 2 v 1sin 2 v 4 5 2 1sin 2 9v 2 5sin v 3 5 (Quadrant I)sin 2v 2 sin v cos v 2 3 5 4 5 2 425 cos 2v cos 2 v sin 2 v 4 5 2 3 5 2tan 2v 127 2 5tanv tan2 v2 3 4 1 3 4 2 3 2 —71 6 or 2 7422. sin 2 v cos 2 v 1 1 3 2 cos 2 v 1cos 2 v 8 9 cos v 2 23(Quadrant I)sin 2v 2 sin v cos v2 2 1 3 2 3 4 29cos 2v cos 2 v sin 2 v 2 23 2 1 3 2 7 9 tan 2v 12tanv tan2 v2 24 1 24 2 22 or 4 2— 7 141 6tan v csinvosv 3 5 — 4 5 3 4 tan v csinvosv 1 3 —2 231 or 24 2223. tan 2 v 1 sec 2 v(2) 2 1 sec 2 v5 sec 2 v5 sec v (Quadrant II)cos v sec v1 or 5 5 5sin 2 v cos 2 v 1sin 2 v 55 2 1sin 2 v 2 025 sin v 2 55(Quadrant II)sin 2v 2 sin v cos v5 5 2 2 5 5 4 5 cos 2v cos 2 v sin 2 v 55 2 2 55 2tan 2v 1224. cos v se1 3 5 tanv tan2 v2(2) 1 (2) 2 43or 4 3 c v1 4 3 3 4 sin 2 v cos 2 v 1sin 2 v 3 4 2 1sin 2 7v 1 6sin v 74sintan v vcosv 74— 3 4 73sin 2v 2 sin v cos v7 2 4 3 4 3 78cos 2v cos 2 v sin 2 v7 2 3 4 2 42 1 6or 1 8 2 tanvtan 2v 1 tan2 v2 73 7 21 32 73— 2 9 or 37(Quadrant II)Chapter 7 222
25. 1 cot 2 v csc 2 12 tanvv tan v co t vtan 2v 1 tan2 v1 3 2 2 csc 2 1v 3 2 2 2 2121 1 34 csc 2 v 2 3 1 2 21221 132 csc v (Quadrant III) 4 21211sin v cs c vsin 2 v cos 2 v 1——or 4 21 1 7172 1 1 2 1313 2 cos 2 v 1 1327. sin 2 a cos 2 sina 1 tan a a2c osa2sin 2 a 7 or 2 13cos 2 1v 17 131 6913 23 2 3 1 cos v 3 1313 23(Quadrant III)sin 2v 2 sin v cos vsin 2 a 7 2 2 1313 3 9 7 or 14 2 21313 sin a 7 1 23(Quadrant II)2 tana13tan 2a 1cos 2v cos 2 v sin 2 tan2 av 3 13 213 2 13 22 1413 2 5 1 141 32 22 tanv14 tan 2v or 2 141 tan2 v5 52 2 3 228. csc 2v 1 2 3 2 1 2 sec v csc v1 sin 2v 1 2 sec v csc v 4 3 1 2sinor 1 v cos v 1 2 sec v csc v2 5 59 1 2 1 1 si n v co s v 1 2 sec v csc v1 126. sin v 2 csc v sec v 1 2 sec v csc vcs c vsin 2 v cos 2 v 11 1 — 5 2 sec v csc v 1 2 sec v csc v 2 5 2 cos 2 v 12 cos2A29. cos A sin A cos 2 v 2 1cos A sin A 2 25 5 cos A sin A c os2 A sin2Acos v 215cosA sinA(cos A sin A)(cos A sin A)(Quadrant IV)cos A sin A cos A sin Asintan v vccos A sin A cos A sin Aosv 2 5 30. (sin v cos v) 2 1 sin 2v—sin 2 v 2 sin v cos v cos 2 v 1 sin 2v 212 sin v cos v 1 1 sin 2v52 sin v cos v sin 2v2 or 2 21 2121 sin 2v sin 2vcos2x 1sin 2v 2 sin v cos v31. cos x 1 2 ( cosx 1)cos x 1 2c os2 2 2 x 1 15 21 5 2(cosx 1)cos x 1 2 cos2x 2 4 21252( cosx 1)cos x 1 2 ( cos2cos 2v cos 2 v sin 2 vx 1)2(cosx 1) 21 2(cos x 1)(cos x 1)5 2 2 5 2cos x 1 2(cos x 1) 1 725 cos x 1 cos x 132. sec 2v c os2v sin2vcos2v sin2v1sec 2v cos 2vsec 2v sec 2v33. tan A 2 sin A 1 cos A223 Chapter 7
tan A 2 sin 2A 2 1 cos 2 A 2 tan A 2 2 sin A 2 cos A 2 1 2 cos 2 A 2 1tan A 2 2 sin A 2 cos A 2 2 cos 2 A 2 tan A 2 tan A 2 tan A 2 34. sin 3x 3 sin x 4 sin 3 xsin(2x x) 3 sin x 4 sin 3 xsin 2x cos x cos 2x sin x 3 sin x 4 sin 3 x2 sin x cos 2 x (1 2 sin 2 x) sin x 3 sin x 4 sin 3 x2sinx(1 sin 2 x) (1 2sin 2 x)sinx 3 sin x 4 sin 3 x2sinx 2sin 3 x sin x 2sin 3 x 3 sin x 4 sin 3 x3 sin x 4 sin 3 x 3 sin x 4 sin 3 x35. cos 3x 4 cos 3 x 3 cos xcos(2x x) 4 cos 3 x 3 cos x(2 cos 2 x 1)cos x 2 sin 2 x cos x 4 cos 3 x 3 cos x(2 cos 2 x 1)cos x 2(1 cos 2 x)cos x 4 cos 3 x 3 cos x2 cos 3 x cos x 2 cos x 2 cos 3 x 4 cos 3 x 3 cos x4 cos 3 x 3 cos x 4 cos 3 x 3 cos xv2 2 gsin 2 2v36. s 2sv 2 gsin 2 vsin A 2 cos A 2 in22vin2v (2 sin v cos v) 2sin2v 4sin 2 v cos 2 vsin2v 4 cos 2 v37. ∠PBD is an inscribed angle that subtends thesame arc as the central angle ∠POD, so m∠PBD 1 2 v. By right triangle trigonometry, tan 1 2 PA v BPA sin v 1 OA 1 cos v.38. R 2v2 cos v sin(v a)g cos 2 aR 2v2 cos v sin(v 45°)g cos 2 45°R 2v2 cos v(sin v cos 45° cos v sin 45°)g cos 2 45°2v 2 cos v(sin v) 22 (cos v) 22 R 2 2R g 22 22v 2 cos v(sin v cos v)g 1 2 2v 2 (2 cos v sin v 2 cos 2 v)gR R v2 2g(2 cos v sin v (2 cos 2 v 1) 1)R v2 2(sin 2v cos 2v 1)gA39a. tan45° 1 2 1 1 c osL1 cosL1 1 c osL1 cosL39b. tan 45° tan L 2 1 tan 45° tan L 2 1 tan L 2 1 1 tan L 2 1 1 c osL1 cosL1 1 c osL1 cosL1 1 c os60˚1 cos60˚1 1 c os60˚1 cos60˚ 1 1 12 1 1 2 1 1 12 1 1 2 1 1 3 1 1 3 3 33 3 3340. tan (a 30°) 2 17tan (a 30°) 3tan a tan 30°1 tan a tan 30° 12 6 36or 2 3 3tan a 33 3 3 tan atan a 3 tan a 3 33(1 3) tan a 3 3331 3tan a 3 341. cos 12 cos 3 4 tan a 9 33 33tan a 6 533 cos 3 cos 4 sin 3 sin 4 1 2 22 32 22 2 64sec 1 2 1 cos 1 2 1 2 64 42 464or 6 242. Sample answer:sin() 2 cos() 2 sin cos 0 (1)1 1Chapter 7 224
43. s rv 1 10 1 1017 10 v 97.4°7 v 1 1077 18 0°44. Let x the distance from A to the point beneaththe mountain peak.htan 21°10 570 xh (570 x) tan 21°10tan 36°40 h x h x tan 36°40(570 x) tan 21°10 x tan 36°40570 tan 21°10 x tan 36°40 x tan 21°10570 tan 21°10 x(tan 36°40 tan 21°10)570 tan 21°10tan 36°40 tan 21°10 x617.7646751 xtan 36°40 h x htan 36°40 61 7.8h 460 ft45. (x (3))(x 0.5)(x 6)(x 2) 0(x 3)(x 0.5)(x 6)(x 2) 0(x 2 2.5x 1.5)(x 2 8x 12) 0x 4 5.5x 3 9.5x 2 42x 18 02x 4 11x 3 19x 2 84x 36 046. y 2x 5yx 2y 5x 5 2y x 52 yy 2x 57-5Page 4581.2.Solving Trigonometric EquationsGraphing Calculator Exploration3. Exercise 1: (1.1071, 0.8944), (4.2487, 0.8944)Exercise 2: (5.2872, 0.5437), (0.9960, 0.5437)4. The x-coordinates are the solutions of theequations. Substitute the x-coordinates and seethat the two sides of the equation are equal.5.Oy x 52x47. x 2y 11 3x 5y 11x 11 2y 3(11 2y) 5y 1133 6y 5y 1111y 22y 2x 2y 11x 2(2) 11x 7 (7, 2)48. ab 3b a3 (a b) 2 64a 2 2ab b 2 643 2 64a 2 2a 3a aa 2 6 3a 2 64a 2 3a 2 70a 2 b 2 70The correct answer is 70.[0, 2] sc1 4 by [3, 3] sc115a. The x-intercepts of the graph are the solutions ofthe equation sin x 2 cos x. They are the same.5b. y tan 0.5x cos x or y cos x tan 0.5xPage 459 Check for Understanding1. A trigonometric identity is an equation that istrue for all values of the variable for which eachside of the equation is defined. A trigonometricequation that is not an identity is only true forcertain values of the variable.2. All trigonometric functions are periodic. Addingthe least common multiple of the periods of thefunctions that appear to any solution to theequation will always produce another solution.3. 45° 360x° and 135° 360x°, where x is anyinteger225 Chapter 7
4. Each type of equation may require adding,subtracting, multiplying, or dividing each side bythe same number. Quadratic and trigonometricequations can often be solved by factoring. Linearand quadratic equations do not require identities.All linear and quadratic equations can be solvedalgebraically, whereas some trigonometricequations require a graphing calculator. A linearequation has at most one solution. A quadraticequation has at most two solutions. Atrigonometric equation usually has infinitelymany solutions unless the values of the variableare restricted.5. 2 sin x 1 0 6. 2 cos x 3 02 sin x 1 2 cos x 3sin x 1 2 cos x 32x 30° x 30°7. sin x cot x 32sin x c osxsinx 32cos x 32x 30° or x 330°8. cos 2x sin 2 x 22 cos 2 x 1 (1 cos 2 x) 22 cos 2 x 1 cos 2 x 13 cos 2 x 0cos 2 x 0cos x 0x 90° or x 270°9. 3 tan 2 x 1 03 tan 2 x 1tan 2 x 1 3 tan x 33x 6 or x 5 6 or x 7 6 or x 11 610. 2 sin 2 x 5 sin x 32 sin 2 x 5 sin x 3 0(2 sin x 1)(sin x 3) 02 sin x 1 0 or sin x 3 0sin x 1 2 sin x 3x 7 6 or x 11 6no solutions11. sin 2 2x cos 2 x 01 cos 2 2x cos 2 x 01 (2 cos 2 x 1) 2 cos 2 x 01 (4 cos 4 x 4 cos 2 x 1) cos 2 x 04 cos 4 x 5 cos 2 x 0cos 2 x(4 cos 2 x 5) 0cos 2 x 0 or 4 cos 2 x 5 0cos x 0cos 2 x 5 4 x 2 k cos x 52no solutions12. tan 2 x 2 tan x 1 0(tan x 1)(tan x 1) 0tan x 1 0tan x 1x 3 4 k13. cos 2 x 3 cos x 2cos 2 x 3 cos x 2 0(cos x 1)(cos x 2) 0cos x 1 0 or cos x 2 0cos x 1cos x 2x (2k 1) no solutions14. sin 2x cos x 02 sin x cos x cos x 0cos x (2 sin x 1) 0cos x 0 or 2 sin x 1 0x 2 k sin x 1 2 15. 2 cos v 1 02 cos v 1cos v 1 2 cos v 1 2 at 2 3 and 4 3 2 3 v 4 316. W Fd cos v1500 100 20 cos v0.75 cos vv 41.41°x 6 2kor x 5 6 2kPages 459–461 Exercises17. 2 sin x 1 0 18. 2 cos x 1 02 sin x 12 cos x 11sin x 2cos x 1 2 sin x 22x 120°x 45°19. sin 2x 1 02 sin x cos x 1 0sin 2 x cos 2 x 1 4 sin 2 x (1 sin 2 x) 1 4 sin 2 x sin 4 x 1 4 0sin 4 x sin 2 x 1 4 0sin 2 x 1 2 sin 2 x 1 2 0sin 2 x 1 2 0sin 2 x 1 2 1sin x or 2 2 2x 45°Chapter 7 226
20. tan 2x 3 0tan 2x 32 tanx 1 tan2 x 32 tan x 3 (1 tan 2 x)2 tan x 3 3 tan 2 x3 tan 2 x 2 tan x 3 0(3 tan x 1)(tan x 3) 03 tan x 1 0 tan x 3 01tan x tan x 333tan x 3x 30°21. cos 2 x cos xcos 2 x cos x 0cos x(cos x 1) 0cos x 0 or cos x 1 0x 90° cos x 1x 0°22. sin x 1 cos 2 xsin x 1 1 sin 2 xsin 2 x sin x 2 0(sin x 1)(sin x 2) 0sin x 1 0 or sin x 2 0sin x 1x 90°23. 2 cos x 1 02 cos x 1cos x 22x 135° or x 225°24. cos x tan x 1 2 sinxcos x c 1 2 osxx 60°sin x 2no solutionsin x 1 2 x 30° or x 150°25. sin x tan x sin x 0sin x (tan x 1) 0sin x 0 or tan x 1 0x 0° or x 180° tan x 1x 45° or x 225°26. 2 cos 2 x 3 cos x 2 0(2 cos x 1)(cos x 2) 02 cos x 1 0 or cos x 2 02 cos x 1 cos x 2cos x 1 2 no solutionx 60° or x 300°27. sin 2x sin x2 sin x cos x –sin x2 sin x cos x sin x 0sin x (2 cos x 1) 0sin x 0 or 2 cos x 1 0x 0° or x 180°2 cos x 1cos x 1 2 x 120°or x 240°28. cos (x 45°) cos (x 45°) 2cos x cos 45° sin x sin 45° cos x cos 45° sin x sin 45° 2cos x 22 sin x 22 cos x 22 sin x 22 22 cos x 2cos x 1x 0°29. 2 sin v cos v 3 sin v 0sin v (2 cos v 3) 0sin v 0 or2 cos v + 3 0v 0° or v 180°2 cos v 3cos v 32v 150°or v 210°30. (2 sin x 1)(2 cos 2 x 1) 02 sin x 1 0 or 2 cos 2 x 1 02 sin x 1 2 cos 2 x 1sin x 1 2 cos 2 x 1 2 x 6 cos x 22or x 5 6x 4 or x 3 4or x 5 4 or x 7 431. 4 sin 2 x 1 4 sin x4 sin 2 x 4 sin x 1 0(2 sin x 1)(2 sin x 1) 02 sin x 1 02 sin x 1sin x 1 2 x 7 6 or x 11 632. 2 tan x 2 sin xsinx2 c osx 2 sin x2 2 cos x 22 cos xx 4 or x 7 42 tan x 2 sin x would also be true if both tan xsinxand sin x equal 0. Since tan x c os,xtan x equals0 when sin x 0. Therefore x can also equal 0 and.0, 4 , , 7 433. sin x cos 2x 1sin x 1 2 sin 2 x 12 sin 2 x sin x 0sin x(2 sin x 1) 0sin x 0 or 2 sin x 1 0x 0 or x 2 sin x 1sin x 1 2 x 7 6 orx 11 6227 Chapter 7
34. cot 2 x csc x 1csc 2 x 1 csc x 1csc 2 x csc x 2 0(csc x 2)(csc x 1) 0csc x 2 0 or csc x 1 0csc x 2 or csc x 1sin x 1 2 sin x 1x 6 or x 5 6x 3 235. sin x cos x 0sin x cos xsin 2 x cos 2 xsin 2 x cos 2 x 0sin 2 x 1 sin 2 x 02 sin 2 x 1 0sin 2 x 1 2 1sin x or 2 2 2sin x and cos x must be opposites, so x 3 4or x 7 .436. 1 3 sin v cos 2v1 3 sin v 1 2 sin 2 v2 sin 2 v 3 sin v 2 0(2 sin v 1)(sin v 2) 02 sin v 1 0 or sin v 2 02 sin v 1 sin v 2sin v 1 2 no solutionv 7 6 or v 11 637. sin x 1 2 141. cos x sec x 1 2 sin 2 x 1 2 0sin 2 x 1 2 sin x 22x 4 k 2k3 tan x 3tan x sin32 x 13sin x 1x 6 kx 2 k46. sec 2 x 2 sec x 0cos x 1 0 no solution 2k or 2kx 7 6 2 k or x 11 6 2k38. cos x tan x 2 cos 2 x 1sinxcos x c 2 cos 2 x 1osxsin x 2(1 sin 2 x) 12 sin 2 x sin x 1 0(2 sin x 1)(sin x 1) 02 sin x 1 0 or sin x 1 02 sin x 1 sin x 1sin x 1 2 x 3 2x 6 2k or x 5 6 2k39. 3 tan 2 x 3 tan x3 tan 2 x 3 tan x 0tan x(3 tan x 3) 0tan x 0 or 3 tan x 3 0x k40. 2(1 sin 2 x) 3 sin x2 2 sin 2 x 3 sin x2 sin 2 x 3 sin x 2 0(2 sin x 1)(sin x 2) 02 sin x 1 0 or sin x 2 02 sin x 1 sin x 2sin x 1 2 no solutionx 6 x 5 6Chapter 7 228sin x cos x sin x(cos x sin x)(cos x sin x) 1cos 2 x sin 2 x 1cos 2 x (1 cos 2 x) 12 cos 2 x 1 12 cos 2 x 2cos 2 x 1cos x 1x k42. 2 tan 2 x 3 sec x 02(sec 2 x 1) 3 sec x 0(2 sec x 1)(sec x 2) 02 sec 2 x 3 sec x 2 02 sec x 1 0 or sec x 2 02 sec x 1 sec x 2cos x 1 2 cos x 2 x 3 2k orno solution x 5 3 2k43. sin x cos x 1 2 sin 2 x cos 2 x 1 4 sin 2 x(1 sin 2 x) 1 4 sin 2 x sin 4 x) 1 4 sin 4 x sin 2 x 1 4 0sin 2 x 1 2 sin 2 x 1 2 044. cos 2 x sin 2 x 32cos 2 x (1 cos 2 x) 322 cos 2 x 1 322 cos 2 x 2 32cos 2 x 2 3423 cos x 2x 1 2 k or x 1 112 k45. sin 4 x 1 0(sin 2 x 1)(sin 2 x 1) 0sin 2 x 1 0 or sin 2 x 1 0sin 2 x 1no solutionssec x(sec x 2) 0sec x 0 or sec x 2 0sec x 2cos x 1 2 x 2 3 2k orx 4 3 2k
47. sin x cos x 1sin 2 x 2 sin x cos x cos 2 x 1sin 2 x 2 sin x cos x 1 sin 2 x 12 sin x cos x 0sin x cos x 0sin 2 x cos 2 x 0sin 2 x (1 sin 2 x) 0sin 2 x 0 or 1 sin 2 x 0sin x 0 sin 2 x 1x 2ksin x 1x 2 2k48. 2 sin x csc x 32 sin 2 x 1 3 sin x2 sin 2 x 3 sin x 1 0(2 sin x 1)(sin x 1) 02 sin x 1 0 or sin x 1 02 sin x 1 sin x 1sin x 1 2 x 2 2k49. cos v 32x 6 2k orx 5 6 2kcos v 32at 5 6 and 7 6 5 6 v 7 650. cos v 1 2 0cos v 1 2 cos v 1 2 at 3 and 5 30 v 3 or 5 3 v 251. 2 sin v 1 02 sin v 1sin v 22sin v 22at 4 and 3 40 v 4 or 3 4 v 252. 0.4636, 3.6052 53. 0, 1.895554. 0.3218, 3.463355. sin v D 10 7sin v 5.5 0.003sin v 0.0001833333333v 0.01°56. sin 2x sin x2 sin x cos x sin x2 sin x cos x – sin x 0sin x(2 cos x 1) 0The product on the left side of the inequality isequal to 0 when x is 0, 3 , , or 5 .3 For the productto be negative, one factor must be positive and theother negative. This occurs if 3 x or 5 3 x 2.57. R v 2gsin 2v20 1 529.8sin 2v0.8711111111 sin 2v2v 60.5880156 or 2v 119.4119844v 30.29° v 59.71°58a. n 1 sin i n 2 sin r1.00 sin 35° 2.42 sin rsin r 1.00 sin35°2.42sin r 0.2370150563r 13.71°58b. Measure the angles of incidence and refractionto determine the index of refraction. If the indexis 2.42, the diamond is genuine.59. D 0.5 sin (6.5 x) sin (2500t)0.01 0.5 sin (6.5(0.5)) sin (2500t)0.02 sin 3.25 sin 2500t0.1848511958 sin 2500t0.1859549654 2500tThe first positive angle with sine equivalent tosin (0.1859549654) is 0.1859549654 or3.326477773.t 3.32 64777732500t 0.0013 s60. a sin(bx c) d d a 2 a sin(bx c) a 2 sin(bx c) 1 2 The period of the function sin(bx c) is 36 0°b, so360°the given interval consists of b periods. 36 0°bThe equation sin (bx c) 1 2 has two solutionsper period, so the total number of solutions is 2b.x 3y 4cos v sin v 3 17 sin v cos v 4223 cos v 4 sin v 17 3 sin v 4 cos v223 cos v 4 sin v 173 sin v 4 cos v 22↓9 cos v 12 sin v 31716 cos v 12 sin v 8225 cos v 82 317cos v 82 317 25v 18.68020037360 v 341.32°61. 229 Chapter 7
62. cot 67.5° cot 13 5°12cot v tatan 13 5°223yO23 1cot 67.5 63. t asec xsinxc osx—1co s x2y 3 cos90˚1 –c os135° cos135°1 22 180˚270˚360˚n v (Quadrant 1)1 22 2 222 2 – 2(2 2)(2 2) (2 2)(2 2) (2 4 2) 2 2 2221 2 22 22 22(2 2)(2 2)(2 2)22 24 2 2 1n x 25 25sin x 25Sample answer: sin x 2564. A 2 3 , 267. 2 ⏐ 1 0 3 22 4 21 2 1 ⏐ 0x 2 2x 1 0(x 1)(x 1) 0x 1 0 x 1 0x 1x 1(x 2)(x 1(x 1)68.[5, 5] sc11 by [2, 8] sc11max: (1, 7), min: (1, 3)69. 3x 4 16 6 2yx 4 y 3 (4, 3)70. x y z 1 x y z 12x y 3z 5 x y z 113x 4z 6 2x 12x 63x 4z 6 x y z 113(6) 4z 6 6 y (3) 114z 12 y 2z 3(6, 2, 3)71.x g(x)g(x)7 45 23 0g(x) |x 3|1 21 4O72. A 1 2 bhA 1 2 (6)(1)A 3The correct choice is C.x65. 45 mihou lesr 52 80m i ftle 12 in ches 1 hourft 3 600secv r v t 792 7 1v 79 27 79 27radians 2 18 rps66. undefined 792 in/secChapter 7 230
Page 462 History of Mathematics1. x 2 5 2 5 2 2(5)(5) cos 10°x 0.87x 2 5 2 5 2 2(5)(5) cos 20°x 1.74x 2 5 2 5 2 2(5)(5) cos 30°x 2.59x 2 5 2 5 2 2(5)(5) cos 40°x 3.42x 2 5 2 5 2 2(5)(5) cos 50°x 4.23x 2 5 2 5 2 2(5)(5) cos 60°x 5x 2 5 2 5 2 2(5)(5) cos 70°x 5.74x 2 5 2 5 2 2(5)(5) cos 80°x 6.43x 2 5 2 5 2 2(5)(5) cos 90°x 7.07Angle Length ofMeasure Chord (cm)10° 0.8720° 1.7430° 2.5940° 3.4250° 4.2360° 5.0070° 5.7480° 6.4390° 7.074.Slope-Intercept Form: y mx b, displaysslope and y-interceptPoint-Slope Form: y y 1 m(x x 1 ),displays slope and a point on the lineStandard Form: Ax by C 0, displaysno informationNormal Form: x cos f y sin f p 0,displays length of the normal and the anglethe normal makes with the x-axisSee students’ work for sample problems.5. x cos f y sin f p 0x cos 30° y sin 30° 10 0 32x 1 2 y 10 03x y 20 06. x cos f y sin f p 0x cos 150° y sin 150° 3 0 32x 1 2 y 3 03x y 23 07. x cos f y sin f p 0x cos 7 4 y sin 7 4 52 0 22x 22 y 52 02x 2y 102 0x y 10 08. 4x 3y 10 A B 2 2 4 2 3 2 or 54 3 104x 3y 10 0 x 5 y 5 5 0 4 5 x 3 5 y 2 0sin f 3 5 , cos f 4 5 , p 2; Quadrant III7-6Normal Form of a Linear Equation 3 5 tan f or 3 4 4 5 fPage 467 Check for Understanding1. Normal means perpendicular2. Compute cos 30° and sin 30°. Use these as thecoefficients of x and y, respectively, in the normalform. The normal form is 32x 1 2 y 10 0.3. The statement is true. The given line is tangent tothe circle centered at the origin with radius p.9. y 3x 23x y 2 0AB 2 2 3 2 1 2 or 103 1 2x y 0 10 10 10 3 1010x y 0 110 10 5 0sin f , 0 110cos f 3 10, 10 p 10;5Quadrant Itan f f 18° 0 1—10or 1 3 3 1010231 Chapter 7
10. 2x 2y 62x 2y 6 0A B 2 2 2 2 (2) 2 or 2 22x 22y 6 2 0 22x 22y 3 0sin f 22, cos f 22, p 3; Quadrant IVtan f 22—or 1 22f 315°11a. 3x 4y 8y 3 4 x 211b. 3x 4y 83x 4y 8 0AB 2 2 34) 2 ( 2 or 5 3 5 x 4 5 y 8 5 0p 8 5 or 1.6 miles3x 4y 8yOx17. x cos f y sin f p 0x cos 210° y sin 210° 5 0 32x 1 2 y 5 03x y 10 018. x cos f y sin f p 0x cos 4 3 y sin 4 3 5 0 1 2 x 32y 5 0x 3y 10 019. x cos f y sin f p 0x cos 300° y sin 300° 3 2 0 1 2 x 32y 3 2 0x 3y 3 020. x cos f y sin f p 0x cos 11 6 y sin 11 6 43 0 32x 1 2 y 43 03x y 83 021. AB 2 2 52 2 1 2 or 135 13 2 65x 11y3 13 05 1x3 1 21y3 5 0sinf 1 251,3cos f 1, p 5; Quadrant III– 1 2132tan f — or 1 55–1 33Pages 467–469 Exercises12. x cos f y sin f p 0x cos 60° y sin 60° 15 0 1 2 x 32y 15 0x 3y 30 013. x cos f y sin v p 0x cos 4 y sin 4 12 0 22x 22y 12 02x 2y 24 014. x cos f y sin f p 0x cos 135° y sin 135° 32 0 22x 22y 32 02x 2y 62 0x y 6 015. x cos f y sin f p 0x cos 5 6 y sin 5 6 23 0 32x 1 2 y 23 03x y 43 016. x cos f y sin f p 0x cos 2 y sin 2 2 00x 1y 2 0y 2 0f 247°22. x y 1x y 1 0AB 2 2 1 2 1 2 or 21 1 1x y 0 2 2 2 22x 22y 22 0sin f 22, cos f 22, p 22; Quadrant I 22tan f — or 1 22f 45°23. 3x 4y 153x 4y 15 0AB 2 2 34) 2 ( 2 or 5 3 5 x 4 5 y 1 5 05 3 5 x 4 5 y 3 0sin f 4 5 , cos f 3 5 , p 3; Quadrant IVtan f 4 5 — 3 5 f 307°or 4 3 Chapter 7 232
24. y 2x – 42x y 4 0AB 2 2 (2) 2 1 2 or 52 1 4x 5 y 5 05 2 55x 55y 4 55 0sin f 55, cos f 2 55, p 4 55; Quadrant IV– 55tan f — or 1 2 2 55f 333°25. x 3x 3 0AB 2 2 1 2 0 2 or 1 1 1 x 3 1 0x 3 0sin f 0, sin f 1, p 3tan f 0 1 or 0f 0°26. 3x y 23x y 2 0AB 2 2 (3) 1) 2 ( 2 or 2 32x 1 2 y 2 2 0 32x 1 2 y 1 0sin f 1 2 , cos f 32, p 1; Quadrant III– 1 2 tan f — or 33 3– 2f 210°27. y 2 1 4 (x 20)y 2 1 4 x 5x 4y 28 0AB 2 2 (1) 2 4 2 or 171 4 28x y 0 17 17 17 x 7 117 4 17y 17 28 17 17 0sin f 4 1717, cos f 7 11,7p 28 17 17; Quadrant II4 1717tan f — or 4– 7 117f 104°28. 3x y 4x 3 y 4 0x 3y 12 0AB 2 2 13) 2 ( 2 or 101 3 12x 10 y 10 010 x 0 110 3 10y 10 6 10 5 0sin f 3 1010, cos f , 0 110p 6 10 ; 5 Quadrant II3 1010tan f — or 3– 0 110f 108°29. 2x0 2y4 1x 2 0 y 2 4 1 06x 5y 124 0AB 2 2 6 2 5 2 or 616 5 124x y 0 61 61 1 6 6 6161x 5 6161y 120 6161 0sin f 5 6161, cos f 6 6161, p 120 61;61Quadrant I5 6161tan f —or 5 6 6 6161f 40°30. AB 2 2 6 2 8 2 or 10; p 106cos f 1 0or 3 5 8, sin f 1 0or 4 5 x cos f y sin f p 0 3 5 x 4 5 y 10 03x 4y 50 031. AB 2 2 (4) 2 4 2 or 42; p 424cos f 4or 242 2, sin f 4or 22 2x cos f y sin f p 0 22x 22y 42 0x y 8 032. 22x y 1822x y 18 0AB 2 2 2(2) 2 (1) 2 9 3 2 23x 1 3 y 1 83 0p 1 83 6 units33a. yO45˚x1.25 ft233 Chapter 7
33b. p 1.25, f 45°x cos (45°) y sin (45°) 1.25 0 22x 22y 1.25 02x 2y 2.5 034a. y36a.y 2O 1xOf and the supplement of v are complementaryangles of a right triangle, so f 180° v 90°.Simplifying this equation gives v f 90°.34b. tan v. The slope of a line is the tangent of theangle the line makes with the positive x-axis34c. Since the normal line is perpendicular to , theslope of the normal line is the negative1reciprocal of the slope of . That is, ta n vcot v.34d. The slope of is the negative reciprocal of the1slope of the normal, or ta n fcot f.35a. A B 2 2 52 2 1 2 or 135 1x3 1 21y3 3 913 05 1x3 1 21y3 3 035b. sin f 1 251,3cos f 1; Quadrant I 1 213—51 3tan f or 1 52f 67°f 90° 67° 90° or 157°x cos 157° y sin 157° 3 01213 1 213y513 x y 3 0O3x5x 1y 3 0335c. See students’ work.35d. The line with normal form x cos f y sin f p 0 makes an angle of f with the positivex-axis and has a normal of length p. The graphof Armando’s equation is a line whose normalmakes an angle of f d with the x-axis and alsohas length p. Therefore, the graph of Armando’sequation is the graph of the original line rotatedd° counterclockwise about the origin. Armando iscorrect. See students’ graphs.xThe angles of the quadrilateral are 180° a,90°, f 2 f 1 , and 90°. Then 180° a 90° f 2 f 1 90° 360°, which simplifies tof 2 f 1 a. If the lines intersect so that a is aninterior angle of the quadrilateral, the equationworks out to be f 2 180° f 1 a.36b. tan f 2 tan(f 1 a)tanf1 tana1 tanf 1 tanaIf the lines intersect so that a is an interiorangle of the quadrilateral, the equation workstanf1 tanaout to be tan f 2 1 tan.f 1 tana37. 5x y 155x y 15 0AB 2 2 51) 2 ( 2 or 265 1 15x y 0 26 26 26 5 2626x y 2626 15 26 26 0, p 1526263x 4y 363x 4y – 36 0A 2 B 2 3 2 4 2 or 56 3 5 x 4 5 y 3 5 0, p 3 55x 2y 205x 2y 20 052) 2 ( 2 25 4 or 295 2 20x y 0 29 29 29 5 2929x 2 2929y 20 29 29 0, p 20292915 26 26 3 65 20 29 296 13.8556487913.85564879 500 6927.824395; $6927.8238. 2 cos 2 x 7 cos x 4 0(2 cos x 1)(cos x 4) 02 cos x 1 0 or cos x 4 02 cos x 1 cos x 4cos x 1 2 no solutionx 3 or x 5 339. sin x 1 s co 2 xsin y 1 s co2 y 1 1 6 2 1 2 3 2 3 636 or 356sin(x y) sin x cos y cos x sin y 35 6 2 3 1 6 53 5 9 or 3 235 5185Chapter 7 234
40. A 1, 2 4 2 or 90°41. r d 2 r 13 .42or 6.7x 2 6.7 2 6.7 2 2(6.7)(6.7) cos 26°20x 2 9.316604344x 3.05 cmx 17142. x 5 25 x 2 x 5x 171 x 5 x 2 25 x 5(x 5)(x 5) xx 5 (x 5)(x 5) 17(x 5) (x 5)Oy1O1(2, 6)(2, 3)y sin 445˚x y 8(5, 3)y 3 (x 5)(x 5) x x(x 5) 17 x 5x 2 5x 17 x 5x 2 4x 12 0(x 6)(x 2) 0x 6 0 or x 2 0x 6 x 243. original box: V wh 4 6 2 48new box: V wh1.5(48) (4 x)(6 x)(2 x)72 x 3 12x 2 44x 480 x 3 12x 2 44x 24x V(x)0.4 4.4160.5 1.125V(0.5) is closer to zero, so x 0.5.4 x 4 0.5 or 4.56 x 6 0.5 or 6.52 x 2 0.5 or 2.54.5 in. by 6.5 in. by 2.5 in.44. y x 2f(x, y) 3x y 4f(2, 3) 3(2) 3 4 or 7f(2, 6) 3(2) 6 4 or 4f(5, 3) 3(5) 3 4 or 1616, 4x90˚15 13 245. 1 5 3 2 1 2 4 34 1 4 11 2 x 03 4 1 5 3 2 1 2 1 5 043151 4 1 5 y 15x 3 2y 4 1x 30y 15x 6y 3(6, 3)46. The value of 2a b cannot be determined fromthe given information. The correct choice is E.7-7Distance From a Point to a LinePage 474 Check for Understanding1. The distance from a point to a line is the distancefrom that point to the closest point on the line.2. The sign should be chosen opposite the sign of Cwhere Ax By C 0 is the standard form ofthe equation of the line.3. In the figure, P and Q are any points on the lines.The right triangles are congruent by AAS. Thecorresponding congruent sides of the trianglesshow that the same distance is always obtainedbetween the two lines.4. The formula is valid in either case. Examples willvary. For a vertical line, x a, the formulasubtracts a from the x-coordinate of the point. Fora horizontal line, y b, the formula subtracts bfrom the y-coordinate of the point.5. 2x 3y 2 → 2x 3y 2 0d Ax 1 By 1 CA B 2 2d d 2PQ2(1) (3)(2) 223) 2 ( 2 or 2 1313136. 6x y 3 → 6x y 3 0d Ax 1 By 1 CA B 2 2d d 6(2) (1)(3) 361) 2 ( 212 or 123773 37235 Chapter 7
7. 3x 5y 1 When x 2, y 1. Use (2, 1).3x 5y 3 → 3x 5y 3 0d Ax 1 By 1 CAB 2 23(2) (5)(1) 3d 35) 2 ( 24d or 2 34 34172 34178. y 1 3 x 3 Use (0, 3).y 1 3 x 7 → x 3y 21 0d Ax 1 By 1 CA B 2 21(0) 3(3) 21d 1 2 3230d or 310103109. d 1 6xd 2 2x 1 3y 1 41 8y 1 56 2 8 2 2 3) 2 ( 2 6x 1 81 y 1 50 2x1 3y1 43 1613x 813y 513 20x 30y 40(20 613)x (30 813)y 40 513 0; 6x 1 8y 1 5103y1 43 1 2x1 613x 813y 513 20x 30y 40(20 613)x (813 30)y 40 513 010. (2000, 0)d Ax 1 By 1 Cd A B 2 25(2000) (3)(0) 053) 2 ( 20,0004 3d 1 or about 1715 ftPages 475–47611. d Ax 1 By 1 CAB 2 2d d 21 2 153(2) (4)(0) 1534) 2 ( 2512. d Ax 1 By 1 CA B 2 2d 5(3) (3)(5) 1053) 2 ( 210d or 5 34 17 53417Exercises34 13. 2x y 3 → 2x y 3 0d Ax 1 By 1 CAB 2 22(0) (1)(0) 3d d 35(2) 2 ) (12 or 3 5514. y 4 2 3 x → 2x 3y 12 0d Ax 1 By 1 CA B 2 22(2) 3(3) (12)d 2 2 3 2 25d 1 or 25 13 3 1325131315. y 2x 5 → 2x y 5 0d Ax 1 By 1 CA B 2 2d 2(3) 3(1)(1) (5)21) 2 ( 20d or 0 516. y 4 3 x 6 → 4x 3y 18 0d Ax 1 By 1 CA B 2 24(1) 3(2) (18)d 4 2 3 2 d 16 5 or 1 65 1 6517. d Ax 1 By 1 CA B 2 23(0) (1)(0) 1d 31) 2 ( 21d or 0 1 1010 0 11018. 6x 8y 3 When x 0, y 3 8 . Use 0, 3 8 .6x 8y 5 → 6x 8y 5 0d Ax 1 By 1 CA B 2 2d 6 2 (8d 10 or 4 5 4 5 6(0) (8) 3 8 18) 219. 4x 5y 12 When x 3, y 0. Use (3, 0).4x 5y 6 → 4x 5y 6 0d Ax 1 By 1 CA B 2 24(3) (5)(0) (6)d 45) 2 ( 26d or 6 41 414120. y 2x 1 Use (0, 1).2x y 2 → 2x y 2 0d Ax 1 By 1 CA B 2 2d d 3 552(0) (1)(1) (2)21) 2 ( 23 5or 35 5Chapter 7 236
21. y 3x 6 Use (0, 6).27. y 2 3 x 1 → 2x 3y 3 03x y 4 → 3x y 4 0d Ax 1 By 1 Cy 3x 2 → 3x y 2 02xA B 2 21 3y 1 33x 1 y 1 2d3(0) 1(6)(1) (4)1 d 2 2 3) 2 ( 23 2 1 2 d 2x3 2 1 2 1 3y 1 3 3x 1 y 1 2 2d or 101310 105210 x 310 y 310 313 x 13 y 213 22. y 8 5 x 1 Use (0, 1).(210 313 )x (13 310 )y 310 213 02x8x 15 5y → 8x 5y 15 01 3y 1 3 3x 1 y 1 2 d Ax 1 By 1 C1310A B 2 2210 x 310 y 310 313 x 13 y 213 8(0) (5)(1) 15d (210 313 )x (13 310 )y 310 213 085) 2 ( 228a. Linda: (19, 112)20d or 20898989d Ax 1 By 1 CA 2089B2 24(19) (3)(112) 22889d 23. y 3 2 43) 2 ( 2x Use (0, 0).y 3 d 322 5or 6.4x 4 → 3x 2y 8 0Father: (45, 120)d Ax 1 By 1 CA B 2 2d Ax 1 By 1 CA B 2 23(0) 2(0) 8d 4(45) (3)(120) 2283 2 2 2 d 43) 2 ( 28d or 81348 1313d 8 135or 9.613Linda24. y x 6 Use (0, 6).28b. 4x 3y 228 0x y 1 04x 3(140) 228 0d Ax 1 By 1 C4x 192A B 2 2x 481(0) 1(6) (1)d 29. Let x 1.1 2 1 2 5d or 5 tan v y2x 2 2ytan 40° 13x25. d 1 d 2 5x 1 12y 1 261 4y 1 10y 0.83909963123 2 4 2 512) 2 ( 2 3x 1 4 y 1 105 5x 1 1 2y 1 26m 0.8 39 01 0y y 1 m(x x 1 )13m 0.839 y 0.839 0.839(x 1)39x 52y 130 25x 60y 130y 0.839x14x 112y 00.839x y 0x 8y 0 3x 1 4 y 1 105 5x 1 1 2y 1 26d Ax 1 By 1 C13A B 2 20.839(16) 1(12) 039x 52y 130 25x 60y 130d 0.83964x 8y 260 0 2 1 2d 1.09206843816x 2y 65 04x26. d 1 d 2 15x 1 8y 1 681 y 1 61.09 m4 2 1 2 (15) 2 8 230. The radius of the circle is [(5) (2)] 2 (6 2) 24x 1 y 1 6 15x 1 8y 1 68or 5. Now find the distance from the center of the 17 17circle to the line.68x 17y 102 1512x 817 y 6817 d Ax 1 By 1 C(68 1517 )x (17 817 )y 102 6817 0AB 2 24x 1 y 1 6 15x 1 8y 1 685(5) (12)(6) 32 d 17 17512) 2 ( 268x 17y 102 1512 x 817 y 6817 d 6513(68 1517 )x (17 817 )y 102 6817 0d 5Since the distance from the center of the circle tothe line is the same as the radius of the circle, theline can only intersect the circle in one point. Thatis, the line is tangent to the circle.237 Chapter 7
31. m 1 432. 73 1or 3 4 y 7 3 4 (x 1)3x 4y 25 0a 1 Ax 1 By 1 CA B 2 2a 1 3(1) (4)(3) 2534) 2 ( 24a 1 3 53 4m 2 or 71 (3)2 y 4 7 2 (x (3))7x 2y 13 0a 2 Ax 1 By 1 CA B 2 27(1) 2(7) 13a 2 7 2 2 2 34a 2 or 34535353m 3 7 ( 3)1 ( 1)or 5y 7 5(x 1)5x y 2 0a 3 Ax 1 By 1 CA B 2 25(3) (1)(4) 2a 3 51) 2 ( 217 or 17262626 3 45, 34 53 53, 172626a 3 y108642OThe standard form of the equation of the linethrough (0, 0) and (4, 12) is 3x y 0. Thestandard form of the equation of the line through(4, 12) and (10, 0) is 2x y 20 0. Thestandard form for the x-axis is y 0. To find thebisector of the angle at the origin, set 3 x y y310 and solve to obtain y 1 x. To find the 10bisector of the angle of the triangle at (10, 0), set 2x y 2052 4 6 8 y and solve to obtain 2x (1 5)y 20 0. The intersection of these two bisectorsis the center of the inscribed circle. To solve the3system of equations, substitute y 1 x into 10the equation of the other bisector and solve for x to20(1 10 )20(1 10 )get x .Then y 5 3 5 210 3601 105 35 210 x5 3 5 210 . This y-coordinate is theinradius of the triangle. The approximate value is3.33.33. 2x 7y 52x 7y 5 0AB 2 2 27) 2 ( 2 or 532 7 5x 53 y 53 053 2 5353x 7 5353y 5 5353 034. cos 2A 1 2 sin 2 A3 2 1 2 6 5 6 35. 2 1 2, 6 0° 1 60°y1y csc ( 60˚)O1 120˚ 300˚ 480˚36. 110 3 330 180° (60° 40°) 80°x 2 330 2 330 2 2(330)(330) cos 80°x 2 179979.4269x 424.24 miles37. T 2 gT 2 29.8 T 2.8 s38. 2 ⏐ 1 8 k2 201 10 ⏐20 k20 k 0k 2039. 2x y z 9 2x y z 92(x 3y 2z) 2(10) → 2x 6y 4z 207y 5z 11x 2y z 7x 3y 2z 10y 2z 35(y z) 5(3)5y 5z 15→7y 5z 11 7y 5z 112y 4y 2y z 3x 2y z 72 z 3x 2(2) (5) 75 zx 6(6, 2, 5)40. square: A s 2 triangle: A 1 2 bh16 s 2 6 1 2 (4)h4 s 3 hAE s hAE 4 3 or 7EF AEEF 7The correct choice is C.Chapter 7 238
Chapter 7 Study Guide and Assessmentx cos 4 x1 1 21. cos 15° cos (45° 30°)2 cos 45° cos 30° sin 45° sin 30°19. sin4 sin2 1 cot 2 xxPage 477 Understanding and Using the(sin 2 x cos 2 x)(sin 2 x cos 2 x)sin 1 cot 2 xxVocabulary sin2 x cos xsin2 1 cot 2 xx1. b 2. g 3. d 4. a1 c os2x5. i 6. j 7. h 8. fsin2 1 cot 2 xx9. e 10. c1 cot 2 x 1 cot 2 x20. cos 195° cos (150° 45°) cos 150° cos 45° sin 150° sin 45°Pages 478–480 Skills and Concepts 32 22 1 2 22111. csc v si n v 6 24 2 212. tan 2 v 1 sec 2 2 32 22 1 2 v4 2 1 sec 2 v 6 2417 sec 2 v22. sin 1 7117 sec v271 2113. sin v cs c vsin 2 v cos 2 v 1sin 4 7 6 1 3 5 2 cos 2 v 1sin 5 4 cos 7 6 cos 4 sin 7 6 3 3 2 3 5 114. sec v co s vcos 2 v 1 2cos v 4 5 tan 2 v 1 sec 2 v1 tan 2 v 1 5 4 2 4 5 5 4 tan 2 9v 1 6tan v 3 4 15. csc x cos 2 1x csc x sin (1 sin 2 x)x sin sin1x sin x16. cos 2 x tan 2 x cos 2 x 1cos 2 sin2xx c os2cos 2 x 1xcos 2 x sin 2 x 11 1 cosv (csc v cot v) 217. 1 1 cosv 1 cosv1 cosv 1sin v c osv 2sinv 1 cosv (1 cosv)21 cosv sin2v 1 cosv (1 cos)1 cosv 1 cos vv22 1 cosv (1 cos v) 21 cosv (1 cos v)(1 cos v) 1 cosv 1 cosv1 cosv 1 cosv18. se c v 1 tanv tanv se c v 1 se c v 1tan v tan v(secv 1)sec2v 1 se c v 1tan v tan v ( secv 1)tan2v se c v 1tan v se c v 1t anv1 sin x sin x651x 22 2 2 1 2 6 24 6 2423. tan 1 11 2 tan 2 3 4 tan 2 tan 3 4 1 tan 2 tan 3 4 3 11 (3)(1) 1 31 3 4 232or 2 324. cos x 1 n si2 xsin y 1 s co2 x 1 7 2 5 2 1 2 3 2 5 76625or 2 42cos (x y) cos x cos y sin x sin y 2 425 2 3 725 53 75 48 755 5 9 or 35239 Chapter 7
125. cos y se c y1sin y 1 s co2 y 2 1 2 3 3 2 1 cos 6 (Quadrant I) 2 3 5 9 or 51 cos 6 3sintan y y1 3c osy2 53 or 51 3 2 2 23 tanx tanytan (x y) 2 2 3 31 tanx tany 5 4 (2 3)(2 3)52(2 3)(2 3) 1 5 4 52 (2 3) 5 2 52 4 3 2 34 8 5 530. sin 2 v cos 2 v 18 1 0 45sin 2 v 3 5 2 18 55sin 2 v 1 6 180 8256 1or 180 25 25561 sin v 426. cos 75° cos 15 5 0°2sin 2v 2 sin v cos v 1 cos150 °2 (Quadrant I) 2 4 5 3 5 1 32 2 425 231. cos 2v 2 cos 2 v 1 2 3 2 3 5 2 1722 527. sin 7 7 sinv2 t 432. tan v c osvtan 2v 8 sin 1 21 cos 7 45 4 or 4 3 (Quadrant II) 3 25 2 1 7 22233. sin 4v sin 2(2v)2 2 sin 2v cos 2v 2 228. sin 22.5° sin 4 5° 2os 45° (Quadrant I) 1 c 21 222 2 2 229. tan 12 tan 2 2 2 6 245 275 3625anvtan2 v2 4 3 1 4 3 2 3 634. tan x 1 sec x(tan x 1) 2 sec 2 xtan 2 x 2 tan x 1 tan 2 x 12 tan x 0tan x 0x 0°35. sin 2 x cos 2x cos x 01 cos 2 x 2 cos 2 x 1 cos x 0cos 2 x cos x 0cos x (cos x 1) 0cos x 0 or cos x 1 0x 90° or x 270° cos x 1x 0°4 Chapter 7 240
36. cos 2x sin x 11 2 sin 2 x sin x 12 sin 2 x sin x 0sin x (2 sin x 1) 0sin x 0 or 2 sin x 1 0x 0° or x 180° sin x 1 2 x 30° orx 150°37. sin x tan x 2 tan x 02tan x sin x 22 0tan x 0 or sin x 2 02x ksin x 22x 4 2k or 3 4 2k38. sin 2x sin x 02 sin x cos x sin x 0sin x (2 cosx 1) 0sin x 0 or 2 cos x 1 0x kcos x 1 2 x 2 3 2kor x 4 3 2k39. cos 2 x 2 cos xcos 2 x cos x 2 0(cos x 1)(cos x 2) 0cos x 1 0 or cos x 2 0cos x 1cos x 2x 2k no solution40. x cos f y sin f p 0x cos 3 y sin 3 23 0 1 2 x 32y 23 0x 3y 43 041. x cos f y sin f p 0x cos 90° y sin 90° 5 00x 1y 5 0y 5 042. x cos f y sin f p 0x cos 2 3 y sin 2 3 3 0 1 2 x 32y 3 0x 3y 6 043. x cos f y sin f p 0x cos 225° y sin 225° 42 0 22x 22 y 42 0x y 8 044. AB 2 2 7 2 3 2 or 587 3 8x y 0 58 58 58 7 58 5858 58 x 3 y 4 058 29sin f 3 58 , cosf 7 58 , p 4 58; Quadrant I585829tan f f 23° 35858 or 3 7 7585845. 6x 4y 56x 4y 5 0AB 2 2 64) 2 ( 2 or 2136452x 132y 132 013 3 1313x 2 1313y 5 1326 0sin f 2 1313, cos f 3 1313, p 5 1326; Quadrant IItan f 21313—— or 2 3 3 1313f 146°46. 9x 5y 39x 5y 3 0AB 2 2 9 2 5 2 or 106 91 x 06 1 506 y 1 306 0 9 1 x06106 5 1 y06106 3 106 106 0sin f 5 1 ,06106cos f 9 1 ,06106p 3 1 ;Quadrant06106Itan f 106 or 5 9 9 1 06 106f 29°47. x 7y 5x 7y 5 0AB 2 2 17) 2 ( 2 or 5275x y 05 515 2 2102202022x 7 y 01 2sin f 7 , cos f 2, p 2; Quadrant II110 2tan f 7 210— 210f 98°48. d A x 1 By1A2 d CB 2or 72(5) (3)(6) 223) 2 ( 26d or 6 13131349. 2y 3x 6 → 3x 2y 6 0d Ax 1 By 1 CAB2 23(3) 2(4) (6)d 3 2 2 2 23d or 23 13 3 1 1323131350. 4y 3x 1 → 3x 4y 1 0d Ax 1 By 1 CAB2 23(2) (4)(4) (1)d 3 4) 2 ( 2d 23 5 or 2 35 2 35241 Chapter 7
51. y 1 3 x 6 → x 3y 18 0d A x 1 By1 CA2 B 21(21) (3)(20) 18d 13) 2 (221d or 2110 01 1052. y 3x 6 Use (0, 6).y 3x 2 → x 3y 6 0d A x 1 By1 CA2 B 21(0) (3)(6) 6d 1 3) 2 ( 224d or 12 10 10510d 12 553. y 3 4 x 3 Use (0, 3).y 3 4 x 1 2 → 3x 4y 2 0d Axd 1 By 1 CAB2 23(0) (4)(3) (2)34) 2 ( 24d 14 5 or 1 5d 1 4554. x y 1 Use (0, 1).x y 5 → x y 5 0d Ax 1 By 1 CAB2 21(0) 1(1) (5)d 1 2 1 2 4d or 22 2d 2255. y 2 3 x 2 Use (0, 2).d Ax 1 By 1 CAB2 22(0) (3)(2) 3d 2 3) 2 ( 29d or 9 13 131356. y 3x 2 → 3x y 2 0xy 2 3 2 → x 2y 3 03xd 1 d 2 x 1 2y 1 31 y 1 23 2 1 2 1 2 2 2 3x 1 y1 2 x 1 2y1 301535x 5y 25 10x 210y 310(35 10)x (5 210)y 25 310 03x 1 y1 2 x 1 2y1 301535x 5y 25 10x 210y 310(35 10)x (5 210)y 25 310 057. x 3y 2 0y 3 5 x 3 → 3x 5y 15 0xd 1 d 2 3x 1 5y 1 151 3y 1 2(1) 23 235) 2 ( 2x 1 3y 1 210 3x 1 5y 1 153434x 334y 234 310x 510y 1510(34 310)x (334 510)y 234 1510 0x 1 3y 1 210310x 510y 1510 34x 334y 234(34 310)x (334 510)y 234 1510 0Page 481 Applications and Problem Solving58. The formulas are equivalent.sin2v v v 220 tan2v 0 c os2v2gsec2 v12g cos 2 vsin2vv 20 c os2v c os2v 1 cos2v2g cos 2 v v 0 2 sin 2 v2g1 By 1 CAB2 24(1600) (2)(0) 0d 42) 2 ( 26400d 0 259. d Ax3x 1 5y 1 1534d 1431 ftx60. sin 30° 10 030° 45° v 90°100 sin 30° x v 15°50 xcos v x y cos 15° 5 y0 Page 48150y cos 15°y 51.76 ydOpen-Ended Assessment1. Sample answer: 15°; 15° 3 2sin 3 2cos 3 2os 30°0° 1 c 21 32 2 2 32os 30°0° 1 c 21 32 2 2 320° Chapter 7 242
tan 3 20° 1 1c os30° cos° 301 321 322 323 2 2(2 3)(2 3) (2 3)(2 3) (2 4 3) 2 3 2 32. Sample answer: sin x tan x 1 cocos 2 xs xosx s xcosx s in2cos xx s in2xc osxsin x tan x 1 cosinx in2sin x cSAT & ACT Preparationcos 2 xs xPage 483 SAT and ACT Practice1. The problem states that the measure of ∠A is 80°.Since the measure of ∠B is half the measure of∠A, the measure of ∠B must be 40°. Because ∠A,∠B, and ∠C are interior angles of a triangle, thesum of their measures must equal 180°.m∠A m∠B m∠C 18080 40 m∠C 180120 m∠C 180m∠C 60The correct choice is B.2. To find the point of intersection, you need to solvea system of two linear equations. Substitution orelimination by addition or subtraction can be usedto solve a system of equations. To solve thissystem of equations, use substitution. Substitute2x 2 for y in the second equation.7x 3y 117x 3(2x 2) 117x 6x 6 11x 5Then use this value for x to calculate the valuefor y.y 2x 2y 2(5) 2 or 8The point of intersection is (5, 8). The correctchoice is A.3. One way to solve this problem is to label the threeinterior angles of the triangle, a, b, and c. Thenwrite equations using these angles and theexterior angles.a b c 180x a 180y b 180z c 180Add the last three equations.x a y b z c 180 180 180x y z a b c 180 180 180Replace a b c with 180.x y z 180 180 180 180x y z 180 180 or 360The correct choice is D.4. Since x y 90°, x 90° y.Then sin x sin (90° y).sin (90° y) cos ysinx c osy sin( 90°cos y)y c osyc osy 1The correct choice is D.Another solution is to draw a diagram and noticethat sin x b c and cos y b c .sinx c osy b c — 1 b c 5. In order to represent the slopes, you need thecoordinates of point A. Since A lies on the y-axis,let its coordinates be (0, y). Then calculate the twoy 0 yslopes. The slope of AB is 0 (3) . 3 The slopey 0 yof ADis 0 3 . 3 The sum of the slopes isy y 3 3 0.The correct choice is B.6. Since PQRS is a rectangle, its angles measure 90°.The triangles that include the marked angles areright triangles. Write an equation for the measureof ∠PSR, using expressions for the unmarkedangles on either side of the angle of x°.90 (90 a) x (90 b)0 90 a b xa b 90 xThe correct choice is A.ax˚bcy˚243 Chapter 7
7. Simplify the fraction. One method is to multiplyboth numerator and denominator by y 2.y2y 1 y 2 y 1 y y y21 2 y 11 2 y 1 y 2y 2y 3 yy 2 2y 1y(y2 1) (y 1)(y 1) y ( y 1)( y 1)( y 1)( y 1) y 2 yy 1Another method is to write both the numeratorand denominator as fractions, and then simplify.y 1 y 1 2 y 1 y 2y 2 y 1 —— y2 2y 1y2 y 2 y2 1 y2y 2y 1 y ( y 1)( y 1)( y 1)( y 1) y 2 yy 1The correct choice is A.8. Since the triangles are similar, use a proportionwith corresponding sides of the two triangles. B C AC B ADE 2 2 3 A4E 2AE 4(2 3)AE 10The correct choice is E.9. Since the volume V varies directly with thetemperature T, the volume and temperaturesatisfy the equation V kT, where k is a constant.When V 12, T 60. So 12 60k, or k 1 5 .The relationship is V 1 5 T.To find the volume when the temperature is 70°,substitute 70 for T in the equation V 1 5 T.V 1 5 (70) or 14. The volume of the balloon is14 in 3 .The correct choice is C.10. Two sides have the same length. The lengths of allsides are integers. The third side is 13. FromTriangle Inequality, the sum of the lengths of anytwo sides must be greater than the length of thethird side. Let s be the length of the other twosides. Write and solve an inequality.2s 13s 6.5The length of the sides must be greater than 6.5.But the length of the sides must be an integer.The smallest integer greater than 6.5 is 7. Theanswer is 7. If you answered 6.5, you did not findan integer. If you answered 6, you found a numberthat is less than 6.5.Chapter 7 244
Chapter 8 Vectors and Parametric Equations8-1Geometric VectorsPage 490 Check for Understanding1. Sample answer:aa bbDraw u a. Then draw u b so that its initial point (tip)is on the terminal point (tail) of u a . Draw a dashedline from the initial point of u a to the terminalpoint of u b. The dashed line is the resultant.2. Sample answer: A vector has magnitude anddirection. A line segment has only length. A vectorcan be represented by a directed line segment.3. Sample answer: the velocities of an airplane and awind current4. No, they are opposites.5-11. Answers may vary slightly.5. 1.2 cm, 120° 6. 2.9 cm, 55° 7. 1.4 cm, 20°8.x y11.12˚2.9 cm, 12°12. h 2.9 cos 55° v 2.9 sin 55°h 1.66 cmv 2.38 cm13a.5 10013b. Use the Pythagorean Theorem.c 2 a 2 b 2c 2 (100) 2 (5) 2c 2 10,025c 10,02 5 or about 100.12 m/sPages 491–492 Exercises14. 2.6 cm, 128° 15. 1.4 cm, 45°16. 2.1 cm, 14° 17. 3.0 cm, 340°18-30. Answers may vary slightly.18.r s3 cm2z2.9 cm1 3 1 3x2z x3.5 cmyrs101˚9.x2.6 cm70˚x210˚3.5 cm, 70°3 cm, 101°19.s3.4 cm s tt25˚x yy3.4 cm, 25°20.su10.z4y z2.6 cm, 210°359˚3.8 cm, 359°3.8 cms u4y12.9 cm51˚12.9 cm, 51°245 Chapter 8
21. 324˚u26.tur5.5 cmr3.5 cm22˚r t u22.5.5 cm, 324°r t3.9 cmr155˚u r27.3.5 cm, 22°r s uu5.4 cmsr3.9 cm, 155°23.t5.4 cm, 133°28.2s133˚u2r5.2 cm5.2 cm, 128°24.s45˚4.2 cm, 45°25.322˚8.2 cm, 322°r3s4.2 cm3u8.2 cm128˚3u 2s2s358˚5.5 cm, 358°29. Draw to scale:2ts3u3.4 cm, 301°30. Draw to scale:3t301˚5.5 cm12s u 2r3.4 cm1 2r3t 2u357˚11.7 cm, 357°31. h ⏐2.6 cos 128°⏐ v ⏐2.6 sin 128°⏐h 1.60 cmv 2.05 cm32. h ⏐1.4 cos 45°⏐ v ⏐1.4 sin 45°⏐h 0.99 cmv 0.99 cm33. h ⏐2.1 cos 14°⏐ v ⏐2.1 sin 14°⏐h 2.04 cmv 0.51 cmrr 2t s 3u 11.7 cm2uChapter 8 246
34. h ⏐3.0 cos 340°⏐ v ⏐3.0 sin 340°⏐h 2.82 cmv 1.03 cm35. c 2 a 2 b 2c 2 (29.2) 2 (35.2) 2c 2 2091.68c 2091. 68 or about 45.73 m36. The difference of the vectors; sample answer: Theother diagonal would be the sum of one of thevectors and the opposite of the other vector, so itwould be the difference.37. Yes; sample answer:s41. h 47 cos 40° v 47 sin 40°h 36 mphv 30 mph42. It is true when k 1 or when u a is the zero vector.43. c 2 a 2 b 2c 2 (50) 2 (50) 2c 5000 or about 71 lb44.a60˚ 60˚ b60˚2.4 cm60˚ 60˚60˚r38.60 Nsr ss r35 Nr40 Nua u b 24equilateral triangle u a u b 24 lb45. The origin is not in the interior of the acute angle.d 1 d 2x y 2d 1 or x y 2 121) ( 2 2y 5d 2 or y 5 0 2 1 2 x y 22 (y 5)x y 2 2(y 5)x y 2 2y 52x y 2 2y 52 0x (1 2)y 2 52 01s46. csc v cos v tan v si cos v cn v s invsinv c osvcosv 1invosv6.147. 4 n where n is an integer48.23˚61 N, 23° north of east39. Sometimes;aba b 5a b 2.3 2.7 5 2 , 5 2 scl 2 by [3, 3] scl1x , 3 2 for 0 x 2549. tan 18°29 0. 5b 5b 0.5 tan 18°29b 29.9 cmsin 18°29 h5 a ba a b 1.5a b 2.3 2.7 55h sin 1 8°29h 15.8 cmb40a. v 1.5 sin 52° h 1.5 cos 52°v 1.18 Nh 0.92 N40b. h 1.5 cos 78° v 1.5 sin 78°h 0.31 Nv 1.47 N247 Chapter 8
50. v o volume of original boxv n volume of new boxv o o w o h o(w 1) w 2w(w 1)2w 2 2w 3 2w 2v n n w n h n (w 2) (w 1) (2w 2) (w 2 3w 2)(2w 2) 2w 3 8w 2 10w 42w 3 8w 2 10w 4 160w 2 8 10 1561 2 6 4 1601 2 10 20 1362 2 12 34 883 2 14 52 0w o 3 o 2w 2 3 or 6h o w 1 3 1 or 4So, the dimensions of the original box are3 ft 4 ft 6 ftx 251. g(x) (x 1)(x 3)vertical: As x approaches 1 and 3, the expressionapproaches or . So, x 1 and x 3 arevertical asymptotes.x 2horizontal: y x 2 2x 3y y As x increases positively or negatively, theexpression approaches 0. So, y 0 is a horizontalasymptote.52. Let x, x 2, and x 4 be 3 consecutive oddintergers.3x 2(x 4) 3 x 4 153x 2x 8 33x 2x 11x 11The correct answer 15.8-2x 2 x 2 x 2 x 2x2 2 x 3x2 x 2 1 x 2 x 21 2 x 3 x 2Algebraic VectorsPages 496–497 Check for Understanding1. Sample answer: u a 8, 6, u b 6, 8; equalvectors have the same magnitude and direction.u2. Use ⏐XY ⏐ (x2 x 1 ) 2 (y 2 y 1 ) 2 and replacethe values for x and y.x(5, 6), y(3, 4)u⏐XY ⏐ [3 5)] ( 2[46)] (2 (8) 2 (2) 2 64 4 or 683. Jacqui is correct. The representation is incorrect.2, 0 0, 5 is not equal to 51, 0 (2)0, 1.The correct expression is 2i u 5j u .u4. MP 3 2, 4 (1) or 5, 5u⏐MP ⏐ (5) 2(5) 2 50 or 52 unitsu5. MP 0 5, 5 6 or 5, 1u⏐MP ⏐ (5) 2) (12 26 unitsu6. MP 4 (19), 0 4 or 23, 4u⏐MP ⏐ (23) 2 (4) 2 545 units7. u t u u v 1, 4 3, 2 1 3, 4 (2) or 2, 28. u t 1 uu u v2 1 1, 4 3, 22 1 , 2 3, 22 1 2 3, 2 (2) or 31 , 429. u t 4u u 6 u v 4 1, 4 63, 2 4, 16 18, 12 4 18, 16 (12) or 14, 410. u t 8u u81, 4 8(1), 8(4) or 8, 3211. ⏐8, 6⏐ 86) 2 ( 2 100 or 108i u 6j u12. ⏐7, 5⏐ (7) 2 ) (52 747i u 5j u13. Let u T represent the force Terrell exerts.Let u W represent the force Mr. Walker exerts.⏐T u x ⏐ 400 cos 65° ⏐Wu x⏐ 600 cos 110° 169.05 205.21⏐T u y ⏐ 400 sin 65° ⏐Wu y⏐ 600 sin 110° 362.52 563.82u uT 169.05, 362.52, W 205.21, 563.82u uT W 36.16, 926.34⏐T u u W⏐ (36. 16) 2 4) (926.32 927 NPages 497–499 Exercisesu14. YZ 2 4, 8 2 or 2, 6u⏐YZ ⏐ (2) 26 2 40 or 210Chapter 8 248
u15. YZ 1 (5), 2 7 or 4, 5u⏐YZ ⏐ 4 5) 2 ( 2 41u16. YZ 1 (2), 3 5 or 3, 2u⏐YZ ⏐ 3 2) 2 ( 2 13u17. YZ 0 5, 3 4 or 5, 7u⏐YZ ⏐ (5) 2) (72 74u18. YZ 0 3, 4 1 or 3, 3u⏐YZ ⏐ (3) 23 2 18 or 32u19. YZ 1 (4), 19 12 or 5, 7u⏐YZ ⏐ 5 2 7 2 74u20. YZ 7 5, 6 0 or 2, 6u⏐YZ ⏐ 2 2 6 2 40 or 210u21. YZ 23 14, 14 (23) or 9, 9u⏐YZ ⏐ 9 2 9 2 162 or 92u22. AB 36 31, 45 (33) or 5, 12u⏐AB ⏐ 5 12) 2 ( 2 169 or 1323. u a u b u c 6, 3 4, 8 6 (4), 3 8 or 2, 1124. u a 2b u u c 26, 3 4, 8 12, 6 4, 8 12 (4), 6 8 or 8, 1425. u a u b 2 u c 6, 3 24, 8 6, 3 8, 16 6 (8), 3 16 or 2, 1926. u a 2b u 3 u c 26, 3 34, 8 12, 6 12, 24 12 (12), 6 24 or 0, 3027. u a b u 4c u6, 3 44, 8 6, 3 16, 32 6 (16), 3 32 or 22, 2928. u a u b 2 u c 6, 3 24, 8 6, 3 8, 16 6 (8), 3 (16) or 14, 1329. u a 3b u 36, 3 3 6, 3 3 or 18, 930. u 1a c u2 1 4, 82 1 2 (4), 1 2 8 or 2, 431. u a 6b u 4c u 66, 3 44, 8 36, 18 16, 32 36 (16), 18 32 or 20, 5032. u a 0.4b u 1.2c u 0.46, 3 1.24, 8 2.4, 1.2 4.8, 9.6 2.4 (4.8), 1.2 9.6 or 7.2, 8.433. u a 1 3 (2b u 5 u c ) 1 3 (26, 3 54, 8) 1 3 (12, 6 20, 40) 1 3 (12 (20), 6 40) 1 3 32, 34 or 3 23, 3 34 34. u a (3b u u c ) 5b u 36, 3 4, 8 56, 3 18, 9 (4, 8 30, 15 18 (4) 30, 9 8 15 or 44, 3235. 3m u 2.5n u 35, 6 2.56 9 [15, 18 15, 22.5 [15 (15), 18 (22.5 30, 4.536. ⏐3, 4⏐ 3 2 4 2 25 or 53i u 4j u37. ⏐2, 3⏐ 23) 2 ( 2 132i u 3j u38. ⏐6, 11⏐ (6) 2 1) (12 157 6i u 11j u39. ⏐3.5, 12⏐ (3.5) 2 12 2 156.2 5 or 12.53.5i u 12j u40. ⏐4, 1⏐ (4) 2 1 2 174i u u j41. ⏐16, 34⏐ (16) 2 34) (2 1412 or 2353 16i u 34j uu42. ST 4 (9), 3 2 or 5, 55i u 5j u43. Student needs to show that( u v 1 vu 2 ) vu 3 vu 1 ( vu 2 vu 3 )( u v 1 vu 2 ) vu 3 [a, b c, d] e, f a c, b d e, f a c e, b d f a c e, b d f a, b c e, d f a, b [c, d e, f] u v 1 ( vu 2 vu 3 )44a.100 N20˚F x44b. sin 20° ⏐ 1uFy⏐00⏐F u y⏐ 100 sin 20° 34 NF y249 Chapter 8
45a.30˚Shore1545b. sin 30° ⏐V u k ⏐⏐V u k ⏐ 15sin 30° 30 mphu u u u46a. Since QR ST 0, QR ST .So, they are opposites.u u46b. QR and ST have the same magnitude, butopposite direction. So, they are parallel.Quadrilateral QRST is a parallelogram.47a. t d r 1 50m5 m/sV k or 30 s47b. d rt (1.0 m/s)(30 s) or 30 m47c. ⏐V u B Vu C⏐ ⏐0.5 1.0⏐ 1 2 5 2 26 or about 5.1 m/s48. cos v (x 2 x 1 ) → (x 2 x 1 ) u v cos v⏐v⏐ u y 1 )V xsin v (y 2⏐v⏐ u → (y 2 y 1 ) u v sin vu49. PQ 2 8, 5 (7) 10, 12u⏐PQ ⏐ (10) 2 12 2 244 uRS 7 8, 0 (7) 1, 7u⏐RS ⏐ (1) 27 2 50none50. d Ax 1 By 1 CA B 2 23(1) 7(4) 1SurferV s 15d 3 7) 2 ( 23 322d 5 or about 4.2851. sin 255° sin (225° 30°) sin 225° cos 30° cos 225° sin 30° 22 32 22 1 2 6 2 452. y A sin (kx c)A: ⏐A⏐ 17A 17 or 17k: 2 k 4 k 8cc: 60° k53. Let a 400, b 600, C 46.3°c 2 400 2 600 2 2(400)(600) cos 46.3°c 2 18.8578.39c 434P a b c 400 600 434 1434 fts 1 2 (a b c)s 1 2 (1434) or 717k s(s a)(s b)(s c)k 717(7 17 400)(7 17 600)(7 17 434) k 7,525 ,766,0 79k 86,751 sq ft54. Sample answer:55.56.f(x) 3x 2 2x 1r 3 2 11 3 1 22 3 4 9f(x) 3x 2 2x 1r 3 2 11 3 5 62 3 8 17[4, 4] scl1 by [4, 4] scl1max: (0, 3), min: (0.67, 2.85)f(x) x 2 3x 1xf(x)10,000 99,970,0011000 997,001100 970110 710 110 131100 10,3011000 1,003,00110,000 100,030,001An upper bound is 2.Alower boundis 1.y → as x → , y → as x → 57. 7x 1 7x 11 1This statement is true regardless of the value of x,so it is true for all real values of x.The correct choice is A.c 60° 8c 480°y 17 sin (8x 480°)Chapter 8 250
8-3Vectors in Three-DimensionalSpacePages 502–503 Check for Understanding1. Sample answer: sketch a coordinate system withthe xy-axes on the horizontal, and the z-axispointing up. Then, vector 2i u is two units along thex-axis, vector 3j u is three units along the y-axis,and vector 4k u is four units along the z-axis. Drawbroken lines to represent three planes.xz2. Sample answer: To find the components of thevector, you will need the direction (angle) with thehorizontal axis. Using trigonometry, you canobtain the components of the vector.3. Sample answer: Neither is correct. The sign forthe u j -term must be the same (), and thecoefficient for the u k -term is 0, so the correct wayto express the vector as a sum of unit vectors isui 4ju .4.G(4,1, 7)yu⏐OG ⏐ 4 1) 2 ( 2 7 2 66u5. RS 3 (2), 9 5, 3 8 or 5, 4, 11u⏐RS ⏐ 5 2 4 2 11) (2 162 or 92u6. RS 10 3, 4 7, 0 (1) or 7, 11, 1u⏐RS ⏐ 7 11) 2 ( 2 1 2 171 or 3197. u a 3f u u g 31, 3, 8 3, 9, 1 3, 9, 24 3, 9, 1 3 3, 9 9, 24 (1) or 6, 0, 258. u a 2 u g 5f u 23, 9, 1 51, 3, 8 6, 18, 2 5, 15, 40 6 5, 18 (15), 2 (40) or 1, 33, 38u9. EF 6 (5), 6 (2), 6 4 11, 4, 211i u 4j u 2k uu10. EF 12 (12), 17 15, 22 (9) 0, 2, 132j u 13k u11. ⏐132, 3454, 0⏐ 132 2 34542 0 2 11,947 ,540 3457 NPages 503–50412.13.ExercisesxB(4,1,3)u⏐OB ⏐ 4 2 1 2 3) (2 26xzB(7, 2, 4)zyyu⏐OB ⏐ 7 2 2 2 4 2 6914.z161412B(10,3,15)108642 412108 642 y4 2 2 426 48xu⏐OB ⏐ 10 2 (3) 2 15 2 334 15. u TM 3 2, 1 5, 4 4 or 1, 4, 8⏐TM u ⏐ 14) 2 ( 2 (8) 2 81 or 916. u TM 3 (2), 5 4, 2 7 or 1, 1, 5⏐TM u ⏐ (1) 2 1 2 (5) 2 27 or 3317. u TM 3 2, 1 5, 0 4 or 1, 4, 4⏐TM u ⏐ 14) 2 ( 2 (4) 2 3318. u TM 1 3, 1 (5), 2 6 or 4, 6, 4⏐TM u ⏐ (4) 2 6 2 (4) 2 68 or 21719. u TM 2 (5), 1 8, 6 3 or 3, 9, 9⏐TM u ⏐ 39) 2 ( 2 (9) 2 171 or 319251 Chapter 8
20. u TM 1 0, 4 6, 3 3 or 1, 2, 6⏐TM u ⏐ 12) 2 ( 2 (6) 2 41u21. ⏐CJ ⏐ 3 (1), 5 3, 4 10or 4, 8, 14)u⏐CJ ⏐ 4 8) 2 ( 2 ) (142 276 or 26922. u 6w u 2 u z 62, 6, 1 23, 0, 4 12, 36, 6 6, 0, 8 18, 36, 223. u 1 2 u v u w 2 u zuu 1 2 4, 3, 5 2, 6, 1 23, 0, 4 2, 3 2 , 5 2 2, 6, 1 6, 0, 8 6, 7 1 2 , 11 1 2 24. u 3 4 u v u w 3 4 4, 3, 5 2, 6, 1 3, 9 4 , 1 54 2, 6, 1 1, 8 1 4 , 4 3 4 25. u 3 u v 2 3 u w 2 u z 34, 3, 5 2 3 2, 6, 1 23, 0, 4 12, 9, 15 4 3 , 4, 2 3 6, 0, 8 16 2 3 , 13, 23 2 3 26. u 0.75v u 0.25 u w 0.754, 3, 5 0.252, 6, 1 3, 2.25, 3.75 0.5, 1.5, 0.25 3.5, 0.75, 3.527. u 4w u u z42, 6, 1 3, 0, 4 8, 24, 4 (3, 0, 4 5, 24, 828. 2 3 u f 3 u g 2 5 uh 2 3 3, 4.5, 1 32, 1, 6 2 5 6, 3, 3 2, 3, 2 3 6, 3, 18 1 256 78 5 25, 3 5, 2 15, 6 5 , 6 5 u29. LB 5 2, 6 2, 2 7 or 3, 8, 53i u 8j u 5k uu30. LB 4 (6), 5 1, 1 0 or 2, 4, 12i u 4j u u ku31. LB 7 9, 3 7, 2 (11) or 2, 4, 92i u 4j u 9k uu32. LB 8 12, 7 2, 5 6 or 20, 5, 1120i u 5j u 11k uu33. LB 8 (1), 5 2, 10 (4)or 7, 3, 67i u 3j u 6 u ku34. LB 6 (9), 5 12, 5 (5)or 15, 7, 015i u 7j uu35. ⏐G 1 G2 ⏐ (x x 2 1 )2 (y 2 y 1 ) 2 (z 2 z 1 ) 2 (x x 1 2 ) 2 (y 1 y 2 ) 2 (z 1 z 2 ) 2 u ⏐G 1 G2 ⏐because (x y) 2 (y x) 2 for all real numbers xand y.36. If u m m 1 , m 2 , m 3 , then⏐m u ⏐ (m 1 ) 2 ) (m 2 2 (m 3 . ) 2 If m u m 1 , m 2 , m 3 , then ⏐m u ⏐ (m 1 ) 2 m ( 2 ) 2 ) (m 2 3 .Since m 2 1 (m 1 ) 2 , m 2 2 (m 2 ) 2 , and m 2 3 (m 3 ) 2 , ⏐m u ⏐ ⏐m u ⏐.37. 3, 2, 4 6, 2, 5 u F u O9, 0, 9 u F u OuF 9, 0, 9 or9, 0, 938. m 1 2 (x 1 x 2 , y 1 y 2 , z 1 z 2 ) 1 2 (2 4, 3 5, 6 2) 1 2 (6, 8, 8) (3, 4, 4)u39a. OK 1 0, 4 0, 0 0 or 1, 4, 0ui 4juu39b. TK 1 2, 4 4, 0 0 or 1, 0, 0i u40. u c u b u acu 3, 1, 5 1, 3, 1cu 2, 2, 441a. z12x126O661241b. Find distance between (0, 0, 0) and (15, 15, 15).d (15 0) 2 (15 0) 2 (15 0) 2 675 or about 26 feet41c. sin v 1 526 v sin 1 156 75v 35.25°u42. ⏐AB ⏐ (1 ) 2 3 2 () 0 2 (0 0) 2 4 or 2u⏐BC ⏐ (1 ) 1 2 3 2 2 1 3 36 63 3 or 1.69u⏐AC ⏐ (1 ) 2 2 1 3 0y 2 3 0 2 2 2 0 2 2 3 2 or 1.41No, the distances between the points are notequal. A and B are 2 units apart, B and C are 1.69units apart, and A and C are 1.41 units apart.43. 3, 5 1, 2 3 (1), 5 2 2, 7Chapter 8 252
u44. AB 3 5, 3 2 or 8, 1u i u j u kuCD d1 0, d 2 0 or d 1 , d 2 2. u a au uAB CD u a x a y a z a x a y a z8, 1 d 1 , d 2 a y a z a x a z a x a yD (8, 1)a y a z a x a z a x a ysin 2X (ay a z a y a z )i u (a x a z a x a z )j u45. 1 cos 2X cot X (a x a y a x a y ) u k2 sin X cos X1 cos 2 X sin 2 X46. cos v 2 3 2 sin X cos X2 sin 2 X c osXsinX cot X cot X cot Xcot X cot Xsin 2 v 1 cos 2 vsin 2 v 1 4 9 sin 2 v 5 9 sin v 5347. y 6 sin 2v amplitude ⏐6⏐ or 6period 2 k rador 4rev48. 16 m in 2 1 min1 rev 6 0 sec 8 15 radians per second49. Yes, because substituting 7 for x and 2 for yresults in the inequality 2 180 which is true.y 4x 2 3x 52 4(7) 2 3(7) 52 18050. 3 2 3 12 1 4 3 3 2 4 3 So, A, C, and D are not correct. 2 3 2 13 1 3 4 2 3 3 4 So, B is not correct.The correct choice is E.8-42 1 2 Perpendicular VectorsPages 508–509 Check for Understanding1. Sample answer: Vector v w is the negative ofvector w v .uiu ju kvu w u 1 0 3 1 2 40 3 1 32 4 1 46i u 7j u 3k uuiu ju k u i u j 1 0 u kvu w u 1 2 41 0 32 4 1 4 1 2 6i u 0 3 7j u 1 3k u 3 1 0 i u j u k u02 i u j u k u 0i u 0j u 0k u 0, 0, 0 u 03. Sample answer: No, because a vector cannot beperpendicular to itself.4. 5, 2 3, 7 5(3) 2(7)15 141, no5. 8, 2 4.5, 18 8(4.5) 2(18)36 36 0, yes6. 4, 9, 8 3, 2, 2 4(3) 9(2) 8(2)12 18 1610, nouiujuk7. 1, 3, 2 2, 1, 5 1 3 22 1 53 2 1 2 11 5 2 5 2 13i u u j 5k u or 13, 1, 5, yes13, 1, 5 1, 3, 213(1) 1(3) (5)(2)13 3 10 013, 1, 5 2, 1, 513(2) 1(1) (5)(5)26 1 25 0uiu ju k i u j u k u8. 6, 2, 10 4, 1, 9 6 2 10 4 1 9 2 10 u i 6 10 u j 6 2 u k1 9 4 9 4 1 8i u 14j u 2k u or 8, 14, 2, yes8, 14, 2 6, 2, 108(6) (14)(2) (2)(10)48 28 20 08, 14, 2 4, 1, 98(4) (14)(1) (2)(9)32 14 18 09. Sample answer: Let T(0, 1, 2), U(2, 2, 4), andV(1, 1, 1)uTU 2, 1, 2uUV 1, 3, 5u uTU UV uiu ju k 2 1 2 1 3 51 2 2 23 5 1 5 u i 8j u 5k u or 1, 8, 5 u i u j 2 1 u k1331253 Chapter 8
u10. AB (0.65, 0, 0.3) (0, 0, 0) 0.65, 0, 0.3uF 0, 0, 32u u uT AB F uiu juk 0.65 0 0.30 0 320 0.3 0.650 32 0 0i u 20.8j u 0k u0.332 i u j u k u⏐T u ⏐ 020.82 ( ) 2 0 2 20.8 foot-pounds0.6500.30Pages 509–511 Exercises11. 4.8 6, 3 4(6) 8(3) 24 24 0, yes12. 3, 5 4, 2 3(4) 5(2) 12 10 2, no13. 5, 1 3, 6 5(3) (1)(6)15 621, no14. 7, 2 0, 2 7(0) 2(2) 0 44, no15. 8, 4 (2, 4 8(2) 4(4) 16 16 32, no16. 4, 9, 3 6, 7, 5 4(6) 9(7) (3)(5)24 63 15 24, no17. 3, 1, 4 2, 8, 2 3(2) 1(8) 4(2) 6 8 8 6, no18. 2, 4, 8 16, 4, 2 2(16) 4(4) 8(2)32 16 16 0, yes19. 7, 2, 4 3, 8, 1 7(3) (2)(8) 4(1) 21 16 4 9, no20. u a u b 3, 12 8, 2 24 24 0, yesub u c 8, 2 3, 2 24 4 28, noau u c 3, 12 3, 2 9 2415, nouiu ju k21. 0, 1, 2 1, 1, 4 0 1 21 1 41 21 4 i u j u k u 2i u 2j u u k or 2, 2, 1, yes2, 2, 1 0, 1, 22(0) 2(1) (1)(2)2 2 2 02, 2, 1 1, 1, 42(1) 2(1) (1)(4)2 2 4 0uiu ju k22. 5, 2, 3 2, 5, 0 5 2 32 5 0 2 3 u i 5 3 u j 5 2 u k5 0 2 0 2 5 15i u 6j u 29k u or 15, 6, 29, yes15, 6, 29 5, 2, 3(15)(5) (6)(2) 29(3)75 12 87 015, 6, 29 2, 5, 0(15)(2) (6)(5) 29(0)30 30 0 0uiu ju k23. 3, 2, 0 1, 4, 0 3 2 0 1 4 0 2 0 u i 3 0 u j 3 2 u k4 0 1 0 1 4 0i u 0j u 10k u or 0, 0, 10, yes0, 0, 10 3, 2, 00(3) 0(2) 10(0)0 0 0 00, 0, 10 1, 4, 00(1) 0(4) 10(0)0 0 0 0uiujuk24. 1, 3, 2 5, 1, 2 1 3 2 5 1 23 2 1 2 1 31 2 5 2 5 1 4i u 12j u 16k u or 4, 12, 16, yes4, 12, 16 1, 3, 24(1) 12(3) 16(2)4 36 32 04, 12, 16 5, 1, 24(5) 12(1) 16(2)20 12 32 0uiuju k i u j u k u25. 3, 1, 2 4, 4, 0 3 1 2 4 4 01 2 3 2 3 14 0 4 0 4 4 8i u 8j u 16k u or 8, 8, 16, yes8, 8, 16 3, 1, 28(3) 8(1) 16(2)24 8 32 08, 8, 16 4, 4, 08(4) 8(4) 16(0)32 32 0 001240111 i u j u k uChapter 8 254
uiu juk26. 4, 0, 2 7, 1, 0 4 027 1 00 2 4 2 4 01 0 7 0 7 1 2i u 14j u 4k u or 2, 14, 4, yes2, 14, 4 4, 0, 22(4) 14(0) 4(2)8 0 8 02, 14, 4 7, 1, 02(7) 14(1) 4(0)14 14 0 027. Sample answer:Let u v v 1 , v 2 , v 3 and v u v 1 , v 2 , v 3 u ui jukvu (v u ) v 1 v 2 vv 1 v 2 v 33v 2 v 3 v 1 v 3 v 1 v 2v 2 v 3 v 1 v 3 v 1 v 2 0i u 0j u 0k u 0uiu uj k28. u a (b u c ) a 1 a 2 a 3(b 1 c 1 ) (b 2 c 2 ) (b 3 c 3 ) a a a a 2313(b 2 c 2 ) (b 3 c 3 ) (b 1 c 1 ) (b 3 c 3 )a [aa 12(b ku2 (b(b 3 c1 3 ) a1 3 (b2 2 cc 2 ) 2 )]i u [a 1 (b 3 c 3 ) a 3 (b 1 c 1 )]j u [a 1 (b 2 c 2 ) a 2 (b 1 c 1 )]k u [(a 2 b 3 a 2 c 3 ) (a 3 b 2 a 3 c 2 )]i u [(a 1 b 3 a 1 c 3 ) (a 3 b 1 a 3 c 1 )]j u [(a 1 b 2 a 1 c 2 ) (a 2 b 1 a 2 c 1 )]k u [(a 2 b 3 a 3 b 2 ) (a 2 c 3 a 3 c 2 )]i u [(a 1 b 3 a 3 b 1 ) (a 1 c 3 a 3 c 1 )]j u [(a 1 b 2 a 2 b 1 ) (a 1 c 2 a 2 c 1 )]k u (a 2 b 3 a 3 b 2 )i u (a 2 c 3 a 3 c 2 )i u (a 1 b 3 a 3 b 1 )j u (a 1 c 3 a 3 c 1 ) u j (a 1 b 2 a 2 b 1 )k u (a 1 c 2 a 2 c 1 )k u [(a 2 b 3 a 3 b 2 )i u (a 1 b 3 a 3 b 1 )j u (a 1 b 2 a 2 b 1 )k u ] [(a 2 c 3 a 3 c 2 )i u (a 1 c 3 a 3 c 1 )j u (a 1 c 2 a 2 c 1 )k u ]a2 a 3 iu a1 a 3 ju a a1 2 ku b 2 b 3 b 1 b 3 b 1 b 2 i u j u k u iu ju ku iu jua 2 a 3 a 1 a 3 a 1 a 2c 2 c 3 c 1 c 3 c 1 c 2b 2 b 3 b 1 b 3 b 1 b 2 iu ju ku a2 a 3 iu a1 a 3 ju a a1 2 ku ( u a u b ) ( u a u c )29. Sample answer:Let T(0, 2, 2), U(1, 2, 3), and V(4, 0, 1)uTU 1, 4, 5uUV 3, 2, 2 u i u j u kuTU UV3 2 2 4 5 u i 1 5 u j 1 4 u k2 2 3 2 3 22i u 17j u 14k u or 2, 17, 14u 1 4 5 30. Sample answer:Let T(2, 1, 0), U(3, 0, 0), and V(5, 2, 0).uTU 1, 1, 0uUV 8, 2, 0 u i u j u kuTU UVu 1 1 0 8 2 01 0 12 0 8 u i 0 u j 1 1 u k 0i u 0j u 6k u or 0, 0, 631. Sample answer:Let T(0, 0, 1), U(1, 0, 1), and V(1, 1, 1).uTU 1, 0, 0uUV 2, 1, 2uTU UVuiujuku 1 0 02 1 20 0 1 01 2 2 2 u i u j 1 0 u k 0i u 2j u k u or 0, 2, 132. The expression is false. m u n u and n u m u havethe same magnitude but are opposite in direction.33a.forearm33b. u u uT AB FuAB 0.04 cos (30°), 0, 0.04 sin (30°) 0.02(3), 0, 0.02uF 0, 0, 600uiujuku uAB F 0.023 0 0.02 0 0 600 0i u 123 u j 0k uu u uT ⏐AB F ⏐ 123 or about 21 N-muiu ju k34. x u y u 2 3 01 1 43 0 21 4 1 4 12i u 8j u 5k uA 1 2 ⏐x u u y ⏐ 1 2 12 2 (8) (5) 2 u i 0 u j 2 3 u k 1 2 233 30˚600 N0.04 melbow135a. o u 120, 310, 60cu 29, 18, 2135b. o u c u 120(29) 310(18) 60(21) $10,320021821255 Chapter 8
36a.F45˚36b. W ⏐F u ⏐⏐d u ⏐ cos vW 120 4 cos 45°W 339 ft-lb37a. u X 2 1, 5 0, 0 3 or 1, 5, 3)uY 3 2, 1 5, 4 0 or 1, 4, 4u u ui j ku uX Y 1 5 3 1 4 4 5 3 u i 1 3 u j 1 5 u k4 4 1 4 1 4 8i u 7j u 9k u or 8, 7, 937b. The cross product of two vectors is always avector perpendicular to the two vectors and theplane in which they lie.38a. v u p (q u u r )u u ui j kqu u r 2 1 4 3 1 5 1 4 u i 2 4 u j 2 1 u k1 5 3 5 3 1i u 22j u 5k u or 1, 22, 5pu (q u r u ) 0, 0, 1 1, 22, 5 0(1) 0(22) (1)(5)5 or 5 units 30 0 138b. 2 1 4 3111 54523452311 0 0 (1)5 or 5 units 3They are the same.39. Need (kv u u w ) u 0.[k1, 2 1, 2] 5, 12 0[k, 2k 1, 2] 5, 12 0k 1, 2k 2 5, 12 0(k 1)5 (2k 2)12 05k 5 24k 24 029k 19 0k 1 299 u40. ⏐BA ⏐ 2 ⏐a u ⏐ 2 ⏐b u ⏐ 2 2⏐a u ⏐⏐b u ⏐ cos v(a b 1 1 ) 2 (a 2 b 2 ) 2 2 a 2 1 a 2 2 2 b 2 1 b 2 22 2a 2 1 b a 22 2 1 b 2 2 cos v(a 1 b 1 ) 2 (a 2 b 2 ) 2 a 2 1 a 2 2 b 2 1 b 2 2 2a 2 1 a 2 2b 2 1 b 2 2 cos v a 2 1 2a 1 b 1 b 21 a 2 2 2a 2 b 2 b 22 a 2 1 a 2 2 b 2 1 b 22 2a 2 1 b a 22 2 1 b 2 2 cos v 2a 1 b 1 2a 2 b 22a 2 1 b a 22 2 1 b 2 2 cos va 1 b 1 a 2 b 2 a 2 1 a 22 b 2 1 b 22 cos va 1 b 1 a 2 b 2 ⏐a u ⏐⏐b u ⏐ cos v u a u b ⏐a u ⏐⏐b u ⏐ cos vu41. AB 5 3, 3 3, 2 (1) or 2, 0, 342. D(8, 3)E(0, 2)uDE 0 8, 2 3) or 8, 5u⏐DE ⏐ (8) 2) (52 8943. 4x y 6 0A B 2 2 4 2 1 2 or 17 4 17 x 1y717 17 6 17 017p 6 17 1.46 units17sin f 17 17cos f 4 1717tan f 1 4 f 14°44. A 36°, b 13, and c 6a 2 b 2 c 2 2 bc cos Aa 2 13 2 6 2 2(13)(6) cos 36°a 8.9 sin 38. 6°9 si n B1 3in36° B sin 1 13 s 8.9B 59.41° or 59°25C 180° 36° 59°25C 84.59° or 84°3545. tan 73° h 4 cos 73° 4 44 tan 73° h cos 73°13.1 h; 13.1 m 13.7 m46. 3 3x 4 103x 4 73x 4 49x 17.6747. 81 3 464 2 6 (2 2 ) 3 or (2 3 ) 24 2 22 2 19 3 2So 64 4 3 8 2The correct choice is B.Chapter 8 256
Page 5111.Mid-Chapter Quiz8-4BGraphing Calculator Exploration:Finding Cross Products2.3 cm46˚F x 2.3 cos 46° F y 2.3 sin 46° 1.6 cm 1.7 cm2.115˚2.7 cmF x 27 cos 245° F y 27 sin 245° 11.4 mm 24.5 mmu3. CD 4 (9), 3 2 or 5, 5u⏐CD ⏐ 5 5) 2 ( 2 50 or 52u4. CD 5 3, 7 7, 2 (1) or 2, 0, 3u⏐CD ⏐ 2 2 0 2 3 2 135. u r u t 2s u 6, 2 24, 3 6, 2 8, 6 6 8, 2 6 or 14, 86. u r 3u u u v 31, 3, 8 3, 9, 1 3, 9, 24 3, 9, 1 3 3, 9 9, 24 (1) or 6, 0, 257. 3, 6 4, 2 3(4) 6(2)12 12 0; yes8. 3, 2, 4 1, 4, 0 3(1) (2)(4) 4(0) 3 8 11; no9. 1, 3, 2 2, 1, 1 u u j u k312112211 3 22 1 11 32 1 i u j u k ui u 5j u 7k u or 1, 5, 7, yes1, 5, 7 1, 3, 2(1)(1) 5(3) (7)(2)1 15 14 01, 5, 7 2, 1, 1(1)(2) 5(1) (7)(1)2 5 7 010. Let X(2, 0, 4) and Y(7, 4, 6).⏐XY⏐ (7 ) 2 2 (4 0) 2 (64) 2 45 or about 6.7 mPage 5121. 49, 32 55 2. 168, 96, 763. 0, 0, 0 4. 11, 15, 35. 0, 0, 7 6. 0, 40, 07. u u x 6, 6, 12⏐u u u v ⏐ 6 2 6 2 12) (2 216 8. u u v 1, 13, 20⏐u u u v ⏐ 113) 2 ( 2 20) (2 570 9. Sample answer: Insert the following lines afterthe last line of the given program.:Disp “LENGTH IS”:Disp ((BZ CY) 2 (CX AZ) 2 (AY BX) 2 )8-5Applications with VectorsPages 516–517 Check for Understanding1. Sample answer: Pushing an object up the sloperequires less force because the component of theweight of the object in the direction of motion ismg sin v. This is less than the weight mg of theobject, which is the force that must be exerted tolift the object straight up.2. The tension increases.3. Sample answer: Forces are in equilibrium if theresultant force is u O .4.Current23 knots5. u F1 300iuuF2 (170 cos 55°) iu (170 sin 55°)j u⏐F u 1 Fu 2 ⏐ (300 os 170 c)255° 70 (1sin55°) 2 421.19 Ntan v 17˚170 sin 55°300 170 cos 55°v tan 1 170 sin 55°300 170 cos 55°6. u F1 50iuuF2 100ju⏐F u 1 Fu 2 ⏐ 502 100 2 111.8 Ntan v 1 0050or 2v tan 1 2 63.43°7. horizontal 18 cos 40° 13.79 Nvertical 18 sin 40° 11.57 N257 Chapter 8
8. u F1 (33 cos 90°)iu (33 sin 90°)j u or 33j uuF2 (44 cos 60°)iu (44 sin 60°)j uor 22i u 22 3 u j⏐F u 1 Fu 2 ⏐ 222 (33 223) 2 74 Ntan v 33 22322 or 3 232 v tan 1 3 2 73°Aforce with magnitude 74 N and direction 73° 180° or 253° will produce equilibrium.9a. 4 mph94˚454 lb342 lb23 9b. If v is the angle between the resultant path ofthe ferry and the line between the landings,4then sin v 1 2or 1 3 . So v sin 1 1 3 , or about19.5°.Pages 517–519Exercises10. 11.12.42 N12 mph53˚13. u F1 425iuuF2 390ju⏐F u 1 Fu 2 ⏐ 4252 390 2 576.82 Ntan v 3 90425or 7 885 v tan 1 7 885 42.5°14. u v1 65iuvu 2 (50 cos 300°)i u (50 sin 300°)j u or 25i u 253 u j⏐v u 1 vu 2 ⏐ 902 (253) 2 99.87 mphtan v 25 9 30or 5 31 838 Wind27˚256 mphv tan 1 5 1Apositive value for v is about 334.3°.15. u v1 (115 cos 60°)iu (115 sin 60°)j uor 57.5i u 57.53 u jvu 2 (115 cos 120°)i u (115 sin 120°)j uor 57.5i u 57.53 u j⏐v u 1 vu 2 ⏐ 02 (1153) 2 1153 199.19 km/hSince tan v is undefined and the verticalcomponent is positive, v 90°.16. The force must be at least as great as thecomponent of the weight of the object in thedirection of the ramp. This is 100 sin 10°, or about17.36 lb.17. u F1 105iuuF2 (110 cos 50°)iu (110 sin 50°)j u⏐F u 1 Fu 2⏐ (105 os 110 c)250° 10 (1sin50°) 2 194.87 Ntan v 110 sin 50°105 110 cos 50°110 sin 50°105 110 cos 50°v tan 1 16.7°20. u F1 (70 cos 330°)iu (70 sin 330°)j u or 353 u i 35j uuF2 (40 cos 45°)iu (40 sin 45°)j u or 202 u i 202 u juF3 (60 cos 135°)iu (60 sin 135°)j u or 30 2 302 u jtan v 35 502353 10235 502353 102250 sin 25° 45 sin 250°250 cos 25° 45 cos 250°v tan 1 37.5°⏐F u 1 Fu 2 Fu 3⏐ (353 2) 10(35 2 502) 2 58.6 lb21. u F1 (23 cos 60°)iu (23 sin 60°)j uuFv tan 1 25.62°18. F w sin v52.1 75 sin v2.1 sin v 5 752.1sin 1 5 75 v44° v19. u F1 (250 cos 25°)iu (250 sin 25°)j uuF2 (45 cos 250°)iu (45 sin 250°)j u⏐F u 1 Fu 2 ⏐(250 os c25°cos 450°) 25 2 in (250 s25°sin 450°) 252 220.5 lb250 sin 25° 45 sin 250°tan v 250 cos 25° 45 cos 250°or 11.5i u 11.53j u2 (23 cos 120°)iu (23 sin 120°)j uor 11.5i u 11.53j u⏐F u 1 Fu 2 ⏐ 02 233) ( 2 233 39.8 NSince tan v is undefined and the verticalcomponent is positive , v 90°. A force withmagnitude 39.8 N and direction 90° 180° or270° will produce equilibrium.Chapter 8 258
22. a g sin 40° 32 sin 40° 20.6 ft/s 223. u F1 (36 cos 20°)iu (36 sin 20°)j uuF2 (48 cos 222° iu (48 sin 222°)j u⏐F u 1 Fu 2 ⏐(36 s co20° s 48 co) 222° 6 2 (30° sin 28 422°) sin 22 19.9 N36 sin 20° 48 sin 222°tan v 36 cos 20° 48 cos 222°v tan 1 36 sin 20° 48 sin 222°36 cos 20° 48 cos 222° 264.7° or 5.3° west of south24a. 135 lb165˚115 lb75˚120˚24b. u F1 70iuuF2 (135 cos 165°)iu (135 sin 165°)j u⏐F u 1 Fu 2 Fu 3 ⏐ (70 os 135 c° 1655 1140°) cos 2 2 in (135 s° 1655 1140°) sin 22 134.5 lb135 sin 165° 115 sin 240°tan v 70 135 cos 165° 115 cos 240°v tan 1 135 sin 165° 115 sin 240° 70 135 cos 165° 115 cos 240° 208.7° or 28.7° south of westSince u F1 Fu 2 Fu 3 0, the vectors are not inequilibrium.25. W u F u d [(1600 cos 50°)i u (1600 sin 50°)j u ] 1500i u (1600 cos 50°)(1500) (1600 sin 50°)(0) 1,542,690 N-m26a. Sample answer: The horizontal forward force isuF cos v. You can increase the horizontal forwardforce by decreasing the angle v between thehandle and the lawn.26b. Sample answer: Pushing the lawnmower at alower angle may cause back pain.327a. tan v 1 8or 1 6 v tan 1 1 6 9.5° south of east27b. s 18 2 3 2 18.2 mph28. F cos v 100 cos 25° 90.63 N29. F 1 cos 174.5° F 2 cos 6.2° 0F 1 sin 174.5° F 2 sin 6.2° 155 0The first equation gives F 2 co s 174.5°cos6.2°F 1 .Substitute into the second equation.cos 174.5° sin 6.2°cos 6.2°70 lbF 1 sin 174.5° F 1 155 0F 1 760 lbF 2 co s 174.5°cos6.2°F 1 761 lb30. Sample answer: Method b is better. Let F be theforce exerted by the tractor, T be the tension inthe two halves of the rope, and v be the anglebetween the original line of the rope and half ofthe rope after it is pulled. At equilibrium,F2T sin v F 0, or T 2s in.vSo, if 0° v 30°,the force applied to the stump using method b isgreater than the force exerted by the tractor.31. Let T be the tension in each towline and supposethe axis of the ship is the vertical direction.2T sin 70° 6000 0T 2s6000in70° 3192.5 tons32. Let T be the tension in each wire. The halves ofthe wire make angles of 30° and 150° with thehorizontal.T sin 30° T sin 150° 25 0 1 2 T 1 2 T 25 0T 25 lb33. u u u v 9(3) 5(2) 3(5)2The vectors are not perpendicular since u u v 0.u34. AB 0 12, 11 (5), 21 18 12, 6, 335. d 2v g0 2 sin v cos v 2 100 232 sin 65° cos 65° 239.4 ft36. Sample answer: A plot of the data suggests aquadratic function. Performing a quadraticregression and rounding the coefficients givesy 1.4x 2 2x 3.9.37. b 0.3p 0.2b p 0.66b pb0.50.40.30.20.1O 0.1155sin 174.5° cos 174.5° tan 6.2°(0.2, 0.3)(0.33, 0.33)(0.2, 0.46)(0.3, 0.3)The vertices are at (0.2, 0.3), (0.3, 0.3), (0.33, 0.33)and (0.2, 0.46).cost function C(p, b) 90p 140b 32(1 p b) 32 58p 108bC(0.2, 0.3) 32 58(0.2) 108(0.3) or 76C(0.3, 0.3) 32 58(0.3) 108(0.3) or 81.8C(0.33, 0.33) 32 58(0.33) 108(0.33) or 86.78C(0.2, 0.46) 32 58(0.2) 108(0.46) or 93.28The minimum cost is $76, using 30% beef and20% pork.p0.2 0.3 0.4F 1 (sin 174.5° cos 174.5° tan 6.2°) 155259 Chapter 8
38. *4 *(3) (4 3 4) [(3) 3 (3)] 60 (24) 84The correct choice is A.Vectors and Parametric EquationsPages 523–524 Check for Understanding1. When t 0, x 3 and y 1. When t 1, x 7and y 1. The graph is a line through (3, 1) and(7, 1).2. Sample answer: For every single unit increment oft, x increases 1 unit and y increases 2 units. Then,the parametric equations of the line are x 3 t,y 6 2t.3. When t 0, x 1 and y 0, so the line passesthrough (1, 0). When t 1, x 0 and y 1, sothe line passes through (0, 1), its y-intercept. Theslope of the line is 1 00 1or 1.4. x (4), y 11 t3, 8x 4, y 11 t3, 8x 4 3ty 11 8tx 3t 4 y 8t 115. x 1, y 5 t7, 2x 1 7ty 5 2tx 1 7ty 5 2t6. 3x 2y 5 7. 4x 6y 122y 3x 5 6y 4x 12y 3 2 x 5 2 y 2 3 x 2x tx ty 3 2 t 5 2 y 2 3 t 28. x 4t 3 9. x 9txx 3 4t 9 t 1 4 x 3 4 t y 4t 210.8-6y 5t 3 y 4 x9 2y 5 1 4 x 3 4 3 y 4 9 x 2y 5 4 x 3 4 t x y1 2 20 2 11 6 02 10 1y11a. receiver:x 5 0t y 50 10tx 5y 50 10tOdefensive player:x 10 0.9ty 54 10.72tx 10 0.9t y 54 10.72tx11b. 50 10t 050 10t5 tWhen t 5, the coordinates of the defensiveplayer are (10 0.9(5), 54 10.72(5)) or (5.5,0.4), so the defensive player has not yet caughtthe receiver.524–525 Exercises12. x 5, y 7 t2, 0x 5 2ty 7 0tx 5 2t y 713. x (1), y 4 t6, 10x 1, y 4 t6, 10x 1 6ty 4 10tx 1 6ty 4 10t14. x (6), y 10 t3, 2x 6, y 10 t3, 2x 6 3ty 10 2tx 6 3ty 10 2t15. x 1, y 5 t7, 2x 1 7ty 5 2tx 1 7ty 5 2t16. x 1, y 0 t2, 4x 1, y t2, 4x 1 2ty 4tx 1 2t17. x 3, y (5) t2, 5x 3, y 5 t2, 5x 3 2ty 5 5tx 3 2ty 5 5t18. x ty 4t 519. 3x 4y 720. 2x y 34y 3x 7y 2x 3y 3 4 x 7 4 y 2x 3x tx ty 3 4 t 7 y 2t 34 21. 9x y 1y 9x 1x ty 9t 123. 4x y 2y 4x 2x ty 4t 222. 2x 3y 113y 2x 11y 2 3 x 1 31 x ty 2 3 t 1 31 24. 3x 6y 86y 3x 8y 1 2 x 4 3 The slope is 1 2 .y 5 1 2 (x 2)y 1 2 x 6x ty 1 2 t 6Chapter 8 260
25. x 2tx 2 ty 1 txy 1 2y 1 2 x 127. x 4t 11x 11 4t 1 4 x 1 41 ty t 3y 1 4 x 1 41 3y 1 4 x 2 3429. x 3 2tx 3 2t 1 2 x 3 2 ty 1 5ty 1 5 1 2 x 3 2 y 5 2 x 1 7230. Regardless of the value of t, x is always 8, so theparametric equations represent the vertical linewith equation x 8.31a. x 11, y (4) t3, 7x 11, y 4 t3, 731b. x 11 3t y 4 7tx 3t 11 y 7t 431c. x 3t 11x 11 3t32. 1 3 x 1 31 ty 7t 4y 7 1 3 x 1 31 4y 7 3 x 8 93[5, 5] Tstep1[10, 10] Xscl1[10, 10] Yscl126. x 7 1 2 tx 7 1 2 t2x 14 ty 3ty 3(2x 14)y 6x 4228. x 4t 8x 8 4t 1 4 x 2 ty 3 ty 3 1 4 x 2y 1 4 x 533.[10, 10] Tstep1[20, 20] Xscl2[20, 20] Yscl234.[10, 10] Tstep1[10, 10] Xscl1[10, 10] Yscl135a. x 2 3t and y 4 7tIf t 0, then x 2 and y 4, so the part of theline to the right of point (2, 4) is obtained.35b. x 02 3t 03t 2t 2 3 36. x y cos 2 t sin 2 t 10 cos 2 t 1 and 0 sin 2 t 1, so the graph isthe segment of the line with equation x y 1from (1, 0) to (0, 1).y112 1 2( , )x1261 Chapter 8
37a. target drone:x 3 (1)t y 4 0tx 3 t y 4missile:x 2 t y 2 2t37b. 3 t 2 t1 2t 1 2 tWhen t 1 2 , the missile has a y-coordinate of 3,not 4, so it does not intercept the drone.38a. Ceres: x 1 t, y 4 t, z 1 2tPallas: x 7 2t, y 6 2t, z 1 t38b. Adding the equations for x and y for Ceres givesx y 3. Subtracting the equations for x and yfor Pallas results in x y 1. The solution ofthis system is x 1 and y 2. Eliminating tfrom the equations for y and z results in thesystem 2y z 7, y 2z 4 which hassolution y 2 and z 3. Hence, the paths crossat (1, 2, 3).38c. 1 t 1 v t 27 2t 1 v t 4Ceres is at (1, 2, 3) when t 2 but Pallas is at(1, 2, 3) when t 4. The asteroids will notcollide.39. The line is parallel to the vector 0 1 3 , 5 1,8 1 or 1 3 , 4, 9. The vector equation of theline is x 1 3 , y 1, z 1 t 1 3 , 4, 9 orx 1 3 , y 1, z 1 t 1 3 , 4, 9.x 1 3 1 3 t y 1 4tx 1 3 1 3 t y 1 4tz 1 9tz 1 9t40. u v 1 (150 cos 330°)iu (150 sin 330°)j uvu 2 (50 cos 245°)i u (50 sin 245°)j u⏐v u 1 vu 2 ⏐ (150 os c° 330 5045°) cos 2 2 0 (15330° sin50 sin 245°) 162.2 km/h150 sin 330° 50 sin 245°tan v 150 cos 330° 50 cos 245°v tan 1 150 sin 330° 50 sin 245° 150 cos 330° 50 cos 245° 47°534 or 47°534 south of east41. 1, 3 3, 2 1(3) 3(2)3Since the inner product is not 0, the vectors arenot perpendicular.42. Since A 90°, a b, and a b sin A, no solutionexists.43. A graphing calculator indicates that there is onereal zero and that it is close to 1. f(1) 0, so thezero is exactly 1.44. x 3 2 y 2x 2 3 2 y 2 3 x 4 3 yy 2 3 x 4 3 2 45. The slope is 1.y 1 1[x (3)]y 1 x 3x y 4 046. The linear velocity of the belt around the largerpulley is (120 rpm)2 9 2 in./rev 1080in./min. The linear velocity around the smallerpulley must be the same, so its angular velocity is(1080 in./min) 1 rev2 3 in. 180 rpm. The correctchoice is D.8-6BGraphing Calculator Exploration:Modeling with ParametricEquationsPage 5261. 408.7t 418.3(t 0.0083)408.7t 418.3t 3.471899.6t 3.47189t 3.4 71899.6 0.362 hr or 21.7 min2. d rt 408.73.4 71899 . 6 147.8 mi5003. The time for plane 1 to fly 500 miles is 4 08..7The500time for plane 2 is 4 18.3 0.0083. Suppose thespeed of plane 1 is increased by a mph.500500 408 4 0.00838-7.7 a 408 .7 a50018.31500 4 18.3 0.0083500a 500 408.7 4 18.3 0.0083 6.7 mphModeling Motion UsingParametric EquationsPage 531 Check for Understanding1. Sample answer: a rocket launched at 90° to thehorizontal; tip-off in basketball2. Equal magnitude with opposite direction.3. The greater the angle of the head of the golf club,the greater the angle of initial velocity of the ball.4. ⏐v u y ⏐ ⏐vu ⏐ sin v 5. ⏐v u x ⏐ ⏐vu ⏐ cos v 50 sin 40° 20 cos 50° 32.14 ft/s 12.86 m/s6. ⏐v u x ⏐ ⏐vu ⏐ cos v ⏐v u y ⏐ ⏐vu ⏐ sin v 45 cos 32° 45 sin 32° 38.16 ft/s 23.85 ft/s7. ⏐v u x ⏐ ⏐vu ⏐ cos v ⏐v u y ⏐ ⏐vu ⏐ sin v 7.5 cos 20° 7.5 sin 20° 7.05 m/s 2.57 m/sChapter 8 262
80ft h 3608a. 300 mph 52 mile0s 440 ft/sx t⏐v u ⏐ cos vx t(440) cos 0 °x 440ty t⏐v u ⏐ sin v 1 2 gt 2 hy t(440) sin 0° 1 2 (32)t 2 3500y 16t 2 35008b. Sample graph:y40003000Height(feet) 20001000x2000 4000 6000Horizontal Distance(feet)8c. 16t 2 3500 016t 2 3500t 2 35001600t 35 16t 14.8 s8d. x 440t 440(14.8) 6512 ftPages 531–533 Exercises9. ⏐v u x ⏐ ⏐vu ⏐ cos v ⏐v u y ⏐ ⏐vu ⏐ sin v 65 cos 60° 65 sin 60° 32.5 ft/s 56.29 ft/s10. ⏐v u x ⏐ ⏐vu ⏐ cos v ⏐v u y ⏐ ⏐vu ⏐ sin v 47 cos 10.7° 47 sin 10.7° 46.18 m/s 8.73 m/s11. ⏐v u x ⏐ ⏐vu ⏐ cos v ⏐v u y ⏐ ⏐vu ⏐ sin v 1200 cos 42° 1200 sin 42° 891.77 ft/s 802.96 ft/s12. ⏐v u x ⏐ ⏐vu ⏐ cos v ⏐v u y ⏐ ⏐vu ⏐ sin v 17 cos 28° 17 sin 28° 15.01 ft/s 7.98 ft/s13. ⏐v u x ⏐ ⏐vu ⏐ cos v ⏐v u y ⏐ ⏐vu ⏐ sin v 69 cos 37° 69 sin 37° 55.11 yd/s 41.53 yd/s14. ⏐v u x ⏐ ⏐vu ⏐ cos v ⏐v u y ⏐ ⏐vu ⏐ sin v 46 cos 19° 46 sin 19° 43.49 km/h 14.98 km/h15a. x t⏐v u ⏐ cos v y t⏐v u ⏐ sin v 1 2 gt 2x 175t cos 35° y 175t sin 35° 16t 215b. y 0175t sin 35° 16t 2 0t(175 sin 35° 16t) 0175 sin 35° 16t 0175 sin 35° 16t 175 s in35°16 tx 175t cos 35° 175 175 s in35°16 cos 35° 899.32 ft or 299.77 yd16. To find the time the projectile stays in the air, sety 0 and solve for t.t⏐v u ⏐ sin v 1 2 gt 2 0t(⏐v u ⏐ sin v 1 2 gt) 0⏐v u ⏐ sin v 1 2 gt 0⏐v u ⏐ sin v 1 2 gt 2⏐vu ⏐ sin vg tThe greater the angle, the greater the time theprojectile stays in the air. To find the horizontaldistance covered, substitute the expression for t inthe equation for x.x t⏐v u ⏐ cos v 2⏐vu ⏐ sin vg ⏐v u ⏐ cos v ⏐vu ⏐ 2g sin 2vgAs the angle increases from 0° to 45°, thehorizontal distance increases. As the angleincreases from 45° to 90°, the horizontal distancedecreases.17a. y 300 when t 77⏐v u ⏐ sin 78° 1 2 (32)7 2 3007⏐v u ⏐ sin 78° 784 3007⏐v u ⏐ sin 78° 1084⏐v u 1084⏐ 7sin78°⏐v u ⏐ 158.32 ft/s17b. x 1 3 t⏐v u ⏐ cos v 50 yd 1 3 (7)(158.32) cos 78° 50 127 yd18. x t⏐v u ⏐ cos vx t⏐v u ⏐ cos vy t⏐v u ⏐ sin v 1 2 gt 2y x ⏐v u ⏐ sin v 1 2 g ⏐v u ⏐⏐v u ⏐ cos vx 2cos v y x tan v 2⏐v u ⏐gcos 2 2 vThe presence of the x 2 -term (due to the force ofgravity) means that y is a quadratic function of x.Therefore, the path of a projectile is a parabolicarc.19. To find the time the projectile stays in the air ifthe initial velocity is u v, set y 0 and solve for t.t⏐v u ⏐ sin v 1 2 gt 2 0t⏐v u ⏐ sin v 1 2 gt 0⏐v u ⏐ sin v 1 2 gt 0⏐v u ⏐ sin v 1 2 gt 2⏐vu ⏐ sin vg t tTo find the range, substitute this expression for tin the equation for x.x t⏐v u ⏐ cos v 2⏐vu ⏐ sin vg ⏐vu ⏐ 2 sin 2vg ⏐v u ⏐ cos v263 Chapter 8
If the magnitude of the initial velocity is doubledto 2⏐v u ⏐, the range becomes (2⏐vu ⏐) 2 sin 2vgor4 ⏐vu ⏐ 2 sin 2vg. The projectile will travel four times asfar.00m h 36020a. 800 km/h 10 km0s 222.2 m/sx t⏐v u ⏐ cos vx 222.2 t cos 45°y t⏐v u ⏐ sin v 1 2 gt 2y 222.2t sin 45° 1 2 (9.8)t 2y 222.2t sin 45° 4.9t 2The negative coefficient in the t-term in theequation for y indicates that the aircraft isdescending. The negative coefficient in theequation for x is arbitrary.20b. y 222.2t sin 45° 4.9t 2222.2(2.5) sin 45° 4.9 (2.5) 2 423.4The aircraft has descended about 423.4 m.20c. 42 3.4m2.5s 169 m/sor169 m/s km1000m 360 0sh80 fti h 360 0s 608.4 km/h21a. 70 mph 52 m 30 3ft/sy 0 sin 35° 16t 2 10 0t 30 83 3 083 sin 35° 30 83sin 35° 2 4 (16) 10 t t 3.84 sx t⏐v u ⏐ cos v 323.2 ft21b. y 88 sin 35° 16t 2 10 8t 30 316t 2 t 30 83 sin 35° 2 0t t 3.71 sx t⏐v u ⏐ cos v 312.4 ft21c. From the calculations in part b, the time isabout 3.71s. y22a.8 30 83sin 35° 30 8n 3si35° 26)2 4(122b.23a. y 300 when t 4.84.8⏐v u ⏐ sin 82° 1 2 (32)(4.8) 2 3004.8⏐v u ⏐ sin 82° 368.64 668.64⏐v u 668.64⏐ 4.8sin82°⏐v u ⏐ 140.7 ft/s23b. x 1 3 t⏐v u ⏐ cos v 100 131.3 yd24a. x t⏐v u ⏐ cos v y t⏐v u ⏐ sin v 1 2 gt 2 hx 155t cos 22° y 155t sin 22° 16t 2 324b. x 420155t cos 22° 420420t 155 cos22°y 155t sin 22° 16t 2 3 155 420155 cos22° sin 22° 16 4202155 cos22° 3 36.04 ftSince 36.04 15, the ball will clear the fence.24c. y 0155t sin 22° 16t 2 3 0155 sin 22° (155 in s) 22° (16)t 2 43t 3.68 sx t⏐v u ⏐ cos v 528.86 ft25. x 11 tx 11 tx 11 ty 8 6ty 8 6(x 11)y 6x 5826a. mg sin v 300(9.8) sin 22° 1101.3 N26b. mg cos v 300(9.8) cos 22° 2725.9 N27. cos A 1 7.421.9A cos 1 1 7.421.9A 37°OtxChapter 8 264
28. 2(2x y z) 2(2)x 3y 2z 3.255x y 0.751(2x y z) 1(2)4x 5y z 2.56x 4y 0.54(5x y) 4(0.75)6x 4y 0.514x 3.5x 3 . 51 4x 0.255x y 0.75 2x y z 25(0.25) y 0.75 2(0.25) 0.5 z 2y 0.5 z 129. 5 2 3 2 25 9 16The correct choice is B.Page 5341. 7 ° 127.2°360° 3 60°1 5 01 5 0 5000 stadiaxHistory of Mathematicsx 50(5000)x 250,000 stadia250,000(500) 125,000,000 ft125,000,000 5280 23,674 miThe actual circumference of Earth is about24,901.55 miles.2. See students’ work. No solution exists.3. See students’ work.8-8Transformation Matrices inThree-Dimensional SpacePages 539–540 Check for Understanding1. Matrix T multiplies x-coordinates by 2 andy- and z-coordinates by 2, so it produces areflection over the yz-plane and increases thedimensions two-fold.u2. CC 8 6, 8 7, 2 3 or 2, 1, 12 2 2 2 2 2The matrix is 1 1 1 1 1 1 .1 1 1 1 1 11 0 03. VU 0 1 0 0 0are the same.1 T, so the transformationsu5a. BE 0 5, 2 5, 4 0 or 5, 3, 4A(5, 5 (3), 0) A(5, 2, 0)C(5 (5), 5, 0) C(0, 5, 0)D(5 (5), 5 (3), 0) D(0, 2, 0)F(5, 5 (3), 0 4) F(5, 2, 4)G(5, 5, 0 4) G(5, 5, 4)H(5 (5), 5, 0 4) H(0, 5, 4)5 5 0 0 0 5 5 0The matrix is 2 5 5 2 2 2 5 5 .0 0 0 0 4 4 4 45 45b. 2 (1)0 25 45 (1)0 20 42 (1)4 20 45 (1)0 25 42 (1)4 20 42 (1)0 25 45 (1)4 29 9 4 4 4 9 9 4 1 4 4 1 1 1 4 4 2 2 2 2 6 6 6 61 0 0 5 5 0 0 0 5 5 05c. 0 1 0 2 5 5 2 2 2 5 5 0 05 20GBx A1550HFC0050E0OD0020zy002444524455440 45 (1)4 2054The image is the reflection over the xz-plane.5d. The dimensions of the resulting figure are halfthe original.6a. The scale factor of the dilation is 4. Thetranslation increases x-coordinates by 2. Thematrices are4 0 0D 0 4 0 and0 0 42 2 2 2 2 2 2 2T 0 0 0 0 0 0 0 0 .0 0 0 0 0 0 0 04a-c.Transformation Orientation Site ShapeReflection yes no noTranslation no no noDilation no yes no265 Chapter 8
6b. Sample answer: If the original prism hasvertices A(3, 3, 0) B(3, 3, 3), C(3, 3, 3),D(3, 3, 0), E(5, 3, 0), F(5, 3, 3), G(5, 3, 3),and H(5, 3, 0), then the image has verticesA(10, 12, 0), B(10, 12, 12), C(10, 12, 12),D(10, 12, 0), E(22, 12, 0), F(22, 12, 12),G(22, 12, 12), and H(22, 12, 0).GHxCD CG DF FH EEPages 540–542 Exercisesu7. FB 3 3, 1 7, 4 4 or 0, 6, 0A(2, 3, (6), 2) A(2, 3, 2C(4, 7 (6), 1) C(4, 1, 1)2 3 4 4 2 3The matrix is 3 1 1 7 3 7 .2 4 1 1 2 4u8. AH 4 (3), 1 (2), 2 2 or 7, 3, 4B(3, 2 3, 2) B(3, 1, 2)C(3, 2 3, 2 (4)) C(3, 1, 2)D(3, 2, 2 (4)) D(3, 2, 2)E(3 7, 2, 2 (4)) E(4, 2, 2)F(3 7, 2, 2) F(4, 2, 2)G(3 7, 2 3, 2) G(4, 1, 2)3 3 3 3 4 4 4 4The matrix is 2 1 1 2 2 2 1 12 2 2 2 2 2 2 2u9. CF 6 4, 0 (1), 0 2 or 2, 1, 2D(2 2, 2 1, 3 (2)) D(4, 1, 1)E(1 2, 0 1, 4 (2)) E(3, 1, 2)2 1 4 4 3 6The matrix is 2 0 1 1 1 0 .3 4 2 1 2 00 0 0 0 0 0 0 010. 0 (2) 3 (2) 3 (2) 0 (2)1 4 2 4 5 4 4 42 0 2 0 2 0 2 00 (2) 0 (2) 3 (2) 3 (2)1 4 4 4 5 4 2 40 0 0 0 2 2 2 2 2 1 1 2 2 2 1 15 6 9 8 5 8 9 6zBABADF HAExThe result is a translation of 2 units along the y-axis and 4 units along the z-axis.0 111. 0 (2)1 (2)1 21ExGOzCB0 13 (2)2 (2)2 10 (2)1 (2)110y0 13 (2)5 (2)2 10 (2)4 (2)0 10 (2)4 (2)2 13 (2)5 (2)2 13 (2)2 (2)The result is a translation of 1 unit along the x-axis, 2 units along the y-axis, and 2 unitsalong the z-axis.0 112. 0 51 (3)1 52xDF GAH1131220 13 52 (3)2 10 51 (3)181zzCB182151DFAE321y3223130 13 55 (3)2 10 54 (3)352CGyBH3100 10 54 (3)2 13 55 (3)3812 13 52 (3)The results is a translation of 1 unit along the x-axis, 5 units along the y-axis, and 3 units alongthe z-axis.351382Chapter 8 266
1 0 0 0 0 0 0 2 2 2 213. 0 1 0 0 3 3 0 0 0 3 3 0 0 1 1 2 5 4 1 4 5 202 03 x1032035004201The transformation does not change the figure.1 0 0 0 0 0 0 2 2 2 20 1 0 0 3 3 0 0 0 3 30 0 1 1 2 5 4 1 4 5 20 0 0 0 2 2 2 2 0 3 3 0 0 0 3 31 2 5 4 1 4 5 2z3 696 3O3 6 99 A y312 6xBED H14. F204235The transformation results in reflections over thexy and xz-planes.1 0 0 0 0 0 0 2 2 2 20 1 0 0 3 3 0 0 0 3 315. 0 0 1 1 2 5 4 1 4 5 20 02 0 33xC1DFEAHBGzG2zG035EAFDCBHyC004y201The transformation results in reflections over allthree coordinate planes.16. The matrix results in a dilation of scale factor 2,so the figure is twice the original size.3 0 0 1 0 0 3 0 017. 0 3 0 0 1 0 0 3 0 , so the0 0 3 0 0 1 0 0 3figure is three times the original size and reflectedover the xy-plane.220423520.75 0 0 1 0 018. 0 0.75 0 0 1 00 0 0.75 0 0 10.75 0 00 0.75 0 , so the figure is three-fourths0 0 0.75the original site and reflected over all threecoordinate planes.2x 2 0 0 x2y 0 2 0 y , so the transformation can5z 0 0 5 z19a. 2 0 0be represented by the matrix 0 2 0 .0 0 519b. The transformation will magnify the x- andy-dimensions two-fold, and the z-dimension5-fold.23.6 23.6 23.6 23.6 23.620a. 72 72 72 72 72 0 0 0 0 020 23.6 136 23.6 247 23.620b. 58 72 71 72 74 7227 0 53 0 59 0302 23.6 351 23.683 72 62 7237 0 52 043.6 159.6 270.6 325.6 374.6 14 1 2 11 1027 53 59 37 5220c. The result is a translation 23.6 units along thex-axis and 72 units along the y-axis.1 0 021. The matrix 0 1 0 would reflect the prism 0 0 10.5 0 0over the yz-plane. The matrix 0 0.5 0 0 0 0.5would reduce its dimensions by half.0.5 0 0 1 0 0 0.5 0 0 0 0.5 0 0 1 0 0 0.5 00 0 0.5 0 0 1 0 0 0.522a. Placing a non zero element in the first row andthird column will skew the cube so that the topis no longer directly above the bottom.1 0 1Sample answer: 0 1 0 0 0 122b. Sample graphs:zxCDGHBAFEyCGBzHFEyxDA267 Chapter 8
23. The first transformation reflects the figure overall three coordinate planes. The secondtransformation stretches the dimensions along they- and z-axes and skews it along the xy-plane.(The first row of T changes the x-coordinate of(x, y, z) to x 2z.)24. To multiply the x-coordinate by 3, the first row ofthe matrix must be 3 0 0. Since the y-coordinate ismultiplied by 2, the second row is 0 2 0. To converta z-coordinate to x 4z, use a third row of 1 0 4.3 0 0The matrix is 0 2 0 .1 0 425a. The x-coordinates are unchanged, they-coordinates increase, and the z-coordinatesdecrease, so the movement is dip-slip.123.9 41.3 201.7 73.8 129.4 36.425b. 88.0205.3145.8246.628.3261.582.6212.097.1166.4123.985.3123.986.4206.541.3144.2247.8201.729.9262.773.884.2213.2129.495.5165.20 0 0 0 0 0 1.6 1.6 1.6 1.6 1.6 1.61.2 1.2 1.2 1.2 1.2 1.226a. La Shawna Jaimiex 0x 35ty 16t 2 150 y 16t 2 15016t 2 150 0150 16t 2 1 5016 1 5016 t 236.4125.584.1 t3.06 tx 35t 35 1 5016 107 ft26b. Since the stones have the same parametricequations for y, they land at the same time. Inpart a, it was calculated that the elapsed time isabout 3.06 seconds.27. x 5t 1x 1 5t x 15 ty 2t 10y 2 x 15y 2 5 x 4 5 10829. 80x 3 80x 2 80x 24.280x 3 80x 2 80x 24.2 0A graphing calculator indicates that there is asolution between 0 and 1. By Descartes’ Rule ofSigns, it is the only solution. When x 0.2, 80x 3 80x 2 80x 24.2 4.36 and when x 0.3,80x 3 80x 2 80x 24.2 9.16. So the solutionto the nearest tenth is 0.2.30. Divide each side of the equations by 2, 3, 4, and 6,respectively, so that the left side is x 2y.I. x 2y 4 II. y 4III. x 2y 2 IV. x 2y 8 3 Only I and II are equivalent, so the correct choiceis A.Chapter 8 Study Guide and AssessmentPage 543 Understanding and Using theVocabulary1. resultant 2. unit3. magnitude 4. cross5. inner 6. vector7. parallel 8. standard9. direction 10. components544–546 Skills and Concepts11. 1.3 cm, 50° 12. 2.9 cm, 10°13.qp qp4.1 cm, 23°14.2p5.3 cm, 25°15.3p q23˚25˚4.1 cmqq5.3 cm2p q28. sec cos 1 2 5 1 2 5 1coscos 1 2 5 2.5 cm98˚3p 5 2 2.5 cm, 98°Chapter 8 268
16.4p q3.5 cm82˚q4p3.5 cm, 82°17. h 1.3 cos 50° v 1.3 sin 50°h 0.8 cmv 1 cm18. h 2.9 cos 10° v 2.9 sin 10°h 2.9 cmv 0.5 cmu19. CD 7 2, 15 3 or 5, 12u⏐CD ⏐ 5 2 2 12 169 or 13u20. CD 4 (2), 12 8 or 6, 4u⏐CD ⏐ 6 2 4 2 52 or 213u21. CD 0 2, 9 (3) or 2, 12u⏐CD ⏐ (2) 212 2148 or 237u22. CD 5 (6), 4 4 or 1, 8u⏐CD ⏐ 1 8) 2 ( 2 6523. u u v u wuu 2, 5 3, 1uu 2 3, 5 (1) or 5, 624. u u v u wuu 2, 5 3, 1uu 2 3, 5 (1) or 1, 425. u 3v u 2w uuu 32, 5 23, 1uu 6, 15 6, 2uu 6 6, 15 (2) or 12, 1726. u 3v u 2w uuu 32, 5 23, 1uu 6, 15 6, 2uu 6 6, 15 (2) or 0, 13u27. EF 6 2, 2 (1), 1 4 or 4, 1, 3u⏐EF ⏐ 4 1) 2 ( 2 (3) 2 26u28. EF 1 9, 5 8, 11 5 or 10, 3, 6u⏐EF ⏐ (10) 2 3) (6 2 2 145 u29. EF 2 (4), 1 (3), 7 0) or 6, 2, 7u⏐EF ⏐ 6 2 2 2 7 2 89u30. EF 4 3, 0 7, 5 (8) or 7, 7, 13u⏐EF ⏐ (7) 2) (73 2 12 267 31. u 2w u 5v uuu 24, 1, 5 51, 7, 4uu 8, 2, 10 (5, 35, 20uu 8 (5), 2 35, 10 (20)uu 13, 37, 3032. u 0.25v u 0.4w uuu 0.251, 7, 4 0.44, 1, 5uu 0.25, 1.75, 1 1.6, 0.4, 2uu 0.25 1.6, 1.75 (0.4), 1 2uu 1.35, 1.35, 133. 5, 1 2, 6 5(2) (1)610 616; no34. 2, 6 3, 4 2(3) 6(4)6 2418; no35. 4, 1, 2 3, 4, 4 4(3) 1(4) (2)4 12 4 8 0; yes36. 2, 1, 4 6, 2, 1 2(6) (1)(2) 4(1)12 2 418; no37. 5, 2, 10 2, 4, 4 5(2) 2(4) (10)(4) 10 8 40 42; nouiu ju k38. 5, 2, 5 1, 0, 3 5 2 5 2 50 351531 0 35 21 0 u u j u k u 6i u 10j u 2k u or 6, 10, 26, 10, 2 5, 2, 56(5) 10(2) (2)(5)30 20 10 06, 10, 2 1, 0, 36(1) 10(0) (2)(3)6 0 6 039. 2, 3, 1 2, 3, 4 ui3 1 2 13 4 2 4 9i u 6j u 0k u or 9, 6, 09, 6, 0 2, 3, 19(2) (6)(3) 0(1)18 18 0 09, 6, 0 2, 3, 49(2) (6)(3) 0(4)18 18 0 0ujuk2 3 12 3 42233 i u j u k u269 Chapter 8
40. 1, 0, 4 5, 2, 1 0uiu juk1 0 45 2 1 i u j u k u28i u 1 19j u 5 2k u 1 5 2or 8, 19, 28, 19, 2 1, 0, 4(8)(1) 19(0)(2)(4)8 0 8 08, 19, 2 5, 2, 1(8)(5) 19(2) (2)(1)40 38 2 041. 7, 2, 1 2, 5, 3 2513472113 iu u j u k7 2 12 5 372251 i u j u k u u i 19j u 31k u or 1, 19, 311, 19, 31 7, 2, 11(7) (19)2 31(1)7 (38) 31 01, 19, 31 2, 5, 31(2) (19)5 31(3)2 (95) 93 042. Sample answer:Let x(1, 2, 3), y(4, 2, 1) and z(5, 3, 0)uxy 4 1, 2 2, 1 3 or 5, 0, 4uyz 5 (4), 3 2, 0 (1) or 9, 5, 15, 0, 4 9, 5, 1 uiujuk5 0 4 9 5 1 0 4 u i 5 4 u j 5 0 u k520i u 1 31j u 9 25k u 1 9 5or 20, 31, 2543. u F1 320iuuF2 260ju⏐F u 1 Fu 2 ⏐ 3202 260 2412.31 Ntan v 2 60320or 1 316v tan 1 1 316 39.09°44. u v1 12juvu 2 (30 cos 116°)i u (30 sin 116°)j u⏐v u 1 vu 2⏐ (30 s co )2 116° 2 (10 3116°) sin2 41 m/stan v 12 30sin116°30cos116°4 30sin116° v tan 1 12 30cos116° 108.65°45. x 3, y (5) t4, 2x 3, y 5 t4, 2x 3 4ty 5 2tx 3 4ty 5 2t46. x (1), y 9 t7, 5x 1, y 9 t7, 5x 1 7ty 9 5tx 1 7ty 9 5t047. x 4, y 0 t3, 6x 4, y t3, 6x 4 3tx 4 3t48. x ty 8t 77 8ty 6t49. x ty 1 2 t 5 2 50. ⏐v u x ⏐ ⏐vu ⏐ cos v ⏐v u y ⏐ ⏐vu ⏐ sin v 15 cos 55° 15 sin 55° 8.60 ft/s 12.29 ft/s51. ⏐v u x ⏐ ⏐vu ⏐ cos v ⏐v u y ⏐ ⏐vu ⏐ sin v 13.2 cos 66° 13.2 sin 66° 5.37 ft/s 12.06 ft/s52. ⏐v u x ⏐ ⏐vu ⏐ cos v ⏐v u y ⏐ ⏐vu ⏐ sin v 18 cos 28° 18 sin 28° 15.89 m/s 8.45 m/su53. CH 4 3, 2 4, 2 (1) or 7, 6, 3A(3, 4 (6), 1 3) A(3, 2, 2)B(3, 4 (6), 1) B(3, 2, 1)D(3, 4, 1 3) D(3, 4, 2)E(3 (7), 4, 1 3) E(4, 4, 2)F(3 (7), 4, 1) F(4, 4, 1)G(3 (7), 4 (6), 1) G(4, 2, 1)The matrix for the figure is3 3 3 3 4 4 4 4 2 2 4 4 4 4 2 2 .2 1 1 2 2 1 1 2The matrix for the translated figure is5 5 5 5 2 2 2 2 2 2 4 4 4 4 2 2 .5 2 2 5 5 2 2 5ABxHGzDCThe figure moves 2 units along the x-axis and 3units along the z-axis.EFyChapter 8 270
1 0 054. 0 1 0 03223 22DC03211321x341341E250 m342442441421422The figure is reflected over the xz-plane.Page 547 Applications and Problem Solvingu55. AB 1 cos 120°, 0, 1 sin 120° or 1 2 , 0, 23uF 0, 0, 50u u uT AB F 1 ftu u ui j k 1 2 0 230 0 50 32 0 u i u j u k0 5050 0i u 25j u 0k u or 0, 25, 0⏐T u ⏐ 025) 2 ( 2 0 2 25 lb-ft56. y t⏐v u ⏐ sin v 1 2 gt 2 h 0.5(38) sin 40° 1 2 (32)(0.5) 2 2 10.2 ft57a.16 km/h 3 km/h35˚ub (16 cos 55°)c u (16 sin 55°)jucu 3ju⏐b u u c ⏐ (16 s co55°)6 2 (155° sin3) 2 13.7 km/hu sin55° 357b. 250 16 16cos55°sin55° 36 cos55°zFAOBu 250 16 1u 275.3 m342 1 2 0442441HGy 3242150 lb 1 2 04220060˚58. u F1 90iuF2 (70 cos 30°)iu (70 sin 30°)j u or353 u i 35j u⏐F u 1 Fu 2 ⏐ (90 353)2 5 32 154.6 N357tan v or 90 353 18 73v tan 1 7 13.1°18 73Page 547 Open-Ended Assessment1a. Sample answer: X(4, 1), Y(1, 1)uXY 1 4, 1 (1) or 3, 2u1b. ⏐XY ⏐ (3) 22 2 or 13uThe magnitude of XY only depends on thedifferences of the coordinates of X and Y, not theactual coordinates.2a. Sample answer: P(1, 1), Q(3, 3), R(3, 1), S(5, 3)uPQ 3 1, 3 1 or 2, 2uRS 5 3, 3 1 or 2, 2u uPQ and RS are parallel because they have thesame direction. In fact, they are the same vector.2b. Sample answer: u a 8, 4, u b 3, 6au u b 8(3) (4)6 or 0au andu b are perpendicular because their innerproduct is 0.Chapter 8 SAT & ACT PreparationPage 549 SAT and ACT Practice1. Recall that the formula for the area of aparallelogram is base times height. You know thebase is 5, but you don’t know the height. Don’t befooled by the segment BD; it is not the height ofthe parallelogram. Try another method to findthe area. The parallelogram is made up of twotriangles. Find the area of each triangle. SinceABCD is a parallelogram, AB DC and AD BC.The two triangles are both right triangles, andthey share a common side, BD. By SAS, the twotriangles are congruent. So you can find the areaof one triangle and multiply by 2. The hypotenuseof the triangle is 5 and one side is 3. Use thePythagorean Theorem to find the other side.5 2 3 2 b 225 9 b 216 b 24 bThe height is 4.Use the formula for the area of a triangle.A 1 2 bhA 1 2 (4)(3) or 6Since the parallelogram consists of two triangles,the area of the parallelogram is 2 6 or 12. Thecorrect choice is A.271 Chapter 8
2. In order to write the equation of a circle, you needto know the coordinates of the center and thelength of the radius. The general equation for acircle is (x h) 2 (y k) 2 r 2 , where the centeris (h, k) and the radius is r. From the coordinatesof points A and B, you know the length of the sideis 4. So the center Q, has coordinates (0, 4).To calculate the length of the radius, draw theradius OB. This creates a 45°-45°-90° righttriangle. The two legs each have length 2. Thehypotenuse has length 22.(x 4) 2 (y 0) 2 (22) 2(x 4) 2 y 2 4(2)(x 4) 2 y 2 8The correct choice is B.3. Write the equation for the perimeter of arectangle. then replace x with its value in termsof y. Solve the equation for y.p 2x 2yp 2 2 3 y 2yp 4 3 y 2yp 1 03y 3 p10 yThe correct choice is B.4. Recall the triangle Inequality Theorem: the sumof the lengths of any two sides of a triangle isgreater than the length of the third side. Let xrepresent the length of the third side.40 80 x120 x40 x 80x 40Since x must be greater than 40, x cannot be equalto 40. The correct choice is A. To check youranswer, notice that the other answer choices aregreater than 40 and less than 120, so they are allpossible values for x.5. Since the answer choices have fractionalexponents of x, start by rewriting the expressionwith fractional exponents. Simplify the fractionsand use the rules for exponents to combine terms. 3 x 2 9 x 3 x 2 3 x 3 9 x 2 3 x 1 3 x (2 3 1 3 )x 1 or xThe correct choice is E.6. This figure looks more complex than it is. A semicircleis just one half of a circle. Notice that theanswer choices include , so don’t convert todecimals. Find the radius of each semi-circle.Calculate the area of each semi-circle.The area of the shaded region is the area of thelarge semi-circle minus the area of the mediumsemi-circle plus the area of the small semi-circle.Large semi-circle area 1 2 3 2 9 2Medium semi-circle area 1 2 2 2 4 2Small semi-circle area 1 2 1 2 1 2Shaded area 9 2 4 2 1 2 6 2 3The correct choice is A.7. The only values for which a rational function isundefined are values which make thedenominator 0. Since f(x) x – 1,the denominator is only 0 when x – 1 0 or x 1.The correct choice is D.8. Start by sketching a diagram of the counterx 2 –3x 230 28 3840–3 – 6m 4 – –2Use your calculator to find the area of the wholecounter and then subtract the area of the whitetiles in the center. The white tiles cover an area of(30 2)(40 2) or (28)(38).(30)(40) 1200(28)(38) 1064Red tiles 1200 1064 136The correct choice is B.9. First, find the slope of the line containing thepoints (–2, 6) and (4, –3).m – 9 36or – 2The point-slope form of the line isy – 6 – 3 2 (x – –2).y – 6 – 3 2 x – 3y – 3 2 x 3So the y-intercept of the line is 3.The correct choice is B.Chapter 8 272
10. Write an expression for the sum of the areas of thetwo triangles. Recall the area of a triangle is onehalf the base times the height. 1 2 (AC)(AB) 1 2 (CE)(ED)From the figure, you know that ABC and CDEare both isosceles, because of the angles marked x°and because BCD is a line segment. These twotriangles have equal corresponding angles.Since they are isosceles triangles, AC AB andCE ED. Use these equivalent lengths in theexpressions for the area sum. 1 2 (AC)(AB) 1 2 (CE)(ED) 1 2 (AC) 2 1 2 (CE) 2 1 2 [(AC) 2 (CE) 2 ]Using the Pythagorean Theorem for ACE, youknow that (AC) 2 (CE) 2 (AE) 2 or 1.So the sum of the two areas is 1 2 (1) 1 2 . You cangrid the answer either as .5 or as 12.273 Chapter 8
Chapter 9 Polar Coordinates and Complex Numbers9-1Pages 557–5581. There are infinitely many ways to represent theangle v. Also, r can be positive or negative.2. Draw the angle v in standard position. Extend theterminal side of the angle in the oppositedirection. Locate the point that is ⏐r⏐ units fromthe pole along this extension.3. Sample answer: 60° and 300°Plot (4, 120) such that v is in standard positionand ⏐r⏐ is 4 units from the pole. Extend theterminal side of the angle in the oppositedirection. Locate the point that is 4 units from thepole along this extension.r 4v 120 180 or v 120 18060 3004. The points 3 units from the origin in the oppositedirection are on the circle where r 3.5. All ordered pairs of the form (r, v) where r 0.90˚6. 7.120˚ 60˚2 2 33180˚150˚210˚240˚8. 90˚ 9.120˚ 60˚C150˚30˚180˚210˚240˚Polar CoordinatesA270˚1 2 3 41 2 3 4270˚Check for Understanding300˚300˚30˚0˚330˚0˚330˚567656764323D43322321 2 3 4B5331 2 3 410. Sample answer: 2, 13 7 , 2, 25 6 , 2, 7 ,62, 19 6 (r, v 2k)→2, 6 2(1) → 2, 13 6 →2, 6 2(2)) → 2, 25 6(r, v (2k 1))→ 2, 6 (1) → 2, 7 6 → 2, 6 (3) → 2, 19 6 53660116011611. 90˚ 12.120˚ 60˚13.180˚150˚210˚240˚180˚150˚120˚210˚240˚180˚180˚150˚210˚240˚270˚90˚270˚1 2 3 4300˚60˚1 2 3 4270˚300˚30˚0˚330˚30˚0˚330˚5101520300˚1 2 3 430˚0˚330˚16. 90˚ 17.120˚ 60˚150˚210˚240˚270˚E300˚30˚0˚330˚567656762343234323223231 2 3 414. P 1 P 2 2.5 2 (3)2 2(2.5)(3 ) cos 4 6.25 9 15 cos5 2 1 5.25 15 co s 5 12 4.3715a.90˚120˚ 60˚15b. 210 (30) 240NA 3 60 (r 2 ) 2 40360 (20 2 ) 838 ft 2Pages 558–560ExercisesF5331 2 3 45360116 6 60116Chapter 9 274
18. 90˚ 19.120˚ 60˚180˚150˚G210˚240˚270˚2 4 6 8300˚30˚0˚330˚20.2 2 3321.56676433201 2 3 4J1165322.2 2 3323.56676180˚43321 2 3 4L53011624. 90˚ 25.120˚ 60˚150˚210˚240˚30˚0˚1 2 3 4N330˚270˚300˚26. 2 2 27.335667643Q321 2 3 45301165676180˚150˚2343120˚K210˚240˚180˚150˚120˚210˚240˚5676180˚2343150˚ RP120˚210˚240˚H23290˚31 2 3 4270˚5360˚1 2 3 490˚270˚2300˚60˚1 2 3 4M3290˚300˚31 2 3 4270˚28. Sample answer: 2, 3 , 2, 7 ,3 (2, 240°),(2, 600°)(r, v 2k)→ 2, 3 2(0) → 2, 3 → 2, 3 2(1) → 2, 7 3(r, v (2k 1)180°)→ (2, 60° (1)180°) → (2, 240°)→ (2, 60° (3)180°) → (2, 600°)5360˚1 2 3 4300˚6011630˚0˚330˚30˚0˚330˚6011630˚0˚330˚29. Sample answer: (1.5, 540°), (1.5, 900°), (1.5, 0°),(1.5, 360°)(r, v 360k°)→ (1.5, 180° 360(1)°) → (1.5, 540°)→ (1.5, 180° 360(2)°) → (1.5, 900°)(r, v (2k 1)180°)→ (1.5, 180° (1)180°) → (1.5, 0°)→ (1.5, 180° (1)180°) → (1.5, 360°)30. Sample answer: 1, 7 , 133 1, ,3 1, 4 3 ,1, 10 3(r, v 2k)→ 1, 3 2(1) → 1, 7 3 → 1, 3 2(2) → 1, 13 3 (r, v (2k 1))→ 1, 3 (1) → 1, 4 3 → 1, 3 (3) → 1, 10 3 31. Sample answer: (4, 675°), (4, 1035°), (4, 135°),(4, 495°)(r, v 360k°)→ (4, 315 360(1)°) → (4, 675°)→ (4, 315 360(2)°) → (4, 1035°)(r, v (2k 1)180°)→ (4, 315 (1)180°) → (4, 135°)→ (4, 315 (1)180°) → (4, 495°)32. 90˚ 33.120˚ 60˚180˚150˚210˚240˚270˚1 2 3 4300˚30˚0˚330˚34. 2 2 35.3356676180˚43321 2 3 453011636. 90˚ 37.120˚ 60˚150˚30˚210˚240˚270˚1 2 3 4300˚0˚330˚56180˚76150˚2343120˚210˚240˚5676234323290˚270˚23231 2 3 45360˚1 2 3 4300˚31 2 3 4536011630˚0˚330˚60116275 Chapter 9
38. 120˚90˚ 39. 60˚120˚150˚30˚180˚1 2 3 4210˚330˚ 210˚240˚ 270˚300˚240˚40.90˚120˚ 60˚0˚150˚r 0180˚90˚60˚30˚1 2 3 40˚330˚270˚300˚49a. When v 120°, r 17. The maximum speed at120° is 17 knots.49b. When v 150°, r 13. The maximum speed at150° is 13 knots.50a.562323601 2 3 4180˚150˚1 2 3 430˚0˚76433253116210˚240˚270˚300˚330˚41. r 2 or r 2 for any v.42. P 1 P 2 4 2 6 2 4)(6) 2( os c5° (1070°) 1 16 36 48 cos°)(65 52 48 cos)(65 5.6343. P 1 P 2 1 2 5 2 2( 1)(5) cos3 4 6 1 5 20 1 cos 26 10 cos 5.35 7 1 2 7 1 2 44. P 1 P 2 (2.5) 2 .75) (1 2 )(1.752(2.5)cos 2 5 8 6.25 25 30.068.75 os c2 140 9.3125 8.75cos 2 140 3.1645. P 1 P 2 1.3 2 ) (3.6(1.3)(2 23.6) s co° (6247°)) ( 1.69 12.966 9.3(62°cos°) 47 14.65 6 9.3(15°cos) 4.8746. r (3) 2 4 2 5sin v 4 5 , v 53°180° 53° 127°Sample answer: (5, 127°)47. There are 360° in a circle. If the circle is cut into 6equal pieces, each slice measures 36 06or 60°.Beginning at the origin, the equation of the firstline is v 0°. The equation of the next line,rotating counterclockwise, is v 0 60 or 60°.The equation of the last line is v 60 60 or120°. Note that lines extend through the origin, so3 lines create 6 pieces.48. P 1 P 2 rr 2 1 2r 2 2 cos 1 r 2 (v v) rr 2 1 2r 2 2 cos 1 r 20 rr 2 1 2r 2 2 1 r 2 (r 1 r2 ) 2 r 1 r 250b. 3 3 2 3 or 120°Let R 3 100 or 300 and let r 0.25 100or 25.NA 3 60 (R 2 N) 3 60 (r 2 ) 1 20360 1 20360180˚150˚210˚240˚3 mph120˚90˚270˚60˚2 4 6 8300˚20 ((300 2 ) 1 360((25) 2 ) (90,000 625) 93,593 ft 2If each person’s seat requires 6 ft 2 of space,there are 93, 5936or 15,599 seats.51. The distance formula is symmetric with respect to(r 1 , v 1 ) and (r 2 , v 2 ). That is,52a.r r 2 2 2r 2 1 cos 2 r 1(v 1 v 2 ) rr 2 1 2r 2 2 cos[1 r 2 (v v 2 1 )] rr 2 1 2r 2 2 cos(v 1 r 2 2 v 1 )30˚0˚330˚52b. P 1 P 2 5 2 6 2 5)(6) 2( os c5° (3410°) 3 25 36 60 cos(35°) 61 60 cos(35°) 3.44No; the planes are 3.44 miles apart.53. Draw a picture.Boatsin v 3 8 sin 1 (sin v) sin 1 3 8 v 22.0° 8 mph54. 3, 2, 4 1, 4, 0 (3)(1) (2)(4) (4)(0) 3 8 0 11No, the vectors are not perpendicular becausetheir inner product is not 0.Chapter 9 276
55. Rewrite y 9x 3 as 9x y 3 0.d Ax 1 By 1 CAB 2 29(3) (1)(2) (3)91) 2 ( 23282 32 82 82 82 24 1 3282 821682 41 Distance is always positive.56. 1 sina 1sin2 a sin 1 2 a csc 2 a 1 cot 2 a357. Arc cos 2 30°In a 30°-60°-90° right12triangle, the angle oppositethe smallest leg is 30°. 358. y 5 cos 4vAmplitude 5; Period 2 4 or 2 59. b sin A 18.6 sin 30°9-2 9.3Since a b sin A, there is one solution.Page 565Find B. Find C.18.69. 3 s inB sin 30°C 180° 90° 30°18.6 sin 30° 9.3 sin B 60° 18.6 sin 30° sin B9.390° BFind c.c 9. 3 sin 60° sin 30°c sin 30° 9.3 sin 60°c 9.3 sin 60°sin 30°c 16.160. 3 or 1 positivef(x) x 3 4x 2 4x 10 negativeP Q 1Since there are only positive real zeros, the onlyrational real zero is 1.61. x 3considered.x 5x2 2x 390˚x120˚ 60˚ 2 5 x 3x 3150˚30˚ 3 x 1 5 10180˚1 210As x → , x→ 0. Therefore, the slant 5210˚330˚240˚ 270˚300˚asymptote is y x 3.62. y-axis:For x: f(x) x 4 3x 2 2For x: f(x) (x) 4 3(x) 2 2 x 4 3x 2 2So, in general, point (x, y) is on the graph if andonly if (x, y) is on the graph.63. 2 0 4 4 0 (1) 11 1101 5 3 5 33 44542(5) 4(5) 1(1)1164. 11 (3) 1411 (2) 1311 (1) 1211 0 11{(3, 14), (2, 13), (1, 12), (0, 11)}For each x-value, there ia a unique v-value.Yes, the relation is a function.65. Since the two triangles formed are right triangles,the side opposite the right angles, AB, interceptan arc measuring 180°, or half the circle. AB is adiameter.C d50 d50 dThe correct choice is E.Graphs of Polar EquationsCheck for Understanding1. Sample answer: r sin 2vThe graph of a polar equation whose form isr a cos nv or a sin nv, where n is a positiveinteger, is a rose.2. 1 sin v 1 for any value of v. Therefore, themaximum value of r 3 5 sin v is r 3 5(1)or 8. Likewise, the minimum value of r 3 5sin v is r 3 5(1) or 2.3. The polar coordinates of a point are not unique. Apoint of intersection may have one representationthat satisfies one equation in a system, anotherrepresentation that satisfies the other equation,but no representation that satisfies bothsimultaneously.4. Barbara is correct. The interval 0 v is notalways sufficient. For example, the interval 0 v only generates two of the four petals for thevrose r sin 2v. r sin 2 is an example wherevalues of v from 0 to 4 would have to be5. 6.cardioid0˚150˚180˚120˚210˚240˚limaçon90˚60˚2 4 6 8270˚300˚30˚0˚330˚277 Chapter 9
7.2 2 338.5665623236Pages 565–567 Exercises11. 90˚ 12.120˚ 60˚120˚90˚60˚76rose9.564323322533011166764333269 1253spiral of Archimedes0116150˚180˚1 2 3 430˚0˚150˚180˚210˚330˚ 210˚240˚ 270˚300˚240˚circlecardioid13. 2 2 14.33120˚566150˚2 4 6 830˚0˚330˚270˚300˚90˚60˚30˚7656764323433243215338121653021162 sin 2 cos 2sin cos 2sin 1 2 sin 2 2 sin 2 sin 1 0(2 sin 1)(sin 1) 02 sin 1 0orsin 1 0sin 1 2 sin 1 6 or 5 6 or 3 2If 6 or 5 6 is substituted in either originalequation, r 1. If 3 2 is substituted in eitheroriginal equation, r 2. The solutions are1, 6 , 1, 5 6 , and 2, 3 2 .10a.26011610b. Sample answer: 0 v 14 3Begin at the origin and “spiral” twice around it,or through 4 radians. Move straight upthrough 4 2 or 9 2 radians. Now move to theleft slightly, through approximately 9 2 6 or 14 radians.376180˚43325 10 15 205301161 20˚330˚180˚210˚240˚270˚spiral of Archimedes lemniscate15. 90˚ 16.90˚120˚ 60˚120˚150˚210˚240˚270˚300˚30˚150˚180˚210˚240˚roserose17. 2 2 18.33566 150˚764332270˚1 2 3 4300˚60˚1 2300˚0180˚4 812162 4 6 853116Spiral of Archimedes19.2 2 3320.5667643cardioid1322 3 4530116120˚210˚240˚limaçon56762343lemniscate90˚270˚213260˚300˚32 3 4530˚330˚30˚0˚330˚30˚0˚330˚60116Chapter 9 278
21. 90˚ 22.120˚ 60˚150˚30˚150˚120˚90˚60˚30˚28. (1, 0.5), (1, 1.0), (1, 2.1), (1, 2.6), (1, 3.7), (1, 4.2),(1, 5.2), (1, 5.8)180˚10˚180˚12340˚210˚330˚210˚330˚240˚ 270˚300˚240˚ 270˚300˚rosecardioid23. Sample answer: r sin 3vThe graph of a polar equation of the form r a cos3v or r a sin 3v is a rose with 3 petals.v24. Sample answer: r 2 4 a 2 r 1 2 1 2 a r 2 25. 3 2 cos 1 cos 0The solution is (3, 0)26. 1 cos 1 cos 2 cos 0cos 0 2 or 3 227.Substituting each angleinto either of the originalequations gives r 1, sothe solutions of the systemare 1, 2 and 1, 3 22567623433231 253 .60116567656762 sin 2 sin 2sin sin 2sin 2 cos sin 0 2 cos sin sin 0 sin (2 cos 1)sin 0 or 2 cos 1 0cos 1 2 0 or or 3 or 5 323432343213223232 3 45331 2536011660116If 0 or is substituted in either originalequation, r 0. If 3 or 5 3 is substituted ineither original equation, r 3 or r 3,respectively. The solutions are (0, 0), (0, ),3, 3 , and 3, 5 3 .[2, 2] scl1 by [2, 2] scl129. (2, 3.5), (2, 5.9)[6, 6] scl:1 by [6, 6] scl130. (3.6, 0.6), (2.0, 4.7)[4, 4] scl1 by [4, 4] scl131a. If the lemniscate is 6 units from end to end, thena 1 (6) or 3.2r 2 9 cos 2 or r 2 9 sin 231b. If the lemniscate is 8 units from end to end, thena 1 (8) or 4.2r 2 16 cos 2 or r 2 16 sin 232.90˚120˚ 60˚180˚150˚210˚240˚270˚2 4 6 8300˚30˚0˚330˚This microphone will pick up more sounds frombehind than the cardioid microphone.33. 0 v 4: Begin at the origin and curl aroundonce, or through 2 radians. Curl around a secondtime and go through 2 2 or 4 radians.34. All screens are [1, 1] scl1 by [1, 1] scl1279 Chapter 9
34a. r cos 2 r cos 7 r cos 4 r cos 9 r cos 6r cos 8When n 10, two more outer rings will appear.34b. r cos 3r cos 5When n 11, the innermost loop will be on theleft and there will be an additional outer ring.35. Sample answer: r 1 sin vAheart resembles the shape of a cardioid. Thesine function orients the heart so that the axis ofsymmetry is along the y-axis. If a 1, the heartpoints in the right direction.36a. For a limaçon to go back on itself and have aninner loop, r must change sign. This will happenif ⏐b⏐ ⏐a⏐.36b. For the other two cases, ⏐a⏐ ⏐b⏐.Experimentation shows that the dimpledisappears when ⏐a⏐ ⏐2b⏐, so there is adimple if ⏐b⏐ ⏐a⏐ ⏐2b⏐.36c. For this remaining case, there is neither aninner loop nor a dimple if ⏐a⏐ ⏐2b⏐.37a. Subtracting a from v rotates the graphcounterclockwise by an angle of a.37b. Multiplying v by 1 reflects the graph about thepolar axis or x-axis.37c. Multiplying the function by 1 changes r to itsopposite, so the graph is reflected about theorigin.37d. Multiplying the function by c results in adilation by a factor of c. (Points on the graphmove closer to the origin if c 1, or fartheraway from the origin if c 1.)38. Sample answer: (4, 405°), (4, 765°), (4, 135°),(4, 225°)(r, 360k°)→ (4, 45° 360(1)°) → (4, 405°)→ (4, 45° 360(2)°) → (4, 765°)(r, (2k 1)180°)→ (4, 45° (1)180°) → (4, 135°)→ (4, 45° (1)180°) → (4, 225°)Chapter 9 280
39. v w i j 2 3k 041 23 0 2 0 2 3 i j k2 4 1 4 1 2 12i 8j 7k 12, 8, 72, 3, 0 12, 8, 7 24 (24) 0 or 01, 2, 4 12, 8, 7 12 (16) 28 or 040. 3.5 cm, 87°sin 2 x41. cos 4 x cos 2 x sin 2 x tan 2 xsin 2 xcos 2 x (cos 2 x sin 2 x)sin 2 x (cos 2 x)(1) tan 2 x tan 2 x tan 2 xtan 2 x tan 2 x42. Find C.C 180° 21°15 49°40 109°5Find b.b28.9 sin 4 9°40 sin 109°5b sin 109°5 28.9 sin 49°40b b 23.3Find a.a28.9 sin 2 1°15 sin 109°5a sin 109°5 28.9 sin 21°159-3sin 2 x cos 2 x28.9 sin 49°40sin 109°528.9 sin 21°15sin 109°5a a 11.143. NY LA MiamiBus $240 $199 $260Train $254$322 $42644. 1 8 6 4 1 8 1 2 138 8 1 3 1 8So 6 4 8 1 3 1 63 3 8 31 6 1 6 2 63The correct choice is A.Polar and RectangularCoordinatesPage 571 Check for Understanding1. Sample answer: (22, 45°)r 2 2 22 Arctan 2 2 8 45° 222. The quadrant that the point lies in determineswhether v is given by Arctan y x or Arctan y x .3. x 2r cos 22r co s r 2 sec 4. To convert from polar coordinates to rectangularcoordinates, substitute r and v into the equationsx r cos v and y r sin v. To convert fromrectangular coordinates to polar coordinates, usethe equation r x 2 y 2 to find r. If x 0, v Arctan y x . If x 0, v Arctan y x . If x 0, youcan use or any coterminal angle for v.2yO5. r (2) 2) 2 ( 2 v Arctan 22 4 or 2 3 42, 3 46. r (2) 2 ) (5 2 v Arctan 5 2 29 5.39 4.33(5.39, 4.33)7. x 2 cos 4 3 y 2 sin 4 3 1 3(1, 3)8. x 2.5 cos 250° y 2.5 sin 250° 0.86(0.86, 2.35)9. y 2r sin v 22.352r si n vr 2 csc v10. x 2 y 2 16(r cos v) 2 (r sin v) 2 16r 2 (cos 2 v sin 2 v) 16r 2 16r 4 or r 411. r 6x 2 y 2 6x 2 y 2 3612. r sec v r r 1r c os v1 1xx 1rx r cos y r sin x281 Chapter 9
13a.150˚180˚120˚90˚210˚330˚240˚ 270˚300˚13b. No. The given point is on the negative x-axis,directly behind the microphone. The polarpattern indicates that the microphone does notpick up any sound from this direction.Pages 572–573 Exercises14. r 22) 2 ( 2 v Arctan 2 8 or 22Add 2 to obtain v 7 4 .22, 7 415. r 0 2 1 2 4 1 or 1Since x 0 when y 1, v 2 .1, 2 16. r 13) 2 ( 2 4 or 22, 3 17. r 1 4 2 3424 1 6 2 4 or 1 2 1 2 , 4 31260˚3430˚0˚v Arctan 3 3v Arctan 4 1 4 Arctan 118. r 3 2 82 v Arctan 8 3 73 8.54 1.21(8.54, 1.21)19. r 47) 2 ( 2 v Arctan 74 65 8.06 1.05Add 2 to obtain v 5.23.(8.06, 5.23)20. x 3 cos 2 y 3 sin 2 0 3(0, 3) 1 2 22 1 2 22 2 44221. x 1 2 cos 3 4 y 1 2 sin 3 4 2, 24 43 122. x 1 cos 6 y 1 sin 6 1 321 1 2 32 1 2 3, 1 2 2 23. x 2 cos 270° y 2 sin 270° 0 2(0, 2)24. x 4 cos 210° y 4 sin 210° 4 32 1 42 232(23, 2)25. x 14 cos 130° y 14 sin 130° 9.00 10.722 (9.00, 10.72)26. x 7r cos v 77r c os vr 7 sec v27. y 5r sin v 55r si n vr 5 csc v3 or 4 328. x 2 y 2 25(r cos v) 2 (r sin v) 2 25r 2 (cos 2 v sin 2 v) 25r 2 25r 5 or r 529. x 2 y 2 2y(r cos v) 2 (r sin v) 2 2r sin vr 2 (cos 2 v sin 2 v) 2r sin vr 2 2r sin vr 2 sin v30. x 2 y 2 1(r cos v) 2 (r sin v) 2 1r 2 (cos 2 v sin 2 v) 1r 2 (cos 2v) 1r 2 1 cos2vr 2 sec 2v31. x 2 (y 2) 2 4x 2 y 2 4y 4 4(r cos v) 2 (r sin v) 2 4r sin v 0r 2 (cos 2 v sin 2 v) 4r sin v 0r 2 4r sin v 0r 2 4r sin vr 4 sin v32. r 2x 2 y 2 2x 2 y 2 433. r 3x 2 y 2 3x 2 y 2 9Chapter 9 282
34. v 3 Arctan y x 3 35. r 2 csc v r r 2r s in v y x 13y 3x1 2 y y 236. r 3 cos vr 2 3r cos vx 2 y 2 3x37. r 2 sin 2v 8r 2 2 sin v cos v 82r sin v r cos v 82yx 8xy 438. y x y x 1Arctan y x Arctan 1v 4 39. r sin vr 2 r sin vx 2 y 2 y40. x 325 cos 70° y 325 sin 70° 111.16 305.40(111.16, 305.40)y2 6 331 5 4 4 41. — — 5 24 6 624 4 24 0.52 unit42. Drop a perpendicular from the point with polarcoordinates (r, v) to the x-axis. r is the length ofthe hypotenuse in the resulting right triangle.x is the length of the side adjacent to angle v, socos v x . Solving for x gives x r cos v. y is therlength of the side opposite angle v, so sin v y r .Solving for y gives y r sin v. (The figure is drawnfor a point in the first quadrant, but the signswork out correctly regardless of where in theplane the point is located.)yOr(r, )xx43. horizontal distance:25(4 2 cos 120°) 75 m eastvertical distance:25(3 2 sin 120°) 118.30 m northyO2 120˚44a. x 4 cos 20° y 4 sin 20° 3.76 1.373.76, 1.37x 5 cos 70° y 5 sin 70° 1.71 4.701.71, 4.7044b. 3.76, 1.37 1.71, 4.70 3.76 1.71, 1.37 4.70 5.47, 6.0744c. 5.47 r cos v; 6.07 r sin v 6 . 07r sinv 5.47r cosv 6 . 07 tan v5.4747.98 v; 47.98°5.47 r cos 47.98°5.47 cos r47.98°8.17 r8.17∠47.98°44d. 8.17 sin (3.14t 47.98°)45. r 2a sin v 2a cos vr 2 2ar sin v ar cos vx 2 y 2 2ay 2axx 2 2ax y 2 2ay 0(x a) 2 (y a) 2 2a 2The graph of the equation is the circle centered at(a, a) with radius 2⏐a⏐.46.90˚120˚ 60˚150˚180˚123430˚0˚210˚330˚240˚ 270˚300˚47. Sample answer: (2, 405°), (2, 765°), (2, 225°),(2, 585°)(r, v 360k°)→ (2, 45° 360(1)°) → (2, 405°)→ (2, 45° 360(2)°) → (2, 765°)(r, v (2k 1)180°)→ (2, 45° (1)180°) → (2, 225°)→ (2, 45° (3)180°) → (2, 585°)48. ⏐r ⏐ 2 50 2 425 2 2 50 425 cos 30°⏐r ⏐ 382.52 mph50 382.52si n v rs in30°50 sin 30° 382.52 sin v425 mph30˚ 50 sin30° sin v382.5250 mph3°45 vThe direction is 3°45 west of south.x283 Chapter 9
49. sin 2 A cos A 11 cos 2 A cos A 10 cos 2 A cos A 20 (cos A 2)(cos A 1)cos A 2 0 or cos A 1 0cos A 2 cos A 1A 0°50. y21O1290˚y 2 cos 180˚ 270˚ 360˚51. The terminal side is in thethird quadrant and thereference angle is210 180 or 30°.cos 210° 32( )3 12,230˚52. Enter the x-values in L1 and the f(x)-values in L2of your graphing calculator. Make a scatter plot.The data points are in the shape of parabola.Perform a quadratic regression.a 0.07, b 0.73, c 1.36Sample answer:y 0.07x 2 0.73x 1.3653. 2⏐ 1003 0 20 2 4 8 1 0 2 0 124 510⏐ 0x 4 2x 3 4x 2 5x 1054. m 62 5 145 (y 145) 60(x 17)25 17 60 y 60x 87555. x y and y z, so x z.zIf x z, then 0 x 1.The correct choice is C.9-4Polar Form of a Linear EquationPages 577–578 Check for Understanding1. The polar equation of a line is p r cos (v f).r and v are the variables. p is the length of thenormal segment from the line to the origin andf is the angle the normal makes with the positivex-axis.2. For r to be equal to p, we must have cos (v f) 1. The first positive value of v for which this istrue is v f.yx3. The graph of the equation x k is a vertical line.Since the line is vertical, the x-axis is the normalline through the origin. Therefore, f 0° or f 180°, depending on whether k is positive ornegative, respectively. The origin is ⏐k⏐ unitsfrom the given vertical line, so p ⏐k⏐. The polarform of the given line is k r cos (v 0°) if k ispositive or k r cos (v 180°) if k is negative.Both equations simplify to k r cos v.4. You can use the extra ordered pairs as a check onyour work. If all the ordered pairs you plot are notcollinear, then you have made a mistake.5. AB 2 2 34) 2 ( 25Since C is negative, use 5. 3 5 x 4 5 y 2 0cos f 3 5 , sin f 4 5 , p 2f Arctan 4 3 53° or 307°p r cos (v f)2 r cos (v 307°)6. AB 2 2 (2) 2 4 225Since C is negative, use 25.2 4 9 2x 5 2y 5 2 05cos f 5, sin f 2 5, p 9 555 10f Arctan(2) 63°Since cos f 0, but sin f 0, the normal lies inthe second quadrant.f 180° 63° or 117°p r cos (v f)9 5 r cos (v 117°)107. 3 r cos (v 60°)0 r cos (v 60°) 30 r (cos v cos 60° sin v sin 60°) 30 1 2 r cos v 3 r sin v 320 1 2 x 3y 320 x 3y 6 orx 3y 6 08. r 2 sec v 4 r cos (v 4 ) 2rcos v cos 4 sin v sin 4 2 0 222 r cos v r sin v 2 02 222x y 2 022x 2y 4 0Chapter 9 284
9. 2 2 3310.566765676432343322 4 6 82325335301162 4 6 860116180˚150˚120˚210˚240˚90˚270˚60˚1 2 3 4300˚11a. p r cos (v f) → 5 r cos v 5 6Since the shortest distance is along the normal,the answer is (p, f) or 5, 5 6 .11b.Pages 578–579 Exercises12. AB 2 2 724) 2 ( 225Since C is positive, use 25.7 2x5 2 425 y 4 07cos f 2,5sin f 2 245 , p 4f Arctan 2 4 7 74°Since cos f 0, but sin f 0, the normal lies inthe second quadrant.f 180° 74° or 106°p r cos (v f)4 r cos (v 106°)13. AB 2 2 21 2 20 229Since C is negative, use 29. 2 129 x 2 029 y 8 729 0cos f 2 129 , sin f 2 029 , p 3f Arctan 2 021 44°p r cos (v f)3 r cos (v 44°)30˚0˚330˚14. AB 2 2 6 8) 2 ( 210Since C is negative, use 10.6 8 1x 0 1y 2 1 00 10cos f 3 5 , sin f 4 5 , p 2.1f Arctan 4 3 53°Since cos f 0, but sin f 0, the normal lies inthe fourth quadrant.f 360° 53° or 307°p r cos (v f)2.1 r cos (v 307°)15. AB 2 2 3 2 2 2 13Sinc C is negative, use 13. x 3 13 2 5 y 13 013cos f 3 13, sin f 2 13 , p 5 131313 13f Arctan 2 3 34°p r cos (v f)5 13 r cos (v 34°)1316. AB 2 2 4 5) 2 ( 214Since C is negative, use 41. x 4 41 y 5 41 1 0 0 41cos f 4 41, sin f 5 41 , p 1041414141f Arctan 5 4 51°Since cos f 0, but sin f 0, the normal lies inthe fourth quadrant.f 360° 51° or 309°p r cos (v f) 10 41 41 r cos (v 309°)17. AB 2 2 (–1) 2 3 2 10Since C is negative, use 10.1 10 3x 10 7y 10 0cos f 10 10sin f 310 , p 7 1010 10f Arctan (3) 72°Since cos f < 0, but sin f > 0, the normal lies inthe second quadrant.f 180° 72° or 108°p r cos (v f) 7 1010 r cos (v 108°)18. 6 r cos (v 120°)0 r (cos v cos 120° sin v sin 120°) 60 1 2 r cos v 3 r sin v 620 1 2 x 3y 620 x 3y 12 orx 3y 12 0285 Chapter 9
19. 4 r cos v 4 0 r cos v cos 4 sin v sin 4 40 2 r cos v 2 r sin v 4220 2x 2y 42 20 2x 2y 8 or2x 2y 8 020. 2 r cos (v )0 r (cos v cos sin v sin ) 20 r cos v 20 x 2x 221. 1 r cos (v 330°)0 r (cos v cos 330° sin v sin 330°) 10 3 r cos v 1 r sin v 1220 3x 1 2 2 y 10 3x y 2 or3x y 2 022. r 11 sec v 7 6r cos v 7 6 11rcos v cos 7 6 sin v sin7 6 11 0 3 r cos v 1 r sin v 11 022 3x 1 y 11 02 23x y 22 023. r 5 sec (v 60°)r cos (v 60°) 5r (cos v cos 60° sin v sin 60°) 5 0 1 2 r cos v 3 r sin v 5 02 1 2 x 3y 5 02x 3y 10 024. 25.90˚120˚ 60˚180˚150˚210˚240˚180˚3 6 9 12270˚300˚1 2 3 430˚0˚330˚26. 90˚27.120˚ 60˚150˚210˚240˚270˚300˚30˚0˚330˚5676180˚150˚2343120˚210˚240˚23290˚31 2 3 4270˚5360˚2 4 6 8300˚6011630˚0˚330˚28. 2 2 3329.5667643322 4 6 853011656762343430. m or 2 6 3 (y 1) 2 3 (x 4) → 2x 3y 5 0AB 2 2 2 2 3 2 13Since C is negative, use 13.2 3 5 13 13 13 0cos f 2 13, sin f 3 13, p 5 131313 13f Arctan 3 2 56°p r cos (v f)5 13 r cos (v 56°)1331. p r cos (v f)→ p 3 cos 4 f f23232 4 6 85360116→ p 2 cos 7 6Use a graphing calculator and the INTERJECTfeature to find solutions to the system at (2.25,0.31) and (5.39, 0.31). Since p, the length of thenormal, must be positive, use f 2.25 and p 0.31.0.31 r cos (v 2.25)32a. p r cos (v f) → 6 r cos (v 16°)Since the shortest distance is along the normal,the closest the fly came was p or 6 cm.32b. (p, f) or (6, 15°)33. Since both normal segments have length 2, p mustbe 2 in both equations. Since the two linesintersect at right angles, their normals alsointersect at right angles. This can be achieved byhaving the two f-values differ by 90°. To makesure neither line is vertical, neither f-valueshould be a multiple of 90°. Therefore, a sampleanswer is 2 r cos (v 45°) and 2 r cos (v 135°).34. m 0(y 4) 0(x 5) → y 4 0cos f 0, sin f 1, p 4Since cos f 0 when sin f 1, f 90°.p r cos (v f)4 r cos (v 90°)Chapter 9 286
35a.180˚150˚120˚210˚240˚90˚270˚60˚30˚125250375500300˚35b. p r cos (v f)→ p 125 cos (130 f)→ p 300 cos (70 f)Use a graphing calculator and the INTERSECTfeature to find the solutions to the system at(45, 124.43) and (135, 124.43). Sinc p, thelength of the normal, must be positive, use f 135° and p 124.43.124.43 r cos (v 135°)36. k r sin (v a)k r [sin v cos a cos v sin a]k r sin v cos a r cos v sin ak y cos a x sin aThis is the equation of a line in rectangularcoordinates. Solving the last equation for y yieldsky (tan a)x co. The slope of the line showss athat a is the angle the line makes with the x-axis.To find the length of the normal segment in thefigure, observe that the complementary angle to ain the right triangle is 90° a, so the v-coordinateof P in polar coordinates is 180° (90° a) a 90°. Substitute into the original polarequation to find the r-coordinate of P:k r sin (a 90° a)k r sin 90°k rTherefore, k is the length of the normal segment.PyO37. p r cos (v f)→ p 40 cos (0° f)→ p 40 cos (72° f)Use a graphing calculator and the INTERSECTfeature to find the solutions of the system at(144, 32.36) and (36, 32.36). Since p, the lengthof the normal, must be positive, use f 36° andp 32.36.32.36 r cos (v 36°)38. r 6x 2 y 2 6x 2 y 2 3639. The graph of a polar equation of the formr a sin nv is a rose.0˚330˚x40. x 3y 6y x 631. 90˚2.120˚ 60˚180˚150˚210˚240˚180˚270˚22 4 6 84300˚6830˚0˚330˚3. 120˚90˚4.60˚150˚210˚240˚y 1 3 x 2x t, y 1 3 t 2N41. A 3 60 r 2 65 3 606 2 20.42 ft 242. Since 360° lies on the x-axisof the unit circle at (1, 0),sin 360° y or 0.43. 2x 3 5x 2 12x 0x(2x 2 5x 12) 0x(2x 3)(x 4) 0x 0 or x 3 2 or x 444. c 2 d 2 48(c d)(c d) 4812(c d) 48c d 4Page 579Mid-Chapter Quiz270˚300˚30˚0˚330˚5. r (2) 2) 2 ( 25676567623432343y23223231 2 3 4(1, 0)x601165335362 v Arctan 4 or 2 4 Since (2, 2) is in the third quadrant,v 4 or 5 4 .2, 5 46. r 04) 2 ( 2 16 or 4Since x 0 when y 4, v 3 2 .4, 3 201116287 Chapter 9
7. x 2 y 2 36x 2 y 2 36r 6 or r 68. r 2 csc vr sin v 2y 29. AB 2 2 5 12) 2 ( 213Since C is positive, use 13.5 1x3 12 1 3y 3 1 3 05cos f 1,3sin f 12 1 3, p 3 1 3f Arctan 1 25 67°Since cos f 0, but sin f 0, the normal lies inthe second quadrant.f 180° 67° or 113°p r cos ( f)3 1 r cos (v 113°)310. AB 2 2 (2) 2 ) (62210Since C is negative, use 210.2 6 2 y 02 2Reals(b 0)1010 cos f 1,010sin f 3 10 , p 0 110 10f Arctan 31 72°Since cos f 0 and sin f 0, the normal lies inthe third quadrant.f 180° 72° or 252°p r cos (v f) 0 1 r cos (v 252°)109-5210 xSimplifying Complex NumbersPage 583 Check for Understanding1. Find the (positive) remainder when the exponentis divided by 4. If the remainder is 0, the answeris 1; if the remainder is 1, the answer is i; if theremainder is 2, the answer is 1; and if theremainder is 3, the answer is i.2. Complex Numbers (a bi)Imaginary(b 0)PureImaginary(a 0)3. When you multiply the denominators, you will bemultiplying a complex number and its conjugate.This makes the denominator of the product a realnumber, so you can then write the answer in theform a bi.4. Sample answer: x 2 1 0(x i)(x i) 0, where the solutions are x i.x 2 xi xi i 2 0x 2 (1) 0x 2 1 05. i 6 (i 4 ) 2 i 2 1 2 (1)16. i 10 i 2 (i 4 ) 2 i 2 i 2 (1) 2 i 2 i 21 (1) or 27. (2 3i) (6 i) (2 (6)) (3i i)4 4i8. (2.3 4.1i) (1.2 6.3i) (2.3 (1.2)) (4.1i (6.3i)) 3.5 10.4i9. (2 4i) (1 5i) (2 (1)) (4i 5i) 1 9i10. (2 i) 2 (2 i)(2 i) 4 4i i 2 3 4ii i11. 1 2i 1 1 2i2i 1 2ii 2i2 1 4i2 i 25 2 5 1 5 i12. (2.5 3.1i) (6.2 4.3i) (2.5 (6.2)) (3.1i 4.3i)3.7 7.4i NPages 583–585 Exercises13. i 6 i 4 i 2 1 1114. i 19 (i 4 ) 4 i 3 1 4 ii15. i 1776 (i 4 ) 444 1 444 116. i 9 i 5 (i 4 ) 2 i (i 4 ) 2 i 3 1 2 i 1 2 i i (i) or 017. (3 2i) (4 6i) (3 (4)) (2i 6i)1 8i18. (7 4i) (2 3i) (7 2) (4i 3i) 9 7i19. 1 2 i (2 i) 1 (2) (i (i))2 3 2 2i20. (3 i) (4 5i) (3 (4)) (i (5i))7 4i21. (2 i)(4 3i) 8 10i 3i 2 5 10i22. (1 4i) 2 (1 4i)(1 4i) 1 8i 16i 215 8iChapter 9 288
23. (1 7i)(2 5i)334. 2 5i 27i 35i 2(2 (2 35) (27 5)i24. (2 3 )(1 12 ) (2 3i)(1 12i)2 212i 3i 36i 22 43i 3i 68 33i2 i 2 i25. 1 2i1 1 2i 2i1 2i 2 3i 2i21 4i 2 4 3i5 4 5 3 5 i3 2i 3 2i26. 4 i 4 i i44 i 12 1 2i161ii 22 10 11i17 1 0 11 i171 727. 5 i 5 i 5 i5 i 5 i 5 i 25 10i i225 i2 24 10i26 1 2 5 131 3 i28. (x i)(x i) 0x 2 i 2 0x 2 1 029. (x (2 i))(x (2 i)) 0(x 2 i)(x 2 i) 0x 2 2x xi 2x 4 2i xi 2i i 2 0x 2 4x 4 1 0x 2 4x 5 030. (2 i)(3 2i)(1 4i) (6 i 2i 2 )(1 4i) (8 i)(1 4i) 8 31i 4i 2 12 31i31. (1 3i)(2 2i)(1 2i) (2 8i 6i 2 )(1 2i) (4 8i)(1 2i) 4 16i 16i 212 16i 1 1 2 3i2 3i32. 1 2i1 2i 1 2i 1 2i 1 2 22i 3i 6i 21 2i 2 1 2 6 22 3i3 1 6 63 3 23 633. 2 2i 2 2i 3 6i3 6i 3 6i 3 6ii6 26i 32i 12i29 6i 2(6 23) (26 32)i15 2 5 2 315 52 2 615 i i 3 i i) 2 (2 i) (2 i)3 i 4 4i i 23 i 3 4i3 i 3 3 4i 4i3 4i 9 9i 4i 29 16i2 13 9i25 1 325 925 i(1 i)35.2(3 2i) 2 (1 i)(1 i)(3 2i)(3 2i)1 2i i2 9 12i 4i 22i 5 12i2i 5 5 12i12i 5 12i10i 24i2 2 5 144i2 24 10i1692410 1 691i6936a. Z R (X L X C )j→ Z 10 (1 2)j → Z 10 j ohms→ Z 3 (1 1)j → Z 3 0j ohms36b. (10 j) (3 0j) (10 3) (1j 0j) 13 j ohms136c. S Z → S 1 6 3j1 6 6 3j3j 6 3j6 3j 3 6 9j 2 6 3j45 0.13 0.07j siemens8i (8i) 2 25)4(1)(37a. x 2(1) 8i 36 23 4i37b. No37c. The solutions need not be complex conjugatesbecause the coefficients in the equation are notall real.37d. (3 4i) 2 8i(3 4i) 25 07 24i 24i 32 25 00 0(3 4i) 2 8i(3 4i) 25 07 24i 24i 32 25 00 038. f(x yi) (x yi) 2 x 2 2xyi y 2 (x 2 y 2 ) 2xyi39a. z 0 2 iz 1 i(2 i) i 2 or 1 2iz 2 i(2i 1) 2i 2 i or 2 iz 3 i(2 i) 2i i 2 or 1 2iz 4 i(1 2i) i 2i 2 or 2 iz 5 i(2 i) 2i i 2 or 1 2i289 Chapter 9
39b. z 0 1 0iz 1 (0.5 0.866i)(1 0i) 0.5 0.866iz 2 (0.5 0.866i)(0.5 0.866i) 0.25 0.866i 0.750.500 0.866iz 3 (0.5 0.866i)(0.500 0.866i)0.250 0.7501.000 0.000iz 4 (0.5 0.866i)(1.000) 0.500 0.866iz 5 (0.5 0.866i)(0.500 0.866i)0.250 0.866i 0.75 0.500 0.866i40. (1 2i) 3 (1 12i) 31 (3 4i)(1 2i)1 11 2i1 11 11 2i 2i 11 2i 1 1 2i12511 21 25 1 25 i41. c 1 (cos 2t i sin 2t) c 2 (cos 2t i sin 2t) c 1 cos 2t c 1 i sin 2t c 2 cos 2t c 2 i sin 2t (c 1 c 2 )(cos 2t) (c 1 c 2 )(i sin 2t) (c 1 c 2 )(cos 2t) only if c 1 c 242. Ab 2 2 62) 2 ( 2210Since C is positive, use 210. 6x 2y 3 02 10 2 10 2 10 cos f 3 10 , sin f 01011, 0p 3 1020 f Arctan 1 3 18°Since cos f 0, but sin f 0, the normal lies inthe second quadrant.f 180° 18° or 162°p r cos (v f)3 10 r cos (v 162°)2043.5676234323237 14 212853644. x (3), y 6 t1, 4x 3, y 6 t1, 445. u 1 8, 6, 4 22, 6, 34 2, 3 2 , 1 4, 12, 6 6, 2 72 , 5 011646. tan a 4 3 cot B 512 tan 2 a 1 sec 2 a 1 cot 2 B csc 2 B5 2 csc 2 B 4 3 2 1 sec 2 a 1 1 2 599 2 526944446923 sec 2 a 1 1 csc 2 B cos 2 a 1 1 sin 2 B 3 5 cos a 1 1 sin Bsin 2 a cos 2 a 1 sin 2 B cos 2 B 1sin 2 a 3 5 2 1 1 213 2 cos 2 B 1sin 2 a 1 625 cos 2 B 126sin a 4 5 5cos B 1 3cos (a B) cos a cos B sin a sin B 3 5 513 4 5 1 1 3 365 47. amplitude 1 2 (7) or 3.5period 2 12or 6 y 3.5 cos 6 t48. h x3tan 52° xx3 4523 x tan 52° 45 tan 52° x3x tan 52° x3 45 tan 52°45tan52°x t an 52° 3x 127.40h x3 127.40(3) 221 ft120˚8˚52˚ 60˚45 ft x30˚h49. Enter the x-values in L1 and the f(x)-values in L2of your graphing calculator. Make a scatter plot.The data points are in the shape of a parabola, soa quadratic function would best model the set ofdata.50. Let d depth of the original pool.The second pool’s width 5d 4, the length 10d 6, and the depth d 2.(5d 4)(10d 6)(d 2) 3420(50d 2 70d 24)(d 2) 342050d 3 100d 2 70d 2 140d 24d 48 342050d 3 170d 2 164d 3372 025d 3 85d 2 82d 1686 0Use a graphing calculator to find the solutiond 3.The dimensions of the original pool are 15 ft by30 ft by 3 ft.51. 80 k(5)(8)2 ky 2(16)(2) 6459Chapter 9 290
52. y 7 x 2x 7 y 2x 7 y 2x 7 y 2x 7 yf 1 (x) 7 x53.yO (1, 1)(6, 8)(6, 1)xf(x, y) 2x yf(1, 1) 2(1) 1 or 3f(6, 1) 2(6) 1 or 11f(6, 8) 2(6) 8 or 4The maximum value is 3 and the minimum valueis 11.54. x 2y 7z 14x 3y 5z 21y 2z 7x 3y 5z 21 → 5x 15y 25z 1055x y 2z 7 5x y 2z 716y 27z 112y 2z 7 → 16y 32z 11216y 27z 112 1 6 y 2 7 z 1 1 2 59z 0z 0y 2(0) 7 → y 7x 2(7) 7(0) 14 → x 0(0, 7, 0)55. Since BC BD, m∠BDC m ∠DCB xm ∠DBC 180 120 or 60.x x 60 1802x 120x 60x 40 60 40 or 100The correct choice is A.9-6The Complex and Polar Formof Complex NumbersPages 589–590 Check for Understanding1. To find the absolute value of a bi, square a andb, add the squares, then take the square root ofthe sum.2. i 0 i; cos 2 0 and sin 2 1i cos 2 i sin 2 3. Sample answer: z 1 i, z 2 i⏐z 1 z 2 ⏐ ⏐z 1 ⏐ ⏐z 2 ⏐⏐i (i)⏐ ⏐i⏐ ⏐i⏐⏐0⏐ i i0 2i4. The conjugate of a bi is a bi.(a i)(a bbi) a 2 b 2 , so the friend’smethod gives the same answer.Sample answer: The absolute value of 2 3i is2 2 3 2 13. Using the friend’s method, theabsolute value is (2 i)(2 33i) 4 9 13.5. 2x y (x y)i 5 4i2x y 5 x y 42x (x 4) 5 y x 4x 1y (1) 4 or 36. 7.21i21O1(2, 1)21211. 2 2 12.3 i 35 6 (4,3) 6764302 4 6 8 325311625676i211 O143223(2, 3)i2(1, 2)⏐z⏐ (2 2 (1) 2 ⏐z⏐ 12) 2 ( 2 5 38. r 22) 2 ( 2 v Arctan 22 2 8 or 22 7 4v is in the fourth quadrant.2 2i 22cos 7 4 i sin 7 4 9. r 4 2 5 2 v Arctan 5 4 41 0.904 5i 41(cos 0.90 i sin 0.90)10. r (2) 2 0 20v Arctan 2 4 or 2 is on the x-axis at 2.2 2 (cos i sin )13253601 2 3 4 324cos 3 i sin 3 2(cos 3 i sin 3) 4 1 2 i — 3—2 2(0.99 i(0.14)) 2 23i 1.98 0.28i116291 Chapter 9
13.5623i2363(2, 2)01 2 3 4 19. 20.i(2, 3)i76433253116OO(3, 4) 3 2 (cos 2 i sin 2)14. 3 2 3 2 (1 i(0))i11 O0.38i10.630.90 0.36 1 1.3015a. magnitude 10 2 15 2 325 18.03 N15b. Arctan 1 510 56.31°Pages 590–591 Exercises16. 2x 5yi 12 15i2x 12 5y 15x 6 y 317. 1 (x y)i y 3xi1 y x y 3xx (1) 3x1 2x 1 2 x18. 4x (y 5)i 2x y (x 7)iy 5 x 74x 2x yy x 12 4x 2x (x 12)3x 12x 4y (4) 12 or 8⏐z⏐ 2 2 3 ⏐z⏐ 34) 2 ( 2 13 25 or 521. 22.(1, 5)⏐z⏐ (1) 2 ) (5 ⏐z⏐ 03) 2 ( 2 26 9 or 323. 24.(1, 5)2⏐z⏐ (1) 2 (5) 2 ⏐z⏐ 42) 2 ( 2 6 18 or 3225. r (4) 2 6 2 52 or 21326. r 3 2 3 2 iOi211 O121v Arctan 3 3 18 or 32 4 3 3i 32 cos 4 i sin 4 27. r (1) 2 3) (2 v Arctan 31 4 or 2 4 3v is in the third quadrant.1 3i 2cos 4 3 i sin 4 3228. r 68) 2 ( 2 v Arctan 6 2 100 or 10 5.36v is in the fourth quadrant.6 8i 10(cos 5.36 i sin 5.36) iOiO(0, 3)8(4, 2)Chapter 9 292
29. r (4) 2 1 2 v Arctan 4 17 2.90v is in the second quadrant.4 i 17(cos 2.90 i sin 2.90)30. r 20 2 ) (21 2 v Arctan 2120 2 841 or 29 5.47v is in the fourth quadrant.20 21i 29(cos 5.47 i sin 5.47)31. r (2) 2 4 2 v Arctan 42 20 or 25 2.03v is in the second quadrant.2 4i 25(cos 2.03 i sin 2.03)32. r 3 2 0 2 v Arctan 0 3 9 or 3 0v is on the x-axis at 3.3 3(cos 0 i sin 0)33. r (4 2) 0 2 20v Arctan 42 32 or 42 v is on the x-axis at 42.42 42 (cos i sin )34. r 02) 2 ( 2 4 or 2Since x 0 when y 2, v 3 .22i 2cos 3 2 i sin 3 2 35. 2 2 36.3i35(3, ) 64 6764301 2 3 4 325311637. 38.2 2 3 i 356676(2,4)34301 2 3 4 32531165676567623432343ii1232353601 2 3 4 (1, )6 1163cos 4 i sin 4 cos 6 i sin 6 3 22 i 22 32 1 2 i 3 22 3 22i2cos 4 3 i sin 4 3 10(cos 6 i sin 6) 2 1 2 i 32 23603 6 9 12 (10, 6)11653 32 10(0.960 i(0.279))1 3i 9.60 2.79i39.2 2 3 i 340.566(2,5)476434301 2 3 4 325311611656762343i2 3(2.5, 1)53601 2 3 4 321162cos 5 4 i sin 5 4 2.5(cos 1 i sin 1) 2 22 i 22 2.5(0.54 i(0.84))2 2i 1.35 2.10i41.2 2 42. 2 2 3 i 33 i 35 56666760.44 0.5i1O0.25 i11i1O1i10.5 0.5i10.5 0.5i1(5, 0) 02 4 6 8 32i530.44 0.44i1 760.50 0.39i0.60 0.39iO 11 (3, )4301 2 3 4 325(cos 0 i sin 0) 3(cos i sin ) 5(1 0) 3(1 0) 5343.i144.45.53116293 Chapter 9
46a. 40∠30° 40(cos 30° j sin 30°) 40 32 j 1 2 34.64 20j60∠60° 60(cos 60° j sin 60°) 60 1 2 j 32 30 51.96j46b. (34.64 20j) (30 51.96j) (34.64 30) (20j 51.96j) 64.64 71.96j46c. v(t) r sin (250t v°)r 64.64 2 .96 71 2 v Arctan 7 1.9664.64 96.73 48°v(t) 96.73 sin (250t 48°)47. The graph of the conjugate of a complex number isobtained by reflecting the original number aboutthe real axis. This reflection does not change themodulus. Since the amplitude is reflected, we canwrite the amplitude of the conjugate as theopposite of the original amplitude. In other words,the conjugate of r(cos v i sin v) can be written asr(cos (v) i sin (v)), or r(cos v i sin v).48a. 10(cos 0.7 j sin 0.7) 7.65 6.44j16(cos 0.5 j sin 0.5) 14.04 7.67j48b. (7.65 6.44j) (14.04 7.67j) (7.65 14.04) (6.44j 7.67j) 21.69 14.11j ohms48c. r 21.69 2 .11 14 214.11 Arctan 21.69 25.88 0.5821.69 14.11j 25.88 (cos 0.58 j sin 0.58) ohms49a. Translate 2 units to the right and down 3 units.49b. Rotate 90° counterclockwise about the origin.49c. Dilate by a factor of 3.49d. Reflect about the real axis.50a. Sample answer: let z 1 1 i and z 2 3 4i.z 1 z 2 (1 i)(3 4i)1 7i50b. z 1 2cos 4 i sin 4 1.41(cos 0.79 i sin 0.79)z 2 5(cos 0.93 i sin 0.93)z 1 z 2 52 (cos 1.71 i sin 1.71) 7.07(cos 1.71 i sin 1.71)50c. Sample answer: Let z 1 2 4i andz 2 1 3i. Thenz 1 25(cos 5.18 i sin 5.18) 4.47(cos 5.18 i sin 5.18),z 2 10(cos 1.89 i sin 1.89) 3.16(cos 1.89 i sin 1.89), andz 1 z 2 (2 4i)(1 3i) 10 10i 102cos 4 i sin 4 14.14(cos 0.79 i sin 0.79).50d. To multiply two complex numbers in polar form,multiply the moduli and add the amplitudes. (Inthe sample answer for 50c, note that 5.18 1.89 7.07, which is coterminal with 0.79.)51. (6 2i)(2 3i) 12 22i 6i 26 22i52. x 3 cos 135° y 3 sin 135°3 22 23 2 3 22 3 22 3 22, 3 22 53. magnitude (3) 2 7 2 583, 7 3i u 7j u54. tan 105° tan (60° 45°)tan 60° tan 45°1 tan 60° tan 45°3 1 1 (3)(1) 3 1 1 31 31 33 3 1 3 4 2322 355. w v t 12( 2)1or 24v rq 18(24) or 432 cm/s432 cm/s 4.32 m/s 13.57 m/s56. sin A 1 218A sin 1 2 3 A 41.8°47. 2a 1 3a 5 2a 1 3a 54 a58. as x → , y → ; as x → , y → y 2x 2 2xy1000 2 10 610 2020 210 2021000 2 10 659. In the fourth month, the person will have received3 pay raises.$500(1.10) 3 $665.50The correct choice is D.Chapter 9 294
9-6BGraphing Calculator Exploration:Geometry in the Complex PlanePage 5921. They are collinear.2. Yes. M is the point obtained when T 0, and N isthe point obtained when T 1.3. The points are again collinear, but closer together.5. r 3 4 v 6 2 3 3 6 or 2 3 4 cos 2 i sin 2 3 4 (0 (1)i) 3 4 i 2 6. r 4 2 or 2 v 9 4 9 4 2 4 or 11 42cos 11 42 i 22 i sin 11 4 222 2i7. r 1 2 (6) or 3 v 3 5 63cos 7 6 i sin 7 6 323 2 6 5 6 or 7 6 i 1 2 3 32 3 2 i8. r 1 223) 2 ( 2 r 2 (3) 2 (3) 2 16 or 4 12 or 23r 4(23) or 83v 1 Arctan 2 23 v2 Arctan 3 3 3 5 64. The points are on the line through M and N.5. If one of a, b, or c equals 0, then aK bM cN ison KMN. If none of a, b, or c equals 0, thenaK bM cN lies on or inside KMN.6. M is the point obtained when T 0 and N is thepoint obtained when T 1. Thus, a point betweenM and N is obtained when 0 T 1.7. The distance between z and 1 i is 5. Thisdefines a circle of radius 5 centered at 1 i.8. The distance between a point z and a point at2 3i is 2.⏐z (2 3i)⏐ 2v 3 5 6 2 6 5 6 or 7 683cos 7 6 i sin 7 6 83 23 i 1 2 12 43i9. E IZ 2cos 11 6 j sin 11 6 3cos 3 j sin 3 r 2(3) or 6 v 11 6 3 11 6 2 6 13 6or 6 V 6cos 6 j sin 6 volts9-7Products and Quotients ofComplex Numbers in Polar FormPages 596 Check for Understanding1. The modulus of the quotient is the quotient of themoduli of the two complex numbers. Theamplitude of the quotient is the difference of theamplitudes of the two complex numbers.2. Square the modulus of the given complex numberand double its amplitude.3. Addition and subtraction are easier in rectangularform. Multiplication and division are easier inpolar form. See students’ work for examples.4. r 2 2 or 4 v 2 3 2 or 2 4 24(cos 2 i sin 2) 4(1 i(0)) 4Pages 596–598Exercises10. r 4(7) or 28 v 3 2 3 3 3 or 28(cos i sin ) 28(1 i(0))2811. r 6 2 or 3 v 3 4 4 2 4 or 2 3cos 2 i sin 2 3(0 i(1)) 3i12. r 1 2 or 1 6 3v 3 6 2 6 6 or 6 1 6 cos 6 i sin 6 1 6 32 3 112 1 2 i 1 2 i 295 Chapter 9
13. r 5(2) or 10 v 3 4 4 4 3 4 or 7 410cos 7 4 i sin 7 4 10 22 i 22 52 52i14. r 6(3) or 18 v 3 5 6 2 6 5 6 3 6 or 2 18cos 2 i sin 2 18(0 i(1)) 18i15. r 3 1 or 3 v 7 3 2 14 6 3 6 or 11 63cos 11 63 i 1 2 i sin 11 6 3 2 3 32 3 2 i16. r 2(3) or 6 v 240° 60° 300°6(cos 300° i sin 300°) 6 1 2 i 22 2 22 1 2 3 33i 3 417. r v 7 4 22 or or 2 4 43 2(cos i sin ) 2(1 i(0))218. r 3(0.5) or 1.5 v 4 2.5 or 6.51.5(cos 6.5 i sin 6.5) 1.46 0.32i19. r 4 1 or 4 v 2 3.6 or 5.64[cos (5.6) i sin (5.6)] 3.10 2.53i20. r 2 015 or 4 3 v 7 6 11 3 7 6 22 6 15 6or 2 4 3 cos 2 i sin 2 4 (0 i (1))3 4 3 i21. r 2(2) or 22 v 3 4 2 3 4 2 4 or 5 422cos 5 4 i sin 5 4 22 22 i 22 2i22. r 2(6) or 12 v 3 6 2 6 6 or 6 12cos 6 i sin 6 12 32 i 1 2 63 6i423. r or 8 v 5 3 3 4 38cos 4 3 i sin 4 3 1 82 i 234 43i2 24. r 1 22) 2 ( 2 r 2 (3) 2 3 2 8 or 22 18 or 32r 22(32) or 12v 1 Arctan 2 2 2 v 2 Arctan 33 7 4 3 4v 7 4 3 4 10 4or 2 12cos 2 i sin 2 12(0 i(1)) 12i25. r 1 (2) 2 2) ( 2 r 2 (3 2) 32) 2 ( 2 4 or 2 36 or 6r 2 6 or 12v 1 Arctan 2 2 v22 Arctan 3232 7 4 5 4v 7 4 5 4 12 4or 12(cos i sin ) 12(1 i(0))1226. r 1 (3) 2 1) ( 2 r 2 2 23) 2 ( 2 4 or 2 16 or 4r 2 4 or 1 2 1v 1 Arctan 2 v32 Arctan 23 2 11 6 5 3v 11 6 5 3 11 6 10 6or 6 1 2 cos 6 i sin 6 1 2 32 i 1 2 34 1 4 i27. r 1 (4 2) (42)2 2 r 2 6 2 6 2 64 or 8 72 or 628r 624 23 2 2 4 2 or 2 26 3242 3 44v 1 Arctan v 2 Arctan 6 6 4 v 3 4 4 2 4 or 2 2 23 cos 2 i sin 2 2 23(0 i(1)) 2 23iChapter 9 296
28. I E Z 13 3 2jr 1 13 r 2 32) 2 ( 2 1313r or 13 13v 1 0v 2 Arctan 2 3 0.59v 0 (0.59) or 0.59I 13(cos 0.59 j sin 0.59) 3 2j amps29. Z E I 100 4 3jr 1 100 r 2 43) 2 ( 2 25 or 5r 10 05or 20v 1 0 v 2 Arctan 34 0.64v 0 (0.64) or 0.64z 20(cos 0.64 j sin 0.64) 16 12j ohms30. Start at z 1 in the complexplane. Since the modulusof z 2 is 1, z 1 z 2 and z zwill2both have the samemodulus as z 1 . Then z 1 z12 can be located byand z z2rotating z 1 by 6 counterclockwise andclockwise, respectively.156762343i2 z 1z 1 z 2 z 2353601 2 3 4 3211631a. The point is rotated counterclockwise about theorigin by an angle of v.31b. The point is rotated 60° counterclockwise aboutthe origin.32. Since a 1, the equation will be the form z 2 bz c 0. The coefficient c is the product of thesolutions, which is 6cos 7 6 i sin 7 ,6 or33 3i in rectangular form. The coefficient bis the opposite of the sum of the solutions, soconvert the solutions to rectangular form to do theaddition.b 3cos 3 i sin 3 2cos 5 6 i sin 5 6 3 2 3 32i (3 i) 3 2 3 3 3 22 i 3 Therefore, the equation is z 2 3 2 3 3 22 iz (33 3i) 0.33. r 512) 2 ( 2 Arctan 125 2 169 or 13 5.115 12i 13(cos 5.11 i sin 5.11)34. r 5 secv 5 635.r cos v 5 6 5rcos v cos 5 6 sin v sin 5 6 5 0 32r cos v 1 2 r sin v 5 0 32x 1 2 y 5 03x y 10 0Since the triangle is isosceles, the base angles arecongruent. Each measures 180 502or 65°.23 x sin 50° sin 65°23 sin 65° x sin 50° 23 sin65°sin50° x27.21 x; 27.21 lb36. cos 2x sin x 11 2 sin 2 x sin x 12 sin 2 x sin x 0sin x (2 sin x 1) 0sin x 0or2 sin x 1 0x 0° sin x 1 2 x 30°37. y cos xx cos yarccos x y38. BC ED BE AF CD 3AB FE 2AC AB BC 2 3 or 5FD FE ED 2 3 or 5perimeter of rectangle ACDF 3 5 3 5or 16perimeter of square BCDE 4(3) or 1216 12 4The correct choice is C.9-8x lb130˚Prop23 lbx lbx lb50˚x lb23 lbPowers and Roots of ComplexNumbersPage 602 Graphing Calculator Exploration1. Rewrite 1 in polar form as 1(cos 0 i sin 0).Follow the keystrokes to find the roots at 1,0.5 0.87i, and 0.5 0.87i.2. Rewrite i in polar form as 1cos 2 i sin 2 .Follow the keystrokes to find the roots at0.92 0.38i, 0.38 0.92i, 0.92 0.38i, and0.38 0.92i. 297 Chapter 9
3. Rewrite 1 i in polar form as 2cos 4 i sin 4 .Follow the keystrokes to find the roots at 1.06 0.17i, 0.17 1.06i, 0.95 0.49i, 0.76 0.76i, and 0.49 0.95i.4. equilateral triangle5. regular pentagon6. If a 0 and b 0, then a bi a. The principalroots of a positive real number is a positive realnumber which would lie on the real axis in acomplex plane.Pages 604-605 Check for Understanding1. Same results, 4 4i; answers may vary.(1 i)(1 i)(1 i)(1 i)(1 i) (1 2i i 2 )(1 2i i 2 )(1 i) (2i)(2i)(1 i)4(1 i)4 4i(1 i) 5→ r 2, v 4 (2) 5 cos (5) 4 i sin (5) 4 42cos 5 4 i sin 5 42 42 22 i 24 4i2. Finding a reciprocal is the same as raising anumber to the 1 power, so take the reciprocal ofthe modulus and multiply the amplitude by 1.3.a aiaa aiiaOa a ai4. Shembala is correct. The polar form of a ai isa2cos 4 i sin 4 . By De Moivre’s Theorem,the polar form of (a ai) 2 is 2a 2 cos 2 i sin 2 .Since cos 2 0, this is a pure imaginary number.5. r (3) 2 1) ( 21or 2 v Arctan or 3 6 2 3 cos (3) 6 i sin (3) 6 8 cos 2 i sin 2 8(0 i (1))8i6. r 35) 2 ( 2 or 34 v Arctan 53 1.03037682734 4 (cos (4)() i sin (4)(v))644 960iaa ai7. r 0 2 1 2 or 1v 2 1 1 6 cos 1 6 2 i sin 1 6 2 1cos 1 2 i sin 1 2 0.97 0.26i8. r (2) 2 ) (1 2 or 5 v Arctan 12 2.6779450455 1 3 cos 1 3 () i sin (3)() 0.82 1.02i9. x 4 i 0 → x 4 iFind the fourth roots of i.r 01) 2 ( 2 1 v 3 2(i) 1 4 1 (cos 3 2 2n i sin 3 2 2n 1 4 cos 3 4n8 i sin 3 8 4nx 1 cos 3 8 i sin 3 8 0.38 0.92ix 2 cos 7 8 i sin 7 8 0.92 0.38ix 3 cos 11 8 i sin 11 8 0.38 0.92ix 4 cos 15 8 i sin 15 8 0.92 0.38i0.92 0.38i1i10.38 0.92i10.38 0.92iO 1 0.92 0.38i10. 2x 3 4 2i 0 → x 3 2 iFind the third roots of 2 i.r (2) 2 ) (1 2 5 v Arctan 12 3.605240263(2 i) 1 3 [5(cos (v 2i) i sin (v 2n))] 1 3 (5) 1 3 cos v 2n3 i sin v 2n3 x 1 (5) 1 3 v vcos 3 i sin 3 0.47 1.22ix 2 (5) 1 3 cos v 23 i sin v 23 1.29 0.20ix 3 (5) 1 3 cos v 43 i sin v 43 0.81 1.02i1 O 11.29 0.20ii110.47 1.22i0.82 1.02iChapter 9 298
11. For w 1 , the modulus (0.8 2 ) (0.7) 2 2 or 1.13.For w 2 , the modulus 1.13 2 or 1.28.For w 3 , the modulus 1.28 2 or 1.64.This moduli will approach infinity as the numberof iterations increases. Thus, it is an escape set.Pages 605–606Exercises12. 3 3 cos (3) 6 i sin (3) 6 27 cos 2 i sin 2 27(0 i(1)) 27i13. 2 5 cos (5) 4 i sin (5) 4 32 cos 5 4 i sin 5 42 32 22162 162i i 2 14. r (2) 2 2 2 22 v Arctan 2 7 4(22) 3 cos (3) 7 9 i sin (3) 7 4 162cos 21 4 i sin 21 4 162 22 i 22 16 16i15. r 13) 2 ( 2 2 v Arctan 31 3 2 4 cos (4) 3 i sin (4) 3 16 cos 4 3 i sin 4 3 16 1 2 i 328 83i16. r 36) 2 ( 2 35 v Arctan 63 1.107148718(35) 4 (cos (4)(v) i sin (4)(v))567 1944i17. r 2 2 3 2 13 v Arctan 3 2 0.9827937232(13) 2 (cos (2)(v) i sin (2)(v))0.03 0.07i18. r 2 2 4 2 25 v Arctan 4 2 1.107148718(25) 4 (cos (4)(v) i sin (4)(v))112 384i 19. 32 1 5 cos 1 5 2 3 i sin 1 5 2 3 2cos 2 15 i sin 2 1 1.83 0.81i20. r (1) 2 0 2 15 1 1 4 cos 1 4 () i sin 1 4 () cos 4 i sin 4 0.71 0.71iv 2 21. r (2) 2 1 2 5(5) 1 4 cos 1 4 (v) i sin 1 4 (v)v Arctan 2 2.677945045 0.96 0.76i22. r 41) 2 ( 2 17 v Arctan 14 0.2449786631(17) 1 3 cos 1 3 (v) i sin 1 3 (v) 1.60 0.13i23. r 2 2 2 2 22v Arctan 2 2 4 cos 1 3 4 i sin 1 3 4 (22) 1 3 1.37 0.37i24. r (1) 2 ) (1 2 2 v Arctan 1 3 4(2) 1 4 cos 1 4 3 4 i sin 1 4 3 4 0.91 0.61i25. r 0 2 1 2 11 1 2 cos 1 2 2 i sin 1 2 2 v 2 1 0.71 0.71i26. x 3 1 0 → x 3 1Find the third roots of 1.r 1 2 0 2 1 v 01 1 3 [1 (cos (0 2n) i sin (0 2n))] 1 3 cos 2n 3 i sin 2n 3x 1 cos 0 i sin 0 1x 2 cos 2 3 i sin 2 3 1 2 23ix 3 cos 4 3 i sin 4 3 1 2 23i 1 3i2 21i1 1 3i2 21O 11299 Chapter 9
27. x 5 1 → x 5 1Find the fifth roots of 1.r (1) 2 0 2 1 v (1) 1 5 [1 (cos ( 2n) i sin ( 2n))] 1 5 cos 2n5 i sin 2n5x 1 cos 5 i sin 5 0.81 0.59ix 2 cos 3 5 i sin 3 5 0.31 0.95ix 3 cos 5 5 i sin 5 5 1x 4 cos 7 5 i sin 7 5 0.31 0.95ix 5 cos 9 5 i sin 9 5 0.81 0.59ii10.31 0.95i10.81 0.59iO 1 0.31 0.95i 0.81 0.59i128. 2x 4 128 0 → x 4 64Find the fourth roots of 64.r 64 2 0 2 64 v 0i32 2 i212 22 23 2 1 O 1 2 31232 2 i64 1 4 [64 (cos (0 2n) i sin (0 2n))] 1 4 22cos n 2 i sin n 2 x 1 22 (cos 0 i sin 0) 22x 2 22cos 2 i sin 2 22ix 3 22 (cos i sin ) 22x 4 22cos 3 2 i sin 3 2 22i29. 3x 4 48 0 → x 4 16Find the fourth roots of 16.r (16) 2 0 2 16 v (16) 1 4 [16 (cos ( 2n) i sin ( 2n))] 1 4 2 cos 2n4 i sin 2n4x 1 2 cos 4 i sin 4 2 2ix 2 2 cos 3 4 i sin 3 4 2 2ix 3 2 cos 5 4 i sin 5 4 2 2ix 4 2 cos 7 4 i sin 7 4 2 2i i 2 2 i1 2 2 i30. x 4 (1 i) 0 → x 4 1 iFind the fourth roots of 1 i.r 1 2 1 2 2v Arctan 1 1 4 (1 i) 1 4 2cos 4 2n i sin 4 2n 1 4 (2) 1 4 cos 8n16 i sin 168n x 1 (2) 1 4 cos 1 6 i sin 1 6 1.07 0.21ix 2 (2) 1 4 cos 9 16 i sin 9 16 0.21 1.07ix 3 (2) 1 4 cos 1 71 6 i sin 1 71 6 1.07 0.21ix 4 (2) 1 4 cos 2 516 i sin 2 516 0.21 1.07i31. 2x 4 2 23i 0 → x 4 1 3i.Find the fourth roots of 1 3i.r (1) 2 3) ( 2 2v Arctan 31 3 or 4 3(1 3i) 1 4 1O 11i10.21 1.07i 2 2 i 2cos 4 3 2n i sin 4 3 2n 1 4 2 1 4 cos 4 6n12 i sin 4 6n12x 1 2 1 4 cos 4 12 i sin 4 12 0.59 1.03ix 2 2 1 4 cos 1 01 2 i sin 1 01 2 1.03 0.59ix 3 2 1 4 cos 2 212 i sin 2 212 0.59 1.03ix 4 2 1 4 cos 2 812 i sin 2 812 1.03 0.59i 2 2 i1.07 0.21i1 O 11.07 0.21i1i11.03 0.59i10.59 1.03i10.21 1.07i0.59 1.03iO 1 1.03 0.59i Chapter 9 300
32. Rewrite 10 9i in polar form as181 cos tan 1 910 i sin tan 1 910 .Use a graphing calculator to find the fifth roots at0.75 1.51i, 1.20 1.18i, 1.49 0.78i,0.28 1.66i, and 1.66 0.25i.33. Rewrite 2 4i in polar form as25[cos (tan 1 (2)) i sin (tan 1 (2))].Use a graphing calculator to find the sixth roots at1.26 0.24i, 0.43 1.21i, 0.83 0.97i,1.26 0.24i, 0.43 1.21i, and 0.83 0.97i.34. Rewrite 36 20i in polar form as4106 cos tan 1 5 9 i sin tan 1 5 9 .Use a graphing calculator to find the eighth rootsat 1.59 0.10i, 1.05 1.19i, 0.10 1.59i,1.19 1.05i, 1.59 0.10i, 1.05 1.19i,0.10 1.59i, and 1.19 1.05i. 3 4 235. For w 1 , the modulus 1 2 2 2 or 0.81.For w 2 , the modulus (0.81) 2 or 0.66.For w 3 , the modulus (0.66) 2 or 0.44.This moduli will approach 0 as the number ofiterations increase. Thus, it is a prisoner set.36a. In polar form the 31st roots of 1 are given bycos 2 n31n i sin 2 3,1n 0, 1, . . . , 30. Thena cos 2 n3.1The maximum value of a cosineexpression is 1, and it is achieved in thissituation when n 0.36b. From the polar form in the solution to part a, weget b sin 2 n. 3 1b will be maximized when 2 n31isas close to 2 as possible. This occurs when n 8,so the maximum value of b is sin 1 6, 3 1or about0.9987.37. x 6 1 0 → x 6 1Find the sixth roots of 1.r 1 2 0 2 1 v 01 1 6 [1 (cos (0 2n) i sin (0 2n))] 1 6 cos n 3 i sin n 3x 1 cos 0 i sin 0 1x 2 cos 3 i sin 3 1 2 32ix 3 cos 2 3 i sin 2 3 1 2 23ix 4 cos 3 3 i sin 3 3 1x 5 cos 4 3 i sin 4 3 1 2 23ix 6 cos 5 3 i sin 5 3 1 2 23i38a. The point at (2, 2) becomes the point at (0, 2).From the origin, the point at (2, 2) had a lengthof 22 and the new point at (0, 2) has a lengthof 2. The dilation factor is 2.222 y (2, 2)(0, 2) 221x1 O 1 212 22(cos 45° i sin 45°) 22 22 i sin 22 0.5 0.5i38b. 22(cos 45° i sin 45°) 2 1 2 (cos 90° i sin 90°)The square is rotated 90° counterclockwise anddilated by a factor of 0.5.39. The roots are the vertices of a regular polygon.Since one of the roots must be a positive realnumber, a vertex of the polygon lies on thepositive real axis and the polygon is symmetricabout the real axis. This means that the non-realcomplex roots occur in conjugate pairs. Since theimaginary part of the sum of two complexconjugates is 0, the imaginary part of the sum ofall the roots must be 0.40. r 2(3) or 6 v 6 5 3 6 10 6or 11 66cos 11 6 i sin 11 6 6 32 i 1 2 33 3i41. (2 5i) (3 6i) (6 2i) (2 (3) (6)) (5i 6i 2i) 5 i42. x t, y 2t 743. cos 22.5° cos 4 5°2 1 c os 45°2 1 2 22 2 2 4 2 2 44. Find B.B 180° 90° 81°15 8°45Find a.atan 81°15 2 828 tan 81°15 a181.9 a2301 Chapter 9
Find c.cos 81°15 2 c8c cos 8281°15c 184.145. Let x the number of large bears produced.Let y the number of small bears produced.x 300y 400x y 1200f(x, y) 9x 5yf(300, 400) 9(300) 5(400) 4700f(300, 900) 9(300) 5(900) 7200f(800, 400) 9(800) 5(400) 9200Producing 800 large bears and 400 small bearsyields the maximum profit.46. 0.20(6) 1.2 quarts of alcohol0.60(4) 2.4 quarts of alcohol 1.2 26 .44 3 . 61 0 or 36% alcoholThe correct choice is A.y12001000800600(300, 900)(800, 400)400200 (300, 400)xO 200 600 1000Chapter 9 Study Guide and AssessmentPages 607 Check for Understanding1. absolute value 2. Polar3. prisoner 4. iteration5. pure imaginary 6. cardioid7. rectangular 8. spiral of Archimedes9. Argand 10. modulusPages 608–610 Skills and Concepts11. 12.90˚90˚120˚ 60˚120˚ 60˚150˚30˚150˚30˚180˚0˚ 180˚0˚1 2 3 41 2 3 4BA210˚330˚210˚330˚240˚ 270˚300˚240˚ 270˚300˚13. 14.2 2 2 2 3333556666C001 2 3 41 2 3 4764332531167643D325311615. Sample answer: (4, 585°), (4, 945°), (4, 45°),(4, 405°)(r, v 360k°)→ (4, 225° 360(1)°) → (4, 585°)→ (4, 225° 360(2)°) → (4, 945°)(r, v (2k 1)180°)→ (4, 225° (1)180°) → (4, 45°)→ (4, 225° (1)180°) → (4, 405°)16. 90˚ 17.120˚ 60˚180˚150˚210˚240˚270˚1 2 3 4300˚30˚18. 19.90˚120˚ 60˚180˚150˚210˚240˚1 2 3 4300˚270˚20. 21.120˚90˚ 60˚150˚180˚210˚240˚180˚210˚240˚2270˚2270˚446300˚6300˚88limaçonrose24. x 6 cos 45° y 6 sin 45° 6 22 6 22 32 32(32, 32)25. x 2 cos 330° y 2 sin 330° 2 32 2 1 2 31(3, 1)0˚330˚30˚0˚330˚30˚0˚330˚0˚330˚56765676567623432343234323223223231 2 3 45331 2 3 453353601166011668 16 24 32circleSpiral of Archimedes22. 23.290˚2 120˚ 60˚335150˚30˚6676432 432653011680116Chapter 9 302
26. x 2 cos 3 4 y 2 sin 3 4 2 22 2 2 22(2, 2)2 27. x 1 cos 2 y 1 sin 2 0 1(0, 1)28. r (3) 3) 2 (2 v Arctan 12 or 23 4 3 33 23, 4 329. r 5 2 5 2 v Arctan 5 5 50 or 52 4 52, 4 30. r (3) 2 1 21v Arctan 3 10 3.16 2.82(3.16, 2.82)31. r 4 2 2 2 v Arctan 2 4 20 4.47 0.46(4.47, 0.46)32. AB 2 2 2 2 125Since C is positive, use 5. 251 3x y 0 55cos f 2 , sin f , p 3 55 5f Arctan 1 2 555 27°Since cos f 0 and sin f 0, the normal lies inthe third quadrant.f 180° 27° or 207°p r cos (v f) 3 5 r cos (v 207°)533. AB 2 2 3 2 1 2 10Since C is positive, use 10.31014 x 10 y 10 0cos f 3 10 , sin f 1,0 21010p 5f Arctan 1 3 10 18°Since cos f 0 and sin f 0, the normal lies inthe third quadrant.f 180° 18° or 198°p r cos (v f) 2 10 5 r cos (v 198°)34. 3 r cos v 3 0 r cos v cos 3 r sin v sin 3 30 1 r 2 cos v 3 r sin v 320 1 x 2 3 y 320 x 3y 6 orx 3y 6 035. 4 r cos v 2 0 r cos v cos 2 r sin v sin 2 40 0 r sin v 40 y 40 y 4 ory 4 036. i 10 i 25 (i 4 ) 2 i 2 (i 4 ) 6 i (1) 2 (1) (1) 6 i1 i37. (2 3i) (4 4i) (2 (4)) (3i (4i))2 7i38. (2 7i) (3 i) (2 (3)) (7i (i))1 6i39. i 3 (4 3i) 4i 3 3i 4 4(i) 3(1)3 4i40. (i 7)(i 7) i 2 14i 49 1 14i 4948 14i41. 4 2i 4 2i 5 2i5 2i5 2i5 2i 20 18i 4i225 4i 2 16 18i29 1 629 1 8i 29 5 i 5 i42. 1 2i1 2i1 2i1 2i5 52i i 2i2 1 2i 2(5 2) (52 1)i3 5 2 1 5233i43. r 2 2 2 2 v Arctan 2 2 8 or 22 4 22 cos 4 i sin 4 44. r 13) 2 ( 2 v Arctan 31 2 10 5.0310 (cos 5.03 i sin 5.03)45. r (1) 2 (3) 2 3 v Arctan 1 4 or 2 2 32cos 2 2 3 i sin 346. r (6) 2 ) (4 2 v Arctan 46 52 or 213 3.73213 (cos 3.73 i sin 3.73)47. r (4) 2 ) (1 2 v Arctan 14 17 3.3917 (cos 3.39 i sin 3.39)303 Chapter 9
48. r 4 2 0 2 v 0 16 or 44(cos 0 i sin 0)49. r (2 2) 0 2 2 v 8 or 2222(cos i sin )50. r 0 2 3 2 v 2 9 or 33cos 2 i sin 2 51. 2 2 52.3 i 356( 2,6 ) 676433201 2 3 4 5311656762343i253,3( )32353601 2 3 4 2cos 6 i sin 6 3cos 5 3 i sin 5 3 2 32 i 1 2 3 1 2 i 32 3 i 3 2 3 3i253. r 4(3) or 12 v 3 3 or 2 312cos 2 3 i sin 2 3 12 1 2 i 326 63i 54. r 8(4) or 32 v 4 2 32cos 3 4 i sin 3 4 4 2 4 or 3 4 32 22 i 22 162 162i55. r 2(5) or 10 v 2 0.5 or 2.510 (cos 2.5 i sin 2.5) 8.01 5.98i56. r 8 2 or 4 v 7 6 5 34cos 2 i sin 2 7 6 10 6or 2 4(0 i(1))4i57. r 6 4 or 3 2 v 2 6 3 2 cos 3 i sin 3 3 6 6 or 3 3 2 1 2 i 32 3 4 3 2i458. r 2 . 24.4or 0.5 v 1.5 0.6 or 0.90.5 (cos 0.9 i sin 0.9) 0.31 0.39i59. r 2 2 2 2 22 v Arctan 2 2 4 (2) 8 cos (8) 4 i sin (8) 4 4096 (cos 2 i sin 2) 409611660. r (3) 2 ) (1 2 2v Arctan 6 1 32 7 cos (7) 6 i sin (7) 6 128 cos 7 6 i sin 7 6 128 32643 64i61. r (1) 2 1 2 2 i 1 2 v Arctan 1 3 4(2) 4 cos (4) 3 4 i sin (4) 3 4 4(cos 3 i sin 3)462. r (2) 2 ) (2 2 22 v Arctan 22 (22) 3 cos (3) 5 4 i sin (3) 5 4 162 cos 15 4 i sin 15 4 162 22 i 22 16 16i63. r 0 2 1 2 1v 2 1 1 4 cos 1 4 2 i sin 1 4 2 cos 8 i sin 8 0.92 0.38i64. r (3) 2 1 2 5 4 1 2 v Arctan 6 32 1 3 cos 1 3 6 i sin 1 3 6 2 1 3 cos 18 i sin 18 1.24 0.22iApplications and Problem Solving v Arctan 1 2575 21,25 0 145.77 59.04°(145.77, 59.04°)r cos cos 2 r sin v sin 2 5 0r sin v 5 0y 5 0y 5 50 180j4 5j 50 180j4 5j 4 5j4 5j200 470j 900j 216 25j 2Page 61165. lemniscate66. r 75 2 125 267. r cos v 2 5 068. I E Z 1100 470j241 26.83 11.46j amps1Chapter 9 304
Page 611 Open-Ended Assessment1a. Sample answer: 4 6i and 3 2i(4 6i) (3 2i) (4 3) (6i 2i) 7 4i1b. No. Sample explanation: 2 3i and 5 i alsohave this sum.(2 3i) (5 i) (2 5) (3i (i)) 7 4i2a. Sample answer: 4 i⏐z⏐ 4 2 1 2 172b. No. Sample explanation: 1 4i also has thisabsolute value.⏐z⏐ 1 2 4 2 17Chapter 9 SAT & ACT PreparationPage 613 SAT and ACT Practice1. ∠a and ∠b form a linear pair, so ∠b issupplementary to ∠a. Since ∠b and ∠d arevertical angles, they are equal in measure. So ∠dis also supplementary to ∠a. Since ∠d and ∠f arealternate interior angles, they are equal. So ∠f issupplementary to ∠a. And since ∠f and ∠h arevertical angles, ∠h is supplementary to ∠a. Theangles supplementary to ∠a are angles b, d, f, andh. The correct choice is A.2. Draw the given triangle and draw the height hfrom point B.A8hB10CThe answer choices include sin x. Write anexpression for the height, using the sine of x.sin x h 8 A 1 2 bh8 sin x h 1 2 (10)(8 sin x) 40 sin xThe correct choice is B.3. Since PQRS is a parallelogram, sides PQ and SRare parallel and m∠Q m∠S b.Mc˚PSx˚Tx˚a˚ OQb˚RNIn SMO, c b a 180 or a 180 (c b).Also, x a 180 or a 180 x since consecutiveinterior angles are supplementary.180 (c b) 180 xx c bThe correct choice is E.4. Volume wh16,500 75 w 1016,500 750w22 wThe correct choice is A.15. 1001100 10 99 10011000 1011009 1 0 1 00The correct choice is A.6. Consider the three unmarked angles at theintersection point. One of these angles, say the topone, is the supplement of the other two unmarkedangles, because of vertical angles. So the sum ofthe measures of the unmarked angles is 180°.The sum of the measures of the marked anglesand the three unmarked angles is 3(180), sincethese angles are the interior angles of threetriangles.m(sum of marked angles) m(sum of unmarked angles) 3(180)m(sum of marked angles) 180 3(180)m(sum of marked angles) 360The correct choice is C.7. Subtract the second equation from the first.5x 2 6x 705x 2 6y 106x 6y 60x y 10, so 10x 10y 100.The correct choice is E.8. Since ∠B is a right angle, ∠C is a right angle also,because they are alternate interior angles.In the triangle containing ∠C, 90 x y 180 orx y 90.The straight angle at D is made up of 3 angles.120 x x 1802x 60 or x 30x y 90(30) y 90y 60The correct choice is B.9. In the slope-intercept form of a line, y mx b,m represents the slope of the line, and brepresents the y-intercept. Since the slope isgiven as 3 2 , the slope-intercept form of the line isy 3 2 x b.Since (–3, 0) is on the line, it satisfies theequation. 0 3 2 (–3) b. So b 9 2 .The correct choice is D.10. Note that consecutive interior angles aresupplementary.110 2x 180 y x 1802x 70 y (35) 180x 35 y 145The answer is 145.wh 10 ft 75 ft305 Chapter 9
Chapter 10 Conics10-1Introduction to Analytic GeometryPages 619–620 Check for Understanding1. negative distances have no meaning2. Use the distance formula to show that themeasure of the distance from the midpoint toeither endpoint is the same.3a. Yes; the distance from B to A is a 5 and the2distance from B to C is also a 5.23b. Yes; the distance from B to A is a 2 b 2 andthe distance from C to A is also a 2 b 2 .3c. No; the distance from A to B is a 2 b 2 , thedistance from A to C is a b, and the distancefrom B to C is b2.4. (1) Show that two pairs of opposite sides areparallel by showing that slopes of the linesthrough each pair of opposite sides are equal.(2) Show that two pairs of opposite sides arecongruent by showing that the distancebetween the vertices forming each pair ofopposite sides are equal.(3) Show that one pair of opposite sides is paralleland congruent by showing that the slopes ofthe lines through that pair of sides are equaland that the distances between the endpointsof each pair of segments are equal.(4) Show that the diagonals bisect each other byshowing that the midpoints of the diagonalscoincide.5. d (x x 2 1 ) 2 (y 2 y 1 ) 2d (5 ) 51 2 (1) 2d 0 0 2 12d 100 or 10 x 1 x 22, y 1 y 22 5 52, 1 112 (5, 6)6. d (x x 2 2 ) 2 (y 2 y 1 ) 2d (4 0) 2 (3 0) 2d (4) 2 ) (32d 25 or 5 x 1 x 22, y 1 2y 2 0 2(4), 0 (3)2 (2, 1.5)7. d [0 2)] ( 2 (4 2) 2d 2 2 2 2 d 8 or 22 x 1 x 22, y 1 2y 2 2 2 (1, 3) 0, 2 42 8. AB (x x 2 1 ) 2 (y 2 y 1 ) 2 (6 ) 3 2 (2 4) 2 13slope of AB2 y12 x1 2 46 3or 2 3 m y xDC (x x 2 1 ) 2 (y 2 y 1 ) 2 (8 ) 5 2 (7 9) 2 13slope of DC2 y12 x1 7 98 5or 2 3 m y xYes; AB DC since AB 13 and DC 13,and AB DC since the slope of AB is 2 3 and theslope of DC is also 2 3 .9. XY (x x 2 1 ) 2 (y 2 y 1 ) 2 [1 (3)]2 6 () 22 68 or 217XZ (x x 2 1 ) 2 (y 2 y 1 ) 2 [5 3)] ( 2 (0 2) 2 68 or 217Yes; XY XZ, since XY 217 and XZ 217,therefore XYZ is isosceles.10a. yA(0, a)B(0, 0)O10b. BD (c ) 0 2 (a 0) 2 c 2 a 2 AC (c ) 0 2 (0 a) 2 c 2 a 2 Thus, AC BD.10c. The midpoint of AC is c , a 2 2 . The midpoint ofBD is c , a 2 2 . Therefore, the diagonals intersectat their common midpoint, E c , a 2 2 . Thus,AE EC and BE ED.10d. The diagonals of a rectangle are congruent andbisect each other.11a. Both players are located along a diagonal of thefield with endpoints (0, 0) and (80, 120). Thekicker’s teammate is located at the midpoint ofthis diagonal. x 1 x 22, y 1 2y 2E 0 2 (40, 60)80, 0 1202 D(c, a)C(c, 0)xChapter 10 306
19. d 11b. d (x x 2 1 ) 2 (y 2 y 1 ) 2(x x 2 1 ) 2 (y 2 y 1 ) 2d 48 or 43AD and BC are one pair of opposite sides. x 1 x 22, y 1 y 22 32 72 5 (1) , slope of ADslope of BC2 2 (52, 3)m y 2 y1x2 xm y 2 y11x2 x1d (40 0) 2 (60 0) d 40 2 60 d 5200 d (c 2 c) 2 (d1 d) d 2 1) 2 ( 2d 5d 2013 or about 72 yards x 1 x 22, y 1 y 22 c c 22, d d 12 2c 22, 2d 12 Pages 620–622 Exercises12. d (x x 2 1 ) 2 (y 2 y 1 ) 2c 1, d 1 2 d [4 1)] ( 2 (13 1) 220. d (x x 2 1 ) 2 (y 2 y 1 ) 2d 5 2 2 1d [w w ( )] 24w ( w) 2d 2 3w)d 169 or 13 ( 2 x 1 x 22, y 1 y 22 1 42, 1 13d 4 w 9 2 or 9w 2 4 (1.5, 7) x 1 x 22, y 1 y 22 w 2 w2, w 4w2 13. d (x x 2 1 ) 2 (y 2 y 1 ) 2d (1 1) 2 (3 3) d (2) 2 ) (621.w 1, 5 2 wd (x x 2 1 ) 2 (y 2 y 1 ) 220 (2a a) 2 [7 (9)]d 40 or 210 x 1 x 22, y 1 20 (3a) 2 16y 22 1 (1)2, 3 (3)2 20 9a 2 256 (0, 0)400 9a 2 25614. d (x x 2 1 ) 2 (y 2 y 1 ) 2144 9a 2d (0 ) 8 (8 0) 2a 2 16d (8) 2 8 2a 16 or 422. Let D have coordinates (x 2 , y 2 ).d 128 or 82 x 1 x 22, y 1 y 22 8 02, 8 02 x 22, 1 y 22 3, 5 2 (4, 4)x 223 1 y 22 5 2 15. d (x x 2 1 ) 2 (y 2 y 1 ) 24 x 2 6 1 y 2 5d [5 1)] ( 2 [36)] (x 2 10 y 2 6Then D has coordinates (10, 6).d 6 2 3 23. Let the verticesyd 45 or 35A(2, 3)of the quadrilateral x 1 x 22, y 1 y 22 1 52, 6 (3)2 be A(2, 3), B(2, 3), (2, 4.5)C(2, 3), and D(3, 2).D(3, 2)16. d (x x 2 1 ) 2 (y 2 y 1 ) 2A quadrilateral is aO xd (72 2) 3[1 (5)]parallelogram if one pairof opposite sides are B(3, 2)d (42) 2 4 2C(2, 3)parallel and congruent.17. d (x x 2 1 ) 2 (y 2 y 1 ) 2 33 (2)d (a ) a9 2 () 72d 0 16) 2 ( 2d 256 or 16 x 1 x 22y 2a, 7 (9)2 , y 1 2 a 2 (a, 1)18. d (x x 2 1 ) 2 (y 2 y 1 ) 2d [r 2 (6r)] 2 (s s) 2d (8) 2 0 2d 64 or 8 x 1 x 22, y 1 2y 2 6 r 2 2r 42, 2 s2 (r 2, s)r 2, s s2 2 3 (2) 2 (3) 1 5 1 5 Their slopes are equal, therefore AD BC.AD (x x 2 1 ) 2 (y 2 y 1 ) 2 [3 2)] ( 2 (2 3) 2 51) 2 ( 2 26BC (x x 2 1 ) 2 (y 2 y 1 ) 2 [2 3)] ( 2 [32)] (2 51) 2 ( 2 26The measures of AD and BC are equal.Therefore AD BC. Since AD BC and AD BC,quadrilateral ABCD is a parallelogram; yes.307 Chapter 10
24. Let the vertices of thequadrilateral beA(4, 11), B(8, 14),C(4, 19), and D(0, 15).A quadrilateral is aparallelogram if bothpairs of opposite sidesare parallel.AB and DC are one pairof opposite sides.slope of ABm y 2 y1x xslope of DCm y 2 y1x x2 12 1 1 4 8 114 19 154 0 3 2 4 4 or 1Since AB ⁄ DC, quadrilateral ABCD is not aparallelogram; no.25. The slope of the line through (15, 1) and (3, 8)should be equal to the slope of the line through(3, 8) and (3, k) since all three points lie on thesame line.slope through (15, 1) slope through (3, 8)and (3, 8) and (3, k)m y 2 y1x xm y 2 y1x x2 12 18 1 3 15 k ( 8)3 ( 3)9 18or 1 2 k 86 k 86 1 2 ⇒ k 526. d (x x 2 1 ) 2 (y 2 y 1 ) 2AB [1 (3)]2 3 (20) 2 16 or 4BC [1 1)] ( 2 (0 23)2 16 or 4CA (3 1) 2 0) (0 2 16 or 4Yes, AB 4, BC 4, and CA 4. Thus,AB BC CA. Therefore the points A, B, and Cform an equilateral triangle.27. EF (x x 2 1 ) 2 (y 2 y 1 ) 2 (4 ) 2 2 (4 5) 2 5HG (x x 2 1 ) 2 (y 2 y 1 ) 2 (2 ) 0 2 (0 1) 2 5slope of EF2 y12 x1 4 54 2or 1 2 m y xslope of FG2 y12 x1 0 42 4or 2 1 m y xslope of HG2 y12 x1 0 12 0or 1 2 m y x4EF HG since EF 5 and HG 5.2016D(0, 15)1284OyC(4, 19)A(4, 11)4B(8, 14)8 12xEF HG since the slope of EF is 1 2 and the slopeof HG is 1 2 . Thus the points form aparallelogram. EF ⊥ FG since the product of theslopes of EF and FG, 1 2 2 1 , is 1. Therefore, thepoints form a rectangle.28. Let A(0, 0), B(b, c), and C(a, 0) be the vertices ofa triangle. Let D be the midpoint of AB and Ebe the midpoint of BC.yDO A(0, 0)The coordinates of D are 0 2The coordinates of E are b 2AC (a ) 0 2 (0 0) 2 a 2 or aDE b a2 b 2 2 c 2 a 4B(b, c)2 or a 2 EC(a, 0) x 2c 2b, 0 c2a, c 02 or b 2 , 2c . or b 2a, 2c .Since DE 1 2 AC, the line segment joining themidpoints of two sides of a triangle is equal inlength to one-half the third side.29. In trapezoid ABCD, let A and B have coordinates(0, 0)and(b,0), respectively. To make the trapezoidisosceles, let C have coordinates (b a, c) and letD have coordinates (a, c).yD(b a, c)O A(0, 0)C(a, c)B(b, 0)AC (a ) 0 2 (c 0) 2 a 2 c 2 BD (b a b) 2 (c 0) 2 a 2 c 2 AC a 2 c 2 a 2 c 2 BD, so thediagonals of an isosceles trapezoid are congruent.xChapter 10 308
30. In ABC, let the vertices be A(0, 0) and B(a, 0).Since AC and BC are congruent sides, let the thirdvertex be C a 2 , b. Let D be the midpoint of ACand let E be the midpoint of BC.y( )aC , b232. Let the vertices of quadrilateral ABCD be A(a, e),B(b, f), C(c, g), and D(d, h). The midpoints of AB,BC, CD, and DA, respectively,are L a b2, e f2 , M b c g2 ,N c d2, g 2h , and P a 2, f 2d, e h2 .A(a, e)yLB(b, f)DEOPOMxO A(0, 0)B(a, 0)The coordinates of D are: a 2 0 , b 2 2The coordinates of E are: a 2 a , b 2 2AE 3 a 4 0 2 0 b 2 2 a 4 2 b 2 2BD a 0 1 2 9 4 1 2 9 40 or a 4 , b 2 .0 or 3 4a 2 b 2a 2 b 2a , b 2 .Since AE BD, the medians to the congruentsides of an isosceles triangle are congruent.31. Let A and B have coordinates (0, 0) and (b, 0)respectively. To make a parallelogram, let C havecoordinates (b a, c) and let D have coordinates(a, c).yD(a, c)O A(0, 0)b, 0 c2, c 02The midpoint of BD is a 2The midpoint of AC is a b 02 or a 2b, 2c . or a c2, . 2Since the diagonals have the same midpoint, thediagonals bisect each other.xC(a b, c)B(b, 0)xbD(d, h) f g2 e f2 g eor c . b ca2 a b2 e h2 g h2 e g or a . a dc2 c d2The slope of LM is The slope of NP isC(c, g)These slopes are equal, so LM NP. f g2 g h2 fhor b c b d2 c d2The slope of MN is The slope of PL is N e h e f2 2 a d2 a b2A.h for d b.These slopes are equal, so MN PL. SinceLM NP, and MN PL, PLMN is a parallelogram.33. Let the vertices of therectangle be A(3, 1),B(1, 3), C(3, 1), andD(1, 3). Since the areaof a rectangle is the lengthtimes the width, find themeasure of two consecutivesides, AD and DC.yBAD (x x 2 1 ) 2 (y 2 y 1 ) 2 [1 3)] ( 2 (3 1) 2 44) 2 ( 2 32 or 42DC (x x 2 1 ) 2 (y 2 y 1 ) 2 (3 ) 11 2 [3)] (2 2 2 2 2 8 or 22Area w (42)(22) 16The area of the rectangle is 16 units 2 .34a.y20ODCx1020 10 O 10 20 3010x309 Chapter 10
34b. The two regions are closest between (12, 12)and (31,0).d (x x 2 1 ) 2 (y 2 y 1 ) 2 [31 )] (120 2 ( 2) 12 43 2 ) (122 1993 or about 44,64The distance between these two points is about44.64 pixels, which is greater than 40 pixels.therefore, the regions meet the criteria.35. Let the vertices of the isosceles trapezoid havethe coordinates A(0, 0), B(2a, 0), C(2a 2c, 2b),D(2c, 2b). The coordinates of the midpoints are:P(a, 0), Q(2a c, b), R(a, 2b), S(c, b).yD(2c, 2b)SO A(0, 0)PQ (2a ) c a 2 (b 0) 2 (a ) c 2 b2QR (2a ) c a 2 (b) 2b2 (a ) c 3 b2RS (a ) cb 2 (2) b2 (a ) c 2 b2PS (a ) c 2 (0 b) 2 (a ) c 2 b2So, all of the sides are congruent andquadrilateral PQRS is a rhombus.36a. distance from fountain to rosebushes:d (x x 2 1 ) 2 (y 2 y 1 ) 2d [1 3)] ( 2 (3 2) 2d 41 or 241 metersdistance from rosebushes to bench:d (x x 2 1 ) 2 (y 2 y 1 ) 2d (3 ) 1 2 [33)] (2d 210 or 410 metersdistance from bench to fountain:d (x x 2 1 ) 2 (y 2 y 1 ) 2d (3 3) 2 3) (2 2d 37 or 237 metersYes; the distance from the fountain to therosebushes is 241 or about 12.81 meters. Thedistance from the rosebushes to the bench is410 or about 12.65 meters. The distance fromthe bench to the fountain is 237 or about12.17 meters.36b. The fountain is located at (3, 2) and therosebushes are located at (1, 3). x 1 x 22, y 1 2y 2R 3 2C(2a 2c, 2b)QP B(2a, 0) x 1, 2 (3)2 1, 1 2 37a. Find a representation for MA and for MB.MA t3t 2 ( 5) 12 tt 2 90t 2 925 2 10t 2 90t 225MB (t ) 9t 2 (3) 122 t8t 2 11 8t 92t 2 744 1 10t 2 90t 225By setting these representations equal to eachother, you find a value for t that would make thetwo distances equal.MA MB10t 2 90t 225 10t 2 90t 225Since the above equation is a true statement,t can take on any real values.37b. A line; this line is the perpendicular bisector ofAB.38. r a 2 b2 bv Arctan a (5) 2 12 212 Arctan 5 169 or 13 1.176005207(5 12i) 2 13 2 [cos 2v i sin 2v]119 120i39. If u v (115, 2018, 0), then⏐v u ⏐ 115 2 20182 0 2 40855 49 or about 2021The magnitude of the force is about 2021 N.40. 2 sec 2 11x 1 1 2 sec 2 x 2 sec 2 2x 1 s in 2 x2 sec 2 2x cos 2 x2 sec 2 x 2 sec 2 x41. s rv11.5 12vv 1 1.512radians 1 1.512 18 0° 54.9°42. sin 390° sin (390° 360°) sin 30° or 1 2 43. z 2 8z 14z 2 8z 16 14 16(z 4) 2 2z 4 2z 4 244. x 2 16x 16 or 4y 2 4y 4 or 2Evaluating (x y) 2 when x 4 and y 2results in the greatest possible value, [4 (2)] 2or 36.10-2sin x sin x(1 sin x) (1 sin x)(1 sin x)(1 sin x)CirclesPage 627 Check for Understanding1. complete the square on each variableChapter 10 310
2. Sample answer: (x 4) 2 (y 9) 2 1,(x 4) 2 (y 9) 2 2, (x 4) 2 (y 9) 2 3,(x 4) 2 (y 9) 2 4, (x 4) 2 (y 9) 2 53. Find the center of the circle, (h, k), by findingthe midpoint of the diameter. Next find theradius of the circle, r, by finding the distancefrom the center to one endpoint. Then writethe equation of the circle in standard form as(x h) 2 (y k) 2 r 2 .4. The equation x 2 y 2 8x 8y 36 0 writtenin standard form is (x 4) 2 (y 4) 2 4.Since a circle cannot have a negative radius, thegraph of the equation is the empty set.5. Ramon; the square root of a sum does not equalthe sum of the square roots.6. (x h) 2 (y k) 2 r 2(x 0) 2 (y 0) 2 9 2x 2 y 2 81yO7. (x h) 2 (y k) 2 r 2[x (1)] 2 (y 4) 2 [3 (1)] 2(x 1) 2 (y 4) 2 16(1, 8) y(1, 4)(0, 9)(9, 0)8. x 2 y 2 4x 14y 47 0x 2 4x 4 y 2 14y 49 47 4 49(x 2) 2 (y 7) 2 100yO(8, 7)O(3, 4)(2, 3)(2, 7)xxx9. 2x 2 2y 2 20x 8y 34 02x 2 20x 2y 2 8y 342(x 2 10x 25) 2(y 3 4y 4) 34 2(25) 2(4)2(x 5) 2 2(y 2) 2 24(x 5) 2 (y 2) 2 12yO(5,2 23)(5, 2)(5 23, 2)x10. x 2 y 2 Dx Ey F 00 2 0 2 D(0) E(0) F 04 2 0 2 D(4) E(0) F 00 2 4 2 D(0) E(4) F 0F 0 F 04D F 16 ⇒ D 44E F 16 E 4x 2 y 2 4x 4y 0x 2 4x 4 y 2 4y 4 0 4 4(x 2) 2 (y 2) 2 8center: (h, k) (2, 2)radius: r 2 8r 8 or 2211. x 2 y 2 Dx Ey F 01 2 3 2 D(1) E(3) F 0 ⇒ D 3E F 105 2 5 2 D(5) E(5) F 0 ⇒ 5D 5E F 505 2 3 2 D(5) E(3) F 0 ⇒ 5D 3E F 34D 3E F 10(1)(5D 5E F) (1)(50)4D 2E 405D 5E F 50(1)(5D 3E F) (1)(34)2E 16E 84D 2(8) 404D 24 (6) 3(8) F 10D 6 F 20x 2 y 2 6x 8y 20 0x 2 6x 9 y 2 8y 16 20 9 16(x 3) 2 (y 4) 2 5center: (h, k) (3, 4)radius: r 2 5r 512. (x h) 2 (y k) 2 r 2[x (2)] 2 (y 1) 2 r 2(x 2) 2 (y 1) 2 r 2(1 2) 2 (5 1) 2 r 225 r 2(x 2) 2 (y 1) 2 25311 Chapter 10
13. midpoint of diameter:18. (x h) 2 (y k) 2 r 2 x 1 x 22, y 1 y 22 2 102, 6 ( 10)2 [x (5)] 2 (y 0) 2 9 2 2 (4, 2)(x 5) 2 y 2 8 1radius: r (x x 2 1 ) 2 (y 2 y 1 ) 24 [4 2)] ( 2 [6 (2)]y9( 6 2 8 5,2) 2 100 or 10(5, 0)(x h) 2 (y k) 2 r 2(x 4) 2 [y (2)] 2 10 219( , 02 )(x 4) 2 (y 2) 2 10014. (x h) 2 (y k) 2 r 2(x 0) 2 (y 0) 2 (1740 185) 2O xx 2 y 2 1925 219. (x h) 2 (y k) 2 r 2(x 6) 2 (y 1) 2 6 2(x 6) 2 (y 1) 2 36Pages 627–630 Exercisesy15. (x h) 2 (y k) 2 r 2(6, 7)(x 0) 2 (y 0) 2 5 2x 2 y 2 25y (0, 5)(0, 1) (6, 1)Ox(5, 0) Ox20. (x h) 2 (y k) 2 r 2(x 3) 2 [y (2)] 2 [2 (2)] 2(x 3) 2 (y 2) 2 1616. (x h) 2 (y k) 2 r 2[x (4)] 2 (y 7) 2 (3) 2(x 4) 2 (y 7) 2 3y (3, 2)yOx(4, 7)(3, 2)(4 3, 7)(7, 2)(4, 7 3)21. 36 x 2 y 2yO xx 2 y 2 (0, 6) 3617. (x h) 2 (y k) 2 r 2[x (1)] 2 [y (3)] 2 2(x 1) 2 (y 3) 2 1 2 26 2( 1, )2(2 2, 32 )(1, 3)yOx22. x 2 y 2 y 3 4 x 2 y 2 y 1 4 3 4 1 4 x 2 y 1 2 2 1O(6, 0)xy1(0, 2)(0, )12O x(1, )12Chapter 10 312
23. x 2 y 2 4x 12y 30 0x 2 4x 4 y 2 12y 36 30 4 36(x 2) 2 (y 6) 2 10yO(2, 6 10)x27. x 2 y 2 14x 24y 157 0x 2 14x 49 y 2 24y 144 157 49 144(x 7) 2 (y 12) 2 36(7, 6)yOx(2, 6)(2 10, 6)(7, 12)(1, 12)24. 2x 2 2y 2 2x 4y 12x 2 2x 2y 2 4y 12x 2 1x 1 4 2(y 2 2y 1) 1 2 1 4 2(1)( 1 , 1)22x 1 2 2 2(y 1) 2 3 2 x 1 2 2 (y 1) 2 3 4 y( 1 3 , 1)2( 1 , 1 3 ) O2 2x25. 6x 2 12x 6y 2 36y 366(x 2 2x 1) 6(y 2 6y 9) 36 6(1) 6(9)6(x 1) 2 6(y 3) 2 96(x 1) 2 (y 3) 2 16yO(3, 3)(1, 3)(1, 7)x26. 16x 2 16y 2 8x 32y 12716x 2 8x 16y 2 32y 12716x 2 1 2 1x 1 6 16(y 2 2y 1) 2 127 16 116 16(1)y( 1 , 14 )O16x 1 4 2 16(y 1) 2 14413( , 1)4( 1 , 24 )x 1 4 2 (y 1) 2 9x 28. x 2 y 2 Dx Ey F 00 2 (1) 2 D(0) E(1) F 0 ⇒E F 1(3) 2 (2) 2 D(3) E(2) F 0 ⇒3D 2E F 13(6) 2 (1) 2 D(6) E(1) F 0 ⇒6D E F 37 E F 1(1)(3D 2E F) (1)(13)3D E 123D 2E F 13(1)(6D 2E F) (1)(37)3D 2E 243D E 12 3(6) E 123D E 246D 36D 6E 6(6) F 1F 7x 2 y 2 6x 6y 7 0x 2 6x 9 y 2 6y 9 7 9 9(x 3) 2 (y 3) 2 25center: (h, k) (3, 3)radius: r 2 25r 25 or 529. x 2 y 2 Dx Ey F 07 2 (1) 2 D(7) E(1) F 0 ⇒7D E F 5011 2 (5) 2 D(11) E(5) F 0 ⇒11D 5E F 1463 2 (5) 2 D(3) E(5) F 0 ⇒3D 5E F 347D E F 50(1)(11D 5E F) 1(146)4D 4E 9611D 5E F 146(1)(3D 5E F) 1(34)8D12D 144(14) 4E 964E 40 7(14) (10) F 50E 10 F 58x 2 y 2 14x 10y 58 0x 2 14x 49 y 2 10y 25 58 49 25(x 7) 2 (y 5) 2 16center: (h, k) (7, 5)radius: r 2 16r 16 or 4313 Chapter 10
30. x 2 y 2 Dx Ey F 0(2) 2 7 2 D(2) E(7) F 0 ⇒2D 7E F 53(9) 2 0 2 D(9) E(0) F 0 ⇒9D F 81(10) 2 (5) 2 D(10) E(5) F 10 ⇒10D 5E F 1252D 7E F 53(1)(9D F) (1)(81)7D 7E 28D E 410D 5E F 125(1)(9D F) (1)(81)D 5E 44D 5E 44D E 44E 40E 10D (10) 4D 69(6) F 81F 135x 2 y 2 6x 10y 135 10x 2 6x 9 y 2 10y 25 135 9 25(x 3) 2 (y 5) 2 169center: (h, k) (3, 5)radius: r 2 169r 169 or 1331. x 2 y 2 Dx Ey F 0(2) 2 3 2 D(2) E(3) F 0 ⇒2D 3E F 136 2 (5) 2 D(6) E(5) F 0 ⇒6D 5E F 610 2 7 2 D(0) E(7) F 0 ⇒7E F 492D 3E F 13(1)(6D 5E F) (1)(61)8D 8E 48D E 66D 5E F 61(1)(7E F) (1)(49)6D 12E 12D 2E 2D E 6D 2E 2E 4E 4D 2(4) 2D 107(4) F 49F 21x 2 y 2 10x 4y 21 0x 2 10x 25 y 2 4y 4 21 25 4(x 5) 2 (y 2) 2 50center: (h, k) (5, 2)radius: r 2 50r 50 or 5232. x 2 y 2 Dx Ey F 04 2 5 2 D(4) E(5) F 0 ⇒4D 5E F 41(2) 2 3 2 D(2) E(3) F 0 ⇒2D 3E F 13(4) 2 (3) 2 D(4) E(3) F 0 ⇒4D 3E F 254D 5E F 41(1)(2D 3E F) (1)(13)6D 2E 283D E 144D 5E F 41(1)(4D 3E F) (1)(25)8D 8E 16D E 23D E 14(1)(D E) (1)(2)2D 12D 66 E 2E 42(6) 3(4) F 13F 37x 2 y 2 6x 4y 37 0x 2 6x 9 y 2 4y 4 37 9 4(x 3) 2 (y 2) 2 50center: (h, k) (3, 2)radius: r 2 50r 50 or 5233. x 2 y 2 Dx Ey F 01 2 4 2 D(1) E(4) F 0 ⇒D 4E F 172 2 (1) 2 D(2) E(1) F 0 ⇒2D E F 5(3) 2 0 2 D(3) E(0) F 0 ⇒3D F 9D 4E F 17(1)(2D E F) (1)(5)D 5E 12D 4E F 17(1)(3D F) (1)(9)4D 4E 8D E 2D 5E 12D E 26E 14E 7 3 D 7 3 2D 1 3 3 1 3 F 9F 8x 2 y 2 Dx Ey F 0x 2 y 2 1 3 x 7 3 y 8 0x 2 1 3 1x 3 6 y 2 7 3 y 4 936 8 316x 1 6 2 y 7 6 2 1 1698 4 936 Chapter 10 314
center: (h, k) 1 6 , 7 6 radius: r 2 1 169869 813 3or 13 226r 1 134. x 2 y 2 Dx Ey F 00 2 0 2 D(0) E(0) F 0 ⇒F 0(2.8) 2 0 2 D(2.8) E(0) F 0 ⇒2.8D F 7.84(5) 2 2 2 D(5) E(2) F 0 ⇒5D 2E F 292.8D 0 7.84 5(2.8) 2E (0) 292.8D 7.84 2E 15D 2.8E 7.5x 2 y 2 2.8x 7.5y 0 0x 2 2.8x 1.96 y 2 7.5y 14.0625 1.96 14.0625(x 1.4) 2 (y 3.75) 2 16.0225or about (x 1.4) 2 (y 3.75) 2 16.0235. (x h) 2 (y k) 2 r 2[x (4)] 2 (y 3) 2 r 2(x 4) 2 (y 3) 2 r 2(0 4) 2 (0 3) 2 r 225 r 2(x 4) 2 (y 3) 2 2536. (x h) 2 (y k) 2 r 2(x 2) 2 (y 3) 2 r 2(5 2) 2 (6 3) 2 r 218 r 2(x 2) 2 (y 3) 2 1837. midpoint of diameter: x 1 x 22y 2(6), 3 (5)2 , y 1 2 2 2 (2, 1)r (x x 2 1 ) 2 (y 2 y 1 ) 2 (2 2) 2 (1 3) 2 (4) 2 ) (42 32(x h) 2 y k) 2 r 2[x (2)] 2 [y (1)] 2 32 2(x 2) 2 (y 1) 2 3238. midpoint of diameter: x 1 x 22y 2 2, 4 12 , y 1 2 3 2 1 2 , 5 2 r (x x 2 1 ) 2 (y 2 y 1 ) 2 1 2 2 2 1 5 2 2 5 2 2 3 2 2 1 72(x h) 2 (y k) 2 r 2x 1 2 2 y 5 2 2 1 7 22x 1 2 2 y 5 2 2 1 2739. (x h) 2 (y k) 2 r 2(x 5) 2 (y 1) 2 r 2x 3y 2x 3y 2 0 ⇒ A 1, B 3, and C 2Ax 1 By 1 Cr A B 2 2 10 10 or 10(x 5) 2 (y 1) 2 10 2(x 5) 2 (y 1) 2 1040. center: (h, 0), radius: r 1(x h) 2 (y k) 2 r 2 22 h 2 20 2 1 22 1 2 2h h 2 1 2 1h 2 2h 1 1h h 2 0h 0 or h 2(x 0) 2 (y 0) 2 1 x 2 2 (y 0) 2 141a. (x h) 2 (y k) 2 r 2x 2 y 2 1 or x 2 2 y 2 12 2(x 0) 2 (y 0) 2 1 2x 2 y 2 3641b. x 2 y 2 36y 2 36 x 2y 36 x 2dimensions of rectangle:2x by 2y ⇒ 2x by 236 x 241c. A(x) 2x 236 x 241d.(1)(5) (3)(1) 21 2 3 2 4x36 x 2[0, 10] scl:1 by [0, 100] scl:2041e. Use 4: maximum on the CALC menu of thecalculator. The x-coordinate of this point is about4.2. The maximum area of the rectangle is thecorresponding y-value of 72, for an area of72 units 2 .42a.[15.16, 15.16] scl:1 by [5, 5] scl:142b. a circle centered at (2, 3) with radius 442c. (x 2) 2 (y 3) 2 16315 Chapter 10
42d. center: (h, k) (4, 2)radius: r 2 36r 36 or 62nd [DRAW] 9:Circle( - 4 , 2 , 6)2(13) (9) F 5F 30x 2 y 2 13x 9y 30 0x 2 13x 42.25 y 2 9y 20.25 30 42.25 20.25(x 6.5) 2 (y 4.5) 2 32.545a. (x h) 2 (y k) 2 r 2(x k) 2 (y k) 2 2 2(x k) 2 (y k) 2 445b. (x 1) 2 (y 1) 2 4y y x(x 0) 2 (y 0) 2 4(x 1) 2 (y 1) 2 4O x[15.16, 15.16] scl:1 by [5, 5] scl:143a. (x h) 2 (y k) 2 r 2(x 0) 2 (y 0) 2 2 4 22 x 2 y 2 14443b. x 2 y 2 6.25 ⇒ r 2 1 6.25r 1 6.25 or 2.5If the circles are equally spaced apart thenradius r 2 of the middle circle is found by addingthe radius of the smallest circle to the radius ofthe largest circle and dividing by two.r 2 12 2.52or 7.25area of area of area ofregion B middle circle smallest circle r 2 2 r 21 (r 2 2 r 2 1 ) (7.25 2 2.5 2 ) (46.3125) or about 145.50The area of region B is about 145.50 in 2 .44.yx 5y 3 0(2, 1) (12, 3)O(5, 1) x2x 3y 7 04x 7y 27x 2 y 2 Dx Ey F 02 2 1 2 D(2) E(1) F 0 ⇒2D E F 55 2 (1) 2 D(5) E(1) F 0 ⇒5D E F 2612 2 3 2 D(12) E(3) F 0 ⇒12D 3E F 1532D E F 5(1)(5D E F) (1)(26)3D 2E 212D E F 5(1)(12D 3E F) (1)(153)10D 2E 1483D 2E 2110D 2E 14813D 169D 133(13) 2E 212E 18E 945c. All of the circles in this family have a radius of 2and centers located on the line y x.46a. (x h) 2 (y k) 2 r 2(x 0) 2 (y 0) 2 14 2x 2 y 2 196y 2 196 x 2y 196 x 2y 196 x 246b. No, if x 7, then y 147 12.1 ft, so thetruck cannot pass.47. x 2 y 2 8x 6y 25 0x 2 8x 16 y 2 6y 9 25 16 9(x 4) 2 (y 3) 2 0radius: r 2 0r 0 or 0center: (h, k) (4, 3)Graph is a point located at (4, 3).48a. (x h) 2 (y k) 2 r 2(x 0) 2 (y 0) 2 r 2x 2 y 2 r 2(475) 2 (1140) 2 r 21,525,225 r 2x 2 y 2 1,525,22548b. r 2 1,525,225r 1,525 ,225 or 1235A r 2 (1235) 2 or approximately 4,792,000 ft 22500 2 4,792,00048c. 2500 2 0.23328about 23%49a. PA has a slope of y 4x 3and PB has slope of y 4x .3If PA ⊥ PB then y 4x 3 y 4x 31. 4 4 1 y x 3 y x 3 y 2 16x2 9 1y 2 16 x 2 9x 2 y 2 2549b. If PA ⊥ PB, then A, P, and B are on the circlex 2 y 2 25.50. d (x x 2 1 ) 2 (y 2 y 1 ) 2d (2 4) 2 (3)] [6 2d (6) 2 9 2d 117 Chapter 10 316
51. (2 i)(3 4i)(1 2i) (6 8i 3i 4i 2 )(1 2i) [6 8i 3i 4(1)](1 2i) (10 5i)(1 2i) 10 20i 5i 10i 2 10 20i 5i 10(1) 20 15i52. x t⏐v u ⏐ cos v y t⏐v u ⏐ sin v 1 2 gt 2x t(60) cos 60° y t(60) sin 60° 1 2 (32)t 2x 60t cos 60° y 60t sin 60° 16t 2x 60(0.5) cos 60° y 60(0.5) sin 60° 16(0.5) 2x 15 y 21.9807621115 ft horizontally, about 22 ft vertically53. A 5 2 or 2.520 2 k and k 1 0y A cos (kt)y 2.5 cos 10 t54. s 1 2 (a b c) 1 2 (15 25 35) 37.5k s(s a)(s b)(s c) 37.5(3 7.5 7.5 15)(37.5 25)(33.5) 26,36 7.187 5 162 units 255. v v4h2 0 695 15 2 64h95 2 15 2 64hh 952 15 6 24h 137.5 ft56. y 6x 4 3x 2 1b 6a 4 3a 2 1 (x, y) (a, b)x-axis: (x, y) (a, b)b 6a 4 3a 2 1; noy-axis: (x, y) (a, b)b 6(a) 4 3(a) 2 1b 6a 4 3a 2 1; yesy x: (x, y) (b, a)a 6b 4 3b 2 1; noorigin: f(x) f(x)f(x)6(x) 4 3(x) 2 1 f(x)(6x 4 3x 2 1)f(x)6x 4 3x 2 1 f(x)6x 4 3x 2 1noThe graph is symmetric with respect to the y-axis.57a. Let x number of cases of drug A.Let y number of cases of drug B.x 10yy 9x 103x 5y 55 12 (0, 9)y 9x 5y 458(5, 8)x 5y 454 (10, 5)3x 5y 55x 0 (0, 0)O 4 8 12 16 20 24xy 0(0, 0), (0, 9), (5, 8), (10, 5)f(x, y) 320x 500yf(0, 0) 320(0) 500(0) 0f(0, 0) 320(0) 500(9) 4500f(5, 8) 320(5) 500(8) 5600f(10, 5) 320(10) 500(5) 5700The maximum profit occurs when 10 cases ofdrug A and 5 cases of drug B are produced.57b. When 10 cases of drug A and 5 cases of drug Bare produced, the profit is $5700.58.yx x x x xx x x x x y y 5x 2xThe correct choice is A10-3EllipsesPages 637–6381. ay 2 2 bx 2 2 1 8y 2 2 5x 2 2 1y2x2 6 4 2 5 1Check For Understanding2. Since the foci lie on the major axis, determinewhether the major axis is horizontal or vertical. Ifthe a 2 is the denominator of the x terms, the majoraxis is horizontal. If the a 2 is the denominator ofthe y terms, the major axis is vertical.3. When the foci and center of an ellipse coincide,c 0.c 2 a 2 b 2ce a0 a 2 b 20b 2 a 2e ab ae 0The figure is a circle.4. In an ellipse, b 2 a 2 c 2 cand a e.c a ec aec 2 a 2 e 25. Shanice; an equation with only one squared termcannot be the equation of an ellipse.6. center: (h, k) (7, 0)a ⏐0 6⏐ or 6b ⏐7 (4)⏐ or 3 (y k) 2a 2 (x h) 2b 2 1 (y 0) 26 2 [x (7)] 232y2 3 6 (x 7) 29c a 2 b2c 6 2 3 2 c 33y 1 1b 2 a 2 c 2b 2 a 2 a 2 e 2b 2 a 2 (1 e 2 )foci: (h, k c) (7, 0 33) (7, 33)317 Chapter 10
7. center: (h, k) (0, 0)a 2 36 b 2 4 c a 2 ba 36 or 6 b 4 or 2 c 36 4 or 42foci: (h c, k) (0 42, 0) or (42, 0)major axis vertices: (h a, k) (0 6, 0) or (6, 0)minor axis vertices: (h, k b) (0, 0 2) or (0, 2)(6, 0)y8. center: (h, k) (0, 4)a 2 81 b 2 49 c a 2 b 2 a 81 or 9 b 49 or 7 c 81 9 4 or 42foci: (h c, k) (0 42, 4) or (42, 4)major axis vertices: (h a, k) (0 9, 4) or(9, 4)minor axis vertices: (h, k b) (0, 4 7) or(0, 11), (0, 3)(9, 4)(0, 2)O(0, 2)yO9. 25x 2 9y 2 100x 18y 11625(x 2 4x ?) 9(y 2 2y ?) 116 ? ?25(x 2 4x 4) 9(y 2 2y 1) 116 25(4) 9(1)25(x 2) 2 9(y 1) 2 225 (x 2) 29 (y 1) 225 1center: (h, k) (2, 1)a 2 25 b 2 9 c a 2 b2a 25 or 5 b 9 or 3 c 25 9 or 4foci: (h, k c) (2, 1 4) or (2, 5), (2, 3)major axis vertices: (h, k a) (2, 1 5) or(2, 6), (2, 4)minor axis vertices: (h b, k) (2 3, 1) or(1, 1), (5, 1)(2, 6)(0, 11)(0, 4)(6, 0)(0, 3)yx(9, 4)x10. 9x 2 4y 2 18x 16y 119(x 2 2x ?) 4(y 2 4y ?) 11 ? ?9(x 2 2x 1) 4(y 2 4y 4) 11 9(1) (4)9(x 1) 2 4(y 2) 2 36 (x 1) 24 (y 2) 29 36center: (h, k) (1, 2)a 2 9 b 2 4 c a 2 b2a 9 or 3 b 4 or 2 c 9 4 or 5foci: (h, k c) 1, 2 5 major axis vertices: (h, k a) (1, 2 3) or(1, 1), (1, 5)minor axis vertices: (h b, k) (1 2, 2) or(3, 2), (1, 2)y(1, 1)Ox(1, 2)(3, 2)(1, 2)11. center: (h, k) (2, 3)a 8 2 or 4b 2 2 or 1 (y a 2 k) 2 (x b 2 h) 2 1 [y (3)] 24 2 [x (2)] 21 2 (y 3) 216 (x 2) 21 1 112. The major axis contains the foci and it is locatedon the x-axis.center: (h, k) 1 12, 0 02 or (0, 0)c 1, a 4c 2 a 2 b 21 2 4 2 b 2b 2 15 (x a 2 h) 2 (y b 2 k) 2 1 (x 0) 24 2 (y 0) 215x2y2 1 6 1 5 1 113. center: (h, k) (1, 2)y(1, 5)(1, 4)The points at (1, 4) and(5, 2) are vertices of theellipse.(5, 1)(2, 1)(1, 1)(5, 2)(1, 2)OxOx(2, 4)a 4, b 2 (x h) 2a 2 (y k) 2b 2 1 (x 1) 24 2 (y 2) 22 2 1 (x 1) 216 (y 2) 24 1Chapter 10 318
14. center: (h, k) (3, 1) foci: (h, k c) 3, 4 39a 62 c19. center: (h, k) (2, 1)ce c 2 a 2 b 2a 2 4 b 2 1 c a 2 b2 aa 4 or 2 b 1 or 1 c 4 1 or 3 1 3 c2 2 6 2 b 2 6 4 36 b 2foci: (h, k c) (2, 1 3)major axis vertices: (h, k a) (2, 1 2) orb 2 32(2, 3), (2, 1) (y k) 2 minor axis vertices: (h b, k) (2 1, 1) ora 2 (h h) 2b 2 1(1, 1), (3, 1) (y 1) 2 3) 2y6 2 1 (x 32 (x 3) 232 (y 1) 236 115. The major axis contains the foci and is located onthe x-axis.center: (h, k) (0, 0)c 0.141732a 1 2 (3.048) or 1.524c 2 a 2 b 2(0.141732) 2 (1.524) 2 b 20.020 2.323 b 2b 2 2.302b 1.517 (x h) 2a 2 (y k) 2b 2 1 ( x 0)21.52 42 ( y 0)21 . 5172x2y2 1.5 24 2 1.5 17 2 1 1Pages 638–641 Exercises16. center: (h, k) (0, 5)a ⏐0 (7)⏐ or 7b ⏐5 0⏐ or 5 (x h) 2a 2 (y k) 2b 2 1 (x 0) 27 2 (y (5)] 252x2 4 9 (y 5) 225 1 1c a 2 b2c 7 2 5 2 or 26foci: (h c, k) (0 26, 5) (26, 5)17. center: (h, k) (2, 0)a ⏐2 2⏐ or 4b ⏐0 2⏐ or 2 (x a 2 h) 2 (y b 2 k) 2 1 [x (2)] 242 (y 0) 22 2 (x 2) 216 y 24 1 1c a 2 b2c 4 2 2 2 or 23foci: (h c, k) (2 23, 0)18. centers: (h, k) (3, 4)a ⏐4 12⏐ or 8b ⏐3 2⏐ or 5 (y a 2 k) 2 (x b 2 h) 2 1 (y 4) 28 2 [x (3)] 252 (y 4) 264 (x 3) 225 1 1c a 2 b2c 8 2 5 2 or 39(3, 1)(2, 3)O(2, 1)(2, 1)(1, 1)x20. center: (h, k) (6, 7)a 2 121 b 2 100a 121 or 11 b 100 or 10c a 2 b2c 121 100or 21foci: (h, k c) (6, 7 21)major axis vertices: (h, k a) (6, 7 11) or(6, 18), (6, 4)minor axis vertices: (h b, k) (6 10, 7) or(16, 7), y (4, 7)(6, 18)(6, 7)(16, 7)21. center: (h, k) (4, 6)a 2 16 b 2 9 c a 2 b2a 16 or 4 b 9 or 3 c 16 9 or 7foci: (h c, k) (4 7, 6)major axis vertices: (h a, k) (4 4, 6) or(8, 6), (0, 6)minor axis vertices: (h, k b) (4, 6 3) or(4, y 3), (4, 9)O(4, 7)O(0, 6)(6, 4)(4, 3)(4, 6)(4, 9)xx(8, 6)319 Chapter 10
22. (h, k) (0, 0)a 2 9 b 2 4 c a 2 ba 9 or 3 b 4 or 2 c 9 4 or 5foci: (h, k c) 0, 0 5 or 0, 5 major axis vertices: (h, k a) (0, 0 3) or(0, 3)minor axis vertices: (h b, k) (0 2, 0) or(2, 0)(2, 0)23. 4x 2 y 2 8x 6y 9 04(x 2 2x ?) (y 2 6y ?) 9 ? ?4(x 2 2x 1) (y 2 6y 9) 9 4(1) 94(x 1) 2 (y 3) 2 4 (x 1) 21 (y 3) 24 1center: (h, k) (1, 3)a 2 4 b 2 1 c a 2 b a 4 or 2 b 1 or 1 c 4 1 or 3foci: (h, k c) 1, 3 3 major axis vertices: (h, k a) (1, 3 2)or (1, 1), (1, 5)minor axis vertices: (h b, k) (1 1, 3)or (2, 3), (0, 3)yO(0, 3)(1, 3)yO(0, 3)(0, 3)(1, 1)(1, 5)(2, 3)(2, 0)xx24. 16x 2 25y 2 96x 200y 14416(x 2 6x ?) 25(y 2 8y ?) 144 ? ?16(x 2 6x 9) 25(y 2 8y 16) 144 16(9) 25(16)16(x 3) 2 25(y 4) 2 400 (x 3) 225 (y 4) 216 1center: (h, k) (3, 4)a 2 25 b 2 16 c a 2 b2a 25 or 5 b 16 or 4 c 25 16 or 3foci: (h c, k) (3 3, 4) or (6, 4), (0, 4)major axis vertices: (h a, k) (3 5, 4) or(8, 4), (2, 4)major axis vertices: (h, k b) (3, 4 4) or(3, 8), (3, 0)(2, 4)yO(3, 8)(3, 4)(3, 0)(8, 4)25. 3x 2 y 2 18x 2y 4 03(x 2 6x ?) (y 2 2 ?) 43(x 2 6x 9) (y 2 2y 1) 4 3(9) 13(x 3) 2 (y 1) 2 24 (x 3) 28 (y 1) 224 1center: (h, k) (3, 1)a 2 24 b 2 8a 24 or 26 b 8 or 22c a 2 b2c 24 8 or 4foci: (h, k c) (3, 1 4) or (3, 5), (3, 3)major axis vertices: (h, k a) (3, 1 26)minor axis vertices: (h b, k) (3 22, 1)(3, 1 26)(3 22, 1)y26. 6x 2 12x 6y 36y 366(x 2 2x ?) 6(y 2 6y ?) 366(x 2 2x 1) 6(y 2 6y 9) 36 6(1) 6(9)6(x 1) 2 6(y 3) 2 96 (x 1) 216 (y 3) 216 1center: (h, k) (1, 3)a 2 16 b 2 16 c a 2 b 2 a 16 or 4 b 16 or 4 c 16 16 or 0foci: (h c, k) or (h, k c) (1, 3)Since a b 4, the vertices are (h 4, k) and(h, k 4) or (5, 3), (3, 3), (1, 1), (1, 7)y(1, 1)(3, 3)xx(3, 1) (3 22, 1)(3, 1 26)O(1, 7)Ox(1, 3) (5, 3)Chapter 10 320
27. 18y 2 12x 2 144y 48x 12018(y 2 8y ?) 12(x 2 4x ?) 120 ? ?18(y 2 8y 16) 12(x 2 4x 4) 120 18(16) 12(4)18(y 4) 2 12(x 2) 2 216 (y 4) 212 (x 2) 218 1center: (h, k) (2, 4)a 2 18 b 2 12a 18 or 32 b 12 or 23c a 2 b2c 18 12 or 6foci: (h c, k) (2 6, 4)major axis vertices: (h g, k) (2 32, 4)minor axis vertices: (h, k b) (2, 4 23)y(2, 4 23)29. 49x 2 16y 2 160y 384 049x 2 16(y 2 10y ?) 384 ?49x 2 16(y 2 10y 25) 384 16(25)49x 2 16(y 5) 2 784x2 1 6 (y 5) 249 1center: (h, k) (0, 5)a 2 49 b 2 16 c a 2 b 2 a 49 or 7 b 16 or 4 c 49 16 or 33foci: (h, k c) (0, 5 33)major axis vertices: (h, k a) (0, 5 7) or(0, 2), (0, 12)minor axis vertices: (h b, k) (0 4, 5) or(4, 5)yO(0, 2)x(2, 4)(2 32, 4) (2 32, 4)(4, 5)(0, 5)(4, 5)O (2, 4 23)28. 4x 2 8y 9x 2 54x 49 04(y 2 2y ?) 9(x 2 6x ?) 49 ? ?4(y 2 2y 1) 9(x 2 6x 9) 49 4(1) 9(9)4(y 1) 2 9(x 3) 2 36x (y 1) 29 (x 3) 24 1center: (h, k) (3, 1)a 2 9 b 2 4 c a 2 b a 9 or 3 b 4 or 2 c 9 4 or 5foci: ( j, k c) (3, 1 5)major axis vertices: (h, k a) (3, 1 3) or (3, 4),(3, 2)minor axis vertices: (h b, k) (3 2, 1) or(5, 1), (1, 1)y(3, 4)30. 9y 2 108y 4x 2 56x 4849(y 2 12y ?) 4(x 2 14x ?) 484 ? ?9(y 2 12y 36) 4(x 2 14x 49) 484 9(36) 4(49)9(y 6) 2 4(x 7) 2 36 (y 6) 24 (x 7) 29 1center: (h, k) (7, 6)a 2 9 b 2 4 c a 2 b 2 a 9 or 3 b 4 or 2 c 9 4 or 5foci: (h c, k) (7 5, 6)major axis vertices: (h a, k) (7 3, 6) or(10, 6)(4, 6)minor axis vertices: (h, k b) (7, 6 2) or(7, 4), (7, 8)yO(0, 12)x(1, 1) (3, 1)O(5, 1)x(4, 6)(7, 4)(10, 6)(7, 6)(3, 2)(7, 8)31. a 7, b 5 (x a 2 h) 2 (y b 2 k) 2 1 [x (3)] 27 2 [y (1)] 25 2 (x 3) 249 (y 1) 225 1 1321 Chapter 10
32. The major axis contains the foci and it is locatedon the x-axis.center: (h, k) 2 20, 0 02 or (0, 0)c 2, a 7c 2 a 2 b 22 2 7 2 b 2b 2 45 (x h) 2a 2 (y k) 2b 2 1 (x 0) 27 2 (y 0) 245x2y2 4 9 4 533. b 3 4 a 1 16 3 4 a8 a (x h) 2a 2 (y k) 2b 2 1 (x 0) 28 2 (y 0) 26 2 1x2y2 6 4 3 6 134. The major axis contains the foci and it is thevertical axis of the ellipse.center: (h, k) 1 (1)2, 1 (5)2 or (1, 2)c ⏐1 k⏐a 213c ⏐1 (2)⏐ or 3c 2 a 2 b 23 2 213 2 b 2b 2 52 9b 2 43 (y a 2 k) 2 (x b 2 h) 2 1 [y ( 2)] 2(2 13) 2 [x ( 1)] 24 3 (y 2) 252 (x 1) 243 1 135. yThe horizontal axis ofthe ellipse is the major12(2, 9)axis.8(11, 5) (7, 5)4(2, 1)1284Oenter: (h, k) 11 72, 5 52 or (2, 5)h a 7 k b 92 a 7 5 b 9a 9 b 4 (x h) 2 a 2 (y k) 2 1b 2 [x (2)] 292 (y 5) 24 2 (x 2) 2 (y 5) 281 16 1 14 8x36. The major axis contains the foci and it is thevertical axis of the ellipse.c 5 (1)2or 3center: (h, k) 1 12, 1 52 or (1, 2) (y k) 2a 2 (x h) 2b 2 1 (2 a 2 2) 2 (y b 2 1) 2 138. ytangent vertices:(4, 0)(4, 0), (0, 7)O 4 8 12 x a ⏐ 7 0⏐ or 74b ⏐4 0⏐ or 4(0, 7)8 (4, 7)1220 2 a 2 3 b2 19 b 2 19 b 2 (y k) 2 a 2 (x h) 2 1b 2 (y 2) 218 (x 1) 2937. 1 2 ac a 2 c 1 1 02 c5 c (x h) 2a 2 (y k) 2 1b 2 (x 0) 2 10 2 (y 0) 275x2y2 1 00 7 5 1 1 (y a 2 k) 2 (x b 2 h) 2 1 [y ( 7)]72 (x 4) 24 2 (y 7) 249 (x 4) 216 1 139. b 2 a 2 (1 e 2 )b 2 2 2 1 3 4 2 b 2 2 816 or 1.75Case 1: Horizontal axis is major axis. (x h) 2 a 2 (y k)b2 (x 0) 2 2 2 (y 0) 21.75 x 2 y24 1 .75 1 1 1Case 2: Vertical axis is major axis. (y k) 2 a 2 (x h) 2 1b 2 (y 0) 2 2 2 (x 0) 21.75 y 2 x24 1 .75 1 1c 2 a 2 b 23 2 a 2 918 a 2b 2 a 2 c 2b 2 10 2 5 2b 2 75Chapter 10 322
40. The major axis contains the foci and it is thehorizontal axis of the ellipse.center: (h, k) 3 12, 5 52 or (2, 5)foci: (3, 5) (h c, k)3 h c3 2 c1 cce ab 2 a 2 (1 e 2 )10.25 b 2 4 2 (1 0.25 2 )a b 2 15a 4 (x a 2 h) 2 (y b 2 k) 2 1 (x 2) 2 4 2 (y 5) 215 (x 2) 216 (y 15) 215 1 141. a 2 02or 10b 2 a 2 (1 e 2 )b 2 10 2 1 710 (y k) 2 a 2 (x h) 2 1b 2 (y 0) 2 10 2 (x 3) 251y2 1 00 (x 3) 251 2 or 51 1 142. focus: (1, 1 5) (h, k c)1 5 k c1 5 1 c5 cce ab 2 a 2 (1 e 2 ) 53 5aa 3major axis: vertical axis (y k) 2 a 2 (x h) 2 1b 2 [y (1)] 23 2 (x 1) 24 (y 1) 29 (x 1) 24 1 1b 2 3 2 1 3b 2 443. x 2 4y 2 6x 24y 41(x 2 6x ?) 4(y 2 6y ?) 41 ? ?(x 2 6x 9) 4(y 2 6y 9) 41 9 4(9)(x 3) 2 4(y 3) 2 44(y 3) 2 4 (x 3) 2(y 3) 2 4 (x 4y 3 y 3) 25 2 (x 3)4 4 2 (x 3)4 4 2 3vertices: (5, 3),(1, 3), (3, 2),(3, 4)44. 4x 2 y 2 8x 2y 14(x 2 2x ?) (y 2 2y ?) 1 ? ?4(x 2 2x 1) (y 2 2y 1) 1 4(1) 14(x 1) 2 (y 1) 2 4(y 1) 2 4 4(x 1) 2y 1 4 4(x 1) 2y 4 (x 4) 1 2 1Vertices: (0, 1),(2, 1), (1, 1),(1, 3)[4.7, 4.7] scl:1 by [3.1, 3.1] scl:145. 4x 2 9y 2 16x 18y 114(x 2 4x ?) 9(y 2 2y ?) 11 ? ?4(x 2 4x 4) 9(y 2 2y 1) 11 4(4) 9(1)4(x 2) 2 9(y 1) 2 369(y 1) 2 36 4(x 2) 2y 1 36 y 36 4 (x 2)29 4 (x 2)29 1Vertices: (1, 1),(5, 1), (2, 3),(2, 1)[7.28, 7.28] scl:1 by [4.8, 4.8] scl:146. 25y 2 16x 2 150y 32x 15925(y 2 6y ?) 16(x 2 2x ?) 159 ? ?25(y 2 6y 9) 16(x 2 2x 1) 159 25(9) 16(1)25(y 3) 2 16(x 1) 2 40025(y 3) 2 400 16(x 1) 2(y 3) 400 y 400 [15.16, 15.16] scl:1 by [10, 10] scl:1 1 6(x 1)225 1 6(x 1)25 3Vertices: (4, 3),(6, 3), (1, 7),(1, 1)[7.28, 7.28] scl:1 by [4.8, 4.8] scl:1323 Chapter 10
47. The target ball should be placed opposite thepocket, 5 feet from the center along the majoraxis of the ellipse. The cue ball can be placedanywhere on the side opposite the pocket. Theellipse has a semi-major axis of length 3 feet and asemi-minor axis of length 2 feet. Using theequation c 2 a 2 b 2 , the focus of the ellipse isfound to be 5 feet from the center of the ellipse.Thus the hole is located at one focus of the ellipse.The reflective properties of an ellipse shouldinsure that a ball placed 5 feet from the centerof the ellipse and hit so that it rebounds once offthe wall should fall into the pocket at the otherfocus of the ellipse.48. Ahorizontal line; see students’ work.49a. a 9 62or 48b 4 62or 23(h, k) (0, 0) (x h) 2 a 2 (y k) 2 1b 2 (x 0) 2 48 2 (y 0) 223 2x2y2 23 04 5 29 1 149b. c a 2 b2c 2304 529c 42.13He could have stood at a focal point, about 42feet on either side of the center along the majoraxis.49c. The distance between the focal points is 2c.2c 2(42) 84about 84 ft50a. x 2 y 2 r 2 x r 2 2 y r 2 2 1 A r 2x 2 ya 2 2 b 2 1A r rA a bA ab50b. x 29 y 24 1a 2 9 b 2 4a 3 b 2A abA (3)(2)A 6 units 251. If (x, y) is a point on the ellipse, then show that(x, y) is also on the ellipse.ax 2 2 by 2 2 1 ( x) 2a 2 ( y) 2b 2 1x 2 ya 2 2 b 2 1Thus, (x, y) is also a point on the ellipse andthe ellipse is symmetric with respect to the origin.52a. a 8 2 or 4b 3c a 2 b 2 c 4 2 3 2 c 7foci: (h c, 0) (0 7, 0) or (7, 0)The thumbtacks should be placed (7, 0) fromthe center of the arch.52b. With the string anchored by thumbtacks at thefoci of the arch and held taunt by a pencil, thesum of the distances from each thumbtack to thepencil will remain constant.53a. GOES 4; its eccentricity is closest to 0.53b. yc a exaOcx 69c55 0.052c 361.66[figure not drawn to scale]x a c Earth’s radiusx 6955 361.66 6357x 959.66x 960 km54. x 2 y 2 Dx Ey F 00 2 (9) 2 D(0) E(9) F 0 ⇒9E F 817 2 (2) 2 D(7) E(2) F 0 ⇒7D 2E F 53(5) 2 (10) 2 D(5) E(10) F 0 ⇒5D 10E F 125 9E F 81(1)(7D 2E F) (1)(53) 7D 7E 28D E 47D 2E F 53(1)(5D 10E F) (1)(125)12D 8E 72(8)(D E) (8)(4)12D 8E 724D 40D 10D E 49E F 8110 E 4 9(6) F 81E 6F 135x 2 y 2 Dx Ey F 0x 2 y 2 10x 6y 135 0(x 2 10x 25) (y 2 6y 9) 135 25 9(x 5) 2 (y 3) 2 169Chapter 10 324
center: (h, k) (5, 3)radius: r 2 169r 13y(5, 16)(5, 3)O(8, 3)55. Graph the quadrilateral with vertices A(1, 2),B(5, 4), C(4, 1), and D(5, 4).A quadrilateral is aparallelogram if onepair of opposite sidesare parallel andcongruent.xD(5, 4) yOC(4, 1)x60. Let h 0.1.x h x 0.11 0.1 or 1.1f(x 0.1) f(1.1) (1.1) 2 4(1.1) 1215.19x h x 0.11 0.1 or 0.9f(x 0.1) f(0.9) (0.9) 2 4(0.9) 1214.79f(x) 16f(x) f(x 0.1) and f(x) f(x 0.1), so the pointis a location of a minimum.61. The graph of the parent function g(x) ⏐x⏐ istranslated 2 units right.g(x)A(1, 2)slope of DAm y xslope of CBm y 2 y1x x2 y12 x12 12 4 1 (5) 4 15 4 64or 3 2 5 1 The slopes are not equal, so DA CB. Thequadrilateral is not a parallelogram; no.56. cos 2v 1 2 sin 2 vcos 2v 1 2 7 8 23 44or 1 73 2 cos 2v 657. ⏐A⏐ 4 2 k 180A 4 k 2c k 20° h 0c 2 20°c 40°y A cos(kx c) hy 4 cos[2x (40°)] 0y 4 cos(2x 40°)58. A 180° (121° 32 42° 5) or 16° 23B(5, 4)a ba c sin A sin Bsin A sin C4.1b4.1c sin 1 6°23 sin 4 2° 5sin 1 6°23 sin 12 1° 32 4ṡ 1 sin42°5in 16°23 b 4.1 sin121°32sin16°23 c9.7 b 12.4 c59. P(x) x 4 4x 3 2x 2 1P(5) 5 4 4(5) 3 2(5) 2 1P(5) 74P(5) 0; no, the binomial is not a factor of thepolynomial.Ox62. Initial location: (2, 0)Rot 90 0 1 2 0or (0, 2)1 0 0 2Rot 80 1 0 2 2 or (2, 0)0 1 0 0Rot 270 0 1 2 0 or (0, 2)1 0 0 263. mQTS mTSR 180a b c d 180 p b c 180b b c c 180 p 90 1802b 2c 180 p 90b c 90The correct choice is C.Page 641 Graphing Calculator Exploration1. Sample answer: The graph will shift 4 units to theright.[15.16, 15.16] scl:2 by [10, 10] scl:2325 Chapter 10
2. Sample answer: The graph will shift 4 units to theleft.[15.16, 15.16] scl:2 by [10, 10] scl:23. Sample answer: The graph will shift 4 units up.[15.16, 15.16] scl:2 by [10, 10] scl:24. Sample answer: The graph will shift 4 units down.2. transverse axis: vertical2a 4a 2An equation in standard form of the hyperbolaymust have 2 2 2 or y 24as the first term; b.c3. e , a so ae c and a 2 e 2 c 2 .Since c 2 a 2 b 2 we havea 2 e 2 a 2 b 2a 2 e 2 a 2 b 2a 2 (e 2 1) b 24. With the equation in standard form, if the firstexpression contains “x”, the transverse axis ishorizontal. If the first expression contains “y”, thetransverse axis is vertical.5. center: (h, k) (0, 0)a 2 25 b 2 4 c a 2 b2a 5 b 2 c 25 4 or 29transverse axis: horizontalfoci: (h c, k) (0 29, 0) or (29, 0)vertices: (h a, k) (0 5, 0) or (5, 0)basymptotes: y k (x a h)8y 0 2 5 (x 0)yy 2 5 x(5, 0) 4 (5, 0)[18.19, 18.19] scl:2 by [12, 12] scl:25. Sample answer: The graph will rotate 90°.8 4 O484 8x[15.16, 15.16] scl:2 by [10, 10] scl:26. For (x c), the graph will shift c units to the left.For (x c), the graph will shift c units to theright.7. For (y c), the graph will shift c units down. For(y c), the graph will shift c units up.8. The graph will rotate 90°.6. center: (h, k) (2, 3)a 2 16 b 2 4 c a 2 b 2 a 16 or 4 b 4 or 2 c 16 4 or 25transverse axis: verticalfoci: (h, k c) 2, 3 25vertices: (h, k a) (2, 3 4) or (2, 7), (2, 1)asymptotes: y k a b (x h)y(2, 7)y 3 4 2 (x 2)y 3 2(x 2)10-4HyperbolasPages 649–650 Check For Understanding1. The equations of both hyperbolas and ellipseshave x 2 terms and y 2 terms. In an ellipse, theterms are added and in a hyperbola these termsare subtracted.O(2, 1)(2, 3)xChapter 10 326
7. y 2 5x 2 20x 50y 2 5(x 2 4x ?) 50 ?y 2 5(x 2 4x 4) 50 (5)(4)y 2 5(x 2) 2 30y2 3 0 (x 2) 26 1x h x 2y k yh 2 k 0center: (h, k) (2, 0)a 2 30 b 2 6 c a 2 ba 30 b 6 c 30 6 or 6transverse axis: verticalfoci: (h, k c) (2, 0 6) or (2, 6)vertices: (h, k a) 2, 0 30 or 2, 30asymptotes: y k a b (x h)y 0 3 (x0 2)6y 5(x 2)y844 O48(2, 30)(2, 0)4 8x(2, 30)8. center: (h, k) (0, 5)transverse axis: horizontala 5, b 3 (x h) 2 a 2 (y k) 2 1b 2 (x 5 2 0) 2 (y 3 2 5) 2 1x2 2 5 (y 5) 29 19. c 9quadrants: II and IVtransverse axis: y xvertices: xy 9xy 93(3) 93(3) 9(3, 3) (3, 3)y(3, 3)11. 2b 6b 3x 2center: x 1 2, y 1 2 3 2 (3, 2)transverse axis: verticala ⏐4 2⏐ or 2y 2 (y k) 2 a 2 (x h) 2 1b 2 (y 2) 2 2 2 (x 3) 2 13 2 (y 2) 2 (x 3) 249x 2 1y 23, 4 02 0, 6 (6)2 12. center: x 1 2, y 1 2 0 2 (0, 0)transverse axis: verticalc distance from center to a focus ⏐0 6⏐ or 6b 2 c 2 a 2 b 2 a 2a 2 c 2 a 2 b 2 182a 2 c 22a 2 c 2a 2 6 22or 18 (y k) 2 a 2 (x h) 2 1b 2 (y 0) 218 (x 0) 218y2x2 1 8 1 8x 2 1 1y 2(10), 0 02 13. center: x 1 2, y 1 2 10 2 (0, 0)transverse axis: horizontalc distance from center to a focus ⏐10 0⏐ or 10ce a b 2 c 2 a 2 5 3 1 0 ab 2 10 2 6 2b 2 64a 6 (x h) 2 a 2 (y k) 2 1b 2 (x 0) 2 6 2 (y 0) 264x2y2 3 6 6 4 1 114a. The origin is located midway between stations Aand B; (h, k) (0, 0). The stations are located atthe foci, so 2c 130 or c 65.yOx10. center: (h, k) (1, 4) (x h) 2 a 2 (y k) 2 1b 2 (x 1) 2 5 2 [y (4)] 222 (x 1) 225 (y 4) 24 1 1(3, 3)A(65, 0) O B(65, 0) xThe difference of the distances from the plane toeach station is 50 miles.50 2a (Definition of hyperbola)25 ab 2 c 2 a 2b 2 65 2 25 2b 2 3600327 Chapter 10
transverse axis: horizontal (x h) 2 a 2 (y k) 2 1b 2 (x 0) 2 25 2 (y 0) 23600x2y2 6 25 36 00 1 114b. Vertices: (h a, k) (0 25, 0) or (25, 0)basymptotes: y k (x a h)y 0 3 6(x0025 0)y 1 25xPlanelocatedon thisbranchStation A(65, 0)60 40 20 O 20 40 6014c. Let y 6.x2y2 6 36 12500x262 6 25 36 00x236 6 25 3 6x2 6 25x2 6 25604020204060 1y00 13 1 36600 Station B(65, 0) 1.01x 2 625(1.01)x 2 631.25x 631.2 5x 25.1Since the phase is closer to station A thanstation B, use the negative value of x to locatethe ship at (25.1, 6).Pages 650–652 Exercises15. center: (h, k) (0, 0)a 2 100 b 2 16a 100 or 10 b 16 or 4c a 2 b 2 c 100 16or 229transverse axis: horizontalfoci: (h c, k) (0 229, 0) or (229, 0)vertices: (h a, k) (0 10, 0) or (10, 0)basymptotes: y k (x a h)4y 0 1(x0 0) yy 2 5 8xx4(10, 0)16 8 O488(10, 0)16x16. center: (h, k) (0, 5)a 2 9 b 2 81a 9 or 3 b 81 or 9c a 2 b2c 9 1 8 or 310transverse axis: horizontalfoci: (h c, k) (0 310, 5) or (310, 5)vertices: (h a, k) (0 3, 5) or (3, 5)basymptotes: y k (x a h)y(0, 5)(3, 5) (3, 5)O8yx4(2, 0) (2, 0)4 2 O42 4x8y 5 9 3 (x 0)y 5 3x17. center: (h, k) (0, 0)a 2 4 b 2 49a 4 or 2 b 49 or 7c a 2 b2c 4 9 4 or 53transverse axis: horizontalfoci: (h c, k) (0 53, 0) or (53, 0)vertices: (h a, k) (0 2, 0) or (2, 0)basymptotes: y k (x a h)y 0 7 2 (x 0)y 7 2 x18. center: (h, k) (1, 7)a 2 64 b 2 4a 64 or 8 b 4 or 2c a 2 b2c 64 4 or 217transverse axis: verticalfoci: (h, k c) (1, 7 217)vertices: (h, k a) y(1, 7 8) or (1, 15), (1, 1)asymptotes: y k a b (x h)(1, 15)y 7 8 2 [x (1)]y 7 4(x 1) (1, 7)(1, 1)OxChapter 10 328
19. x 2 4y 2 6x 8y 11(x 2 6x ?) 4(y 2 2y ?) 11 ? ?(x 2 6x 9) 4(y 2 2y 1) 11 9 (4)(1)(x 3) 2 4(y 1) 2 16 (x 3) 216 (y 1) 24 1center: (h, k) (3, 1)a 2 16 b 2 4 c a 2 b 2 a 16 or 4 b 2 4 or 2 c 16 4 or 25transverse axis: horizontalfoci: (h c, k) (3 25, 1)vertices: (h a, k) (3 4, 1) or (1, 1),(7, 1)basymptotes: y k (x a h)y (1) 2 4 [x (3)]y 1 1 2 (x 3)y21. 16y 2 25x 2 96y100x356016(y 2 6y?)25(x 2 4x?)35616(y 2 6y9)25(x 2 4x4)35616(9)25(4)16(y3) 2 25(x2) 2 400 (y 3) 2 (x 2) 225 161center: (h, k) (2, 3)a 2 25 b 2 16 c a 2 ba 25 b 16 c 25 16 or 5 or 4 or 41transverse axis: verticalfoci: (h, k c) (2, 3 41)vertices: (h, k a) (2, 3 5) or (2, 8), (2, 2)asymptotes: y k a b (x h)yy 3 5 4 (x 2)(2, 8)(7, 1)O x(1, 1)(3, 1)O(2, 3)x(2, 2)20. 4x 9y 2 24x 90y 153 09(y 2 10y ?) 4(x 2 6x ?) 1539(y 2 10y 25) 4(x 2 6x 9) 153 9(25) 4(9)9(y 5) 2 4(x 3) 2 36 (y 5) 24 (x 3) 29 1center: (h, k) (3, 5)a 2 4 b 2 9 c a 2 b 2 a 4 or 2 b 9 or 3 c 4 9 or 13transverse axis: verticalfoci: (h, k c) (3, 5 13)vertices: (h, k a) (3, 5 2) or (3, 7), (3, 3)asymptotes: y k a b (x h)y 5 2 3 [x (3)]y 5 2 3 (x 3)y(3, 7)22. 36x 2 49y 2 72x294y216936(x 2 2x?)49(y 2 6y?)2169??36(x2x1)49(y 2 6y9)216936(1)49(9)36(x1) 2 49(y3) 2 1764 (x 1) 2 (y 3) 249 361center: (h, k) (1, 3)a 2 49 b 2 36 c a 2 ba 49 b 36 c 49 36 or 7 or 6 or 85transverse axis: horizontalfoci: (h c, k) (1 85, 3)vertices: (h a, k) (1 7, 3) or (8, 3),(6, 3)asymptotes: y k ab (x h)y (3) 6 7 (x 1)yy 3 6 7 (x 1)(3, 5)(3, 3)OxO(6, 3) (8, 3)(1, 3)x329 Chapter 10
23. 25y 2 9x 2 100y72x269025(y 2 4y?)9(x 2 8x?)269??25(y 2 4y4)9(x 2 8x16)26925(4)9(16)25(y2) 2 9(x4) 2 225 (y 2) 2 (x 4) 29 251center: (h, k) (4, 2)a 2 9 b 2 25 c a 2 ba 9 b 25 c 9 5 2or 3 or 5 or 34transverse axis: verticalfoci: (h, k c) 4, 2 34vertices: (h, k a) (4, 2 3) or (4, 5), (4, 1)asymptotes: y k a b (x h)y 2 3 5 [x (4)]y 2 3 5 (x 4)24. center: (h, k) (4, 3)transverse axis: verticala 4, b 3 (y k) 2a 2 (x h) 2b 2 1 (y 4 2 3) 2 (x 3 2 4) 2 1 (y 3) 216 (x 4) 29 125. center: (h, k) (0, 0)transverse axis: horizontala 3, b 3 (x h) 2a 2 (y k) 2b 2 1 (x 3 2 0) 2 (y 3 2 0) 2 1 x 29 y 29 126. center: (h, k) (4, 0)transverse axis: verticala 2, b 1 (y k) 2a 2 (x b) 2b 2 1 (y 0) 22 2 [x (4)] 212 y 24 (x 4) 21y(4, 2 34)(4, 2 34) 1 1(4, 5)3y 2 (x 4)5(4, 2)y 2 35(x 4)Ox(4, 1)27. c 49quadrants: I and IIItransverse axis: y xvertices: xy 49 xy 497(7) 49 7(7) 49(7, 7) (7, 7)yO(7, 7)(7, 7)28. c 36quadrants: II and IVtransverse axis: y xvertices: xy 36 xy 366(6) 36 6(6) 36(6, 6) (6, 6)(6, 6)yO(6, 6)29. 4xy 25xy 2 54c 2 54quadrants: II and IVtransverse axis: y xvertices: xy 2 54xy 2 5 4 5 2 5 2 2 54 5 2 5 2 2 54 5 2 , 5 2 5 2 , 5 2 ( 52 , 5 2)Oy5( 2 , 52)xxxChapter 10 330
30. 9xy 16xy 1 69c 1 69quadrants: I and IIItransverse axis: y xvertices: xy 1 69xy 1 69 4 3 4 3 1 69 4 3 , 4 3 31. center: (h, k) (4, 2) (y k) 2a 2 (x h) 2b 2 1yO( 43 , 43) [y (2)] 222 (x 4) 23 2 (y 2) 24 (x 4) 29x 26 4 3 4 3 1 9 4 3 , 4 3 1 1y 20, 3 (3)2 32. center: x 1 2, y 1 2 0 2 (0, 0)transverse axis: verticala distance from center to a vertex ⏐0 3] or 3c distance from center to a focus ⏐0 (9)⏐ or 9b 2 c 2 a 2b 2 9 2 3 2 or 72 (y a 2 k) 2 (x b 2 h) 2 1 (y 0) 23 2 (x 0) 272 y 2 x29 7 2 1 133. 2a 6a 3center: x 1 x 22, y 1 y 22 5 (5)2 (0, 2)transverse axis: horizontalc distance from center to a focus ⏐0 5⏐ or 5b 2 c 2 a 2b 2 5 2 3 2 or 16 (x b) 2a 2 (y k) 2b 2 1 (x 0) 23 2 (y 2) 216 x 29 (y 2) 216( )43 , 4 3 1 1x, 2 22 34. 2b 8b 4center: x 1 x 22, y 1 y 22 3 (3)2 (3, 2)transverse axis: verticala distance from center to a vertex ⏐2 9⏐ or 7 (y k) 2a 2 (x h) 2b 2 1 (y 2) 27 2 [x (3)] 242 (y 2) 249 (x 3) 216x 2 1 1y 2, 9 (5)2 (8), 0 02 35. center: x 1 2, y 1 2 8 2 (0, 0)transverse axis: horizontalc distance from center to a focus ⏐0 8⏐ or 8b 2 c 2 a 2 b 2 a 2a 2 c 2 a 2 b 2 322a 2 c 22a 2 c 2a 2 8 22or 32 (x h) 2a 2 (y k) 2b 2 1 (x 0) 232 (y 0) 232x2y2 3 2 3 2 1 136. centers: (h, k) (3, 1)c distance from center to a focus ⏐3 2⏐ or 5ce a b 2 c 2 a 2 5 4 5 b 2 5 2 4 2a b 2 9a 4transverse axis: horizontal (x h) 2a 2 (y k) 2b 2 1 [x (4 3)] 22 (y 1) 29 (x 3) 216 (y 1) 29 1 137. center: (h, k) (4, 2)a distance from center to a vertex ⏐2 5⏐ or 3transverse axis: vertical4y 4 3x4y 4 12 3x 124y 8 3x 124(y 2) 3(x 4)y 2 3 4 (x 4) a b 3 4 3 b 3 4 b 4 (y a 2 k) 2 (x b 2 h) 2 1 (y 3 2 2) 2 (x 4 2 4) 2 1 (y 2) 29 (x 4) 216 1331 Chapter 10
38. center: (h, k) (3, 1)a distance from center to a vertex ⏐3 5⏐ or 2transverse axis: horizontal3x 11 2y3x 11 4 2y 43x 15 2y 43(x 5) 2(y 2)b a 3 2 b 2 3 2 3 2 (x 5) y 2y 2 3 2 (x 5)b 3 (x a 2 b) 2 (y b 2 k) 2 1 (x 3) 22 2 [y (1)] 232 (x 3) 24 (y 1) 29x 2 1 1y 20, 8 (8)2 39. center: x 1 2, y 1 2 0 2 (0, 0)c distance from center to a focus ⏐0 8⏐ or 8e ac b 2 c 2 a 2 4 3 8 b 2 8 2 6 2a b 2 28a 6transverse axis: vertical (y k) 2a 2 (x h) 2b 2 1 (y 0) 26 2 (x 0) 228y2x2 3 6 2 8x 2 1 140. centers: x 1 2, y 1 2 10 2 (4, 3)c distance from center to a focus ⏐4 10⏐ or 6y 2(2), 3 (3)2 e ac b 2 c 2 a 2 6 5 6 b 2 6 2 5 2a b 2 11a 5transverse axis: horizontal (x h) 2a 2 (y k) 2b 2 1 (x 4) 25 2 [y ( 3)]11 (x 4) 225 (y 3) 211x 2 1 1y 2(9), 0 02 41. center: x 1 2, y 1 2 9 0 (0, 0)c distance from center to a focus ⏐0 9⏐ or 9b 2 c 2 a 2 b 2 a 2a 2 9 2 a 2 b 2 8 22a 2 81a 2 8 121transverse axis: horizontal (x h) 2a 2 (y k) 2b 2 1__ (x 0) 2 8 12 1 2 x281 2 y28 1x 2 1y 21, 5 (3)2 42. center: x 1 2, y 1 2 1 2 (1, 1)c distance from center to a focus ⏐1 5⏐ or 4transverse axis: vertical a b 2a 2bc 2 a 2 b 24 2 (2b) 2 b 216 5b 2 1 65 b 2 (y a 2 k) 2 (x b 2 h) 2 1(y 1) (x 1)22 1 1 6 6 455 5(y 1) 264 5(x 1) 21 6 1V25020015010050O 2 4 6 8 10c 2 a 2 b 2 .c 2 a 2 a 2c 2 2a 2c a2cSince e , a we havece aa2 a43a. quadrants: I and IItransverse axis: y x43b. PV 505(101)V 505V 5.0 dm 343c. PV 505(50.5)V 505V 10.0 dm 343d. If the pressure is halved, then the volume isdoubled, or V 2(original V ).44. In an equilateral hyperbola, a b ande __ (y 0) 2 8 12Pa 2 (2b) 2a 2 4 1 65a 2 6 45a be 2Thus, the eccentricity of any equilateral hyperbolais 2. Chapter 10 332
45a.45b.2a 150a 75ce a 5 3 7c5 125 ctransverse axis: horizontalcenter: (h, k) (0, 0) (x a 2 h) 2 (y b 2 k) 2 1 (x 0) 275 2 (y 0) 21002x2y 7 52 2 10 0 2top: 7xbase: 7xy150 ftOy(0, 100)O(0, 350)2 y52 2 10 0 2x2 7 52 1 0021002x2 7 5x2 7 5 1 1 1 1 (x, y) (x, 100)2 1 12 y52 2 10 0 2x2 7 52 ( 350) 21002x2 7 5x2 7 52 2x 2 11,250x 106.07 ft 1 1 (x, y) (x, 350)2 12.25 12 13.25b 2 c 2 a 2b 2 125 2 75 2b 2 10,000b 100x 2 74,531.25x 273.00 ft46. The origin is located midway between stations Aand B. The stations are located at the foci, so2c 4 or c 2 miles.c 2 mic 2 mi 52 80ft1 mic 10,560 ftd rtyd 1100(2) or 2200 ftx(x, 100)x(x, 350)The lightning is 2200 feet farther from station Bthan from station A. The difference of distancesequals 2a.2200 2a (Definition of hyperbola)1100 ab 2 c 2 a 2b c 2 a 2 b 10,56 0100 2 12b 10,503center: (h, k) (0, 0)transverse axis: horizontal (x h) 2a 2 (y k) 2b 2 1 (x 01100 ) 22 ( y 0)21 0,5032x2y2 11 00 2 10, 503 2 1 147. center: (h, k) (0, 0)c ⏐0 6⏐ or 6⏐PF 1 PF 2 ⏐ 102a 10a 5b 2 c 2 a 2b 2 6 2 5 2b 2 11transverse axis: horizontal (x h) 2a 2 (y k) 2b 2 148a. 1x (x 0) 25 2 (y 0) 211x2y2 2 5 1 126 y 29 1 1 1⏐PF 1 PF 2 ⏐ 2acenter: (h, k) (0, 0)a 2 16 b 2 9a 16 or 4 b 9 or 3transverse axis: horizontalvertices: (h a, k) (0 4, 0) or (4, 0)basymptotes: y k (x a h) y 2 x29 1 6 1y 0 3 4 (x h)y 3 4 xcenter: (h, k) (0, 0)a 2 9 b 2 16a 9 or 3 b 16 or 4transverse axis: verticalvertices: (h, k a) (0, 0 3) or (0, 3)asymptotes: y k a b (x h)yOy 0 3 4 (x 0)y 3 4 xxA(10,560, 0)OB(10,560, 0) x333 Chapter 10
48b. They are the same lines.48c. (y 2) 225 (x 3) 21648d. (x 3) 216 (y 2) 225 1 1center: (h, k) (3, 2)a 2 16 b 2 25a 16 or 4 b 25 or 5transverse axis: horizontalvertices: (h a, k) (3 4, 2) or (7, 2), (1, 2)basymptotes: y k (x a h) (y 2) 225 (x 3) 216y 2 5 4 (x 3) 1center: (h, k) (3, 2)a 2 25 b 2 16a 25 or 5 b 16 or 4transverse axis: verticalvertices: (h, k a) (3, 2 5) or (3, 7), (3, 3)asymptotes: y k a b (x h)yx 249. center: x 1 2, y 1 2844 O 44y 2 5 4 (x 3)y 2 2 22, 3 (3)2 (2, 0)a 4c distance from center to a focus ⏐0 3⏐ or 3b 2 a 2 c 2b 2 4 2 3 2 or 7major axis: vertical (y k) 2 a 2 (x h) 2 1b 2 (y 0) 2 4 2 (x 2) 27y2 1 6 (x 2) 27 1 150. x 2 y 2 4x 14y 28 0(x 2 4x ?) (y 2 14y ?) 28 ? ?(x 2 4x 4) (y 2 14y 49) 28 4 49(x 2) 2 (y 7) 2 81yO8xx51.AB (2 ) 11 2 () 3 2 5BC (2 ) 61 2 () 2 2 5CD (6 ) 3 2 (2 6) 2 5AD (3 ) 1 2 (6 3) 2 5Thus, ABCD is a rhombus. The slope of AD 6 33 1or 3 4 3 1and the slope of AB 1 2or 4 3 .Thus, AD is perpendicular to AB and ABCD is asquare.52. (r, v) (90, 208°)(r, v 360 k°) (90, 208° 360(1)°) (90, 152°)(r, v (2k 1)(180°)) (90, 208 (2(1) 1)(180°)) (90, 28°)53. 4(5) 1(2) 8(2) 6No, the inner product of the two vectors is notzero.54. x cos f y sin f p 0x cos 60 y sin 60 3 055.3 1 2 x 2y 3 0x 3y 6 0xtan 30° 90x00 9000 tan 30° x5196 xd rt 51 9615yA(1, 3)O9000 m30˚ d t r r60˚D(3, 6)346.4 rabout 346 m/s56.C(6, 2)B(2, 1)x(2, 7)(11, 7)(2, 16)Since 0.2506 is closer to zero than 0.6864, thezero is about 1.3.Chapter 10 334
Since 0.0784 is closer to zero than 0.2446, thezero is about 0.6.57. Case 1: r is positive and s is negative.Case 2: r is negative and s is positive.I. r 3 s 3 is false if r is negative.II. r 3 s 2 is false for each case.III. r 4 s 4 is true for each case.The correct choice is C.7. y 2 4x 2y 5 0y 2 2y 4x 5y 2 2y ? 4x 5 ?y 2 2y 1 4x 5 1(y 1) 4(x 1)vertex: (h, k) (1, 1)4p 4p 1focus: (h p, k) (1 1, 1) or (2, 1)directrix: x h px 1 1x 0axis of symmetry: y ky 1yx 010-5ParabolasO(2, 1)xPages 658–659 Check for Understanding1. The equation of a parabola will have only onesquared term, while the equation of a hyperbolawill have two squared terms.2. vertex: (h, k) (2, 1)p 4(x h) 2 4p(y k)(x 2) 2 4(4)(y 1)(x 2) 2 16(y 1)3. The vertex and focus both lie on the axis ofsymmetry. The directrix and axis of symmetry areperpendicular to each other. The focus and thepoint on the directrix collinear with the focus areequidistant from the vertex.4. (h, k) (4, 5)p 5(y k) 2 4p(x h)(y 5) 2 4(5)[x (4)](y 5) 2 20(x 4)5a. ellipse 5b. parabola5c. hyperbola 5d. circle6. vertex: (h, k) (0, 1)4p 12p 3focus: (h, k p) (0, 1 3) or (0, 4)directrix: y k py 1 3y 2axis of symmetry: x hx 0yO(0, 4)(0, 1)y 2x(1, 1)8. x 2 8x 4y 8 0x 2 8x 4y 8x 2 8x ? 4y 8 ?x 2 8x 16 4y 8 16(x 4) 2 4(y 2)vertex: (h, k) (4, 2)4p 4p 1focus: (h, k p) (4, 2 (1)) or (4, 1)directrix: y k py 2 (1)y 3axis of symmetry: x hx 4(4, 2)(4, 1)9. vertex: (h, k) (0, 0)opening: downwardp 4(x h) 2 4p(y k)(x 0) 2 4(4)(y 0)x 2 16yyy 3O xyOx335 Chapter 10
10. (y k) 2 4p(x h)(1 5) 2 4p[2 (7)] (h, k) (7, 5);(6) 2 36p (x, y) (2, 1)1 p(y k) 2 4p(x h)(y 5) 2 4(1)[x (7)](y 5) 2 4(x 7)y12b. The maximum height is s 52 ft.12c. Let s 0.s 56t 16t 2 30 16t 2 56t 3b bac2 4t 2a56 56 2 6)(3)4(1t 2(16)t 3.6 or 0.053.6 s11. vertex: (h, k) (4, 3)opening: upward(x h) 2 4p(y k)(5 4) 2 4p[2 (3)] (h, k) (4, 3);1 2 20p (x, y) (5, 2)1 2 0 p(x h) 2 4p(y k)1 (y 3)(x 4) 2 4 2(x 4) 2 1 5 (y 3)yO(7, 5)0(4, 3)12a. s v 0 t 16t 2 3s 56t 16t 2 316t 2 56t s 3O16t 2 7 5 t ? s 3 ?16t 2 7 2 t 4 916 s 3 (16) 4 116t 7 4 2 s 52t 7 4 2 116(s 52)t 7 4 2 1 4 6 4 (s 52)vertex: (h, k) 7 4 , 52 s55opening: downward 5045403530252015105xxO196 2 3 4xPages 659–661 Exercises13. vertex: (h, k) (0, 0)4p 8p 2focus: (h p, k) (0 2, 0) or (2, 0)directrix: x h px 0 2x 2axis of symmetry: y ky 0yOx 2(2, 0)14. vertex: (h, k) (0, 3)4p 4p 1focus: (h, k p) (0, 3 (1)) or (0, 2)directrix: y k pyy 3 (1)y 4axis of symmetry: x hx 015. vertex: (h, k) (0, 6)4p 4p 1focus: (h p, k) (0 1, 6) or (1, 6)directrix: x h px 0 1x 1axis of symmetry: y kyy 6x 1x(0, 6)OOy 4(0, 3)(0, 2)(1, 6)xxChapter 10 336
16. y 2 12x 2y 13y 2 2y 12x 13y 2 2y ? 12x 13 ?y 2 2y 1 12x 13 1(y 1) 2 12(x 1)vertex: (h, k) (1, 1)4p 12p 3focus: (h p, k) (1 (3), 1) or (4, 1)directrix: x h px 1 (3)x 2axis of symmetry: y ky 117. y 2 x 2 4xx 2 4x y 2x 2 4x ? y 2 ?x 2 4x 4 y 2 4(x 2) 2 y 2vertex: (h, k) (2, 2)4p 1p 1 4 focus: (h, k p) 2, 2 1 4 or 2, 7 4 directrix: y k py 2 1 4 y 9 4 axis of symmetry: x hx 2yyx 2(4, 1) (1, 1)Ox18. x 2 10x 25 8y 24(x 5) 2 8(y 3)vertex: (h, k) (5, 3)4p 8p 2focus: (h, k p) (5, 3 (2)) or (5, 1)directrix: y k py 3 (2)y 5axis of symmetry: x hx 519. y 2 2x 14y 41y 2 14y 2x 41y 2 14y ? 2x 41 ?y 2 14y 49 2x 41 49(y 7) 2 (2x 4)vertex: (h, k) (4, 7)4p 2p 2 4 or 1 2 focus: (h p, k) 4 1 2 , 7 or 7 2 , 7directrix: x h px 4 1 2 x 9 2 axis of symmetry: y ky 7x 29(5, 3)(5, 1)yOy 5yxOxO11 2 37( 2, 4)x7( , 72)(2, 2)y 49(4, 7)337 Chapter 10
20. y 2 2y 12x 13 0y 2 2y 12x 13y 2 2y ? 12x 13 ?y 2 2y 1 12x 13 1(y 1) 2 12(x 1)vertex: (h, k) (1, 1)4p 12p 3focus: (h p, k) (1 3, 1) or (4, 1)directrix: x h px 1 3x 2axis of symmetry: y ky 1x 2yO21. 2x 2 12y 16x 20 02x 2 16x 12y 202(x 2 8x ?) 12y 20 ?2(x 2 8x 16) 12y 20 2(16)2(x 4) 2 12y 12(x 4) 2 6(y 1)vertex: (h, k) (4, 1)4p 6p 6 4 or 3 2 (1, 1)(4, 1)focus: (h, k p) 4, 1 3 2 or 4, 1 2 directrix: y k py 1 3 2 y 5 2 axis of symmetry: x hx 4yO1(4, 2)(4, 1)y 25xxvertex: (h, k) (3, 2)4p 10p 1 04or 5 2 focus: (h, k p) 3, 2 5 2 or 3, 9 2 directrix: y k py 2 5 2 y 1 2 axis of symmetry: x hx 3yO9( 3,2 )(3, 2) y 2123. 2y 2 16y 16x 64 02y 2 16y 16x 642(y 2 8y ?) 16x 64 ?2(y 2 8y 16) 16x 64 2(16)2(y 4) 2 16x 32(y 4) 2 8x 16(y 4) 2 8(x 2)vertex: (h, k) (2, 4)4p 8p 2focus: (h p, k) (2 (2), 4) or (4, 4)directrix: x h pyx 2 (2)x 0O xaxis of symmetry: y ky 4x 0(4, 4) (2, 4)24. vertex: (h, k) (5, 1)opening: righth p 25 p 2p 7(y k) 2 4p(x h)(y 1) 2 4(7)[x (5)](y 1) 2 28(x 5)yx22. 3x 2 30x 18x 87 03x 2 18x 30y 873(x 2 6x ?) 30y 87 ?3(x 2 6x 9) 30y 87 3(9)3(x 3) 2 30y 60(x 3) 2 10y 20(x 3) 2 10(y 2)84(5, 1)12 8 4 O48xChapter 10 338
25. opening: leftfocus: (h p, k) (0, 6)h p 0 k 6h (3) 0h 3(y k) 2 4p(x h)(y 6) 2 4(3)(x 3)(y 6) 2 12(x 3)161284y(3, 6)28. vertex: (h, k) (2, 3)(y k) 2 4p(x h)[1 (3)] 2 4p[3 (2)] (h, k) (2, 3);4 2 4p (x, y) (3, 1)4 p(y k) 2 4p(x h)y(y 3) 2 4(4)(x 2)(y 3) 2 16(x 2)O x(2, 3)26. opening: upwardvertex: (h, k) 4, 5 12 or (4, 3)focus: (h, k p) (4, 1)k p 13 p 1p 2(x b) 2 4p(y k)(x 4) 2 4(2)[y (3)](x 4) 2 8(y 3)y8 4 O 4 84x29. opening: upwardp 2focus: (h, k p) (1, 7)h 1 k p 7k 2 7k 5(x h) 2 4p(y k)[x (1)] 2 4(2)(y 5)(x 1) 2 8(y 5)y(1, 5)O(4, 3)27. opening: downwardvertex: (h, k) (4, 3)(x h) 2 4p(y k)(5 4) 2 4p(2 3) (h, k) (4, 3);1 2 4p (x, y) (5, 2) 1 4 p(x h) 2 4p(y k)(x 4) 2 4 1 4 (y 3)(x 4) 2 (y 3)y(4, 3)x30. opening: downwardvertex: (h, k) (5, 3) (maximum)(x h) 2 4p(y k)(1 5) 2 4p[7 (3)] (h, k) (5, 3);(4) 2 16p (x, y) (1, 7)1 p(x h) 2 4p(y k)(x 5) 2 4(1)(y 3)(x 5) 2 4(y 3)yO(5, 3)xOxOx31. opening: rightvertex: (h, k) (1, 2)(y k) 2 4p(x h)(0 2) 2 4p[0 (1)] (h, k) (1, 2);(2) 2 4p (x, y) (0, 0)1 p(y k) 2 y 4p(x h)(y 2) 2 4(1)[x (1)](y 2) 2 4(x 1)(1, 2)Ox339 Chapter 10
32. opening: upwardh x 1 x 22 1 22or 3 2 vertex: (h, k) 3 2 , 0(x h) 2 4p(y k)1 3 2 2 4p(1 0) (h, k) 3 2 , 0; 1 4 4p (x, y) (1, 1)1 1 6 p(x h) 2 4p(y k)1 (y 0)x 3 2 2 4 1y 3 2 1 4 y33a. vertex: (h, k) (0, 0)depth: x 4p 2yO6(4, y 1 )(2, 0) (4, 0) x(4, y 2 )(y k) 2 4p(x h)(y 0) 2 4(2)(x 0) (h, k) (0, 0);y 2 8x p 2y 8x y 8(4) y 42y 1 42, y 2 42diameter ⏐y 1 y 2 ⏐ ⏐42 (42)⏐ 82 in.33b. depth: x 1.25(4)x 5(y k) 2 4p(x h)(y 0) 2 4(2)(x 0) (h, k) (0, 0);y 2 8x p 2y 8x y 8(5) y 210y 1 210, y 2 210diameter ⏐y 1 y 2 ⏐ ⏐210 (210)⏐ 410 in.y1O1( 2 3 , 0)2x34a. Let y income per flight.Let x the number of $10 price decreases.Income number of passengers cost of a tickety (110 20x) (140 10x)y 15,400 1100x 2800x 200x 2y 200x 2 1 72x 15,400y 15,400 200x 2 1 72xy 15,400 200 1 7 24 200 x 2 1 72 x 1 97 21 200 (y 19,012.5) x 1 7 24 The vertex of the parabola is at 1 74, 19,012.5,and because p is negative it opens downward. Sothe vertex is a maximum and the number of $10price decreases is 1 74 or 4.25.number of passengers 110 20x 110 20(4.25) 195However, the flight can transport only 180people.number of passengers 110 20x180 110 20x3.5 xTherefore, there should be 3.5 $10 pricedecreases.cost of ticket 140 10x 140 10(3.5) $10534b. Let y income per flight.Let x the number of $10 price decreases.Income number of passengers cost of a tickety (110 10x) (140 10x)y 15,400 300x 100x 2y 100(x 2 3x) 15,400y 15,400 100(x 2 3x)y 15,400 100 3 2 2 100 x 2 3x 3 2 2 y 15,625 100x 3 2 21 1 00 (y 15,625) x 3 2 2The vertex of the parabola is at 3 2 , 15,625, andbecause p is negative, it opens downward. So thevertex is a maximum and the number of $10price decreases is 3 2 or 1.5.number of passengers 110 10x 110 10(1.5) 125This is less than 180, so the new ticket price canbe found using 1.5 $10 price decreases.cost of a ticket 140 10x 140 10(1.5) $125Chapter 10 340
35a. Let (h, k) (0, 0).x 2 4pyx 2 4pyx 2 4 1 8 y p 1 8 x 2 4 1 4 y p 1 4 2x 2 yx 2 4pyx 2 1(1)y p 1 1 4 x yThe opening becomesnarrower.35b. The opening becomes wider.36a. Sample answer:opening: upwardvertex: (h, k) (0, 0)y(2100, 490)(0, 0)2100 10 ft O 2100 xroadwayx 2 y500 fty2x 2 yOx 2 y1x 2 y 4x38b. 4p 16p 4focus of parabola center of circlevertex: (h, k) (1, 4)focus: (h, k p) (1, 4 (4)) or (1, 0)diameter latus rectum 16radius 1 2 (16) 8(x h) 2 (y k) 2 r 2(x 1) 2 (y 0) 2 8 2(x 1) 2 y 2 6439. center: (h, k) (2, 3)a 2 25 b 2 16a 25 or 5 b 16 or 4c a 2 b2c 25 16 or 41transverse axis: verticalfoci: (h, k c) (2, 3 41)vertices: (h, k a) (2, 3 5) or (2, 8), (2, 2)asymptotes: y k a b (x h)y(2, 8)y 3 5 4 (x 2)(x h) 2 4p(y k)(2100 0) 2 4p(490 0) (h, k) (0, 0)2250 p (x, y) (2100, 490)(x h) 2 4p(y k)(x 0) 2 4(2250)(y 0)x 2 9000y36b. x 2 9000y(720) 2 9000y57.6 y57.6 10 67.6 ft37. (y k) 2 4p(x h)y 2 2ky k 2 4px 4phy 2 4px 2ky k 2 4ph 0y 2 Dx Ey F 0(x h) 2 4p(y k)x 2 2hx h 2 4py 4pkx 2 4py 2hx h 2 4pk 0x 2 Dx Ey F 038a. 4p 8p 2 or p 2opening: right(y k) 2 4p(x h)(y 2) 2 4(2)[x (3)](y 2) 2 8(x 3)opening: left(y k) 2 4p(x h)(y 2) 2 4(2)[x (3)](y 2) 2 8(x 3)O(2, 3)(2, 2)x40. 4x 2 25y 2 250y 525 04x 2 25(y 2 10y ?) 525 ?4x 2 25(y 2 10y 25) 525 25(25)4x 2 25(y 5) 2 100x2 2 5 (y 5) 24 1center: (h, k) (0, 5)a 2 25 b 2 4a 25 or 5 b 4 or 2c a 2 b2c 25 4 or 21foci: (h c, k) (0 21, 5) or (21, 5)major axis vertices: (h a, k) (0 5, 5) or(5, 5)minor axis vertices: (h, k b) (0, 5 2) or(0, 3), (0, 7)y(5, 5)O(0, 3)(0, 5)(5, 5)x(0, 7)341 Chapter 10
41.5676 12 cos 2 (r, )0 12 (12, 0) 6 6 6, 6 3 6 6, 3 2 12 12, 2 2 36 6, 2 3 5 66 6, 5 6 12 (12, ) 7 66 6, 7 6 4 36 6, 4 3 3 2 12 12, 3 2 5 36 6, 5 3 11 6 6 6, 11 6234323233 6 95360116 42. 2n, where n is any integer43. The measure of a is360° 12 or 30°.acos 30° 6 .4 6.4 cma 6.4 cos 30° a5.5 a; 5.5 cm45. 19 t 14 1919 t 33t 3 3 or 27perimeter 14 19 t 14 19 27 60The correct choice is C.Page 661 Mid-Chapter Quiz1a. AB (x x 2 1 ) 2 (y 2 y 1 ) 2 (6 ) 3 2 (9 3) 2 3 2 6 2 45BC (x x 2 1 ) 2 (y 2 y 1 ) 2 (9 ) 6 2 (3 9) 2 36) 2 ( 2 45Since AB BC, triangle ABC is isosceles.1b. AC (x x 2 1 ) 2 (y 2 y 1 ) 2 (9 ) 3 2 (3 3) 2 6 2 0 2 6perimeter AB BC AC 45 45 6 19.42 units2. Diagonals of a rectangle intersect at theirmidpoint.midpoint of AC x 1 x 22, y 1 y 22 4 52, 9 52 (0.5, 7)3. x 2 y 2 6y 8x 16(x 2 8x ?) (y 2 6y ?) 16 ? ?(x 2 8x 16) (y 2 6y 9) 16 16 9(x 4) 2 (y 3) 2 9center: (4, 3): radius: 9 or 3y(4, 6)44.g(x) x 24 1x y g(x)10,000 4 10 81000 4 10 6100 4 10 410 0.040 410 0.04100 4 10 41000 4 10 610,000 4 10 8y → 0 as x → , y → 0 as x → 4. (x h) 2 (y k) 2 r 2[x (5)] 2 (y 2) 2 7 25a.O(4, 3)(7, 3)(x 5) 2 (y 2) 2 7yx1Odd 2a cxLet d 1 be the greatestdistance from thesatellite to Earth. Letd 2 be the leastdistance from thesatellite to Earth.Chapter 10 342
a 1 2 (10,440)a 5220c a e 52c20 0.16c 835.20radius of Earth 1 2 (7920) 3960d 1 a c Earth radiusd 1 5220 835.20 3960d 1 2095.2 milesd 2 major axis d 1 Earth diameterd 2 10,440 2095.2 7920d 2 424.8 miles5b. (h, k) (0,0)a 5220b 2 a 2 (1 e 2 )b 2 (5220) 2 (1 0.16 2 )b 2 26,550,840.96 (x b) 2a 2 (y k) 2b 2 1 (x 0) 2 (y 0)25220 26,5 50,840.96x2y2 27,24 8,400 26,550 , 840.96 1 16. 9x 2 25y 2 72x 250y 554 09(x 2 8x ?) 25(y 2 10y ?) 544 ? ?9(x 2 8x 16) 25(y 2 10y 25) 544 9(16) 25(25)9(x 4) 2 25(y 5) 2 225 (x 4) 225 (y 5) 29 1center: (h, k) (4, 5)a 2 25 b 2 9 c a 2 b2a 5 b 3 c 25 9 or 4major axis vertices: (h a, k) (4 5, 5) or(9, 5), (1, 5)minor axis vertices: (h, k b) (4, 5 3) or(4, 2), (4, 8)foci: (h c, k) (4 4, 5) or (8, 5), (0, 5)yO(1, 5)(4, 2)(4, 5)(4, 8)(9, 5)x7. 3y 2 24y x 2 2x 41 03(y 2 8y ?) (x 2 2x ?) 41 ? ?3(y 2 8y 16) (x 2 2x 1) 41 3(16) 13(y 4) 2 (x 1) 2 6 (y 4) 22 (x 1) 26 1center: (h, k) (1, 4)a 2 2 b 2 6 c a 2 b2a 2 b 6 c 2 6 or 22transverse axis: verticalvertices: (h, k a) 1, 4 2foci: (h, k c) 1, 4 22asymptotes: y k a b (x h)y (4) [x (1)]6y 4 3 (x 1)3yO(1, 4 2)(1, 4)(1, 4 2)x8. To find the center, find the intersection of theasymptotes.y 2x 42x 2x 44x 4x 1y 2 1y 2The center is at (1, 2).Notice that (4, 2) must be a vertex and a equals4 1 or 3.Point A has an x-coordinate of 4.Since y 2x, the y-coordinate is 2 4 or 8.The value of b is 8 2 or 6.The equation is (x 1) 29 (y 2) 236 1.9. y 2 4x 2y 5 0y 2 2y 4x 5y 2 2y ? 4x 5 ?y 2 2y 1 4x 5 1(y 1) 2 4(x 1)vertex: (h, k) (1, 1)4p 4p 1focus: (h p, k) (1 1, 1) or (2, 1)axis of symmetry: y ky 1directrix: x h pyx 0x 1 1x 0(1, 1)2O(2, 1)x343 Chapter 10
10. vertex: (h, k) (5, 1)(x h) 2 4p(y k) (h, k) (5, 1)(9 5) 2 4p[2 (1)] (x, y) (9, 2)4 2 4p4 p(x h) 2 4p(y k)(x 5) 2 4(4)[y (1)](x 5) 2 16(y 1)10-6Page 6651.Rectangular and ParametricForms of Conic SectionsGraphing Calculator Exploration3. Sample answer:rectangular equation: y 2 xparametric equations: y tyy 2 xt 2 xt 2 xy t, x t 2 , t 4. A 1, c 9; since A and C have the same signand are not equal, the conic is an ellipse.x 2 9y 2 2x 18y 1 0(x 2 2x ?) 9(y 2 2y ?) 1 ? ?(x 2 2x 1) 9(y 2 2y 1) 1 1 9(1)(x 1) 2 9(y 1) 2 9 (x 1) 29 (y 1) 21 1center: (h, k) (1, 1)a 2 9 b 2 1a 3 b 1vertices: (h a, k) (1 3, 1) or (2, 1), (4, 1)(h, k b) (1, 1 1) or (1, 2), (1, 0)Tmin: [0, 6.28] step: 0.1[7.58, 7.58] scl1 by [5, 5] scl11a. (1, 0) 1b. clockwise2.Tmin: [0, 6.28] step: 0.1[7.58, 7.58] scl:1 by [5, 5] scl12a. (0, 1) 2b. clockwise3.Tmin: [0, 6.28] step: 0.1[7.58, 7.58] scl1 by [5, 5] scl1an ellipse4. The value of a determines the length of the radiusof the circle.5. Each graph is traced out twice.Page 667 Check for Understanding1. For the general equation of a conic, A and C havethe same sign and A C for an ellipse. A and Chave opposite signs for a hyperbola. A C for acircle. Either A 0 or C 0 for a parabola.2. t (1, 2)(4, 1)(1, 0) O(2, 1)(1, 1)5. A 1, C 0; since C 0, the conic is a parabola.y 2 8x 8yy 2 8x 8y 2 8(x 1)vertex: (h, k) (1, 0)opening: right(1, 0)Ox6. A 1, C 1; since A and C have different signs,the conic is a hyperbola.x 2 4x y 2 5 4y 0(x 2 4x ?) (y 2 4y ?) 5 ? ?(x 2 4x 4) (y 2 4y 4) 5 4 4(x 2) 2 (y 2) 2 5x (x 2) 25 (y 2) 25 1center: (h, k) (2, 2)a 2 5a 5vertices: (h a, k) (2 5, 2)asymptotes: y k ab (x h)yO(2 5, 2) (2 5, 2)(2, 2)y (2) (x 2)5y 2 (x 2)x5Chapter 10 344
7. A 1, C 1; since A C, the conic is a circle.x 2 6x y 2 12y 41 0(x 2 6x ?) (y 2 12y ?) 41 ? ?(x 2 6x 9) (y 2 12y 36) 41 9 36(x 3) 2 (y 6) 2 4center: (h, k) (3, 6)radius: r 2 4r 2yO(3, 8)(3, 6)8. y t 2 6t 2y x 2 6x 2(5, 6)x10. Sample answer:Let x t.y 2x 2 5xy 2t 2 5tx t, y 2t 2 5t, t 11. Sample answer:x 2 y 2 36x2y2 3 3 166 x6 2 y6 2 1cos 2 t sin 2 t 1 x6 2 cos 2 t y6 2 sin 2 tx 6 cos ty 6 sin tx 6 cos ty 6 sin tx 6 cos t, y 6 sin t, 0 t 212.tx 2 8 0 2yx 8 080x y 2y 2 80xy0 3 (0, 3)t x y (x, y)Pages 667–669 Exercises6 6 2 (6, 2)13. A 1, C 0; since C 0, the conic is a parabola.5 5 7 (5, 7)x 2 4y 6x 9 04 4 10 (4, 10)x 2 6x ? 4y 9 ?3 3 11 (3, 11)x 2 6x 9 4y 9 9(x 3) 2 4y2 2 10 (2, 10)vertex: (h, k) (3, 0)1 1 7 (1, 7)opening: upward0 0 2 (0, 2) yO (3, 0) x14. A 1, C 1; since A C, the conic is a circle.x 2 8x y 2 6y 24 0O x(x 2 8x ?) (y 2 6y ?) 24 ? ?(x 2 8x 16) (y 2 6y 9) 24 16 99. x 2 cos t y 3 sin t(x 4) 2 (y 3) 2 1x y center: (h, k) (4, 3) 2 cos t 3 sin t yradians: r 2 1cos 2 t sin 2 t t 12r 1 x2 2 y3 2 1y x 24 y t t 029 1O xOx(4, 3)3t (3, 3)2(4, 4)t x y (x, y)0 2 0 (2, 0) 2 0 3 (0, 3) 2 0 (2, 0) 3 2345 Chapter 10
15. A 1, C 3; since A and C have different signs,the conic is a hyperbola.x 2 3y 2 2x 24y 41 0x 2 3y 2 2x 24y 41 03(y 2 8y ?) (x 2 2x ?) 41 ? ?3(y 2 8y 16) (x 2 2x 1) 41 3(16) 13(y 4) 2 (x 1) 2 6 (y 4) 22 (x 1) 26 1center: (h, k) (1, 4)a 2 2a 2vertices: (h, k a) (1, 4 2)asymptotes: y k a b (x h)y (4) [x (1)]6y 4 3(x 1)3y(1, 4 2)(1, 4)(1, 4 2)16. A 9, C 25; since A and C have the same signand are not equal, the conic is an ellipse.9x 2 25y 2 54x 50y 119 09(x 2 6x ?) 25(y 2 2y ?) 119 ? ?9(x 2 6x 9) 25(y 2 2y 1) 119 9(9) 25(1)9(x 3) 2 25(y 1) 2 2252 (x 3) 225 (y 1) 29 1center: (h, k) (3, 1)a 2 25 b 2 9a 5 b 3vertices: (h a, k) (3 5, 1) or (8,1), (2, 1)(h, k b) (3, 1 3) or (3, 4), (3, 2)y(2, 1)OO(3, 4)(3, 1)(3, 2)(8, 1)xx17. A 1, C 0; since C 0, the conic is a parabola.x 2 y 8x 16x 2 8x ? y 16 ?x 2 8x 16 y 16 16(x 4) 2 yyvertex: (h, k) (4, 0)opening: upward18. A C D E 0; the conic is a hyperbola.2xy 3xy 3 2 quadrants: I and IIItransverse axis: y xvertices: 3 ,2 3 2 or 6 ,2 6 ,2 3 ,2 3 2 or 626( 6 , )2yO, 62 19. A 5, C 2; since A and C have the same signand are not equal, the conic is an ellipse.5x 2 2y 2 40x 20y 110 05(x 2 8x ?) 2(y 2 10y ?) 110 ? ?5(x 2 8x 16) 2(y 2 10y 25) 110 5(16) 2(25)5(x 4) 2 2(y 5) 2 20 (x 4) 24 (y 5) 210 (y 5) 210 (x 4) 24 1 1(4, 0)center: (h, k) (4, 5)a 2 10 b 2 4a 10 b 2vertices: (h, k a) (4, 5 10)(h b, k) (4 2, 5) or (6, 5), (2, 5)y2(2, 5)6( , 6 )2(4, 5 10)(4, 5)(6, 5)2xOxO(4, 5 10)xChapter 10 346
20. A 1, C 1; since A C, the conic is a circle.x 2 8x 11 y 2(x 2 8x ?) y 2 11 ?(x 2 8x 16) y 2 11 16(x 4) 2 y 2 5center: (h, k) (4, 0)radius: r 2 5r 5yO21. A 9, C 8; since A and C have different signs,the conic is a hyperbola.8y 2 9x 2 16y 36x 100 08(y 2 2y ?) 9(x 2 4x ?) 100 ? ?8(y 2 2y 1) 9(x 2 4x 4) 100 8(1) 9(4)8(y 1) 2 9(x 2) 2 72 (y 1) 29 (x 2) 28 1center: (h, k) (2, 1)a 2 9a 3vertices: (h, k a) (2, 1 3) or (2, 4), (2, 2)asymptotes: y k a b (x h)3y 1 2(x 2)2y 1 3 2(x 2)4Oy(2, 4)(2, 2)22. A 0, C 4; since A 0, the conic is a parabola.4y 2 4y 8x 154(y 2 y ?) 8x 15 ?4y 2 y 1 4 8x 15 4 1 4 4y 1 2 2 8x 16y 1 2 2 2x 4y 1 2 2 2(x 2)vertex: (h, k) 2, 1 2 opening: left(4, 5)(4 5, 0)(4, 0) x(2, 1)xy23. 4y 2 10x 16y x 2 5x 2 4y 2 10x 16y 5 0A 1, C 4; since A and C have different signs,the conic is a hyperbola.x 2 4y 2 10x 16y 5 0(x 2 10x ?) 4(y 2 4y ?) 5 ? ?(x 2 10x 25) 4(y 2 4y 4) 5 25 4(4)(x 5) 2 4(y 2) 2 4(5, 2)(7, 2)yO(3, 2) (x 5) 24 (y 2) 21x 1center: (h, k) (5, 2)a 2 4a 2vertices: (h a, k) (5 2, 2) or (3, 2),(7, 2)basymptotes: y k (x a h)y (2) 1 2 [x (5)]y 2 1 2 (x 5)24. Ax 2 Bxy Cy 2 Dx Ey F 02x 2 0 2y 2 (8)x 12y 6 02x 2 2y 2 8x 12y 6A C; circle2(x 2 4x ?) 2(y 2 6y ?) 6 ? ?2(x 2 4x 4) 2(y 2 6y 6y 9) 6 2(4) 2(9)2(x 2) 2 2(y 3) 2 20(x 2) 2 (y 3) 2 10ycenter: (h, k) (2, 3)radius: r 2 10r 1025. y 2t 2 4t 1y 2x 2 4x 1O(2, 3 10)(2 10, 3)(2, 3)OyxxO( ) 2 , 1 2xt x y (x, y)1 1 7 (1, 7)0 0 1 (0, 1)1 1 1 (1, 1)2 2 7 (2, 7)347 Chapter 10
26. x cos 2t y sin 2tcos 2 2t sin 2 2t 1x 2 y 2 1t x y (x, y)0 1 0 (1, 0) 2 0 1 (0, 1) 1 0 (1, 0) 0 1 (0, 1) 3 2y29. x sin 2t y 2 cos 2tx sin 2ty 2 cos 2tcos 2 2t sin 2 2t 1yt x y (x, y)t 00 0 2 (0, 2)O x 4 1 0 (1, 0) 2 0 2 (0, 2) 3 427. x cos t y sin t30. x 2t 1x cos tx 1 2tycos 2 t sin 2 t 1(x) 2 y 2 1x 2 y 2 1 x 12 ty ty x 1t x y (x, y)2O0 1 0 (1, 0) 2 0 1 (0, 1) 1 0 (1, 0) 3 2 0 1 (0, 1)t x y (x, y)y0 1 0 (1, 0)1 1 1 (1, 1)2 3 2 (3, 2)t 03 5 3 (5, 3)O x4 7 2 (7, 2)31. x 3 cos 2t y 3 sin 2tx 3 cos 2ty 3 sin 2tcos 2 2t sin 2 2t 128. x 3 sin t y 2 cos txx y 3 2 3 sin t y3 2 12 cos tcos 2 t sin 2 x 29 y 29 1t 1y y2 2 x3 2 x 2 y 2 9 1t 032. Sample answer: y 24 x 29 1x 2 y 2 25 x 29 y 2x2y2 4 1O2 5 x2 5 1 x5 2 y5 2 1cos 2 t sin 2 t 1 x5 2 cos 2 t y5 2 sin 2 tt x y (x, y)x 5 cos ty 5 sin t0 2 0 (0, 2)x 5 cos ty 5 sin t 2 3 0 (3, 0)x 5 cos t, y 5 sin t, 0 t 2 0 2 (0, 2) 3 2 2t 0y 2 (x) 2 1 y 24 x 2 1x 2 y 24 1O 1 0 (1, 0)xxChapter 10 348
33. Sample answer:x 2 y 2 16 0x 2 y 2 16x2y2 1 1 166 x4 2 y4 2 1cos 2 t sin 2 t 1 x4 2 cos 2 t y4 2 sin 2 tx y 4 cos t 4 sin tx 4 cos ty 4 sin tx 4 cos t, y 4 sin t, 0 t 234. Sample answer:2 y2 2 1 x 45 x2 2 y5 2 1cos 2 t sin 2 t 1 x2 2 cos 2 t y5 2 sin 2 tx y 2 cos t 5 sin tx 2 cos ty 5 sin tx 2 cos t, y 5 sin t, 0 t 235. Sample answer:y2 1 x 2 16x 2 y4 2 1cos 2 t sin 2 t 1x 2 cos 2 t 4x cos ty 2 sin 2 t 4y sin ty 4 sin tx cos t, y 4 sin t, 0 t 236. Sample answer:Let x t.y x 2 4x 7y t 2 4t 7x t, y t 2 4t 7, t 37. Sample answer:Let y t.x y 2 2y 1x t 2 2t 1x t 2 2t 1, y t, t 38. Sample answer:Let y t.(y 3) 2 4(x 2)(t 3) 2 4(x 2)0.25(t 3) 2 x 20.25(t 3) 2 2 xx 0.25(t 3) 2 2, y t, t 39a. Answers will vary. Sample answers:Let x t.x yt yt 2 yx t, y t 2 , t 0Let y t.x yx tx t, y t, t 039b.Tmin: [0, 5] step: 0.1[7.58, 7.58] scl1 by [5, 5] scl139c. yes39d. There is usually more than one parametricrepresentation for the graph of a rectangularequation.40a. a circle with center (0, 0) and radius 6 feet(x h) 2 (y k) 2 r 2(x 0) 2 (y 0) 2 6 2x 2 y 2 3640b. Sample answer:x 2 y 2 36x2y2 3 3 1 x6 2 y6 2 1sin 2 (qt) cos 2 (qt) 166Since the paddlewheel completes a revolution in2 seconds, the period is 2 q 2, so q .sin 2 (t) cos 2 (t) 1 x6 2 sin 2 (t) y6 2 cos 2 (t)x y 6 sin(t) 6 cos(t)x 6 sin(t)y 6 cos (t)x 6 sin (t), y 6 cos (t), 0 t 240c. C 2rC 26C 37.7 ftThe paddlewheel makes 1 revolution, or moves37.7 ft in 2 seconds. 37 .7 ft2 st x y (x, y)0 0 0 (0, 0)1 1 1 (1, 1)2 2 4 (2, 4)3 3 9 (3, 9)yO 60 s 1131 ftThe paddlewheel moves about 1131 ft in1 minute.x349 Chapter 10
41a. A 2, C 5; since A and C have the same signand A C, the graph is an ellipse.2x 2 5y 2 05y 2 2xy 2 2 5 xy 2 5 xThis equation is true for (x, y) (0, 0).The graph is a point at (0, 0); the equation isthat of a degenerate ellipse.41b. A 1, C 1; since A C, the graph is a circle.x 2 y 2 4x 6y 13 0(x 2 4x ?) (y 2 6y ?) 13(x 2 4x 4) (y 2 6y 9) 13 4 9(x 2) 2 (y 3) 2 0center: (h, k) (2, 3)radius: 0The graph is a point at (2, 3); the equation isthat of a degenerate circle.41c. A 9, C 1; since A and C have differentsigns, the graph is a hyperbola.y 2 9x 2 0y 2 9x 2y 3xThe graph is two intersecting lines y 3x; theequation is that of a degenerate hyperbola.42. The substitution for x must be a function thatallows x to take on all of the values stipulated bythe domain of the rectangular equation. Thedomain of y x 2 5 is all real numbers, butusing a substitution of x t 2 would only allow forvalues of x such that x 0.43a. center: (h, k) (0, 0)(x h) 2 (y k) 2 r 2(x 0) 2 (y 0) 2 6x 2 y 2 3643b. x 2 y 2 36x2y2 3 3 166 x6 2 y6 2 1sin 2 t cos 2 t 1 x6 2 sin 2 t y6 2 cos 2 tx y 6 sin t 6 cos tx 6 sin ty 6 cos tx 6 sin t, y 6 cos tSince the second hand makes 2 revolutions,0 t 4.43c.Tmin: [0, 4] step: 0.1[9.10, 9.10] scl1 by [6, 6] scl144. After drawing a vertical line through (x, y) and ahorizontal line through the endpoint opposite(x, y), two right triangles are formed. Bothtriangles contain an angle t, since correspondingangles are congruent when two parallel lines arecut by a transversal. Using the larger triangle,xcos t a or x a cos t. Using the smaller triangle,ysin t b or y b sin t.45. x 2 12y 10x 25x 2 10x ? 12y 25 ?x 2 10x 25 12y 25 25(x 5) 2 12yvertex: (h, k) (5, 0)4p 12p 3focus: (h, k p) (5, 0 3) or (5, 3)axis of symmetry: x hx 5directrix: y k py 0 3y 3(5, 0)46. c 25quadrants: II and IVtransverse axis: y xvertices: xy 25xy 255(5) 255(5) 25(5, 5) (5, 5)y(5, 5)Ox(5, 5)yO x47. 3x 2 3y 2 18x 12y 93(x 2 6x ?) 3(y 2 4y ?) 9 ? ?3(x 2 6x 9) 3(y 2 4y 4) 9 3(9) 3(4)3(x 3) 2 3(y 2) 2 48(x 3) 2 (y 2) 2 16center: (h, k) (3, 2)radians: r 2 16r 4y (3, 2)O(3, 2)(7, 2)xChapter 10 350
48.x60˚30yxcos 60° 3 0 ysin 60° 3 0 x 30 cos 60° y 30 sin 60°x 15 lby 153 lb49. y 0.13x 37.80.13x y 37.8 0A 0.13, B 1, C 37.8Car 1: (x 1 , y 1 ) (135, 19)d 1 Ax 1 By 1 CA B 2 20.13(135) 1(19) (37.8)d 1 (0.13) 2 1 2d 1 1.24The point (135, 19) is about 1 unit from the liney 0.13x 37.8.Car 2: (x 2 , y 2 ) (245, 16)d 2 Ax 2 By 2 CA B 2 20.13(245) 1(16) (37.8)d 2 (0.13) 2 12d 2 9.97The point (245, 16) is about 10 units from the liney 0.13x 37.8.Car 1: the point (135, 19) is about 9 units closerto the line y 0.13x 37.8 than the point(245, 16).50. Let v Sin 1 1 2 .Sin v 1 2 v 30sin 2 Sin 1 1 2 sin (2v) sin (2 30) sin 6051. s 1 2 (a b c)3 2s 1 2 (48 32 44)s 62K s(s a)(s b)(s c)K 62(62 )(62 482 32)(64) 4K 46872 0K 685 units 252. 2y 3 2y 3 12y 3 2y 3 12y 3 2y 3 22y 3 17 22y 3 7 2 2y 3 4 94 2y 3 3 74 2y 3 78 y53. y kxz y kxz16 k(5)(2) y 1.6(8)(3)1.6 k y 38.454. 5 9 5(3) 7(9) 7 3 78Yes, an inverse exists since the determinant of thematrix 0.55. m y 2 y1x x2 17 4 3 (6) 1 3 y y 1 m(x x 1 )y 4 1 3 (x 6) or y 7 1 3 (x 3)y mx by mx b4 1 3 (6) b y 1 3 x 66 b56. (1 # 4) @ (2 # 3) 1 @ 2 2The correct choice is B.10-7Transformation of ConicsPages 674–675 Check for Understanding1. Sample answers:(h, k) (0, 0)x y 2yO(h, k) (3, 3)(x h) (y k) 2(x 3) (y 3) 22. Replace x with x cos 30° y sin 30° or3 x 1 2 2y.Replace y with x sin 30° y cos 30° or 1 2 x, y.2 2x y 2xyx 3 (y 3) 2Ox351 Chapter 10
3. y90° or 270°y 2 1y 2x 2 125 10090˚O 270˚4. Ebony; B 2 4AC 63 2 4(7)(13) 0and A C5. B 2 4AC 0 4(1)(1)4A C 1; circle(x h) 2 (y k) 2 7(x 3) 2 (y 2) 2 7 (h, k) (3, 2)x 2 6x 9 y 2 4y 4 7x 2 y 2 6x 4y 6 06. B 2 4AC 0 4(2)(0) 0parabolay 2x 2 7x 5y 5 2x 2 7xy 5 2x 2 7 2 xy 5 2 7 4 2 2x 2 7 2 x 7 4 2 y 5 4 98 2x 7 4 2y 9 8 2x 7 4 2y 9 8 k 2x 7 4 h 2y 9 8 5 2x 7 4 4 2 (h, k) (4, 5)y 3 18 2x 9 4 2y 3 18 2x 2 1 84x 8 116y 3 18 2x 2 9x 8 180 2x 2 9x y 142x 2 9x y 14 07. B 2 4AC 0 4(1)(1) 4hyperbolax 2 y 2 9(x cos 60° y sub 60°) 2 (x sin 60° y cos 60°) 2 9 1 2 x 32y 2 3x 1 2 2 y 2 93x 2100 25 1 4 (x) 2 xy 3 2 4 (y) 2 x 3 4 (x) 2 3xy 1 2 4 (y) 2 9 1 2 (x) 2 3xy 1 2 (y) 2 9(x) 2 23xy (y) 2 18(x) 2 23xy (y) 2 18 08. B 2 4AC 0 4(1)(1)4A C 1; circlex 2 5x y 2 3x cos 4 y sin 4 2 5x cos 4 y sin 4 xsin 4 y cos 4 2 3 2x 22 2y 2 5 2x 22 2y 22x 22y 2 3 1 2 (x) 2 xy 1 2 (y) 2 5 2x 5 2y2 2 1 2 (x) 2 xy 1 2 (y) 2 3(x) 2 (y) 2 5 22x 5 22y 32(x) 2 2(y) 2 52x 52y 62(x) 2 2(y) 2 52x 52y 6 09. B 2 4AC 4 2 4(9)(4)128A C; ellipseBtan 2v A C4tan 2v 9 4tan 2v 0.82v 38.65980825°v 19°10. B 2 4AC 5 2 4(8)(4) 153hyperbolaBtan 2v A C5tan 2v 8 (4)tan 2v 0.4166666672v 22.61986495°v 11°11. 3(x 1) 2 4(y 4) 2 03(x 2 2x 1) 4(y 2 8y 16) 03x 2 6x 3 4y 2 32y 64 04y 2 32y (3x 2 6x 67) 0↑ ↑ ↑a b cy b b2 ac 42ax 4(4)(3x 2 6) 672(4)y 32 322 y ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩32 48x2 x 96 48832 48(x 1)2y 8x 1, y 4; pointyO(1, 4)xChapter 10 352
12a. y 1 6 x 214. B 2 4AC 0 4(4)(5)x sin 30° y cos 30° 1 6 (x cos 30° y sin 30°) 280A C; ellipse2a 1 2 x 32y 1 6 32x 1 2 y 2 1 2 x 32y 1 6 3 4 (x) 2 2 34xy 1 4 (y) 2 1 2 x 32y 1 8 (x) 2 3 112xy 2(y)4212x 123y 3(x) 2 23xy (y) 20 3(x) 2 23xy (y) 2 12x 123y3(x) 23xy (y) 2 12x 123y 012b. 3x 2 23xy y 2 12x 123y 01y 2 (23x 123)y (3x 2 12x) 0↑ ↑ ↑a b cb bacy 2 4y ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩⎧⎪⎪⎪⎨⎪⎪⎪⎩y 3x 63 y 3x 63 and y 1 6 x 2Pages 675–677(23x 123) (23x 3) 12 2 )(3x 4(1 2 12x)2(1)Exercises13. B 2 4AC 0 4(3)(0) 0parabolay 3x 2 2x 5y 5 3x 2 2xy 5 3x 2 2 3 xy 5 3 1 3 2 3x 2 2 3 x 1 3 2 y 1 43 3x 1 3 2192 x2 43 2y 1 43 k 3x 1 3 h 212x 2 144x432x 128x2 42y 1 43 3 3x 1 3 2 2 (h, k) (2, 3)y 5 3 3x 7 3 2y 5 3 3x 2 1 43x 4 99y 5 3 3x 2 14x 4 930 3x 2 14x y 183x 2 14x y 18 0 4x 2 5y 2 204(x h) 2 5(y k) 2 204(x 5) 2 5(y 6) 2 20(h, k) (5, 6)4(x 2 10x 25) 5(y 2 12y 36) 204x 2 40x 100 5y 2 60y 180 204x 2 5y 2 40x 60y 260 015. B 2 4AC 0 4(3)(1)12A C; ellipse3x 2 y 2 93(x h) 2 (y k) 2 93(x 1) 2 (y 3) 2 9 (h, k) (1, 3)3(x 2 2x 1) y 2 6y 9 93x 2 6x 3 y 2 6y 9 93x 2 y 2 6x 6y 3 016. B 2 4AC 0 4(12)(4)192A C; ellipse4y 2 12x 2 244(y k) 2 12(x h) 2 244(y 4) 2 12(x 1) 2 24(h, k) (1, 4)4(y 2 8y 16) 12(x 2 2x 1) 244y 2 32y 64 12x 2 24x 12 24y 2 8y 16 3x 2 6x 3 63x 2 y 2 6x 8y 13 017. B 2 4AC 0 4(9)(25) 900hyperbola9x 2 25y 2 2259(x h) 2 25(y k) 2 2259(x 0) 2 25(y 5) 2 225 (h, k) (0, 5)9x 2 25(y 2 10y 25) 2259x 2 25y 2 250y 850 018. (x 3) 2 4yx 2 6x 9 4y 0B 2 4AC 0 4(1)(0) 0parabola(x 3) 2 4y(x 3 h) 2 4(y k)(x 3 7) 2 4(y 2) (h, k) (7, 2)(x 10) 2 4y 8x 2 20x 100 4y 8x 2 20x 4y 108 019. B 2 4AC 0 4(1)(0) 0parabolax 2 8y 0(x cos 90° y sin 90°) 2 8(x sin 90° y cos 90°) 0(y) 2 8(x) 0(y) 2 8x 0353 Chapter 10
20. B 2 4AC 0 4(2)(2)16A C; circle2(xcos30°ysin30°) 22 3 4 (x) 2 322 322x 2 2y 2 82(xsin30°ycos30°) 2 8x 1 2 y 2 2 1 2 x 32y 2 8xy 1 4 (y) 2 2 1 4 (x) 2 32xy 3 4 (y) 2 8 3 2 (x) 2 3xy 1 2 (y) 2 1 2 (x) 2 3xy 3 2 (y) 2 82(x) 2 2(y) 2 8(x) 2 (y) 2 4 021. B 2 4AC 0 4(1)(0) 0A C; parabolay 2 8x 0x sin 6 y cos 6 2 8x cos 6 y sin 6 0 1 2 x 32y 2 8 3x 1 2 2 y 0 1 4 (x) 2 3xy 3 2 4 (y) 2 43x 4y 0(x) 2 23xy 3(y) 2 163x 16y 022. B 2 4AC 1 2 4(0)(0) 1hyperbolaxy 8x cos 4 y sin 4 x sin 4 y cos 4 8 2x 2y 2x 22 2 2 2y 8 1 2 (x) 2 1 2 xy 1 2 xy 1 2 (y) 2 8(x) 2 (y) 2 16(x) 2 (y) 2 16 023. B 2 4AC 0 4(1)(1)4A C; circlex 2 5x y 2 3x cos 3 y sin 3 2 5x cos 3 y sin 3 1 2 x 32y 2 5 1 2 x 2x sin 3 y cos 3 2 33y 3x 1 2 2 y 2 3 1 4 (x) 2 3xy 3 2 4 (y) 2 5 2 x 5 3y2 3 4 (x) 2 3xy 1 2 4 (y) 2 3(x) 2 (y) 2 5 2 x 5 3y 322(x) 2 2(y) 2 5x 53y 6 024. B 2 4AC 0 4(16)(4) 256hyperbola16x 2 4y 2 6416(x cos 60° y sin 60°) 2 4(x sin 60° y cos 60°) 6416 1 2 x 32y 2 4 32x 1 2 y 2 6416 1 4 (x) 2 32xy 3 4 (y) 2 4 3 4 (x) 2 32xy 1 4 (y) 2 644(x) 2 83xy 12(y) 2 3(x) 2 23xy (y) 2 64(x) 2 103xy 11y) 2 64 025. 6x 2 5y 2 306(x cos 30° y sin 30°) 26 326 3 4 (x) 2 32 1 84 5(x sin 30° y cos 30°) 2 30x 1 2 y 2 5 1 2 x 32y 2 30 xy 1 4 (y) 2 3 5 1 4 (x) 2 3xy 3 2 4 (y) 2 30(x) 2 6 xy 6 2 4 (y) 2 5 4 (x) 2 5 xy23 1 5(y)42 30 2 3(x)42 3xy 2 1 (y)242 30 023(x) 2 23xy 21(y) 2 120 026. 3 2 4AC 4 2 4(9)(5)164A C; ellipseBtan 2v A C4tan 2v 9 5tan 2v 12v 45°v 23°27. B 2 4AC (1) 2 4(1)(4) 17hyperbolaBtan 2v A C1tan 2v 1 ( 4)tan 2v 1 5 2v 11.30993247°v 6°28. B 2 4AC 8 2 4(8)(2) 0parabolaBtan 2v A C8tan 2v 8 tan 2v 4 3 2v 53.13010235°v 27°29. B 2 4AC 9 2 4(2)(14)31A C; ellipse2Btan 2v A C9tan 2v 2 14tan 2v 3 4 2v 36.86989765°v 18°Chapter 10 354
30. B 2 4AC 4 2 4(2)(5)24A C; ellipseBtan 2v A C4tan 2v 2 tan 2v 4 3 2v 53.13010235°v 27°31. B 2 4AC 43 2 4(2)(6) 0parabolaBtan 2v A tan 2v tan 2v 32v 60°v 30°32. B 2 4AC 4 2 4(2)(2) 0parabolaA C; v 4 or 45°33. (x 2) 2 (x 3) 2 5(y 2)x 2 4x 4 x 2 6x 9 5(y 2)10x 5 5(y 2)2x 1 y 2y2x 3 yy 2x 3O xy 2x 343 (0, 0)2 6O36. (x 2) 2 (y 2) 2 4(x y) 8x 2 4x 4 y 2 4y 4 4x 4y 8(1)y 2 0y x 2 0↑ ↑ ↑a b c5y b b2 4 2ay 0 0 4 )2(1)4x y 2x 0, y 0; pointyCline37. x 2 2xy y 2 5x 5y 0(1)y 2 (2x 5)y (x 2 5x) 0↑ ↑ ↑a b cy b b2 4y ⎧⎪⎪⎨⎪⎪⎩⎧ ⎪⎪⎨⎪⎪⎩x2a(2x 5) (2x 5)2 4(1)(x2 ) 5x2(1)20x 25 4x 2 20x22x 5 40x 252y 2x 5 4x2 y 34. 2x 2 6y 2 8x 12y 14 0x 2 3y 2 4x 6y 7 03y 2 (6)y (x 2 4x 7) 0↑ ↑ ↑a b cy b b2 4y ⎧⎪⎪⎪⎨⎪⎪⎪⎩2a(6) (6)2 )(x 4(3 2 7)4x 2(3)x 48 486y 6 12x26 12(x 2)y 26x 2, y 1; pointy[6.61, 14.6] scl1 by [2, 12] scl138. 2x 2 9xy 14y 2 514y 2 (9x)y (2x 2 5) 0↑ ↑ ↑a b cy b b2 42a⎧ ⎪⎪⎨⎪⎪⎩(2x 4(14) 2 5)2(14)y 9x (9x)2 y 9x 21x 2 0 2828(2, 1)Ox35. y 2 9x 2 0y 2 9x 2y 9x 2y 3xintersecting linesyOy 3xxy 3x[7.58, 7.58] scl1 by [5, 5] scl1355 Chapter 10
42. 9x 2 4xy 6y 2 2039. 8x 2 5xy 4y 2 2y b b2 42a 2a(4)y 2 (5x)y (8x 2 2) 06y 2 (4x)y (9x 2 20) 0↑ ↑ ↑↑ ↑ ↑a b ca b cy 5x (5x)2 )(8x 4(4 2)y 4x (4x)2 x 4(6)(90) 2y ⎧ ⎪⎪⎨⎪⎪⎩5x 153x 2 32y ⎧ ⎪⎪⎨⎪⎪⎩2(6)4x 212 x802 412[7.58, 7.58] scl1 by [5, 5] scl140. 2x 2 43xy 6y 2 3x y6y 2 (43x 1)y (2x 2 3x) 0↑ ↑ ↑a b cy b b2 4y ⎧⎪⎪⎪⎨⎪⎪⎪⎩⎧⎪⎪⎨⎪⎪⎩2a(43x 1) (43x 1)2 )(2x 4(6 2 3x)2(6)83x 1 48x 2 72x12y 43x 1 48x2 y 43x 1 83x x 72 112[8.31, 2.31] scl1 by [2, 5] scl141. 2x 2 4xy 2y 2 22x 22y 122y 2 (4x 22)y (2x 2 22x 12) 0↑ ↑ ↑a b cy b b2 4ac 2a(4x 22) (4x y )(2x 4(2 22x) 12⎧⎪⎪⎪⎨⎪⎪⎪⎩⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩y 4x 22 16x2 y 2(2) 162x 816x2 2x 169644x 22 32 2x 8 84[10.58, 4.58] scl1 by [2, 8] scl1[7.85, 7.85] scl1 by [5, 5] scl143a.y(0, 5280)O(1320, 1320)(5280, 0)xT (1320, 1320)43b. circlecenter: (h, k) (1320, 1320)radius: r 1320(x h) 2 (y k) 2 r 2(x 1320) 2 (y 1320) 2 1320 2(x 1320) 2 (y 1320) 2 1,742,40044a. B 2 4AC 0 4(1)(0) 0parabola; 360°44b. B 2 4AC 0 4(8)(6)192A C; ellipse; 180°44c. B 2 4AC 4 2 4(0)(0) 16hyperbola; 180°44d. 3 2 4AC 0 4(15)(15)900A C; circle; There is no minimum angle ofrotation, since any degree of rotation will resultin a graph that coincides with the original.45. Let x x cos v y sin v andy x sin v y cos v.x 2 y 2 r 2(x cos v y sin v) 2 (x sin v y cos v) 2 r 2(x) 2 cos 2 v xy cos v sin v (y) 2 sin 2 v (x) 2 sin 2 v xy cos v sin v (y) 2 cos v r 2[(x) 2 (y) 2 ] cos 2 v [(x) 2 (y) 2 ] sin 2 v r 2[(x) 2 (y) 2 ](cos 2 v sin 2 v) r 2[(x) 2 (y) 2 ](1) r 2(x) 2 (y) 2 r 246a. B 2 4AC 103 2 4(31)(21)2304A C; ellipticalChapter 10 356
46b. 31x 2 103xy 21y 2 14448a. center: (h, k) (0, 0)21y 2 (103x)y (31x 2 144) 0major axis: horizontal↑ ↑ ↑a 81 or 9a b cb 36 or 6y b b2 ac 4c a 2 b2 2a c 81 36 (103x) (10c 45 or 35y 3x)2 (31x 4(21) 2 ) 1442(21)y103x 230 4x y 2 612,098424yOx⎧⎪⎪⎪⎨⎪⎪⎪⎩84O484 8x46c. tan 2v A Btan 2v C103 31 21tan 2v 32v 60°v 30°B47a. tan 2v A C23tan 2v 9 11tan 2v 32v 60°v 30°The graph of this equation has been rotated 30°.To transform the graph so the axes are on thex- and y-axes, rotate the graph 30°.47b. B 2 4AC 23 2 4(9)(11)384A C; the graph is an ellipse.9x 2 23xy 11y 2 24 09[x cos (30°) y sin (30°)] 2 23[(x cos (30°) y sin (30°)][x sin (30°) y cos (30°)] 11[x sin (30°) y cos (30°) 2 ] 24 03x 1 2 y9 32x 1 2 y 2 23 2 1 2 x 32y 11 1 2 x 32y 2 24 09 3 4 (x) 2 3xy 1 2 4 (y) 2 23 34 2 74 6 34(x) 2 3 4 xy 1 4 xy 3(y)42 11 1 4 (x) 2 3xy 3 2 4 (y) 2 24 03(x) 2 9 xy 9 24 (y) 2 6 4 (x) 23xy 2 xy 6 44 (y) 2 1 1 (x)42 11 3xy 3 324(y) 2 24 08(x) 2 12(y) 2 24 0 (x ) 2 3 (y ) 2 2 148b. T (35, 0) 8x2 y21 3 6 (x h) 281 (y k) 236(x 35) 2361 1 148c. (y ) 2 8 149. A 3, C 5; since A and C have different signs,the conic is a hyperbola.50. (h, k) (2, 3)e ac 2 6 c 5 1 2 65 cb 2 a 2 c 2b 2 1 2 2 56 2 b 2 1 2 5major axis: horizontal (x h) 2a 2 (y k) 2b 2 1 (x 2) 21 1(x 2) 2 25(y 3) 2 1major axis : vertical (y a 2 k) 2 (x b 2 h) 2 1 [y (3)] 212[y (3)] 212 5(x 2) 2 112 5(y 3) 2 25(x 2) 2 151.v 0 30° 60° 90° 120° 150° 180°r 1 1.4 3.9 3.9 1.4 1 1150˚180˚120˚210˚240˚90˚60˚1 2 3 4300˚270˚30˚0˚330˚357 Chapter 10
52.WN5S53. cos 70° 0.34cos 170° 0.98cos 70°8c54. 5 16 5 16 18 0°55. y 2 56.25 56 1 560 56° 15E2y 5 2y 5 3y 2 (y 2) (y 1)2y 5 A B y 2 3y 2 y 2 y 1c 2 a 2 b 2c 2 8 2 5 2c 89c 9.4about 9.4 m/s2y 5 A(y 1) B(y 2)2(2) 5 A(2 1) B(2 2)1 A1 A2y 5 A(y 1) B(y 2)2(1) 5 A(1 1) B (1 2)3 BA B 13 y 2 y 1 y 2 y 156. x 1y x 22 y1 1 2y 5 4 2y 2 9.657. 8m 3n 4p 68m 6 3n 4pm 3 4 3 8 n 1 2 p4m 9n 2p 4 3 8 n 1 2 p 9n 2p 44 3 4 3 3 2 n 2p 9n 2p 4 2 12n 7n 2 3 8m 3n 4p 6 6m 12n 5p 18m 3 2 3 4p 6 6m 12 2 3 5p 18m 2 4p 6 6m 8 5p 18m 4p 4 6m 5p 72m p 12m p 1 → 10m 5p 56m 5p 7 → () 6m 5p 716m 12m 3 4 8m 3n 4p 68 3 4 3 2 3 4p 66 2 4p 6 3 4 , 2 3 , 1 2 p 1 2 58.5a59. The expression 8b5 1 80a6b 2 simplifies to a 2b . 3Since361 b and 2 a, the expression is always largerthan 22 13 1 . Since b 2 and a 3, the369expression is always less than 32 23 36 7 23 or 2.6The correct choice is B.10-8x h(x) [[x]] 33 x 262 x 151 x 040 x 1 31 x 2 22 x 3 13 x 4 04 x 5 15 x 6 26 x 7 3h(x)Systems of Second-DegreeEquations and InequalitiesPage 682 Check for Understanding1. Possible number of solutions: 0, 1, 2, 3, or 4xChapter 10 358
2. Sample answer: y x 2 , x 2 (y 3) 2 97. x 2 y 2 4100 40y 5y 2 1611.y5y 2 40y 84 08y b b2 4 2a4(40) (40)y 5)(84) 4( 2(5)40 80 8 4 O 4 8 xy 4yy 2 4 x 2y x 29x 2 4y 2 36 x 2 y 2 49x 2 4(4 x 2 ) 36 2 2 y 2 4O x13x 2 16 36 y 2 0x 2 4 y 0x 2 (y 3) 2 9x 2(2, 0)y3. The system contains equation(s) that areequivalent. The graphs coincide.(2, 0) (2, 0)4. Graph each second-degree inequality. The regionO xin which the graphs overlap represents thesolution to the system.5. x y 0x y (x 1) 220 (y 1) 28. xy 1 x5 1 yx(x 2 ) 11 2 y (x 1) 220 (x 1) 25 1x 3 11 y(x 1) 2 4(x 1) 2 20x 1x 2 2x 1 4(x 2 2x 1) 20(1, 1)5x 2 10x 5 20 0y5(x 2 2x 3) 05(x 3)(x 1) 0x 3 0 x 1 0(1, 1)x 3x 1(3, 3), (1, 1)O xy(3, 3)9. y10. yO4x(1, 1)OO x44 8 126. x 2y 108x 10 2yx 2 y 2 1612(10 2y) 2 y 2 16no solution810yxOx359 Chapter 10
12a. Let x side length of flowerbed 1.Let y side length of flowerbed 2.A 1 x x or x 2A 2 y y or y 2Total Area x 2 y 2680 x 2 y 2x 2 y 2 680Difference of Areas x 2 y 2288 x 2 y 2x 2 y 2 28812b. ySince side length40cannot be negative, an20estimated solution is(22, 14)(22, 14).40 20 O 20 40x204012c. x 2 y 2 680y 2 680 x 2x 2 y 2 288 x 2 y 2 680x 2 (680 x 2 ) 288 22 2 y 2 6802x 2 968 y 2 196x 2 484 y 14x 2222 ft and 14 ftPages 682–684 Exercises13. x 1 0x 1y 2 49 x 2y 2 49 (1) 2y 2 48y 6.9(1, 6.9)14. xy 2y 2 x x 2 3 y 2x 2 3 2 x 2x 2 3 x42x 4 3x 2 4x 4 3x 2 4 0(x 2 4)(x 2 1) 0x 2 4 0 x 2 1 0x 2 4x 2xy 2 xy 2x 2 12(y) 2 (2)y 2y 1y 1yO(1, 6.9)(1, 6.9)x(2, 1), (2, 1)(2, 1)15. 1 2x y1 2x y4x 2 y 2 254x 2 (1 2x) 2 254x 2 1 4x 4x 2 258x 2 4x 24 04(2x 2 x 6) 04(2x 3)(x 2) 02x 3 0 x 2 0x 1.5x 21 2x y1 2x y1 2(1.5) y 1 2(2) y4 y3 y(1.5, 4), (2, 3)(2, 3)yOyO(2, 1)16. x y 2x 2 yx 2 100 y 2(2 y) 2 100 y 24 4y y 2 100 y 22y 2 4y 96 02(y 2 2y 48) 02(y 6)(y 8) 0y 6 0 y 8 0y 6y 8x y 2 x y 2x 6 2 x (8) 2x 8x 6(8, 6), (6, 8)yOxx(1.5, 4)(8, 6)x(6, 8)Chapter 10 360
17. x y 0x y (x 1) 29 y 2 1 (y 1) 29 y 2 1(y 1) 2 9y 2 9y 2 2y 1 9y 2 98y 2 2y 8 04y 2 y 4 0y b b2 ac 42ay 1 12 (4)(4) 42(1)1 63 2yy no solution19. x y 1y 1 x(y 1) 2 4 x(1 x 1) 2 4 x(2 x) 2 4 x4 4x x 2 4 xx 2 3x 0x(x 3) 0x 0 x 3 0x 3x y 1x y 10 y 1 3 y 1y 1 y 2(0, 1), (3, 2)(3, 2)yOxO(0, 1)x18. x 2 2y 2 10x 2 10 2y 23x 2 9 y 23(10 2y 2 ) 9 y 230 6y 2 9 y 25y 2 21y 2 4.2y 2.03x 2 9 y 2 3x 2 9 y 23x 2 9 (2.0) 2 3x 2 9 (2.0) 2x 2 1.6 x 2 1.6x 1.3x 1.3(1.3, 2.0), (1.3, 2.0)y(1.3, 2.0) (1.3, 2.0)O(1.3, 2.0)x(1.3, 2.0)20. xy 6 0y 6 x x 2 y 2 13x 2 6 x 2 13x 2 3 6x2 13x 4 36 13x 2x 4 13x 2 36 0(x 2 9)(x 2 4) 0x 2 9 0 x 2 4 0x 2 9 x 2 4x 3x 2xy 6 0 xy 6 03y 6 0 2y 6 0y 2y 3xy 6 0 xy 6 03y 6 0 2y 6 0y 2 y 3(3, 2), (3, 2) (2, 3), (2, 3)(2, 3)(3, 2)yOx(3, 2)(2, 3)361 Chapter 10
21. x 2 y 3 0x 2 y 3x 2 4y 2 36y 3 4y 2 364y 2 y 33 0(y 3)(4y 11) 0y 3 0 4y 11 0y 3 y 2.8x 2 y 3 0 x 2 y 3 0x 2 3 3 0 x 2 (2.8) 3 0x 2 0 x 2 5.8x 0 x 2.4(0, 3), (2.4, 2.8)yO(2.4, 2.8)(0, 3)x(2.4, 2.8)23. xy 4y 4 x x 2 25 9y 2x 2 25 9 4 x 2442x 2 25 1 xx 4 25x 2 144x 4 25x 2 144 0(x 2 9)(x 2 16) 0x 2 9 0 x 2 16 0x 2 9 x 2 16x 3x 4xy 4xy 43y 44y 4y 1.3y 1xy 4xy 43y 44y 4y 1.3 y 1(3, 1.3), (3, 1.3) (4, 1), (4, 1)24. 25.y22. 2y x 3 02y 3 xx 2 16 y 2(2y 3) 2 16 y 24y 2 12y 9 16 y 25y 2 12y 7 0y b b2 ac 4y 2a12 12 2 7)4(5)(26.88 y 4xO4 4 848yOx12 284 y 10y 0.5 or y 2.92y x 3 0 2y x 3 02(0.5) x 3 0 2(2.9) x 3 0x 4.0 x 2.8(4.0, 0.5), (2.8, 2.9)y27.yOx8O(2.8, 2.9)2(5)(4, 0.5)x844 O 4 8x48Chapter 10 362
28. 29.y8y4Ox8 4 O 4 8x4830. 31.y8y37a. 2x 2y P xy A2x 2y 150 xy 80037b. A system of a line and a hyperbola may have 0,1, or 2 solutions.37c.y80 408040O408040 80x844 O4832. 33.x48 4 O 4 88 y4834. parabola:vertex: (1, 3)(y k) 2 4p(x h)(5 3) 2 4p(1 1) 1 2 p(y 3) 2 4 1 2 (x 1)(y 3) 2 2(x 1)line:m 2, b 7y mx by 2x 735. circle:center: (0, 0), radius: 22x 2 y 2 r 2x 2 y 2 8hyperbola:(2)(2) 4xy 436. large ellipse:a 5, b 4, center (0, 0)48x8Oy44 O484xx37d. xy 800y 80 0x2x 2y 1502x 2 80 x0 1502x 16 x00 1502x 2 1600 150xx 2 75x 800 0x b b2 ac 4x x 2a(75) (75) 2 1)(800 4( )2(1)75 2425 2x 12.88 or x 62.12xy 800 xy 80012.88y 800 62.12y 800y 62.11 y 12.8812.9 m by 62.1 m or 62.1 m by 12.9 m38a. (h, k) (0, 4)(x, y) (6, 0)(x h) 2 4p(y k)(6 0) 2 4p (0, 4)36 16p2.25 p(x h) 2 4p(y k)(x 0) 2 4(2.25)(y 4)x 2 9(y 4)x 2 9(y 4), y 038b.yOx ay 2 2 bx 2 2 1y2x2 2 5 1 6 1(interior is shaded)small ellipse:a 3, b 2, center (0, 1) ax 2 2 (y b 2 k) 2 1 x 29 (y 1) 24 1(exterior is shaded)363 Chapter 10
38c. (h, k) (0, 3)(x, y) (6, 0)(x h) 2 4p(y k)(6 0) 2 4p(0 3)36 12p3 p(x h) 2 4p(y k)(x 0) 2 4(3)(y 3)x 2 12(y 3)x 2 12(y 3), y 039. xy 12y 1 2xx y 1x 1 x2 1x 2 12 xx 2 x 12 0(x 4)(x 3) 0x 4 0 x 3 0x 4x 3xy 12xy 124y 123y 12y 3 y 4(4, 3) (3, 4)Check that (4, 3) and (3, 4) are also solutionsof y 2 25 x 2 .y 2 25 x 2 y 2 25 x 2(3) 2 25 (4) 2 (4) 2 25 (3) 29 9 16 16 (4, 3), (3, 4)(3, 4)yO(4, 3)x40b. yestimate: (40, 30)40c. x 2 y 2 2500x 2 2500 y 2x 2 (y 30) 2 16002500 y 2 y 2 60y 900 160060y 1800 0y 30(x 35) 2 (y 18) 2 169x 2 70x 1225 (30 18) 2 169x 2 70x 1200 0(x 30)(x 40) 0x 30 0 or x 40 0x 30 x 40Check (30, 30) and (40, 30):x 2 y 2 2500 x 2 y 2 250030 2 30 2 2500 40 2 30 2 25001800 2500 2500 2500 (40, 30)41. x 3y k2y 2 3y k2y 2 3y k 0y 2 3 2 y 1 2 k 0y 2 3 2 y 3 4 2 0y 3 4 2 0 1 2 k 3 4 2 1 2 9k 1 6k 9 8 42a.8040O80 40 40 80x408yComplete the square.440a. first station: (h, k (0, 0)x 2 y 2 r 2x 2 y 2 50 2x 2 y 2 2500second station: (h, k) (0, 30)(x h) 2 (y k) 2 r 2x 2 (y 30) 2 40 2x 2 (y 30) 2 1600third station: (h, k) (35, 18)(x h) 2 (y k) 2 r 2(x 35) 2 (y 18) 2 13 2(x 35) 2 (y 18) 2 169842b. yes; (6, 2) or (6, 2)42c. Earth’s surface:x 1 2 y 1 2 40 x140 x 2140 y 2140y140 1 2 2 1cos 2 t sin 2 t 1 x1404 O4 2 cos 2 tx 12 10 848x y140 2 sin 2 ty cos t 12 sin t10 x 1 210 cos t y 1 210 sin tChapter 10 364
asteroid:Let y 2 t.x 2 0.25y 2 5x 2 0.25 t 2 542d.49. No; the domain value 4 is mapped to two elementsin the range, 0 and 3.50. area of rectangle q 8(4) or 32area of circles 2 (r 2 ) 2(4p) or 8parea of shaded region 32 8pThe correct choice is E.10-8BGraphing Calculator Exploration:Tmin 8, Tmax 8, Tstep 0.13Shading Areas on a Graph[15.16, 15.16] scl1 by [10, 10] scl 12 y 2 Page 686 11. (x sin 30° y cos 30°) 2 9 143. x 9(x cos 30° y sin 30°) 2 2 9 3 4 (x) 2 3xy 1 2 4 (y) 29 1 2 x 32y 2 1 1 4 (x) 2 3xy 3 2 4 (y) 2 1 3 4 (x) 2 3xy 1 2 4 (y) 2 9 4 (x) 2 9 3xy 2 24(y) 2 93(x) 2 43xy 7(y) 2 9 044. x 4t 1 y 5t 7 x 14 t y 5 t x 14 y 75(x 1)(5) (4)(y 7)5x 5 4y 285x 4y 33 045. 4 csc v cos v tan v 4 si3x 1 2 y 21n v 446. r 10 cm or 0.10 mv rqv (0.10) 5 2p1 v 3.14 m/s47.r 1 0 0 40 1 0 0 41 1 1 1 32 1 2 4 47 (cos v) csinvosv 7[9.1, 9.1] scl1 by [6, 6] scl12.[9.1, 9.1] scl1 by [6, 6] scl13.[9.1, 9.1] scl1 by [6, 6] scl14.between 1 and 248. y (x 2) 2 3x (y 2) 2 3x 3 (y 2) 2x 3 y 2x 3 2 yyy (x 2) 2 3Ox[9.1, 9.1] scl1 by [6, 6] scl15a. 35b. Find the points of intersection for the boundaryequation by using the TRACE function.y x 3 2365 Chapter 10
5c. SHADE(((36X 2 )/9),((36X 2 )/9),6,2,3,4);SHADE(X 2 2,((36X 2 )/9),2,2,3,4,);SHADE(((36X 2 )/9), ((36X 2 )/9),2,6,3,4)14. (x h) 2 (y k) 2 r 2(x 0) 2 (y 0) 2 33 2yx 2 y 2 27(0, 33)O(33, 0)x[9.1, 9.1] scl1 by [6, 6] scl16. See students’ work.Chapter 10 Study Guide and AssessmentPage 687 Understanding and Using theVocabulary1. true 2. false; center3. false; transverse 4. true5. false; hyperbola 6. false; axis or axis ofsymmetry7. true 8. false; parabola9. true 10. false; ellipsePages 688–690 Skills and Concepts11. d (x x 2 1 ) 2 (y 2 y 1 ) 2d (3 1) 2 (6)] [4 2d 20 or 25 x 1 x 22, y 1 2y 2 1 2 (1, 5)12. d x 2 x(y 1 ) 2 y 2 1 ) 2(3), 6 (4)2 d (a 3 a) 2 (b4 b) 2d 25 or 5 x 1 x 22, y 1 y 22 a a 2 (a 1.5, b 2)13. AB (x x 2 1 ) 2 (y 2 y 1 ) 2 3, b b 42 [3 5)] ( 2 [4 (2)]2 100 or 10DC (x x 2 1 ) 2 (y 2 y 1 ) 2 (10 2) 2 (3)] [3 2 100 or 10BC (x x 2 1 ) 2 (y 2 y 1 ) 2 (10 3) 2 4) (3 2 50 or 52AD (x x 2 1 ) 2 (y 2 y 1 ) 2 [2 5)] ( 2 [3)] (22 50 or 52Yes; AB DC 10 and BC AD 52. Sinceopposite sides of quadrilateral ABCD arecongruent, ABCD is a parallelogram.15. (x h) 2 (y k) 2 r 2(x 2) 2 (y 1) 2 2 2(x 2) 2 (y 1) 2 416. x 2 y 2 6yx 2 (y 2 6y ?) 0 ?x 2 (y 2 6y 9) 0 9x 2 (y 3) 2 9y(0, 3)O(2, 1)(2, 1)17. x 2 14x y 2 6y 23(x 2 14x ?) (y 2 6y ?) 23 ? ?(x 2 14x 49) (y 2 6y 9) 23 49 9(x 7) 2 (y 3) 2 81(16, 3)(0, 6)(3, 3)18. 3x 2 3y 2 6x 12y 60 0x 2 y 2 2x 4y 20 0(x 2 2x ?) (y 2 4y ?) 20 ? ?(x 2 2x 1) (y 2 4y 4) 20 1 4(x 1) 2 (y 2) 2 25(1, 3)(7, 6)(7, 3)yyOxO x(1, 2) (4, 2)xyO(4, 1)xChapter 10 366
19. x 2 y 2 Dx Ey F 01 2 1 2 D(1) E(1) F 0 ⇒D E F 2(2) 2 2 2 D(2) E(2) F 0 ⇒2D 2E F 8(5) 2 1 2 D(5) E(1) F 0 ⇒5D E F 26D E F 2(1)(5D E F) (1)(26)6D 24D 42D 2E F 8(1)(D E F) (1)(2)3D E 63(4) E 6E 6D E F 24 (6) F 2F 12x 2 y 2 Dx Ey F 0x 2 y 2 4x 6y 12 0(x 2 4x ?) (y 2 6y ?) 12 ? ?(x 2 4x 4) (y 2 6y 9) 12 4 9(x 2) 2 (y 3) 2 25center: (h, k) (2 3)r 25 or 520. center: (h, k) (5, 2)a 2 36 b 2 16a 36 or 6 b 16 or 4c a 2 b2c 6 4 or 2foci: (h, k c) (5, 2 2)major axis vertices: (h, k a) (5, 2 6) or(5, 8), (5, 4)minor axis vertices: (h b, k) (5 4, 2) or(9, 2), (1, 2)y (5, 8)(1, 2)O(5, 2)(5, 4)(9, 2)x21. 4x 2 25y 2 24x 50y 394(x 2 6x ?) 25(y 2 2y ?) 39 ? ?4(x 2 6x 9) 25(y 2 2y 1) 39 4(9) 25(1)4(x 3) 2 25(y 1) 2 100 (x 3) 225 (y 1) 24 1center: (h, k) (3, 1)a 2 25 b 2 4a 25 or 5 b 4 or 2c a 2 b2c 25 4 or 21foci: (h c, k) (3 21, 1)major axis vertices: (h a, k) (3 5, 1) or(8, 1), (2, 1)minor axis vertices: (h, k b) (3, 1 2) or(3, 1), (3, 3)yO(2, 1)(3, 1)(3, 1)(3, 3)(8, 1)22. 6x 2 4y 2 24x 32y 64 06(x 2 4x ?) 4(y 2 8y ?) 64 ? ?6(x 2 4x 4) 4(y 2 8y 16) 64 6(4) 4(16)6(x 2) 2 4(y 4) 2 24 (x 2) 24 (y 4) 26 24center: (h, k) (2, 4)a 2 6 b 2 4a 6 b 4 or 2c a 2 b2c 6 4 or 2foci: (h, k c) (2, 4 2)major axis vertices: (h, k a) 2, 4 6minor axis vertices: (h b, k) (2 2, 4) or(0, 4), (4, 4)y(2, 4 6)(4, 4)(2, 4)(2, 4 6)O(0, 4)x23. x 2 4y 2 124 8x 48y(x 2 8x ?) 4(y 2 12y ?) 124 ? ?(x 2 8x 16) 4(y 2 12y 36) 124 16 4(36)(x 4) 2 4(y 6) 2 36 (x 4) 236 (y 6) 29 36center: (h, k) (4, 6)a 2 36 b 2 9a 36 or 6 b 9 or 3c a 2 b2c 36 9 or 33foci: (h c, k) (4 33, 6)major axis vertices: (h a, k) (4 6, 6) or(10, 6), (2, 6)minor axis vertices: (h, k b) (4, 6 3) or(4, 9), (4, 3)yx(2, 6)O(4, 9)(4, 3)(4, 6)(10, 6)x367 Chapter 10
24. (h, k) (4, 1)a 9b 6 (y k) 2a 2 (x h) 2b 2 1 (y 1) 29 2 [x (4)] 262 (y 1) 281 (x 4) 236 1 125. center: (h, k) (0, 0)a 2 25 b 2 16a 25 or 5 b 16 or 4c a 2 b2c 25 16 or 41transverse axis: horizontalfoci: (h c, k) (0 41, 0) or (41, 0)vertices: (h a, k) (0 5, 0) or (5, 0), (5, 0)basymptotes: y k a (x h)yy 0 4 5 (x 0)y 4 5 x(5, 0) (5, 0)O x26. center: (h, k) (1, 5)a 2 36 b 2 9a 36 or 6 b 9 or 3c a 2 b2c 36 9 or 35transverse axis: verticalfoci: (h, k c) 1, 5 35vertices: (h, k a) (1, 5 6) or (1, 1), (1, 11)asymptotes: y k a b (x h)y (5) 6 3 (x 1)y 5 2(x 1)y(1, 1)Ox(1, 5)center: (h, k) (0, 2)a 2 4 b 2 1a 4 or 2 b 1 or 1c a 2 b2c 4 1 or 5transverse axis: horizontalfoci: (h c, k) (0 5, 2) or (5, 2)vertices: (h a, k) (0 2, 2) or (2, 2),(2, 2)basymptotes: y k (x a h)(2, 2)y (2) 1 2 (x 0)y 2 1 2 x28. 9x 2 16y 2 36x 96y 36 09(x 2 4x ?) 16(y 2 6y ?) 36 ? ?9(x 2 4x 4) 16(y 2 6y 9) 36 9(4) 16(9)9(x 2) 2 16(y 3) 2 144 (y 3) 29 (x 2) 216 1center: (h, k) (2, 3)a 2 9 b 2 16a 9 or 3 b 16 or 4c a 2 b2c 9 6 1 or 5transverse axis: verticalfoci: (h, k c) (2, 3 5) or (2, 2), (2, 8)vertices: (h, k a) (2, 3 3) or (2, 0), (2, 6)asymptotes: y k a b (x h)yOyO(0, 2)x(2, 2)y (3) 3 4 (x 2)(2, 0)(2, 6)y 3 3 4 (x 2)(2, 3)x(1, 11)27. x 2 4y 2 16y 20x 2 4(y 2 4y ?) 20 ?x 2 4(y 2 4y 4) 20 4(4)x 2 4(y 2) 2 4 x 24 (y 2) 21 1Chapter 10 368
29. c 9quadrants: I and IIItransverse axis: y xvertices: xy 9 xy 93(3) 9 (3)(3) 9(3, 3) (3, 3)(3, 3)yO30. 2b 10b 5center: x 1 x 22, y 1 y 22 1 12, 1 2 (1, 2)transverse axis: verticala distance from center to a vertex ⏐2 (1)⏐ or 3 (y k) 2a 2 (x h) 2b 2 1 (y 3 2 2) 2 (x 5 2 1) 2 1 (y 2) 29 (x 1) 225 1x 2(3, 3)y 2 5 6, 3 (3)2 31. center: x 1 2, y 1 2 2 2 (2, 3)a distance from center to a vertex ⏐2 (2)⏐ or 4c distance from center to a focus ⏐2 (4)⏐ or 6b 2 c 2 a 2b 2 6 2 4 2b 2 20transverse axis: horizontal (x a 2 h) 2 (y b 2 k) 2 1 (x 2) 24 2 [y ( 3)] 220 (x 2) 216 (y 3) 220 1 132. vertex: (h, k) (5, 3)4p 8p 2focus (h, k p) (5, 3 2) or (5, 5)directrix: y k py 3 2y 1axis of symmetry: x hx 5yx33. vertex: (h, k) (1, 2)4p 16p 4focus: (h p, k) (1 (4), 2) or (3, 2)directrix: x h px 1 (4)x 5axis of symmetry: y ky 234. y 2 6y 4x 25y 2 6y ? 4x 25 ?y 2 6y 9 4x 25 9(y 3) 2 4(x 4)vertex: (h, k) (4, 3)4p 4p 1focus: (h p, k) (4 1, 3) or (5, 3)directrix: x h px 4 1x 3axis of symmetry: y ky 3yO(3, 2)(4, 3)x 335. x 2 4x y 8x 2 4x 4 y 8 4(x 2) 2 y 4vertex: (h, k) (2, 4)4p 1p 1 4 yOx(1, 2)x 5(5, 3)xfocus: (h, k p) 2, 4 1 4 or (2, 4.25)directrix: y k py 4 1 4 y 3.75yaxis of symmetry: x hx 2 (2, 4.25)(5, 3)(5, 5)y 1y 3.75(2, 4)OxOx369 Chapter 10
36. vertex: (h, k) (1, 3)(y k) 2 4p(x h)(7 3) 2 4p[3 (1)]16 8p2 pSince parabola opens left, p 2.(y k) 2 4p(x h)(y 3) 2 4(2)[x (1)](y 3) 2 8(x 1)37. vertex: (h, k) 5, 2 42 or (5, 1)focus: (h, k p) (5, 2)k p 21 p 2p 3(x h) 2 4p(y k)(x 5) 2 4(3)[y (1)](x 5) 2 12(y 1)38. A 5, c 2; ellipse39. A C 0; equilateral hyperbola40. A C 5; circle41. C 0; parabola42. y t 2 3y x 2 3t x y (x, y)2 2 1 (2, 1)1 1 2 (1, 2)0 0 3 (0, 3)1 1 2 (1, 2)2 2 1 (2, 1)y44. x 2 sin t y 3 cos tx y 2 sin t 3 cos tsin 2 t cos 2 t 1 2t x 2 y3 2 145. x tx 2 tty 2 12 x 24 y 29y x 2 1 1t x y (x, y)0 0 3 (0, 3) 2 2 0 (2, 0) 0 3 (0, 3) 3 2 2 0 (2, 0)32yOt 0t t 2xt x y (x, y)0 0 1 (0, 1)4 2 1 (2, 1)9 3 3.5 (3, 3.5)Oxyt 943. x cos 4t y sin 4tcos 2 4t sin 2 4t 1x 2 y 2 1t x y (x, y)0 1 0 (1, 0) 8 0 1 (0, 1) 4 1 0 (1, 0) 3 8 0 1 (0, 1)t 4yOt 83t 8t 0xOt 046. Sample answer:Let x t.y 2x 2 4y 2t 2 4, t 47. Sample answer:x 2 y 2 49x2y2 4 4 199 x7 2 y7 2 1sin 2 t cos 2 t 1 x7 2 sin 2 t y7 2 cos 2 txx 7 sin ty 7 cos tx 7 sin ty 7 cos t, 0 t 2pChapter 10 370
48. Sample answer:x2y2 3 8 161 x6 2 y9 2 1cos 2 t sin 2 t 1 x6 2 cos 2 t y9 2 sin 2 tx y 6 cos t 9 sin tx 6 cos ty 9 sin t, 0 t 2p49. Sample answer:Let y t.x y 2x t 2 , t 50. B 2 4AC 0 4(4)(9)144A C; ellipse4x 2 9y 2 364x cos p 6 y sin p 6 2 9x sin p 6 y cos p 6 2 364 324 3 4 (x) 2 32x 1 2 y 2 9 1 2 x 32y 2 36xy 1 4 (y) 2 9 1 4 (x) 3 32xy 3 4 (y) 2 363(x) 2 23xy (y) 2 9 4 (x) 2 9 32xy 2 4(y) 2 36 2 14(x) 2 5 xy 3 24(y) 2 3621(x) 2 103xy 31(y) 2 144 051. B 2 4AC 0 4(0)(1) 0parabolay 2 4x 0(x sin 45° y cos 45°) 2 4(x cos 45° y sin 45°) 0 22x 232y 2 4 2712x 22y 0 1 2 (x) 2 xy 1 2 (y) 2 22x 22y 0(x) 2 2xy (y) 2 42x 42y 052. B 2 4AC 0 4(4)(16) 256hyperbola4x 2 16(y 1) 2 644(x h) 2 16(y k 1) 2 644(x 1) 2 16(y 2 1) 2 644(x 1) 2 16(y 1) 2 644(x 2 2x 1) 16(y 2 2y 1) 644x 2 8x 4 16y 2 32y 16 64 0x 2 4y 2 2x 8y 19 053. B 2 4AC 23 2 4(6)(8)180A C; ellipseBtan 2v A C23tan 2v 6 8tan 2v 32v 60°v 30°54. B 2 4AC (6) 2 4(1)(9) 0parabolaBtan 2v A C6tan 2v 1 9tan 2v 3 4 2v 36.86989765°v 18°55. (x 1) 2 4(y 1) 2 20(y 1) 2 4(y 1) 2 205(y 1) 2 20(y 1) 2 4y 1 2y 3 or 1y 3 x yx 3y 1 x yx 1(3, 3), (1, 1)yO(1, 1)(3, 3)x56. 2x y 02x yy 2 49 x 2(2x) 2 49 x 24x 2 49 x 23x 2 49x 4.042x y 0 2x y 02(4.04) y 0 2(4.04) y 0y 8.08 y 8.08(4.0, 8.1), (4.0, 8.1)y(4.0, 8.1)(4.0, 8.1)O x371 Chapter 10
57. x 2 4x 4y 461. 62.x 2 b b2 c 464. 2a 2,000 e a 2aa 60000.2 60c00x 2 12 12 6)4(1)(12(1)1200 cx 2 10.472 or x 2 1.528b 2 a 2 cx 3.236 x 1.2362b 2 6000 2 1200xy 4xy 42b 2 34,560,0003.236y 4 1.236y 4 (x h) 2a 2 (y k) 2y 1.236 y 3.236b 2 1x 2 4x 4 4y(x 2) 2 4y 0(x 2) 2 x 2 4x 4 0x 2 4x 4 x 2 4x 4 02x 2 8x 02x(x 4) 0y2O 2 xyO2x 0 x 4 0x 0 x 4(x 2) 2 4y 0 (x 2) 2 4y 0(0 2) 2 4y 0 (4 2) 2 4y 0y 1y 1(0, 1), (4, 1)Page 691 Applications and Problem Solvingy63a. r (x x 2 1 ) 2 (y 2 y 1 ) 2r (12 0) 2 (16 0) r 20Ox 2 y 2 r 2(0, 1)xx 2 y 2 20 2(4, 1)x 2 y 2 40063b. area of watered portion pr 2 p20 258. xy 4 1256.6 ft 2area of backyard qy 4 x 50(40)x 2 y 2 2000 ft 12x 2 4 x 2 area of nonwatered portion 2000 1256.6 12 743.4 ft 2x 2 1 6x2 12percent not watered 7 43.42000x 4 16 12x 2 0 0.37(x 2 ) 2 12(x 2 ) 16 0about 37%xy 4xy 4 (x 0) (y 0) 60002 3.236y 41.236y 434 ,560,000x2y2y 1.236 y 3.236 36,00 0,000 34,56 0,000 1(3.2, 1.2), (3.2, 1.2), (1.2, 3.2), (1.2, 3.2)65. a 3.5yb 3(1.2, 3.2)(3.2, 1.2)Oxc a 2 bc 3.5 2 3 c 1.8about 1.8 feet from the center(3.2, 1.2)x(1.2, 3.2)59. 60.yy4Ox4O 44xChapter 10 372
Page 691 Open-Ended Assessment1. Sample answer:ce a 1 9 c aLet a 9. 1 9 c 9c 1 ax 2 2 by 2 2 1x 2 y29 2 8 0x2y2 8 1 8 0 1 12. Sample answer:axis of symmetry: x hx 2, so h 2focus: (h, k p) (2, 5)k p 5Let k 2, p 3.(x h) 2 4p(y k)(x 2) 2 4(3)(y 2)(x 2) 2 12(y 2)SAT & ACT PreparationPage 693 SAT & ACT Practice1. Add the two numbers of parts to get the whole, 8.The fraction of red jelly beans to the whole is 3 8 .The total number of jelly beans is 160. Thenumber of red jelly beans is 3 8 (160) or 60. Thecorrect choice is C.Or you can use a ratio box. Multiply by 20.Green Red Whole5 3 860 160b 2 a 2 c 2b 2 9 2 1 2b 2 802. Notice the capitalized word EXCEPT. You mightwant to try the plug-in method on this problem.Choose a value for b that is an odd integer, say 1.Then substitute that value for b in the equation.a 2 b 12 2a 2 (1) 12 2a 2 12 2a 12Check the answer choices for divisors of this valueof a. 12 is divisible by 3, 4, 6, and 12, but not by 9.The correct choice is D.3. The information in the question confirms theinformation given in the figure. Recall the formulafor the area of a triangle — one half the basetimes the height. The triangle DCB is obtuse, sothe height will lie outside of the triangle.Let DC be the base. The length of the base is 6.The height will be equal to 7, since it is a linesegment parallel to AD through point B.A 1 2 bh 1 2 (6)(7) or 21The correct choice is A.4. The problem asks how many more girls there arethan boys. First find how many girls and howmany boys there are in the class.One method is to find the fraction of girls in thewhole class and the fraction of boys in the wholeclass. Since the ratio of girls to boys is 4 to 3, thefraction of girls in the whole class is 4 7 . Find thenumber of girls in the class by multiplying thisfraction by 35. 4 7 (35) 20 There are 20 girls in the class.Using the same process, the fraction of boys is 3 7 . 3 7 (35) 15 There are 15 boys in the class.So there are 5 more girls than boys. The correctchoice is D.Another method is to use a “Ratio Box.” Firstenter the given information, shown in the darkercells below. Then enter the number for the total ofthe first row, 7. To go from the total of 7 to thetotal of 35, you must multiply by 5. Write a 5 ineach cell in the second row.Girls Boys Total4 3 75 5 520 15 35Then multiply the two numbers in the firstcolumn to get 20 girls, shown with a dark border.Multiply the second column to get 15 boys.Subtract to find there are 5 more girls than boys.5. Set A is the set of all positive integers less than30. Set B is the set of all positive multiples of 5.The intersection of Sets A and B is the set of allelements that are in both Set A and Set B. Theintersection consists of all positive multiples of5 which are also less than 30. The intersectionof the two sets is {5, 10, 15, 20, 25}.The correct choice is A.373 Chapter 10
6. For a quadratic equation in the formy a(x – h) 2 k, the coordinates of the vertexof the graph of the function are given by theordered pair (h, k). So the vertex of the graphof y 1 2 (x – 3) 2 + 4 has coordinates (3, 4).The correct choice is C.7. On the SAT, if you forget the relationships for 45°right triangles, look at the Reference Informationin the gray box at the beginning of eachmathematics section of the exam. The measure ofeach leg of a 45–45–90 triangle is equal to thelength of the hypotenuse divided by 2. Multiplyboth numerator and denominator by 2 andsimplify.8BC 28 2 22 8 22or 42The correct choice is D. You could also use thePythagorean Theorem and the fact that the twolegs must be equal in length, but that methodmight take more time.8. Form a ratio using the given fractions asnumerator and denominator. Write a proportion,using x as the unknown. Multiply the crossproducts.Solve for x.9. Let d represent the number of dimes in the jar.Since there are 4 more nickels than dimes, thereare d 4 nickels in the jar. So, the ratio of dimesdto nickles in the jar is . This ratio is lessd 4than 1. The only answer choice that is less than81 is choice A, 1 0 . If d 8 , then d 16. So,d 4 1 0there are 16 dimes and 16 4 or 20 nickels in thejar, and 1 62 0 8.10The correct choice is A.10. Set up a proportion.totallitersx liters t otalbottles1 bottle8 x 2 0 120x 8x 0.4 or 2 5 The correct answer is .4 or 2/5. 1 7 1 5 10 0x 1 7 x 1 5 (100) 1 x 207x 140The correct choice is E.Chapter 10 374
Chapter 11 Exponential and Logarithmic Function11-1 Real ExponentsPage 695 Graphing Calculator Exploration1.2.8. 32 3 5 (2 5 ) 3 5 2 1 55 2 3 89. (3a 2 ) 3 3a 5 3 3 a 6 3a 5 3 4 a 1 81a 1 or 8 a10. m 3 n 2 m 4 n 5 (m 3 n 2 ) 1 2 (m 4 n 5 ) 1 2 11. 8n 4 n2 7 2 7 8n 4 n 1 2 2 7 2 4 n 2 8n 2 1 m 3 2 n 2 2 m 4 2 n 5 2 m 7 2 n 7 2 or m 3 n 3 mn (23 ) n 2 2 7 2 (2 2 ) n 2 3. a m a n a mn4. (a m ) n a mn5. a b m a b m m, when b 0Page 700 Check for Understanding1. The quantities are not the same. When thenegative is enclosed inside of the parentheses andthe base is raised to an even power, the answer ispositive. When the negative is not enclosed insideof the parentheses and the base is raised to aneven power, the answer is negative.2. If the base were negative and the denominatorwere even, then we would be taking an even rootof a negative number, which is undefined as a realnumber.3. Laura is correct. The negative exponent of 10represents a fraction with a numerator of 1 and adenominator of a positive power of 10. The productof this fraction and a number between 1 and 10 isbetween 0 and 1.4. 5 4 514 6125 6. 216 1 3 (6 3 ) 1 3 6 3 3 65. 196 2 1 919 2 1 6 1 6292 2 56816 2 7. 27 3 27 1 2 3 1 2 (3 3 ) 1 2 3 1 2 3 3 2 3 1 2 3 4 2 3 2 9n 7 23 2 22 2 2 n 2 2 3 n 2 2 7 2 2 2 n 2 2 5n 72 2 2n3 2 n 12 2 2n3 2n112. (2x 4 y 8 ) 1 2 2 1 2 (x 4 ) 1 2 (y 8 ) 1 2 2 1 2 x 4 2 y 8 2 13. 169x 5 (169x 5 ) 1 2 2 1 2 x 2 y 4 or x 2 2 4 2 169 1 2 (x 5 ) 1 2 13x 5 2 14. 4 ad 2 b 3 c 4 5 (a 2 b 3 c 4 d 5 ) 1 4 15. 6 1 4 b 3 4 c 1 4 (6b 3 c) 1 4 4 (a 2 ) 1 4 (b 3 ) 1 4 (c 4 ) 1 4 (d 5 ) 1 4 ⏐a⏐ 1 2 ⏐b⏐ 3 4 ⏐c⏐ ⏐d⏐ 5 4 6b 3 c17. 3 p 4 q 6 r 5 (p 4 q 6 r 5 ) 1 3 18. y 4 5 34y 4 5 5 4 34 5 4 y (34 5 ) 1 4 y 82.1 (p 4 ) 1 3 (q 6 ) 1 3 (r 5 ) 1 3 p 4 3 q 2 r 5 3 pq 2 r 3 pr 216. 15x 1 3 y 1 5 15x 155y 3 15 15 (x 5 y 3 ) 11 5 15 15 x5 y 3375 Chapter 11
19. A r 2 r 3.875 10 7 mA (3.875 10 7 m) 2 (3.875) 2 (10 7 ) 2 m 2 (15.015625 10 14 ) m 2 4.717 10 13 m 2Pages 700–70320. (6) 4 1 ( 6) 422. (5 3) 2 15 224. 7 8 3 1 7 8 325. (3 1 3 3 ) 1 3 1 26. 81 1 2 (9 2 ) 1 2 28. 3329. 2 1 2 12 1 2 2 1 2 (2 6) 1 2 30. 64 1 2 (2 6 ) 132. ( 3 6 729Exercises21. 6 4 16433. 3 216 2 (216 1 3 ) 2 216 2 3 (6 3 ) 2 3 6 2 3634. 81 1 2 81 1 2 (9 2 ) 1 2 1 (9 2 ) 1 2 11 9 1 9 12 9612 962423. 8 8 9 2 1 2 4(1) 225 2 51135. 7 ) (1284 (128) 4 7 32 1 [(2) 7 ] 4 7 8 7 31 ( 2) 41 8 3 1 673 5 1236. (3n 2 ) 3 3 3 (n 2 ) 3343 27n 6137. (y2 ) 4 y 8 1 (y3 12 y 8) 4 1 1 y 8 y 8 1 3 1 9 1 1 38. (4y 4 ) 3 2 4 3 2 (y 4 ) 3 2 4 9 (2 2 ) 3 2 (y 4 ) 3 2 2 6 2 y 1 2 9 24 27. 729 1 3 (9 3 ) 1 3 8y 6 9 939. (27p 3 q 6 r 1 ) 1 3 27 1 3 (p 3 ) 1 3 (q 6 ) 1 3 1 r 1 3 27 (3 3 ) 1 3 (p 3 ) 1 3 (q 6 ) 1 3 1 r 1 3 (3 3 ) 2 327 2 3 3 3 3 3 3 p 3 3 q 6 3 1 32r 1 3 3 3pq 2 r 1 3 40. [(2x 4 ] 2 1 2 1 2 2 1 2 6 1 2 [(2x ) 4 ] 21 2 6 1 2 2 8x 8 2 8 x 8 1or 2625 6x 812 31. 16 1 4 1 41. (36x 6 ) 1 2 36 1 2 (x 6 ) 1 2 42. b2nb2n 1 2 2 1 2 16 1 4 or 2 6⏐x⏐ 3 1(2 4 ) 1 4 43. 2n n1 2 1 2n2 3 7 )(94) 37 (3 2 ) 444n 1 2 27 6(3 3 ) 6 2 3 15 2n1 2 394 n2 (b 2n b 2n ) 1 2 (b 4n ) 1 2 b 2nChapter 11 376
44. 3m 1 2 27n 1 4 4 3 4 m 1 2 4 27 4 14n 4 3 4 m 2 (3 3 ) 4 n 3 16 m 2 n45. f 161256g 4 h44 (f 16 256 1 g 4 h 4 ) 1 4 (f 16 ) 1 4 (256 1 ) 1 4 (g 4 ) 1 4 (h 4 ) 1 4 f 4 256 1 4 ⏐g⏐ ⏐h⏐ 1 4 f 4 ⏐g⏐⏐h⏐ 1 or 4⏐46. 6 x 2 x 3 4 x 3 4 x 2 x 3 4 x 3 4 1 6 x 2 x 3 4 x 6 4 x 5 4 x 3 2 47. 2x 1 4 y 1 3 3x 1 4 y 2 3 6x 2 4 y 3 3 48. m n a a 1n 1 m a 1 1 n m a 1nm 1 a mn mn a50. xy 3 (x) 1 2 (y 3 ) 1 2 ⏐x⏐ 1 2 ⏐y⏐ 3 2 6x 1 2 y51. 3 8x 3 y 6 8 1 3 (x 3 ) 1 3 (y 6 ) 1 3 2xy 2 1 1 6 1 1 6 52. 17 7 x 14 y 7 z 12 17(x 14 ) 1 2 (y 7 ) 1 7 (z 12 ) 1 7 17x 2 yz 1 2753. 5 a 10 b 2 4 c 2 (a 10 ) 1 5 (b 2 ) 1 5 (c 2 ) 1 4 f 4 ⏐g⏐h⏐49. m 6 n (m 6 ) 1 2 n 1 2 ⏐m⏐ 3 n 1 2 a 2 b 2 5 ⏐c⏐ 1 2 54. 60 8 t r 80 s 56 27 60(r 80 ) 1 8 (s 56)1 8 (t 27 ) 1 8 60r 10 ⏐s⏐t 2 7855. 16 1 5 5 16 56. (7a) 5 8 b 3 8 8 7 5 a 5 b 3 57. p 2 3 q 1 2 r 1 3 p 4 6 q 3 6 r 2 6 6 p 4 q 3 r 259. 13a 1 7 b 1 3 13a 231b 7 21 13 21 a3 b 760. (n 3 m 9 ) 1 2 (n 3 ) 1 2 (m 9 ) 1 2 n 3 2 m 9 2 ⏐n⏐m 4 mn 2 2 3 58. 2 1 32 1 3 3 261. x 3 (245 ) 1 5 3 0.33 0.6962. d 3 e 2 f 2 (d 3 ) 1 2 (e 2 ) 1 2 (f 2 ) 1 2 d⏐e⏐⏐f⏐d63. 3 a 5 b 7 c (a 5 ) 1 3 (b 7 ) 1 3 (c) 1 3 ab 2 3a2 bc64. 20x 3 y 6 (20) 1 2 (x 3 ) 1 2 (y 6 ) 1 2 2x⏐y⏐ 3 5x 65. 14.2 x 3 2 (14.2) 2 3 (x 3 2 ) 2 3 0.17 xx 3 x y8 3 8 965 616 3 6 1 7 29 5 3 5 1 2 1 033 3 1 3043 3 1.395 1 2 3 1 2 1.732 2 3 3 2 3 2.0800 1 093 1 93.389 5 3 3 5 3 14.620 7 2 3 7 2 46.76566. 724 15a 5 2 12712 15a 5 2 7 1215 7 1215 2 5 a 5 2 (a 5 2 ) 2 5 4.68 a67. 1 8 x 5 3.5x 5 2 28(x 5 2 ) 2 5 (28) 2 5 x 3.7968. d 6.794 10 3 km so r 3.397 10 3 kmV 4 3 r 3 4 3 (3.397 10 3 km) 3 1.64 10 11 km 369. y 3 x ; x 8, 6, 5, 1 033 , 1 2 , 2 3 , 1 09, 5 3 , 7 2 69a. If x 0then y 0. If x 0then y 1. Sincex 0, y 0 and y 1. So, 0 y 1.69b. If x 0 then y 1. If x 1 then y 3. So,1 y 3.69c. If x 1 then y 3. So, y 3.377 Chapter 11
69d. If the exponent is less than 0, the power is greaterthan 0 and less than 1. If the exponent is greaterthan 0 and less than 1, the power is greater than 1and less than the base. If the exponent is greaterthan 1, the power is greater than the base.Any number to the zero power is 1. Thus, if theexponent is less than zero, the power is less than1. A power of a positive number is nevernegative, so the power is greater than 0.Any number to the zero power is 1 and to thefirst power is itself. Thus, if the exponent isgreater than zero and less than 1, the power isbetween 1 and the base.Any number to the first power is itself. Thus, ifthe exponent is greater than 1, the power isgreater than the base.70. r (1.2 10 15 )A 1 3 If r 2.75 10 15 then2.75 10 5 (1.2 10 15 )A 1 3 2 . 75 10151.2 1015 A 1 3 2.29 A 1 3 12.04 ASince 12.04 12, which is the mass number ofcarbon, the atom is carbon.71. 32 (x2 4x) 16 (x2 4x3)(2 5 ) (x2 4x) (2 4 ) (x2 4x3)2 (5x2 20x) 2 (4x2 16x12)5x 2 20x 4x 2 16x 12x 2 4x 12 0(x 6)(x 2) 0x 6 0 x 2 0x 6 x 272a.72b. A5-mile per hour increase in the wind speedwhen the wind is light has more of an effect onperceived temperature than a 5-mile per hourincrease in the wind speed when the wind isheavy.73a. r 3 G4r 3 Wind Speed Wind Chill5 0.810 12.715 22.620 29.925 35.330 39.2Met 22G 6.67 10 11M t 5.98 10 24t 1 day 86,4000 seconds(6.67 10 11 )(5.98 10 24 )(86,400) 24 2r 42,250,474.31 m73b. 42,250,474.31 m 42,250.47431 km42.250.47431 6380 35870.47431 35,870 kmm factors n factors mn factors⎧⎪⎨⎪⎩74a. a m a n a a ... a a a ... a a a ... a a mnChapter 11 378n factorsm factors m factors m factors⎧ ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩ ⎪⎨⎪⎩ ⎧⎪⎨⎪⎩ ⎧⎪⎨⎪⎩⎧⎧ ⎪⎨⎪⎩ ⎪⎨⎪⎩⎧ ⎧⎧ ⎪⎨⎪⎩⎪⎨⎪⎩ ⎪⎨⎪⎩⎧ ⎧ ⎪⎨⎪⎩ ⎪⎨⎪⎩⎧⎪⎨⎪⎩⎧⎪⎨⎪⎩74b. (a m ) n a a ... a a a ... a ... a a ... a m n factorsm factors m factors m factorsa a ... a a mn74c. (ab) m ab ab ... ab a a ... a b b ... b a m b mm factors74d. a b m a b a b ... a b a b m m74e. a a75.m factorsmn a a ... an factorsa mna a ... a76. y 2 12x(y 0) 2 4(3)(x 0)Vertex is at (0, 0) and p 3. The parabola opensto the right so the focus is at (0 3, 0) or (3, 0).Since the directrix is 3 units to the left of thevertex, the equation of the directrix is x 3.77. (23 2i) 1 5 Convert to polar form r(cos v i sin v).78.y8642O2468r 23 2 2 2 v Arctan 223 16 Arctan 33 4yO 6 So, (23 2i) 4cos 6 i sin 6 .Use De Moivre’s Theorem.4cos 6 i sin 6 1 5 4 1 5 cos 1 5 6 sin 1 5 6 56762343Lemniscate2 4 6232353xx 4 1 5 cos 3 0 4 1 5 i sin 3 0 1.31 0.14i61 2 3 40116
79. Use the equation y t⏐v u ⏐ sin v 1 2 qt 2 h.⏐v u ⏐ 105 g 32 h 3 v 42y t(105)(sin 42°) 1 2 (32)t 2 3 16t 2 (105 sin 42°)t 3Find t when y 0 (i.e., the ball is on the ground).105 sin 42° (105 in s) 42° 16)( 4( 3) t 2(16)t 0.04, 4.43So, the ball hits the ground after about 4.43 s.u80. TC (2 3), (6 (4)), (5 6) 1, 10, 11u⏐TC ⏐ (2 ) 3 2 (64)) ( 2 (5 6) 2 222 81. Sample answer:tan S cos S 1 2 sinS c osS co s S 1 1 2 sin S 1 2 82. cot v 0v 2 Period of cot v is socot v 0v 2 n, where n is an integer.83. r 6h 150 mr 15 0 m6 hr 25 m/h84. 90°, 270°85. x 3 25x 0Degree of 3 so there are 3 complex roots.x 3 25x 0x(x 5)(x 5) 0x 0 x 5 0 x 5 0x 5 x 53; 5, 0, 586.87. The time it takes to paint a building is inverselyproportional to the number of painters.48 k 8 k 384So t 3 8416t 24The correct choice is E.11-2Exponential FunctionsPage 705 Graphing Calculator Exploration1. positive reals2. (0, 1)3. For a 0.5 and 0.75, y → as x → , y → 0 asx → . For a 2 and 5, y → 0 as x → , y → asx → .4. horizontal asymptote at y 0, no verticalasymptotes5. Yes; the range of an exponential function is allpositive reals because the value of any positivereal number raised to any power is positive.6. Any nonzero number raised to the zero power is 1.7. The graph of y b x is decreasing when 0 b 1because multiplying by number between 0 and 1results in a product less than the original number.The graph of y b x is increasing when b 1because multiplying by a number greater than 1results in a product greater than the originalnumber.8. There is a horizontal asymptote at y 0 because apower of a positive real number is never 0 or less.As a number between 0 and 1 is raised to greaterand greater powers, its value approaches 0. As anumber greater than 1 is raised to powersapproaching negative infinity, its valueapproaches 0.[5, 5] scl:0.5 by [5, 5] scl:0.5abs. min; (0.75, 1.88)Page 708 Check for Understanding1. Power; in a power function the variable is thebase, in an exponential function the variable isthe exponent.2. Both graphs are one-to-one, have the domain of allreals, a range of positive reals, a horizontalasymptote of y 0, a y-intercept of (0, 1), and novertical asymptote. The graph of y b x isdecreasing when 0 b 1 and increasing whenb 1.3. If the base is greater than 1, the equationrepresents exponential growth. If base is between0 and 1, the equation represents exponentialdecay.4. The graphs of y 4 x and y 4 x 3 are the sameexcept the graph of y 4 x 3 is shifted downthree units from the graph of y 4 x .379 Chapter 11
5. x y1 1 9 0 11 32 9yy 3 x11. x y1 1 2 0 11 22 4yOxy 2 x6. x y2 91 30 11 1 9 y 3 xOyx12. x y2 41 20 11 1 2 y 2 xyO7. x y2 3 3 4 Use (0, 0) as a test point.1 3 1 2 0 ? 2 0 40 4 0 ? 1 41 2 0 3 2 0 The statement is true so shadethe region containing (0, 0).yOxx13. x y3 12 21 40 814. x y3 12 21 40 8Oyy 2 x 3 OyOy 2 x 3xxxy 2 x 48. Use N N 0 (1 r) t where N 0 3750, r 0.25,and t 2.N 3750 (1 0.25) 2 3750 (0.75) 2 2109.389a. 9,145,219 8,863,052 282,167 282 ,1677 40,309.5715. x y (0, 0)1 1 3 4 0 ? 4 2 20 1 0 ? 1 21 2 0 1; no2 14yOy 4 x 2x 4 0,309.578,863,052 0.0045 0.45%9b. Use N N 0 (1 r) t .N 8,863,052(1 0.0045) 20N 9,695,76616. x y1 50 11 1 5 yy (1)x5Pages 708–71110. x y1 1 2 0 11 22 4ExercisesyOxy 2 x17. x y2 41 20 11 1 2 (0, 0)0 ? 1 2 00 ? 1 y (1)x2OyxOxChapter 11 380
18. x y (0, 0)3 1 2 0 ? 2 044 1 0 ? 2 415 2 0 1 66 4yy 2 x 423b. The graph of y 3 x is a reflection of the graphof y 3 x across the x-axis.Ox19. x y1 1000 111 1 00 Byy 0.01 x(0, 1) ( 1, 1100)[10, 10] scl:1 by [10, 10] scl:123c. The graph of y 7 x is a reflection of the graphof y 7 x across the y-axis.20. x y1 50 11 1 5 O(1, 5)yy 5 xxC21. x y1 490 71 1AyO(0, 7)(0, 1) ( ) 1, 1 5x[10, 10] scl:1 by [1, 9] scl:123d. The graph of y 1 2 x is a reflection of the graphof y 2 x across the y-axis.22.O(1, 1)xy 7 1 x 2[5, 5] scl:1 by [5, 5] scl:122a. The graph of y 5 x is a reflection of y 5 xacross the x-axis. The graph of y 5 x is areflection of y 5 x across the y-axis.22b. The graph of y 5 x 2 is shifted up two units,while the graph of y 5 x 2 is shifted downtwo units.22c. The graph of y 10 x increases more quicklythan the graph of y 5 x . The graphs are not thesame because 5 2x 10 x .23a. The graph of y 6 x 4 is shifted up four unitsfrom the graph of y 6 x .24a.[10, 10] scl:1 by [1, 9] scl:1y323028262422201816141210864O 2 4 6 8 1012 141618 2022x24b. y 9.25(1.06) 50y 170.386427 thousandy $170,40025a. y (0.85) x[10, 10] scl:1 by [1, 9] scl:1381 Chapter 11
25b. y28d. Sample answer: A borrower might choose the1.330-year mortgage in order to have a lower1.21.1monthly payment. A borrower might choose the 2,668,760.458 or 2,668,760Compounded twice:26c. 152,307 139,510(1 r) 10I 1000[1 0. 052 2 1]152,307 (1 r) 10139,510 50.625; $50.631.020-year mortgage in order to have a lower0.9interest expense.0.80.729a. P 4000, n 43, i 0.04750.6(1 0.0475)F n 4000 10.50.04750.4 535,215.918; $535,215.920.30.229b. P 4000, n 43, i 0.05250.1(1 0.0525)F n 4000 10.0525 O 1 2 3 4 5 6 7 8x 611,592.1194 or $611,592.1225c. y (0.85) 12$611,592.12 535,215.92 $76,376.20 0.14 or 14%30. The function y a x is undefined when a 0 and25d. No; the graph has an asymptote at y 0, so thepercent of impurities, y, will never reach 0.26a. N 876,156(1 0.0074) 15 978,612.2261 or 978,61226b. N 2,465,326(1 0.0053) 15the exponent x is a fraction with an evendenominator.31a. Compounded once:I 1000[(1 0.05) 1 1] 50; $50110Compounded four times:152,307 139,510 1 rI 1000[1 0. 054 4 1]N 139,510(1 r) 25 50.9453; $50.94N 173,736.7334 or 173,737Compounded twelve times:26d. 191,701 168,767(1 r) 10I 10001 0 .051212 1191,701 168,767 (1 r) 10110191,701 168,767 1 rN 168,767(1 r) 25N 232,075.6889 or 232,07627a. O 100(3 3 5 5) 100(3 3 ) 2700 units27b. s 4 .2 mi1 h r 52 80ft 1 hr1 mi 3 600s 6.16 ft/sO 100(3 36 .165) 5800.16 5800 units28a. P n 121,000, n 30 12 or 360,i 0.075 12 or 0.00625121,000 P 1 (1 0.00625) 3600.00625P 846.04955; $846.0528b. P n 121,000, n 20 12 or 240,i 0.0725 12 or 0.006041 1 121,000 P 0.0 72512———725 0.0 12 240P 956.35494; $956.3528c. 30 year: I 360(846.05) 121,000 $183.57820 year: I 240(956.35) 121,000 $108,524Chapter 11 382 51.1619; $51.16Compounded 365 times:I 10001 0 . 05365365 1 51.2675; $51.2631b. Let x represent the investment.Statement savings: I x[(1 0.051) 1 1] 0.051xThe return is 5.1%Money Market Savings: I x1 0.0 5051212 1The return is 5.17% 0.517xSuper Saver: I x1 0 . 05365365 1 0.513xThe return is 5.13%Money Market Savingsx 36531c. (1 0.05) 1 36 5 (1.05) 1365 x 1 36 5365[(1.05) 1365 1] x0.04879 x; 4.88%32. 4x 2 (4x) 2 4x 2 (4) 2 (x) 24x2 1 6x2 1 4 33. y 15Use y r sin v.y 15So, 15 r sin v.34. 3, 9 2, 1 (3)(2) (9)(1) 33; no because the inner product does not equal 0.
35. 1 3 cos 7 8 i sin 7 8 33cos4 i sin 4 4 1 3 33cos 7 8 4 i sin 7 8 3cos 5 8 i sin 5 80.66 1.60i36. s 72t 16t 2 4s 4 16t 2 72ts 4 (16)(5.0625) 16(t 2 4.5 5.0625)(s 85) 16(t 2.25) 2Vertex: (2.25, 85)Maximum height: 85 feet.37a.y(12, 8)(7, 6)42.1y x 3yO1y xxThe parent graph is translated 3 units left. Thevertical asymptote is now x 3. The horizontalasymptote, y 0, is unchanged.43. C AC 32 C AB 16 100.53 50.27100.53 50.27 50.26 or about 50.The correct choice is E.37b. (12 (7)), (8 6)2 2 (2.5, 7)38. sin 4 A cos 2 A cos 4 A sin 2 A(sin 2 A) 2 cos 2 A cos 4 A sin 2 A(1 cos 2 A) 2 cos 2 A cos 4 A sin 2 A(1 2cos 2 A cos 4 A) cos 2 A cos 4 A sin 2 Acos 4 A 1 cos 2 A cos 4 A sin 2 Acos 4 A sin 2 A cos 4 A sin 2 A39. 2900 rev 21 1 rev0O 4800V 9.2 480 1 138,732.73 or about 139,000 cm/s40.21˚xtan 69° 20 0200 tan 69° x521.02 xabout 521 feet41.x200 ftx11-3The Number ePage 714 Check for Understanding1. C2. If k is positive, the equation models growth.If k is negative, the equation models decay.3. Amount in an account with a beginning balance of$3000 and interest compounded continuously atan annual rate of 5.5%.4. reals, positive reals5. Sample answer: Continuously compoundedinterest is a continuous function, but interestcompounded monthly is a discrete function.6a. growth6b. 33,4306c. y 33,430e 0.0397(60) 361,931.0414 or 361,9317. A 12,000e 0.064(12) 25,865.412 or $25,865.41Pages 714–717 Exercises8. p (100 18)e 06(2) 18 42.6 or 43%9a. y 84e 0.23(15) 76 78.66 or 78.7°F9b. too cold; After 5 minutes, his coffee will be about90°F.10a.[4, 4] scl:1 by [1, 10] scl:110b. symmetric about y-axisSample answer: y 948.4x 4960.6383 Chapter 11
11a. Annually: I 100[(1 0.08) 1 1] 80 $80.00; 8%Semi-annually: I 10001 0. 082 2 1 81.6 $81.60; 8.16%Quarterly: I 10001 0. 084 4 1 82.4316 $82.43; 8.243%Monthly: I 10001 0 .081212 1 82.9995 $83.00; 8.3%Daily: I 10001 0 . 08365365 1 83.2776 $83.28; 8.328%Continuously: I 1000(e 0.08(1) 1) 83.2871 $83.98; 8.329%EffectiveInterestAnnualCompounded Interest YieldAnnually $80.00 8%Semi-annually $81.60 8.16%Quarterly $82.43 8.243%Monthly $83.00 8.3%Daily $83.28 8.328%Continuously $83.29 8.329%11b. continuously 11c. E 1 nr n 111d. E e r 112a. y 525(1 e 0.038(24) ) 314.097 314 people12b. after about 61h[0, 100] scl:10 by [0, 550] scl:5013a. P 1 e 6(0.5) 0.95021 95%13b.x 0.02; about 0.02 h 0.0 2h1 60 min1 h 1.214b. 2 decimal places; 4 decimal places; 6 decimalplaces14c. always greater15a. 5 days: P 1 e 0.047(5) 0.20943 20.9%20 days: P 1 e 0.047(26) 0.60937 60.9%90 days: P 1 e 0.047(90) 0.98545 98.5%20.9%; 60.9% 98.5%15b.about 29 days15c. Sample answer: The probability that a personwho is going to respond has respondedapproaches 100% as t approaches infinity. Newads may be introduced after a high percentage ofthose who will respond have responded. Thegraph appears to level off after about 50 days.So, new ads can be introduced after an ad hasrun about 50 days.16a. all reals16b. 0 f(x) 116c. c shifts the graph to the right or left17. 120,000 P (1 0 0.. 035) 8 1035120,000 P(9.051687)P 13,257.19725$13,257.2018. x 8 5 y 3 5 z 1 5 x 5 x3 y 3 z19. y 6x 2 v 45°x sin 45° y cos 45° 6(x cos 45° y sin 45°) 22 2 2 2 x y 6 x y 2 22 2 x y 6 1 2 x 2 xy 1 2 y 2 22 22 2 x y 3x 2 6xy 3y 22x 2y 6x 2 12xy 6y 26x 2 12xy 6y 2 2x 2y 020. r (5) 2 ) (1 2 v Arctan 15 26 3.3426(cos 3.34 i sin 3.34) 2 2 2 2[0, 1] scl:0.1 by [0, 1] scl:0.1 about 1.2 min14a. For x 10: 2 ( 10) 1 102(10) 1 2 1 1019 2.720551414For x 100: 2 ( 100) 1 1002(100) 1 2 01100199 2.718304481For x 1000: 2 ( 1000) 1 10002(1000) 1 2 00110001999 2.7182820552.720551414; 2.718304481; 2.718282055Chapter 11 384
21.yO(0, 150)d28˚xu ud x, y F 0, 150xcos 28° 1 0 xsin 28° 1 0 x 10 cos 28° x 10 sin 28°x 8.8295 x 4.6947d 8.8295, 4.6947W u F u dW 0, 150 8.8295, 4.6947 0 704.205 704.2 ft-lb22. y 1.5(10) 2 13.3(14) 19.4 2.423. 2x 3 42x 3 162x 13x 1 3224. ⏐3x 2⏐ 63x 2 63x 4x 4 3 x ⏐ 8 3 x 4 3 (x, y)3x 2 63x 8x 8 3 3 2 2 325. 3 9 6 6 9 y2 6 5 1 6 18 15 3J(9, 6), K(6, 18), L(6, 15), M(9, 3); Thedilated image has sides that are 3 times thelength of the original figure.4x y26. 6x 2y 124x y 6 4(2y 12) y 6 x 2(6) 12x 2y 12 9y 54 x 0y 6(0, 6)27. {4, 2, 5}; {5, 7}; yes28. The correct choice is D.4xPage 717 Mid-Chapter Quiz1. 64 2 82. ( 3 343 ) 2 (343 1 3 ) 2 ((7 3 ) 1 3 ) 2 7 23. 28x3 y67w6 z9 1 3 21 4 98x3z97w6y6 1 3 (23 ) 1 3 (x 3 ) 1 3 (z 9 ) 1 3 (3 3 ) 1 3 (w 6 ) 1 3 (y 6 ) 1 3 2xz3 3 w2y24. a 6 b 3 (a 4 b 3 ) 1 2 or 2 3 w 2 xy 2 z 3 (a 6 ) 1 2 (b 3 ) 1 2 ⏐a⏐ 3 b 3 2 5. (125a 2 b 3 ) 1 3 3 125a b 2 3 35 3 a 2 b 3 5b 3 a26. 1.75 10 2 0.094 A31.75 10 2 0.94 A 3 2 1.75 10 20. 94 1.75 10 2 0. 94 2 3 A 3 2 (A 3 2 ) 2 3 151.34 A1.51 10 2 mm 27. 1,786,691 1,637,859 (1 r) 8 1 , 786,6911,637,859 1 , 786,6911,637,859 1 8 1 , 786,6911,637,859 1 8 (1 r) 8 [(1 r) 8 ] 1 8 1 rr 0.011N 1,637,859 (1 0.011) 24 2,216,156.9792,126,1578. A 3500 1 0.0 52 (4)(3.5)4 3500(1.013) 14 4193.728$4193.738.1Store the exact value inyour calculator’smemory.Use the storedvalue for r.8.19a. y 6.7e 4 159b. y 6.7e 4 50 0.271292 2.560257271,292 ft 3 2,560,257 ft 32010. 2 years: n 1 2 0e 00.35(2) 18.320015 years: n 1 20 181260 years: n 1 20 e 20018.3; 181; 200e0.35(15)000.35(60)385 Chapter 11
11-4Pages 722–723Logarithmic FunctionsCheck for Understanding1. yy 3 x and log 3 x arey 3 x similar in that they areboth continuous, one-toone,increasing andinverses.O xy 3 x and log 3 x are noty log 3 x similar in that they areinverses. The domain ofone is the range ofanother and the rangeof one is the domain ofthe other. y 3 x has ay-intercept and ahorizontal asymptotewhereas y log 3 x has ax-intercept and avertical asymptote.2. Let b x m, then log b m x.(b x ) p m pb xp m plog b b xp log b m pxp log b m pp log b m log b m p3.yOy log 1 x5y log 5 xLog 5 x is an increasing function and log x is a1 5decreasing function.4. Sean is correct. The product property states thatlog b mn log b m log b n.5. In half-life applications r 1 2 . So, (1 r)becomes 1 1 2 or 1 2 . Thus, the formulaN N 0 (1 r) t becomes N N 0 1 2 t .6. 9 3 2 27 7. 25 1 2 58. log 7 y 6 9. log 1 8 4 2 3 110. log 2 1 6 x2 x 1 1 62 x 2 4x 4112. log 7 3 43 x7 x 1 3 43 7 x 7 3x 3x111. log 10 0.01 x10 x 0.0110 x 10 2x 213. log 2 x 52 5 x32 x14. log 7 n 2 3 log 7 8log 7 n log 7 8 2 3 n 8 2 3 n 415. log 6 (4x 4) log 6 644x 4 64x 1516. 2 log 6 4 1 4 log 6 16 log 6 xlog 6 4 2 log 6 16 1 2 log 6 x4 2log 6 log 6 x16 1 4 17. x y1 02 14 218. x y1 06 119. 16 3.3 logt4 10244 216 1 4 xx 8t 16(33 log 4 1024) log 4 1024 xt 16(33 5) 4 x 1024t 264 h 2 2x 2 10Pages 723–725Exercises2x 10x 520. 27 1 3 3 21. 16 1 2 422. 7 4 1 24 0123. 4 5 2 3224. e x 65.98 25. 6 4 3626. log 81 9 1 2 27. log 36 216 3 2 28. log1512 3 29. log 8 6 362yOyOy log 1 x2y log 6 x1xxChapter 11 386
30. log 16 1 0 31. log x 14.36 1.23832. log 8 64 x8 x 648 x 8 2x 233. log 125 5 x2 x 322 x 2 5x 535. log 4 128 x9 x 9 6x 637. log 49 343 x8 x 162 3x 2 43x 4x 4 3 39. log 84096 x10 log 10 24 x2 4 x16 x42. log 3 3x log 3 363x 36x 1243. log 6 x log 6 9 log 6 54log 6 9x log 6 549x 54x 644. log 8 48 log 8 w log 8 6log 4 8 8 w log8 6 4 8 w 66w 48125 x 5(5 3 ) x 5 13x 1x 1 3 34. log 2 32 x4 x 1282 2x 2 72x 7x 7 2 or 3.536. log 9 9 6 x49 x 3437 2x 7 32x 3x 3 2 or 1.538. log 8 16 x8 x 40968 x 2 8 4x 2 4x 840. 10 4 log 10 2 x41. log x 49 2x 2 49x 7w 8x 345. log 6 216 x 46. log 5 0.04 x6 x 2165 x 0.046 x 6 3 5 x 5 2x 247. log 10 3 10 x10 x 3 1010 x 10 1 3 x 1 3 48. log 12 x 1 2 log 12 9 1 3 log 12 27log 12 x log 12 9 1 2 log 12 27 1 3 log 12 x log 12 9 1 2 27 1 3 x 9 1 2 27 1 3 x 3 3x 949. log 5 (x 4) log 5 8 log 5 64log 5 (x 4)(8) log 5 64(x 4)(8) 64x 4 8x 450. log 4 (x 3) log 4 (x 3) 2log 4 (x 3)(x 3) 24 2 (x 3)(x 3)16 x 2 925 x 25 x51. 1 2 (log 7 x log 7 8) log 7 16 1 2 (log 7 8x) log 7 16log 7 (8x) 1 2 log 7 16(8x) 1 2 168x 256x 3252. 2 log 5 (x 2) log 5 36log 5 (x 2) 2 log 5 36(x 2) 2 36x 2 6x 8y53. x y1 0y log 4 x2 1 2 4 1Ox387 Chapter 11
54. x y1 02 34 655. x y2 06 156. x y1 02 14 257. x y1 02 24 458. x y0 09 159. Use N N 0 (1 r) t ; r 1 since the rate of growthis 100% every t time periods.64,000 1000 (1 1) t64 2 tlog 2 2 6 log 2 2 t6 tt 15 90 min60. All powers of 1 are 1, so the inverse of y 1 x isnot a function.61. Let log b m x and log b n y.So, b x m and b y n.xy b xy m n b b m n b xylog b m n x ylog b m n log b m log b n20151055O101520yyyyyy 3 log 2 x510152025303540xy log 5 (x 1)y log 2 xy 2 log 2 xy log 10 (x 1)xxxx62a. 5000 25001 4r 4102 7 1 4r 4062b. 2 1 4r 402 140 r1 4 401.0175 1 4r 0.0699 r6.99%62c. 2 1 4r 282 1 28 1 4r 1.0251 1 4r 0.1004 r10.04%63a. n log 2n log 2 42 n 4n 21 401— 63b.13 log 2 p 1 4 64. Let y log a x, so x a y .x a ylog b x log b a ylog b x y log b alogbxy l65a.log a x ll4ogbaogbxogba2 3 p1 P65b. log 2.72 14 .70.02(1)P 14 .7 2.72 0.02P 14.7 (272 0.02 ) 14.4 psiP65c. log 2.72 14 .70.02(6.8)P 14.7(2.72 0.136 ) 16.84 psi66. 6.8 38 1 2 t 6 . 838log 6 . 838log 6 . 838 1 2 t log 1 2 t t log 1 2 t 1615.51514.51413.51312.52 Olog 6 . 8__ 38log 1 2 Pt 2.52.5 3.82 9.55about 9 days2 4h8 p 1 1 8 plog P 2.72 ( ) 0.02h14.7less light; 1 8 Chapter 11 388
67. 69.61641 1 68. 90,000 P 0. 11512.30______12115 P 891.262$891.2669. ellipse9x 2 18x 4y 2 16y 11 09x 2 18x 4y 2 16y 119(x 2 2x 1) 4(y 2 4y 4) 11 9 169(x 1) 2 4(y 2) 2 36 (x 1) 24 (y 2) 29 170. r 3, v 2(3 cos 2t, 3 sin 2t)71. AB (1 (1)) 2 (33) 2 06) 2 ( 2 6BC (3 1)) ((0 2 3)) (2 4 2 3 2 5AC (3 1)) ((0 2 3) 2 4 2 3 2 572. 5cos 3 4 i sin 3 4 2cos 2 3 i sin 2 3 5 2cos 3 4 2 3 i sin 3 4 2 37 10cos 1 1 2 i sin 1 1772 72 10 cos 1 1 2 10i sin 1 1 2.59 9.66i10cos 1 71 2 i sin 1 71 2 , 2.59 9.66i73. (3 4j)(12 7j) 36 21j 48j 28j 2 64 27j64 27j volts74.Ay50˚yOC 0. 12x50˚75. cos (A B) cos A cos B sin A sin B5cos A 1 3cos B 3 537 x 2 5 2 13 2 x 2 35 2 37 2x 2 144 x 2 144x 12 x 12So, sin A 1 213So, sin B 1 2375 5 2 2cos (A B) 1S3 3 37 1 13 1 37 1 75481 1 4448131 4 8176. y A sin (kt c) hA 90 642h 90 64 2 2k 4 13 77 k 2 y 13 sin 2 t c 7764 13 sin 2 (1) c 77 c c13 13 sin 2 1 sin 2 sin 1 (1) 2 c3.14 cSo, y 13 sin 2 k 3.14 7777. c 2 (6.11) 2 (5.84) 2 2(6.11)(5.84) cos 105.3c 2 37.3321 34.1056 71.3648 cos 105.3c 2 90.2689c 9.5(6.11) 2 (5.84) 2 (9.5) 2 2(5.84)(9.5) cos A37.3321 34.1056 90.25 110.96 cos Acos A 0.7843A 38.34 or 38° 20B 180 (105° 18 38° 20) 36° 22c 9.5, A 38° 20, B 36° 2278. Mc˚ x˚QPb˚c˚NmSMN mQNM alternate interior anglesx° b° c° Exterior Angle TheoremThe correct choice is E.OBDxBoth vectors have the same direction. 50° south ofeast. Therefore, ABu and CDu are parallel.389 Chapter 11
11-5Common LogarithmsPage 730 Check for Understanding1. log 1 0 means log 10 1 0. So, 10° 1.log 10 1 means log 10 10 1. So, 10 1 10.2. Write the number in scientific notation. Theexponent of the power of 10 is the characteristic.3. antilog 2.835 10 2.835 683.91164. log 15 1.1761log 5 0.6990log 3 0.4771log 5 log 3 0.6990 0.4771 1.17615. log 80,000 log (10,000 8) log 10 4 log 8 4 0.9031 4.90316. log 0.003 log (0.001 3) log 10 3 log 33 0.47712.52297. log 0.0081 log (0.0001 3 4 ) log 10 4 4 log 34 4(0.4771)2.09158. 2.6274 9. 74,816.9510. yy log (x 3)og1811. log 12 18 l log12 1.163212. log 8 15 l og15log8 1.302313. 2.2 x 5 9.32(x 5) log 2.2 log 9.32(x 5) l og9.32log2.2x 7.8314. 6 x2 4 x(x 2) log 6 x log 4x log 6 2 log 6 x log 42 log 6 x log 4 x log 62 log 6 x (log 4 log 6)2log 6 log 4 log6 x8.84 x15. 4.3 x 76.2x log 4.3 log 76.2x l og76.2log4.3x 2.97x16. 3 x3 2 4 4x13 x3 24 x 4(x 3) log 3 log 2 x 4log 4(4x 12)log 3 4 log 2 (x 1) log 44x log 3 12 log 3 4 log 2 x log 4 log 44x log 3 x log 4 4 log 2 log 4 12 log 3x(4 log 3 log 4) 4 log 2 log 4 12 log 34 log 2 log 4 12 log 3x 4 log 3 log 4x 4.8417.[10, 10] scl:1 by [20, 100] scl:105.585018a. R log 2 001.6 4.2 6.318b. 10 times; According to the definition oflogarithms, R in the equationR log aT B is an exponent of the base of thelogarithm, 10. 10 5 is ten times greater than 10 4 .1Pages 730–732 Exercises19. log 4000,000 log (100,000 4) log 100,000 log 4 5 0.6021 5.602120. log 0.00009 log (0.00001 9) log 0.00001 log 95 0.95424.045821. log 1.2 log (0.1 12) log 0.1 log 121 1.0792 0.079222. log 0.06 log 0.01 1 22 log 0.01 log log 0.01 log 12 1 2 log 42 1.0792 1 2 (0.6021)1.221823. log 36 log (4 9) log 4 log 9 0.6021 0.9542 1.556324. log 108,000 log (1000 12 9) log 1000 log 12 log 9 3 1.0792 0.9542 5.0334124 1 2 1Chapter 11 390
25. log 0.0048 log (0.0001 12 4)46. 3 x1 2 x7log 1539. log115 2 log 1 40. 2 x 9549. 0.5 2x4 0.1 5x2 x log 2 log 95(2x 4) log 0.5 (5 x) log 0.1 log 0.0001 log 12 log 44 1.0792 0.60212.318826. log 4.096 log (0.001 4 6 )(x 1) log 3 (x 7) log 2x log 3 log 3 x log 2 7 log 2x log 3 x log 2 log 3 7 log 2x(log 3 log 2) log 3 7 log 2 log 0.001 6 log 4x lo g 3 7 log2log3 log23 6(0.6021)x 9.2571 0.612427. log 1800 log 100 9 4 1 2 47. logx 6 1 log 100 log 9 1 2 l og6log 4logx 1 2 0.9542 1 log 6 log x2 (0.6021)6 x 3.2553When x 1, log 1 0, which means l og6logxis28. 1.9921 29. 2.9515undefined. When x 1, l og6logxis negative, which is30. 0.871 31. 2.001not greater than 1. So, x must also be greater than32. 3.2769 33. 2.17451. Therefore, 1 x 6.34. log 2 8 l og8log235. log 5 625 lo g 625log548. 4 2x5 3 x3 3 4(2x 5) log 4 (x 3) log 336. log 6 24 l og24log637. log 7 4 l og42x log 4 5 log 4 x log 3 3 log 3log72x log 4 x log 3 5 log 4 3 log 3 1.7737 0.7124x(2 log 4 log 3) 5 log 4 3 log 338. log 6.5 0.0675 log 0.0675log0.5x 5 log4 3 log32 log4 log3 3.8890x 2.17192x(log 0.5) 4 log 0.5 5 log 0.1 x log 0.1x l og95log22x log 0.5 x log 0.1 5 log 0.1 4 log 0.5x 6.5699x(2 log 0.5 log 0.1) 5 log 0.1 4 log 0.541. 5x 4 x 35 log 0.1 4log 0.5x 2 log 0.5 log 0.1x log 5 (x 3) log 4x log 5 x log 4 3 log 4x(log 5 log 4) 3 log 4Change inequality sign because (2 log 0.5 log 0.1) is negative.log4x 3.87253x log 5 log 4x 18.637742. 1 3 log x log 8x 1 3 8x 51243. 0.16 43x 0.3 8x(43x) log 0.16 (8 x) log 0.34 log 0.16 3x log 0.16 8 log 0.3 x log 0.33x log 0.16 x log 0.3 8 log 0.3 4 log 0.16x(3 log 0.16 log 0.3) 8 log 0.3 4 log 0.168 log 0.3 4 log 0.16x 3 log 0.16 log 0.3x 0.343444. 4 log (x 3) 9log (x 3) 9 4 (x 3) antilog 9 4 50. log 2 x 3x 2 3x 0.125052.[10, 10] scl:1 by [3, 3] scl:153.g 52.751. x lo log3x 3.6087x antilog 9 4 3x 174.829745. 0.25 log 16 x0.25 x log 160.25x lo g 16x 0.2076[1, 10] scl:1 by [1, 3] scl:1391 Chapter 11
54.[10, 1] scl:1 by [2, 10] scl:155. x 0.3210[5, 5] scl:1 by [10, 50] scl:1056. x 0.1975[5, 5] scl:1 by [3, 10] scl:157. x 2[5, 5] scl:1 by [5, 10] scl:158a. h 10 09log 1 0.314.7 1.7 mi58b. 4.3 10 0 P9log 14 .70.3870 log P log 14.70.3870 log 14.7 log P0.7803 log P6.03 P; 6 psi59a. M 5.3 5 5 log 0.018 1.5859b. 5.3 8.6 5 5 log P8.3 5 log P1.66 log P0.0219 P60a. q 1 2 0.89 1 2 0.1342 0.9112$91,11660b. 0.9535 1 2 0.8tloglog 0.9535 0.8 t log 1 2 log 0.9535log 1 2 0.8 tlog 0.9535 t log 0.8 log 1 2 12.0016 t12 years61. Sample answer: x is between 2 and 3 because 372is between 100 and 1000, and log 100 2 and log1000 3.62a. L 10 log 1.0 110 12 10(log 1 log (1.0 10 12 )) 120 dB62b. 20 log 1.0 I10 122 log I log (1.0 10 12 )2 log I 1210 log I1 10 10 I; 1 10 10 W/m 263. Use N N 0 1 2 t .N 630 micrograms 63 10 4 gramN 0 1 milligram 1.0 10 3 gram6.3 10 4 (1.0 10 3 ) 1 2 tlog 6 . 3 1041.0 103 t log 1 2 0.6666 t0.6666 5730 3819 yr64. log a y log a P log a q log a rlog a y log p a q log a rlog a y log p r a qy p rq65. log x 243 5x 5 243x 366.increasing from to 67. (a 4 b 2 ) 1 3 c 2 3 (a 4 ) 1 3 (b 2 ) 1 3 (c 2 ) 1 3 a 3 ab 2 c 268. (5) 2 (0) 2 D(5) E(0) F 05D F 25 0(1) 2 (2) 2 D(1) E(2) F 0D 2E F 5 0(4) 2 (3) 2 D(4) E(3) F 04D 3E F 25 0Chapter 11 392
74. f(x) x 3 2x 2 11x 125D 0E F 25 0 31.6758 or 31.68 cm 2() D 2E F 5 0f(1) 1 2(1) 11(1) 12 Test f(1).4D 2E 20 0f(1) 0(x 1) is a factor.4D 3E F 25 01⏐ 1 2 11 12() D 2E F 5 01 1 123D E 20 01 1 12 ⏐ 04D 2E 20 0x 2 x 12 02(3D E 20 0)(x 4)(x 3) 010D 60 0So, the factors are (x 4)(x 3)(x 1).D 675. y4(6) 2E 20 02E 4 0E 25(6) 0(2) F 25 0F 5 0y 5x 3 2x 5F 5x 2 y 2 6x 2y 5 0O x(x 2 6x 9) (y 2 2y 1) 5 9 1(x 3) 2 (y 1) 2 Neither; the graph of the function is not 569. 65 25, 18 symmetric with respect to either the origin or the 422 (25, 11)y-axis.70. r 6r 2 36x 2 y 2 3676. 7 5 4 1 1717,000,000The correct choice is D.71.120˚90˚60˚150˚30˚11-6 Natural Logarithms180˚2 4 6 0˚Page 735 Check for Understanding210˚330˚1. ln e 1 is the same as log e e 1. And e 1 e. So,240˚ 270˚300˚ln e 1.u 2. The two logarithms have different bases.72. AB (6 5), (5 6)log 17 ⇒ 10 x 17 or x 1.23 1, 1ln 17 ⇒ eu 17 or x 2.83⏐AB ⏐ (6 ) 5 5 2 () 63. ln 64 4.1589 2 1.414ln 16 2.772673.ln 4 1.3863ln 16 ln 4 2.7726 1.3863 4.15894. The two equations represent the same thing,A Pe rt is a special case of the equationN N 0 e kt and is used primarily for computationsinvolving money. 3.65 cm5. 4.7217 6. 1.1394a7. 15.606 8. 0.45709. log 5 132 ln 132ln 5bv 36 10 360° 3.033910. log 3 64 l n 64ln3ạcos 36° 3 65 bsin 36° 3. 65 3.785611. 18 e 3xa 3.65 cos 36° b 3.65 sin 36°ln 18 3x ln e 2.9529 2.1454Use A 1 2 ln 18aP, where P 10(2.1454) 21.454.3 xA 1 0.9635 x2 (2.9529)(21.454)393 Chapter 11
12. 10 5e 5k2 e 5kln 2 5k ln e33. log 8 0.512 ln ln8 0.3219 ln 25 k34. log 6 323 ln 3031.60.1386 k 3.224613. 25e x 10035. log 5 288 ln 288e x ln5 4 1.7593x ln e ln 436. 6x 1.3863x 72 37. 2 x 2714. 4.5 e 0.031tx ln 6 ln 72 x ln 2 ln 27ln 4.5 0.031t ln ex l n 72ln6x l n 27ln2 l n 4.50.031 t 2.3869 4.754938. 948.5186 tx4 7.13(x 4) ln 9 ln 7.1315. x 13.57x ln 9 4 ln 9 ln 7.13x ln 9 ln 7.13 4 ln 9x ln 7.13 4ln9ln 9x 4.894039. 3x 32x ln 3 ln 3 ln 2x ln 3 ln 2ln 3[20, 20] scl:2 by [4, 20] scl:2x 1.315516. x 26.9041. 60.3 e 0.1tln 60.3 0.1t ln e ln 60.30.1 t40.9933 t42. 6.2e 0.64t 3e t1ln 6.2 0.64t ln e ln 3 (t 1) ln eln 6.2 ln 3 1 0.36t[15, 30] scl:5 by [50, 150] scl:1017a. p 760e 0.125(3.3) ln 6.2 ln3 10.36 t 760e 0.41250.7613 t 503.1 torrs43. 22 44 (1 e 2x )17b. 450 760e 0.125a 1 4 2 1 e 2x50760 e 0.125a 1 2 e 2xln 4 50760 0.125a ln eln 1 2 2x ln eln 4 50760ln 120.125 a 2 x4.1926 a; 4.2 km0.3466 x44. 25 e 0.075yPages 736–737 Exercise18. 5.4931 19. 0.270520. 6.8876 21. 0.965722. 0 23. 2.232224. 10.4395 25. 1.213426. 0.0233 27. 0.996628. 146.4963 29. 0.241730. log 12 56 l n 56ln1231. log 5 36 l n 36ln5 1.6199 2.226632. log 4 83 l n 83ln4 3.18750.512ln 25 0.075y ln eln 25 0 .075 yy 42.918340. 25e x 1000e x 40x ln e ln 40x 3.6889Chapter 11 394
45. 5 x 76x ln 5 ln 7 ln 6x ln 7 ln 6ln 5x 1.765746. 12 x4 4 x(x 4) ln 12 x ln 4x ln 12 4 ln 12 x ln 4x ln 12 x ln 4 4 ln 12x(ln 12 ln 4) 4 ln 124 ln 12x ln12 ln4x 9.047447. x 2 3 27.6x (27.6) 3 2 x 144.998548. x 3.76[5, 5] scl:1 by [50, 700] scl:5049. x 7.64[70, 10] scl:1 by [3, 10] scl:150. t 133.14[10, 150] scl:10 by [100, 2000] scl:10051. x 2.14[6, 6] scl:1 by [4, 24] scl:252. x 4.72[10, 10] scl:1 by [10, 75] scl:553. x 0.37[5, 5] scl:1 by [2, 5] scl: 0.554. 0.6 1eln 0.6 20,000(4 10 11 ) ln 0.6 t4.09 10 7 t4.09 10 7 s55. 2.8 9 1 2 tln 2 .8 9 t ln 1 2 ln 2 .8 __ 9ln 1 2 t1.6845 t1.6845 8 24 323.4236324 h56a. ln ⏐180 72⏐ k(0) c4.6821 c56b. ln ⏐150 72⏐ k(2) 4.6821 ln 78 4.68212t20,000(410 11 )t20,000(4 10 11 ) k0.1627 k56c. ln ⏐100 72⏐ (0.1622)t 4.6821 ln 2 8 4.68210.1627 t8.3 t 8.3 2 6.3about 6.3 min57. e 2x 4e x 3 0(e x 3)(e x 1) 0e x 3 0 e x 1 0e x 3 e x 1x ln e ln 3 x ln e ln 1x 1.0986 x 00 or 1.098658a. 2 e 0.063tln 2 0.063t ln eln2 0 . 063 t11.0023 tabout 11 years58b. See students’ work.ln e395 Chapter 11
59. 1800 5000 ln r68. 2x 5y 3 01800 5000 ln rAB 2 2 25) 2 ( 2 292xantiln 18005000 r 2 y 9 52 9 2 0390.6977 r; about 70% 2 29 x 5 29 y 3 29 060a. 1 2 1e k(1622)29 29 29p ln 1 3 29 0.56 units2 29 1622 k ln e5sin ln 1 29 2cos 2 1622 k529___ 29tan 0.000427 k 22960b. 1.7 23e (0.000427)(t)29ln 1 . 7tan 5 2 2.30.000427t ln e 112°ln 1 .72.30.000427 t707.9177 tabout 708 yr61. y is a logarithmic function of x. The pattern in thetable can be determined by 3 y x which can beexpressed as log 3 x y.62. 1.284463. 16 3 4 864. x 2 y 4x 2 4y 2 8(y 4) 4y 2 84y 2 y 4 01 1 4y (4)(4 )865 y 1 8y 0.9, 1.1x 2 0.9 4 x 2 11 4x 4.9 x 2.9 x 2.2, 2.2 x 1.7, 1.765. 52 .4 Nm2yO 146 cm 3 m 100 cx3 cm 30.00765 c; 0.00765 N m66. x 0.25 cos y 0.25 sin 0.25 0(0.25, 0)67. a u 1, 2 34, 3 1, 2 12, 9 13, 73 2 29 2929 29 2929 x 5 y 3 0; 3 0.56; 112°29 292969. y 70 cos 4v70. d 800 (10 55) 250The correct answer is 250.11-6BGraphing Calculator Exploration:Natural Logarithms and AreaPages 738–7391. 0.693147182. 0.6931471806; It is the same value as found inExercise 1 expressed to 10 decimal places.3a. The result is the opposite of the result inExercise 1.3b. Sample answer: a negative value4a. 0.693147184b. 1.09861234c. 1.38629444d. 0.6931471806, 1.098612289, 1.3862943614e. The value for each area is the same as the valueof each natural logarithm.5. 0.5108256238; 0.6931471806; 0.9162907319;These values are equal to the value of ln 0.6,ln 0.5, and ln 0.4.6. If k 1, then the area of the region is equal toln k. If 0 k 1, then the opposite of the area isequal to ln k.7. The value of a should be equal to or very close to1, and the value of b should be very close to e. Thisprediction is confirmed when you display theactual regression equation.8. Sample answer: Define ln k for k 0 to bethe area between the graph of y 1 x , the x-axis,and the vertical lines x 1 and x k if k 1 andto be the opposite of this area if 0 k 1. Definee to be the value of k for which the area of theregion is equal to 1.Chapter 11 396
11-7Modeling Real-World Data withExponential and LogarithmicFunctionsPage 744 Check for Understanding1. Replace N by 4N 0 in the equation N N 0 e kt ,where N 0 is the amount invested and k is theinterest rate. Then solve for t.2. The data should be modeled with an exponentialfunction. The points in the scatter plot approach ahorizontal asymptote. Exponential functions havehorizontal asymptotes, but logarithmic functionsdo not.3. y 2e (ln 4)x or y 2e 1.3863x ; ln y ln 2 (ln 4)x orln y 0.6931 1.3863xn 2ln4. t 0ḷ 01755. t 20 .08 39.61 yr 8.66 yr6a. y 10.0170(0.9703) x6b. y 10.0170(0.9703) xy 10.0170(e ln 0.9703 ) xy 10.0170e (ln 0.9703)xy 10.0170e 0.0301x6c. 5 10.0170e 0.0301x5ln 10.0 170 0.0301x5ln 10.0 1700.0301 xPages 745–7487. t 0ḷ 0225 30.81 yrln 29. t 0.023.08 x; 23.08 minn 27125Exercisesln8. t 02.05 13.86 yr 9.7310. exponential; the graph has a horizontal asymptote11. logarithmic; the graph has a vertical asymptote12. logarithmic; the graph has a vertical asymptote13. exponential; the graph has a horizontal asymptote14a. y 4.7818(1.7687) x14b. y 4.7818(1.7687) xy 4.7818(e ln 1.7687 ) xy 4.7818e (ln 1.7687)xy 4.7818e 0.5702x14c. Use t ln k; k 0.5702.n 2t 0ḷ 5702 1.215 hr15a. y 1.0091(0.9805) x15b. y 1.0091(0.9805) xy 1.0091(e ln 0.9805 ) xy 1.0091e (ln 0.9805)xy 1.0091e 0.0197x215c. 0.415 1.0091e 0.0197x0.415ln 1 . 00910.415ln 1 . 0091 0.0197 0.0197x x45.10 x45.10 10 35.10 min16a. y 2137.5192(1.0534) x16b. y 2137.5192(1.0534) xy 2137.5192(e ln 1.0534 ) xy 2137.5192e (ln 1.0534)xy 2137.5192e 0.0520x16c. 2631.74 2137.52e 4rln 2 631.742137.52 2 631.742137.52 4rln 4 r0.0520 r; 5.2%17. y 40 14.4270 ln x18a. y –826.4217 520.4168 ln x18b. The year 1960 would correspond to x 0 andln 0 is undefined.19. Take the square root of each side.y cx 2y cx 2y cx20a. 1034.34 1000(1 r) 11.03034 1 r0.03034 r; 3.034%20b. y 1000.0006(1.0303) x20c. y 1000.0006(1.0303) xy 1000.0006(e ln 1.0303 ) xy 1000.0006e (ln 1.0303)xy 1000.0006e 0.0299x20d. 1030.34 1000e r21a.ln 10 30.341000 r0.0299 r; 2.99%x 0 50 100 150 190 200ln y 1.81 2.07 3.24 3.75 4.25 4.3821b. ln y 0.0136x 1.688921c. ln y 0.0136x 1.6889y e 0.0136x1.688921d. y e 0.0136(225)1.6889 115.4572115.5 persons per square mile22a. The graph appears to have a horizontalasymptote at y 2, so you must subtract 2 fromeach y-value before a calculator can performexponential regression.22b. y 2 1.0003(2.5710) x23a. ln y is a linear function of ln x.y cx aln y ln(cx a )ln y ln c ln x aln y ln c a ln x397 Chapter 11
23b. The result of part a indicates that we shouldtake the natural logarithms of both the x- andy-values.23c. ln y 0.4994 ln x 1.390123d. ln y 0.4994 ln x 1.3901e ln y e0.4994ln x1.3901y e 0.4994ln x e 1.3901y (e ln x ) 0.4994 4.0153y 4.0153x 0.499424. 2 e k(85)ln 2 85k0.0082 k12 e 0.0082tln 12 0.0082t303 t303.04 min or about 5 h25. 0.0126. log 5 (7x) log 5 (5x 16)7x 5x 162x 16x 827a. y x(400 20(x 3))y x(460 20x)y 20x 2 460xy 2645 20(x 2 23x 132.25)y 2645 20(x 11.5)vertex at (11.5, 2645), maximum at x 11.5$11.5027b. At maximum, y 2645.$264528. 5cos 6 i sin 6 5 32 1 2 i 5 3 5 2 2 i60 b29. sin 24° sin 48°60 ftln x 6.21 6.91 8.52 9.21 9.62ln y 4.49 4.84 5.65 5.99 6.1960 sin 48° b sin 24°b 109.625about 109.6 ft30. 5x 2 8x 12 0Discriminant: (8) 2 4(5)(12) 176The discriminant is negative, so there are2 imaginary roots.x 8 10b176 42˚18˚ 8 41 i110or 4 2 i115 31. 4 units left and 8 units down32.(2, 3)x 2y 4y 1O(2, 1): f(x) 2(2) 8(1) 10 22(4, 1): f(x) 2(4) 8(1) 10 26(2, 8): f(x) 2(2) 8(3) 10 38(4, 4): f(x) 2(4) 8(4) 10 5050; 2233. Circle X contains the regions a, b, d, and e.Circle Z contains the regions d, e, f, and g. Sixregions are contained in one or both of circles Xand Z.The correct choice is C.Chapter 11 Study Guide and AssessmentPage 749 Understanding the Vocabulary1. common logarithm 2. exponential growth3. logarithmic function 4. scientific notation5. mantissa 6. natural logarithm7. linearizing data 8. exponential function9. nonlinear regression 10. exponential equationPages 750–752Skills and Concepts11. 1 4 2 12. (64) 1 2 8 1 4 213. (27) 4 3 (3 3 ) 4 3 3 4 81yx 4x 2(4, 4)x23x) 2 3 x29x2 1 3 15. 3x 2 (3x) 2 (3x(2, 1)17. 1 2 x 4 3 1 2 3 (x 4 ) 3 16 1 8 x 1218. (w 3 ) 4 (4w 2 ) 2 w 12 4 2 w 4 16w 1619. (2a) 1 3 (a 2 b) 1 3 3 (2a) 1 3 3 (a 2 b) 1 3 2 (2a)(a 2 b) 2a 3 b20. 3x 1 2 y 1 4 (4x 2 y 2 ) 12x 5 2 y 9 4 1(4, 1)14. ( 4 256 ) 3 (256) 3 4 (4 4 ) 3 4 4 3 6416. 6a 1 3 3 6 3 a 1 3 3 216aChapter 11 398
21. 22.yy45. 2 log 6 4 1 3 log 6 8 log 6 xlog 6 4 2 log 6 8 1 3 log 6 xlog 64 2 log 6 xy 3 xy (1)x28 1 3 x4 2O x23. 24.yy 2 x 1Oyxy 2 x 246. log 2 x 1 3 log 6 27log 2 x log 2 27 1 3 47.yx 27 1 3 x 38 1 3 8 xOxOxOy log 10 xx25. 26.yyOy 2 x 127. A 2500e 0.065(10) 4788.8520; $4788.8528. A 6000e 0.0725(10) 12,388.3866; $12,388.3929. A 12,000e 0.059(10) 21,647.8610, $21,647.8630. 8 2 3 4 31. 3 4 1 8 1132. log 2 16 4 33. log 5 2 5234. 2 x 32 35. 10 x 0.0012 x 2 5 10 x 10 3x 5x 336. 4 x 1 1 637. 2 x 0.54 x 4 2 2 x 2 1x 2x 138. 6 x 216 39. 9 x 1 9 6 x 6 3 9 x 9 1x 3x 140. 4 x 1024 41. 8 x 5124 x 4 5 8 x 8 3x 5 x 342. x 4 81 43. 1 2 4 xx (81) 1 4 x 344. log 3 3 log 3 x log 3 45log 3 3x log 3 453x 45x 15xy 2 x 2 O16 xx48. log 300,000 log (100,000 3) log 100,000 log 3 5 0.4771 5.477149. log 0.0003 log (0.0001 3) log 0.0001 log 34 0.47713.522950. log 140 log (10 14) log 10 log 14 1 1.1461 2.146151. log 0.014 log (0.001 14) log 0.001 log 143 1.14611.853952. 4x 6 x2x log 4 (x 2) log 6x log 4 x log 6 2 log 6x log 4 x log 6 2 log 6x(log 4 log 6) 2 log 62 log6x log 4 log 6x 8.8453. 12 0.5x 8 0.1x40.5x log 12 (0.1x 4) log 80.5x log 12 0.1x log 8 4 log 80.5x log 12 0.1x log 8 4 log 8x(0.5 log 12 0.1 log 8) 4 log 8x 4 log 80.5 log 12 0.1 log 8x 8.04399 Chapter 11
54. 1 4 3x 6 x23x log 1 4 (x 2) log 63x log 1 4 x log 6 2 log 63x log 1 4 x log 6 2 log 6x(3 log 1 4 log 6) 2 log 6Ox 2 log 63 log 1 4 log 6Change the inequality because 3 log 1 4 log 6 isnegative.x 0.655. 0 1 2x8 7 x4(2x 8) log 0.1 (x 4) log 72x log 0.1 8 log 0.1 x log 7 4 log 72x log 0.1 x log 7 4 log 7 8 log 0.1x(2 log 0.1 log 7) 4 log 7 8 log 0.1x x 456. log (2x 3) log (3 x)log (2x 3) log (3 x) 1(2x 3) (3 x) 1(2x 3)(3 x) 12x 2 3x 8 0x 3 73 4 1.39, 2.8957. yy 3 log (x 2)x4 log 7 7 log 0.12 log 0.1 log 7g 100og12562. log 4 100 lo log963. log 15 125 l log15 2.0959 1.782964. 4x 100x ln 4 ln 100x ln 100ln 4x 3.321965. 6 x2 30(x 2) ln 6 ln 30x 2 l n 30ln6n 30x l ln6 2x 3.898266. 3 x1 4 2x(x 1) ln 3 2x ln 4x ln 3 ln 3 2x ln 4x ln 3 2x ln 4 ln 3x(ln 3 2 ln 4) ln 3ln3x ln 3 2 ln 4x 0.656367. 9 4x 5 x44x ln 9 (x 4) ln 54x ln 9 x ln 5 4 ln 54x ln 9 x ln 5 4 ln 5x(4 ln 9 ln 5) 4 ln 54 ln5x 4ln 9 ln5x 0.896768. 24 e 2x 69. 15e x 200ln 24 2x ex 2 001 5x ln 242x ln 2 0015x 1.5890 x 2.590370. x 3.33358.yOy 7 x 2x[5, 5] scl:1 by [10, 60] scl:1071. x 2.2059. x 3.42[1, 5] scl:1 by [1, 10] scl:1[5, 5] scl:1 by [5, 5] scl:1og15og2460. log 4 15 l log461. log 8 24 l log8 1.9534 1.5283Chapter 11 400
ln2ln 272. t 0 . 02873. t 0.0 51 24.76 13.5274. 18 ln 2kk ln 218 0.0385; 3.85%Page 75375. 0.065 1 2 tlog 0.65 t log 1 2 log 0.65log 1 2 t25Applications and Problem solving0.6215 t0.6215 5730 3561.13 or 3561 yr.76a. 10 log 1.15 10 20.676b. 10 log 9 10 39.5 10 10121092 10 376c. 10 log 8.95 1012 99.577. 200,000 142,000e 0.014t 1 0071ln 1 0071ln 1 0071 0.014 e 0.014t 0.014t t20.6 dB39.5 dB99.5 dB24.4 t1990 24 201478a. N 65 30e 0.20(2) 44.89; 45 words per minute78b. N 65 30e 0.20(15) 63.50; 64 words per minute78c. 50 65 30e 0.20t 1 2 ln 1 2 ln 1 2 0.20Page 753 e 0.20t0.20t t3.47 t; 3.5 weeksOpen-Ended Assessment1. Sample answer: (n 4 ) 1 4 (4m) 12. Sample answer:log 2 log(x 2) 1 2 log 36Chapter SAT & ACT PreparationPage 755 SAT and ACT Practice1. To find the greatest possible value, the other 3values must be as small as possible. Since theyare distinct positive integers, they must be 1, 2,and 3. The sum of all 4 integers is 4(11) or 44. Thesum of the 3 smallest is 1 2 3 or 6, so thefourth integer cannot be more than 44 6 or 38.The correct choice is B.2. Since one root is 1 2 , x 1 2 , 2x 1, and 2x 0.Similarly for the root that is 1 3 ,x 1 3 , 3x 1, and 3x 1 0.To find the quadratic equation, multiply these twofactors and let the product equal zero.(2x 1)(3x 1) 06x 2 5x 1 0The correct choice is E.3. The result of dividing T by 6 is 14 less than thecorrect average. T 6 correct answer 14 T 6 14 correct averageThe correct average is the total divided by thenumber of scores, 5.correct average T 5 T 6 14 T 5 The correct choice is E.4. yA x 2tan A y 2x 2 yxFind the area of ABC.A 1 2 bh 1 2 xy x y2Simplify the ratio. area x y of ABC2tanA 2 yxB (x , y)2 2xy x y2The correct choice is E.5. x y 1 02y2yx 10y2x 10x 5The correct choice is B.yC(x, 0) x 4x2401 Chapter 11
6. C D9. A is the arithmetic mean of three consecutive 2 x (x 2) (x 4)3 Dpositive even integers, so A 3D 2 3 and, therefore, r 1 3 . 3x 6 x 2, where x is a positive even integer.3Now use this value for the radius to calculate half Then A is also a positive even integer. Since A isof the area.even, when A is divided by 6, the remainder must 1 2 A 1 2 r 2 1 2 1 3 2 1 2 1 9 1 8 also be an even integer. The possible evenremainders are 0, 2, and 4.The correct choice is A.The correct choice is C.7. The average of 8 numbers is 20.10. First notice that b must be a prime integer. Nextsum of eight numbers20 notice that 3b is greater than 10. So b could be 5,8since 3(5) 15. (b cannot be 3.) Check to be surethat 5 fits the rest of the inequality.sum of eight numbers 160The average of 5 of the numbers is 14.3(5) 15 5 6 (5) 2 5 4 1 6 6 sum of five numbersSo 5 is one possible answer. You can check to see14 5that 7 and 11 are also valid answers.sum of five numbers The correct answer is 5, 7, or 11.70The sum of the other three numbers must be160 70 or 90. Calculate the average of thesethree numbers.sum of three numbersaverage 9 33 30The correct choice is D.8. The sum of the angles in a triangle is 180°. Since∠B is a right angle, it is 90°. So the sum of theother two angles is 90°. Write and solve anequation using the expressions for the two angles.2x 3x 905x 90x 18The question asks for the measure of ∠A.∠A 2x 2(18) 36The correct choice is C.0Chapter 11 402
Chapter 12 Sequences and Series12-1Arithmetic Sequences and SeriesPages 762–763 Check for Understanding1. a 1 6 4(1) or 2a 2 6 4(2) or 2a 3 6 4(3) or 6a 4 6 4(4) or 10a 5 6 4(5) or 142, 2, 6, 10, 14; yes, there is a commondifference of 4.2a.a n1O1231 2 3 4 5 6n2b. linear2c. The common difference is 1. This is the slope ofthe line through the points of the sequence.3a. The number of houses sold cannot be negative.3b. S n n 2 [2a, (n 1)d]S 10 1 02[2 3750 (10 1)500] $60,0004. Negative; let n and n 1 be two consecutivenumbers in the sequence.d (n 1) n or 15. Neither student is correct, since neither sequencehas a common difference. The differencefluctuates between 1 and 1. The second sequencehas a difference that fluctuates between 2 and 2.6. d 11 6 or 516 5 21, 21 5 26, 26 5 31, 31 5 3621, 26, 31, 367. d 7 (15) or 81 8 9, 9 8 17, 17 8 25, 25 8 339, 17, 25, 338. d (a 2) (a 6) a a 2 6 or 4a 2 4 a 6, a 6 4 a 10,a 10 4 a 14, a 14 4 4 a 18a 6, a 10, a 14, a 189. a n a 1 (n 1)da 17 10 (17 1)(3)3810. a n a 1 (n 1)d37 13 (n 1)550 5(n 1)10 n 111 n11. a n a 1 (n 1)d3 a 1 (7 1)(2)3 a 1 1215 a 112. a n a 1 (n 1)d34 100 (12 1)d66 11d6 d13. a n a 1 (n 1)d24 9 (4 1)d15 3d5 d9 5 14, 14 5 199, 14, 19, 2414. S n n 2 [2a 1 (n 1)d]S 35 3 52[2.7 (35 1) 2] 143515. S n n 2 [2a 1 (n 1)d]210 n 2 [2 30 (n 1)(4)]420 60n 4n 2 4n4n 2 64n 420 04(n 21)(n 5) 0n 21 or n 5Since n cannot be negative, n 21.16. n 19, a 19 27, d 1a n a 1 (n 1)d27 a 1 (19 1)19 a 1S 19 n 2 (a 1 a 19 ) 1 92(9 27) 342 seatsPages 763–765Exercises17. d 1 5 or 67 (6) 13, 13 (6) 19,19 (6) 25, 25 (6) 3113, 19, 25, 3118. d 7 (18) 114 11 15, 15 11 26, 26 11 37, 37 11 4815, 26, 37, 4819. d 4.5 3 or 1.56 1.5 7.5, 7.5 1.5 9, 9 1.5 10.5,10.5 1.5 127.5, 9, 10.5, 1220. d 3.8 5.6 or 1.82 (1.8) 0.2, 0.2 (1.8) 1.6,1.6 (1.8) 3.4, 3.4 (1.8) 5.20.2, 1.6, 3.4, 5.221. d b 4 b or 4b 8 4 b 12, b 12 4 b 16,b 16 4 b 20, b 20 4 b 24b 12, b 16, b 20, b 2422. d 0 (x) or xx x 2x, 2x x 3x, 3x x 4x, 4x x 5x2x, 3x, 4x, 5x403 Chapter 12
23. d n 5n or 6x7n (6n) 13n 13n (6n) 19n,19n (6n) 25n, 25n (6n) 31n13n, 19n, 25n, 31n24. d 5 (5 k) or k5 k (k) 5 2k, 5 2k (k) 5 3k,5 3k (k) 5 4k, 5 4k (k) 5 5k5 2k, 5 3k, 5 4k, 5 5k25. d (2a 2) (2a 5) or 72a 9 7 2a 16, 2a 16 7 2a 23,2a 23 7 2a 30, 2a 30 7 2a 372a 16, 2a 23, 2a 30, 2a 3726. d 5 (3 7) or 2 77 7 2 7 9 27,9 27 2 7 11 37, 11 37 2 7 13 479 27, 11 37, 13 4727. a 25 8 (25 1)3 8028. a 18 1.4 (18 1)(0.5) 9.929. 41 19 (n 1)(5)60 5(n 1)12 n 113 n30. 138 2 (n 1)7140 7(n 1)20 n 121 n31. 38 a 1 (15 1)(3)38 a 1 4280 a 132. 10 2 3 a 1 (7 1) 1 3 10 2 3 a 1 28 2 3 a 133. 58 6 (14 1)d52 13d4 d34. 26 8 (11 1)d18 10d1 4 5 d35. d (1 5) (4 5) or 3a 8 4 5 (8 1)3 17 536. d 6 (5 i) 1 ia 12 5 i (12 1)(1 i) 5 i 11 11i 16 10i37. d 10.5 12.2 or 1.7a 33 12.2 (33 1)(1.7)42.238. d 4 (7) or 3a 79 7 (79 1)3 22739. 21 12 (3 1)d9 2d4.5 d12 4.5 16.512, 16.5, 2140. 4 5 (4 1)d9 3d3 d5 3 2, 2 3 15, 2, 1, 441. 12 3 (4 1)d12 3 3d12 3 3 d3 12 33 12 23,3 12 233 12 33 24 333, 12 2 3, 24 33,31242. 5 2 (5 1)d3 4d0.75 d2 0.75 2.75, 2.75 0.75 3.5,3.5 0.75 4.252, 2.75, 3.5, 4.25, 543. d 1 3 2 or 1 2 a 11 3 2 (11 1) 1 2 3.5S 11 1 21 3 2 3 1 2 1144. d 4.8 (5) or 0.2a 100 5 (100 1)0.2 14.8S 100 10 02(5 14.8) 49045. d 13 (19) or 6a 26 19 (26 1)6 131S 26 2 62(19 131) 145646. 14 n 2 [2(7) (n 1)1.5]28 14n 1.5n 2 1.5n0 1.5n 2 15.5n 28n 15.5 (15. 5) 4(1.5)2 (28)2(1.5)n 8 or n 2 1 3 Since there cannot be a fractional number ofterms, n 8.47. 31.5 n 2 [2(3) (n 1)2.5]63 6n 2.5n 2 25n0 2.5n 2 8.5n 638.5 (8.5 ) 2.5)(n 2 4( 63) 2(2.5)n 7 or n 3.6Since n cannot be negative, n 7.48. d 7 5 or 2a n 5 (n 1)2 2n 349. d 2 6 or 8a n 6 (n 1)(8)8n 14Chapter 12 404
50. 9:00, 9:30, 10:00, 10:30, 11:00, 11:30, 12:00n 7, d 2, a 1 3a 7 3 (7 1)2 15 data items per minute51. Let d be the common difference. Then, y x d,z x 2d, and w x 3d. Substitute thesevalues into the expression x w y and simplify.x (x 3d) (x d) x 2d or z.52. a 1 5, d 4, n 25a 25 5 (25 1)4 101S 25 2 52(5 101) 1325 bricks53. S n n 2 (128° 172°); S n (n 2)180°150n 180n 36030n 360n 1254a. S 4 (4 2)180° or 360°S 5 (5 2)180° or 540°S 6 (6 2)180° or 720°S 7 (7 2)180° or 900°360°, 540°, 720°, 900°54b. The common difference between eachconsecutive term in the sequence is 180,therefore the sequence is arithmetic.54c. a 35 180 (35 1)180 5940°55a. a 1 1, d 2S 5 5 2 [2(1) (5 1)2] 2555b. S 10 1 02[2(1) (10 1)2] 10055c. Conjecture: The sum of the first n terms of thesequence of natural numbers is n 2 .Proof:Let a n 2n 1. The first term of the sequenceof natural numbers is 1, so a 1 1.Then, using the formula for the sum of anarithmetic series,S n n 2 (a 1 a n )S n n 2 [1 (2n 1)] n 2 (2n) or n 256. a 1 5, d 7, n 15S 15 1 52[2(5) (15 1)7] 810 feet57. n 10, S 10 5510, d 1005510 1 02[2a 1 (10 1)100]5510 10a 1 45001010 10a 1101 a 1a 10 101 (10 1)100 1001least: $101, greatest: $100158. S n a 1 a 2 (a 31 a 32 ) (a 41 a 42 ) (a 51 a 52 ) a 1 a 2 a 2 a 1 a 3 a 2 a 4 a 3 a 1 a 2 a 2 a 1 (a 2 a 1 ) a 2 a 3 a 2 a 3 059. A Pe rt 100e 0.07(15) $285.7760. 4x 2 25y 2 250y 525 04x 2 25(y 2 10y) 5254x 2 25(y 5) 2 100x2 2 5 (y 5) 24 1h 0, k 5, a 5, b 2, c 21center: (0, 5)foci: (21, 5)vertices: major → (5, 5)minor → (0, 3) and (0, 7)y661. r 1 2or 0.5v 5 8 2 or 8 0.5cos 8 i sin 8 0.46 0.19i62. 2, 1, 3 5, 3, 0 2(5) (1)(3) (3)(0) 763. x cos 30° y sin 30° 5 0 32x 1 2 y 5 03x y 10 064. y246(5, 5)OO(0, 3)(0, 5)(0, 7)2 x65. Find A.A 90° 19° 32 70° 28Find a.acos 19° 32 4 .5 4.2 aFind b.bsin 19° 32 4 .5 1.5 b66. discriminant (3) 2 4(4)(2)23Since the discriminant is negative, this indicatestwo imaginary roots.x(5, 5)405 Chapter 12
67. x 1x 3x 2 4x 2(x 2 3x)x 2(x 3)1y x 10 1 2 0 1 11 0 1 3 4 2A(1, 2), B(3, 0), C(4, 1)69. a 4b 15 → a 15 4b4a b 154(15 4b) b 1560 16b b 1515b 45b 34a (3) 154a 12a 3a b 3 (3) or 6The correct choice is C.68. 3 412-21Geometric Sequences and SeriesPage 771 Check for Understanding1. Both arithmetic and geometric sequences arerecursive. Each term of an arithmetic sequence isthe sum of a fixed difference and the previousterm. Each term of a geometric sequence is theproduct of a common ratio and the previous term.2. a n (3) 11 or 9a 2 (3) 21 or 27a 3 (3) 31 or 81The expression generates the following sequence:9, 27, 81, . The common ratio is 3, thereforeit is a geometric sequence.3. If the first term in a geometric sequence werezero, then finding the common ratio would meandividing by zero. Division by zero is undefined.4. Sample answer: The first term of the series5 10 20 is 5 and the sum of the first 6terms of the sequence is 105, but 105 is notgreater than 5.5a. No; the ratio between the first two terms is 2,but the ratio between the next two terms is 3.5b. Yes; the common ratio is 3.5c. Yes; the common ratio is x.6a.Beginning of Value ofYear Computers1 27,500.002 15,125.003 8318.754 4575.315 2516.426 1384.03064b.Value27,50025,00022,50020,00017,50015,00012,50010,0007500500025006c. an exponential function47. r — or 6 2 3 24(6) 144, 144(6) 864, 864(6) 5184144, 864, 51848. r 3 2 9 2 3 2 2 74, 2 74 2 74, 8 18, 2 43167.2.800 1 2 3Years 3 2 8 81, 8 18 3 2 2 14369. r 1or 428.8(4) 115.2, 115.2(4) 460.8,460.8(4) 1843.2115.2, 460.8, 1843.210. r 2 .1 7 or 0.3a n a 1 r n1a 7 7(0.3) 71 0.00510311. a n a 1 r n124 a 1 (2) 5124 16a 1 3 2 a 112. a 3 2 .5 2 or 1.25a 2 1. 25 2 or 0.625a 1 0.8 252or 0.31250.3125, 0.625, 1.2513. a n a 1 r n127 1(r) 4127 r 33 r1(3) 3, 3(3) 91, 3, 9, 27114. r 0 . 5or 2S n a 1 a1 rn1 rS 9 0.5 0.5( 2) 91 ( 2) 85.54 5 6Chapter 12 406
15. r 1.035The value of the car after 10, 20, and 40 years willbe the 11th, 21st, and 41st terms of the sequencerespectively.a 11 20,000(1.035) 111 $28,211.98a 21 20,000(1.035) 211 $39,795.78a 42 20,000(1.035) 411 $79,185.1926. a 5 8 3 2 51 8 12 3 8 27. r — or 3 4 1 2 b5, 2 ba 7, 3 a 9 4 7 , 2, 7a 6 1 2 3 4 612432 04828. r 0 . 44.0or 0.01a 7 40(0.01) 71Pages 771–773 Exercises 4 10 11216. r 1 0or 0.229. r 10 or 250.4(0.2) 0.08, 0.08(0.2) 0.016,a0.016(0.2) 0.00329 52 910.08, 0.016, 0.003217. r 20 8 or 2.550(2.5) 125, 125(2.5) 312.5,312.5(2.5) 781.2530. 165192 a 1 (4) 61192 1024a 10.1875 a 1125, 312.5, 781.2531. 322 a 1 (2) 51 2 322 4a 13 82 a18. r or 31 2 9 32. 6 a 1 1 3 5112(3) 6, 6(3) 18, 18(3) 546 8a1 16, 18, 54486 a 13——1 019. r — or 2 a 2 486 15 3 or 162— 3 4 —a 3 162 1 3 or 543 2 5 2 5 6 1 25 6, 1 25 2 5 1212 6 2, 5 6 25 2 5 24 3 125 486, 162, 546 1 25 1224, 6 2, 5 3 125 33. 0.32 a 1 (0.2) 510.32 0.0016a20. r 3 . 517or 0.5200 a 11.75(0.5) 0.875, 0.875(0.5) 0.4375,a 2 200(0.2) or 400.4375(0.5) 0.21875a 3 40(0.2) or 80.875, 0.4375, 0.21875200, 40, 8634. 81 256r 5121. r —— or 2813 2 2 56 r622 12, 122 122, 1222 24 3 4 r12, 122, 2422. r 3 39or 3256 34 192, 192 3 4 144,33 33 3, 3 33 1, 1 33 3144 3 4 10833, 1, 3256, 192, 144, 108, 81323. r 35. 54 2r 411 1i or i27 r 31i(i) 1, 1(i) i, i(i) 13 r1, i, 12(3) 6, 6(3) 1824. r t 5t8 or t 32, 6, 18, 54t 2 (t 3 ) t 1 , t 1 (t 3 ) t 4 , t 4 (t 3 ) t 736. 7 4 7 r 31t 1 , t 4 , t 7 4 94 r 2a25. b 2 b a 2 1 a1b, a b b a 2 1 1 a 3, a 3 b a 2 ab5 , 7 2 rb a 5 ab2 b 2 ba 7, 2 a 7 ba2 b 3 a 9 4 7 7 2 21 1 b a, b a 3, a407 Chapter 12
37. r —5or 343a. r 5 5. 50 3 5 or 1.1S 5 38. r 1 36 5 3 5 3 (3) 51 3 60 535 or 0.25(0.2)6 0.2 81.2448 1S 6 65 6 1 3 2 39. r or 3 2 S 10 1 1 3 2 10——1 3 2 1 1,60551240. r 2 32or 3S 8 2 2(3) 8a 10 5(1.1) 101 $11.79a 20 5(1.1) 201 $30.58a 10 5(1.1) 401 $205.72$11.79, $30.58, $205.725(1.1) 5243b. S 52 5 1 1.1 $7052.1543c. Each payment made is rounded to the nearestpenny, so the sum of the payments will actuallybe more than the sum found in b. 1 5 1 8 44a. 1 x 2 1 380 1 x 8 013 x1 2 1 z 1 344b. 1 8 0160 2 1 3 1 4 1 2 0 1 160z 1 1 3 31 3 1 5 1 z 160(1 3)25 z 80(1 3)45. a 2 3(a 1 ) by definition41a. The population doubles every half-hour, so r 2.After 1 hour, the number of bacteria is the third3(2) 6r a 2term in the sequence and n 1 2. After 2 hours, 3a1it is the fifth term and n 1 4. After 3 hours,Then a n a 1 rit is the seventh term and n 1 6. After tn1So, a n (2)(3)hours it is the 2t 1 term and n 1 2t.n1b 46. a 1 1, r 2.5, n 15t b 0 2 2ta 15 1(2.5)41b. b 151t 30 2 2(5) 372,529 30,72047a. 251 0. 02441c. Sample answer: It is assumed that favorable12 $25.05conditions are maintained for the growth of thebacteria, such as an adequate food and oxygensupply, appropriate surrounding temperature,25.05 25.051 0. 024and adequate room for growth.12S 24 42a. a1 1 0. 0247 a 4 r 31212 4r 33 r 3 $615.23 3 3 ra42b. a 4 a 1 r 411 a 1 1 0. 024122447c. 750 4 a 1 3 3 34 3a 1 4 3 a 1a 28 4 3 3 3 281 26,24447b. No; at the end of two years, she will have only$615.23 in her account. 241.5 a 1 1 1 0. 1$30.54 a 10241 1 0 .24120242 24a 0 1 0. 12 $30.54a 0 $30.48The least monthly deposit is $30.48.Chapter 12 408
2112 7—18 148. r a ar 3a n a 1 r n116561 8 1 3 n1(6561)(81) 3 n1(3 8 )(3 4 ) 3 n13 12 3 n112 n 113 n6561 is the 13th term of the sequence.49. S n n 2 [2a 1 (n 1)d]650 n 2 [2(20) (n 1)5]1300 40n 5n 2 5n0 5n 2 35n 13000 5(n 13)(n 20)n 13 or n 20Since n cannot be negative, n 13 weeks.50. log 11 265 lo g 265log11 2.326951.yOx55. Find the amplitude.A 86 362or 25Find h.h 86 362or 61Find k. 4 2 kk 2 y 25 sin 2 t c 6136 25 sin 2 1 c 611 sin 2 c 2 2 c cy 25 sin 2 t 3.14 6156. Since 43° 90°, consider the following.b sin A 20 sin 43° 13.64Since 11 13.64, no triangle exists.57. (n 2 ) 49n 2 7<strong>Solution</strong> set: {n⏐n 3 or n 3}n m n 1 nm5 5 1 or 4 (5)(4) or 204 4 1 or 3 (4)(3) or 123 3 1 or 2 (3)(2) or 63 3 1 or 4 3(4) or 124 4 1 or 5 4(5) or 205 5 1 or 6 5(6) or 30← leastpossiblevalue52. AB 2 2 35 2 ( 2 )34Since C is positive, use 34. x 3 34 5 5y 34 0345p or 5 34353434, cos f , sin f 345 34tan f 3 3434 tan f 5 3 f 59°Since cosine is negative and sine is positive, 59° 180° or 121°.p r cos(v f)5 3434 r cos(v 121°)53. 3x 4y 5y 3 4 x 5 4 x t, y 3 4 t 5 4 54. csc v 3 si1n v 3 1 3 sin vThe answer is 6.12-3 Infinite Sequences and SeriesPages 780–781 Check for Understanding1a.a n1.00.5O 1 2 3 4 5 6 7 8 9 10n1b. The value of a n approaches 1 as the value of nincreases. n 1n→n1c. lim 1409 Chapter 12
n 1n→n1d. lim limn→ n n 1 0 1 lim n→ n1 The limits are equal.2a. See students’ work. Student’s should draw thefollowing conclusions:limn→ 1 2 n 0limn→ 1 4 n 0limn→ (1)n 1limn→ (2)n no limitlimn→ (5)n no limit2b. If ⏐r⏐ 1, then lim r n 0. If ⏐r⏐ 1, thenn→limn→ rn 1. If ⏐r⏐ 1 then lim r n does not exist.n→3. Sample answer: 2 4 8 4. Zonta is correct. As n approaches infinity theexpression 2n 3 will continue to grow larger andlarger. Tyree applied the method of dividing by thehighest powered term incorrectly. Both thenumerator and the denominator of the expressionmust be divided by the highest-powered term. It isnot appropriate to apply this method here sincethe denominator of the expression 2n 3 is 1.5. 0; as n → , 5 n becomes increasingly large and1thus the value 5 n becomes smaller and smaller,approaching zero. So the sequence has a limit ofzero.n6. lim 25 limn→ 5 2nn→ 2n 1 2 n5lim n→ 2 n lim 5n→2 1 n 5 2 0 0As n approaches infinity, 1 2 n becomes increasinglylarge, so the sequence has no limit.7. 3 7 6, lim lim 3nn→7nn→ 3 7 6 7 n1 lim n→ 3 7 3 7 6 7 0 or 3 7 7 78. 0.7 1 0 1 00 7 1a 1 1,0r 1 0S n 7 9 26126000 1,0 00,0002610,00r 10 00 1261 000——11 10 00 5 1 26999 5 1 411 9. 5.126 5 11a 1 1171 0—11 1 0S n 5 lim 6n→7 1 lim n→n 310. r 6or 1 2 S n 6——1 1 2 411. r or 1 3 1 4 — 3 4 S n 3 4 —1 1 3 1 1 8 312. r or 3 3The sum does not exist since ⏐r⏐ ⏐3⏐ 1.13. a 1 75, r 2 5 S n 125 mPages 781–783 7 2nn→5n14. lim n3 2n→n215. lim limExercisesn→ 7 5 n1 2 5 lim 7n→5 1 lim n→n lim 2n→5 7 5 0 2 5 or 2 5 2 limn .n→n2 1lim n→ n lim 2 n→n 2 0 or 0, but as n approachesinfinity, n becomes increasingly large, so thesequence has no limit.16. lim 6n 2 5n→3n2 lim n→ 6 3 5 3 n 2 lim 2 lim 5n→ n→3 1 lim n→n 2 9n3 5n 2n→2n317. lim (3n 4 )(1 n)n→n218. lim 2 5 3 0 or 2 lim n→ 9 2 5 2 n 2 n13 lim lim 9n→2 lim 5n→2 1 1 lim n→n 2 lim n→n 3 9 2 5 2 0 0 or 9 2 3n2 n 4n→n2 lim n→3 n1 n42 1 lim(3) lim n→ n→n lim 4 limn→3 0 4 0 or 319. Dividing by the highest powered term, n 2 , we findlimn→751 2 5 8 n5 n22 —— n32 n2 which as n approaches infinity0 0simplifies to 8 0 0 8 0 . Since this fraction isundefined, the limit does not exist.1n→n 2Chapter 12 410
420. lim n→23n nn2 3 3n2 5 lim n→ lim n→lim 4 lim 3 lim 3 lim 2 limn→ n→lim 2 lim 3 lim1 1 n→ n→ n→ n lim 5 lim n→ n→ n 3 4 0 3 0 02 3 0 5 0or 021. As n → , 3 n becomes increasingly large and thus1the value 3 n becomes smaller and smaller,approaching zero. So the sequence has a limit ofzero.22. Dividing by the highest powered term, n, we find ( 2) nnlim lim( 2)—n 14 n n→ n→4n 1 n 1 which as n approaches infinity1lim 1, but lim ( 2)nn→n→nhas no limit since4 n 1o2o 1.( 1)23. limnn1) 2 lim 2 limn2n→ 5n n 5 (n→n n→n ( 1) nn→n2 ( 1) nn→n25 lim n→n lim limAs n increases, the value of the numeratoralternates between 1 and 1. As n approachesinfinity, the value of the denominator becomesincreasingly large, causing the value of thefraction to become increasingly small. Thus, theterms of the sequence alternate between smallerand smaller positive and negative values,approaching zero. So the sequence has a limit ofzero.4 424. 0.4 1 0 1 00 …4 1a 1 1,0r 1 041 0S n —11 1 025. 0.51 15 4 9 1 5100 10 ,0001 10,0r 1 00 511 00a 1 15S n … 5 199 or 1 73370370000 1,0 00,0007010,00r 10 003701 00026. 0.370 13a 1 13—1 1100 S n ——11 1000 3 70999or 1 027 4 n 3 3 nn3 n 2n3 2 nn 33 3 n2n 3 n53 n43 n32 n1 2 n3 n53 n→ n1 …n→ n1n→ n1 227. 6.259 6 12591a 1 1, r 10000S n 6 ——11 1028. 0.15 1100 6 2 599997 62 75 1500 10 ,0005 10,0r 1 00 151 00a 1 11S n —1 1159259000 1,0 00,000002591 000 …00 1 599 5or 3 3630 1 000 63 100 , 0003000 1, r 1 00 63 1 000 —11 1 00 1 5 63 9 9029 1 1029. 0.263 12a 1 16S n 1 5 … …30. The series is geometric, having a common ratio of0.1. Since this ratio is less than 1, the sum of theseries exists and is 2 9 .231. r 1 16or 3 4 S n 161 3 4 6432. r 7 .5 5 or 1.5This series is geometric with a common ratio of1.5. Since this ratio is greater than 1, the sum ofthe series does not exist.533. r 1 0or 1 2 S n 101 1 2 2034. The series is arithmetic, having a general term of7 n. Since lim 7 n does not equal zero, thisn→series has no sum. 1 4 35. r — or 2 1 8 This series is geometric with a common ratio of 2.Since this ratio is greater than 1, the sum of theseries does not exist.411 Chapter 12
1 9 36. r — or 1 6 S n 2 3 2 3 ——1 1 6 4 7 4 5 37. r — or 2 3 6 5 S n 3 3 5 6 5 —1 2 3 138. r or 55 5S n 51 555 1 5 5 —1 551 555 151 2 5 4 5 (5 1)39. r 4 38or 328S n 1 32 1 382 ——1 31 322 81 32 ——1 3 4 321 32 32 16340a. a 1 35 r 2 5 a 2 35 2 5 or 14a 3 14a 4 14 2 5 or 5.6a 5 5.635, 14, 14, 5.6, 5.61440b. S n 35 141 2 1 2 5 5 35 2 03 7 03 81 2 3 m or about 82 m41a. The limit of a difference equals the difference ofthe limits only if the two limits exist. Sincen2n2neither lim n→ 2n 1nor lim n→ 2n 1exists, thisproperty of limits does not apply.41b. limn→ n2n22n 1 2n 1 lim n→ lim n→ lim n→ 4n 2 lim n→n 2 (2n 1) n 2 (2n 1)(2n 1)(2n 1)2n 3 n 2 2n 3 n24n 2 12n 2 1 2n 2n2— 4 n21n2 n 22 lim n→ 4 n12 1 2 42a. 12 4 3r 3 3a 4 a 1 r 414 a 1 3 3 34 a 1 (3) 4 3 a 142b. a 28 a 1 r 281 4 3 3 3 27 4 3 (3 9 ) 4(3 8 ) 26,24443. No; if n is even, lim cos n n→2 1 2 , but if n is odd,lim cos n 2 1 2 .n→44a. After 2 hours, 1 2 D exists. After 4 hours, 1 2 1 2 D or 1 4 D exists. After 6 hours and before the seconddose, 1 2 1 2 1 2 D or 1 8 D exists.44b. a 1 D, r 1 8 1 1 8 n S n D ——1 1 8 8 7 D1 1 8 n 44c. lim S n Sn→a1S 1 8 7 rD 1 1 8 D44d. 350 8 7 D306.25 DThe largest possible dose is 306.25 mg.45a. Aside of the original square measures 2 04or5 feet. Half of 5 feet is 5 2 feet. 5 2 2 5 2 2 s 2 5 04 s 25 22 sPerimeter 4 5 22or 102 feet.Chapter 12 412
45b. a 1 20, r 10 220or 22S 201 221 2202 ——1 21 222 20 102 1 2 4 40 202 ft or about 68 ft46a.n 17.3032181(0.864605) n1 15.02 12.93 11.24 9.75 8.46 7.27 6.28 5.49 4.710 4.015.0, 12.9, 11.2, 9.7, 8.4, 7.2, 6.2, 5.4, 4.7, 4.046b. The 2000–2001 school year corresponds to the9 th term of the sequence, 4.7. The model is 0.3below the actual statistic.46c. The 2006–2007 school year would correspond tothe 15 th term of the sequence.17.3032181(0.864605) 15 2.046d. Yes; as n → , 17.3032181(0.864605) n → 0.46e. No, the number of students per computer mustbe greater than zero.47. 3 2 3 2, 2 2 3 1 1 3 , 1 1 3 2 3 8 9 , 8 9 2 3 1 627 2, 1 1 3 , 8 9 , 1 2748. a 16 1.5 (16 1)0.5 949. x 2 4y 2 12x 16y 16(x 6) 2 4(y 2) 2 450.6 (x 6) 24 (y 2) 21 1h 6, k 2, a 2, b 1, c 5center: (6, 2)foci: (6 5, 2)vertices: (8, 2) and (4, 2)562323651. v y 125 sin 20° 42.75 milesv x 125 cos 20° 117.46 miles52. cos 112.5° cos 22 253.180˚5° 1 cos225 °2 1 222 2 2 422 125 miles20˚54. possible values of p: 1, 2possible values of q: 1, 2, 4, 8possible rational zeros, p q : 1, 2, 1 2 , 1 4 , 1 8 55. If b 1, then 4b 26 30 which is divisible by 2,5, and 6.If b 11, then 4b 26 70, which is divisibleby 7.4b 26 is not divisible by 4 since 4 divides 4bevenly, but does not divide 26 evenly. The correctchoice is B.12-3BGraphing Calculator Exploration:Continued FractionsPage 7841. 1.6181818182. N 243. x 1 1 x x 2 x 1x 2 x 1 0y1(1) (1)2 )(1)4(1x 2(1)x 1 52Since the sum 1 1O2180˚111 1 360˚cannot be1negative, the value of 1 is 1 5 .211 1 2 4 6 8076433253116413 Chapter 12
14. Let x 3 , then x 3 1 x .11. Answers will vary. Sample answers: A 1, B 14;3 1 3 A 4, B 1.A 2 B: If A 1 and B 14, 14 2 1 15.Solve for x.x 3 1 x x 2 3x 1x 2 3x 1 0)(1)4(12(1) 3 132x 3 (3)2Since the sum of positive numbers must remainpositive, x 3 13.25. The output and the decimal approximation areequal.6. Let x A , then x A 1 x .1A A1 Solve for x.x A 1 x x 2 Ax 1x 2 Ax 1 0)(1) 4(12(1)x A (A)2A A42 2Since the sum of positive numbers must remainA A4positive, x 2 27. Sample program:Program: CCFRAC:Prompt A :Prompt B:Disp “INPUT TERM”:Disp “NUMBER N, N 3”:Prompt N:1 → K:B 1/(2A) → C:Lbl 1:B 1/(2A C) → C:K 1 → K:If K N 1:Then: Goto 1:Else: Disp C A8. For large values of N, the program output and thedecimal approximation of AB 2 are equal.9. x A 2A BB2A 2A Bx A 2A x A(x A) 2 2A(x A) Bx 2 2Ax A 2 2Ax 2A 2 Bx 2 A 2 Bx AB2 Since the sum A B cannot beB2A 2A Bnegative, the value of A isB2A 2A A B. 2 10. They will be opposites.Chapter 12 41412-4If A 4 and B 1, 41)2 ( 15.Convergent and Divergent SeriesPages 790–791 Check for Understanding1a. See students’ work.1b. See students’ work.1c. See students’ work.1d. In a given trigonometric series where ⏐r⏐ 1,each succeeding term is larger than the onepreceding it. Therefore, the series approaches and thus does not converge.2. As n → , S → 6.3a.s n1.61.41.210.80.60.40.2O3b. convergent3c. n 23 n3d. We can use the ratio test to determine whetherthe series is indeed convergent.a n n 23 and n an1 (n 1) 3n21r lim n→ (n 1) 2 3n→3n 1 n n 2r limr lim n2 n→3r lim n→1 2 3 4 5 6 7 8 (n 1) 3n21— n 3 n 22n 1n21 n2 n1 3r 1 3 Since r 1, the series is convergent.4. Consider the infinite series a n1. If the lim a n 0, the sum of the series does notn→exist, and thus the series is divergent. If thelim a n 0, the sum may or may not exit andn→therefore it cannot be determined from this testif the series is convergent or divergent.2. If the series is arithmetic then it is divergent.3. If the series is geometric then the seriesconverges for ⏐r⏐ 1 and diverges for ⏐r⏐ 1.4. Ratio test: the series converges for r 1 anddiverges for r 1. If r 1, the test fails. Thistest can only be used if all the terms of then
series are positive and if the series can beexpressed in general form.5. Comparison test: may only be used if all theterms in the series are positive.n5. a n 2 n, a n1 n 12 n 1 a n1ann→ n 12 n 1 nn→ 2 nr lim lim (n 1)2n→n2 n n1 (n 1)n→2nn lim lim lim n→ 1 2 21 1 2 convergent6. a n 4n , 14na 4(n n1 4( 4(n 1) 14( n 1)——n→ 4n 14nn 3)(4n) 4)( 4n 1)16n 2 12nn→ 16 n2 12n 4 16 nn 22 1 2nn 2r lim(4 lim n→(4n lim lim n→ 1 616—— 16 nn 22 1 2nn 2 n42 1) 1n 1) or 1test provides no information7. The general term is (n 1)n. n 1 1n nfor all n, so divergent8. The series is arithmetic, so it is divergent.19. The general term is 2 .n 21 1 2 n 2 n for all n, so convergent10. a n n 12, n a n1 (n 11)2 n1r (n 11)2 n1——1 n 2 nn2n n→(n 1)2 n1nn→ 2(n 1) lim lim lim n→1 2(1 0) 1 2 n n ——2 n n n1 convergent11. The series is geometric where r 3 4 .Since 3 4 1, it is convergent.90012a. The series is geometric where r 1or 3 5 .S 10 3727 m12b. S 15001 3 5 1500 1500 3 5 10———1 3 5 500 3750 mNo, the sum of the infinite series modeling thissituation is 3750. Thus, the spill will spread nomore than 3750 meters.Pages 791–793 Exercises4413. a n 3 n, a n 1 3 n14 3 n1—n→4 3 n 3 3 nn 3r lim 4lim n→4 1 3 convergent2n214. a n 5, a n1 5(n1nn 1)2n15( n 1)—n→ 2n5 n2 n 2 5n (5n 5)r lim lim n→ 2 n10nn→ 5n 5 lim lim n→ 10 nn— 5 n 5 n n 1 05or 2divergent215. a n n 2n 2, a n1 (n2n1 (n 1 ) 2——n→2 n n 2r lim lim n→2 n (n2 limn1 1 ) 22 n2 2 n2 2n 1)n2n→n 2 2n 1 lim n→ 2divergent 2 n2n2—— n 2n2 2 nn2 n12 415 Chapter 12
2216. a n (n 1),(n 2)a n1 (n 2)2 (n 2) (n 3)——n→ 2 (n 1) (n 2) 2 ( n2 3n 2)n→2(n2 5n 6) n 2n2 3 nn2 n22 ——n→ n 2n2 5 nn2 n62 r lim lim lim 1test provides no information2n 117. a n , a n1 r lim1 2 ... (2n 1)2(n 1) 11 2 ... [2(n 1) 1]n→ 2n 11 2 ... (2n 1) 0convergent5n18. a n 1 2, a n1 (n 3)2(n 1) 11 2 ... [2(n 1) 1]124. The general term is 2n .11 1 2n 1 n for all n, so convergent25. The series is geometric where r 3 4 .Since 3 4 1, it is convergent.26. The general term is 2 n 12n.1 2 n 1 12n 1 nfor all n, so divergent127. The general term is 5 .n 21 1 5 n 2 n for all n, so convergent128. The general term is .n1 1 n n for all n, so divergent29. The series is arithmetic, so it is divergent30. a n 2 n 12n1, a n1 2(n 1) 12n 2 2(n 1) 1(2n 1)(1 2 ... (2n 1))2 n 2 lim (2n 1)(1 2 ... (2n 1))n→r lim ——n→ 2 n 112n1 lim n→(2n 1)(2n)(2n 1) 21 limn 2 lim n→4n 2 (2n 1) 2 n 2 2 2nn→ lim 2 n 1n→4n 25n1 2 n 1 n n ... n1 2 ... (n 1) lim —n→5 n1 4 n 2 n n1 2 ... (n 1)r lim n→ 5n 24 or 1 2 1 2 ... n5 limn 5(1 2 ... n)5 n (1 2 ... (n 1))n→5 lim n→ n 1 0convergentr 0.70.( n 1)2(n 1) 2(n 2)700,000n, a n1 2 n1S 1 0.702(n 1) 2(n 2) $2.3 millionr lim 2 n1n→ 2n 2 (n 1)2n2(n 1) 2(n 2) 2 lim n2n 2(n 1) 2 n 2n→ lim n 2n→2nn lim n→2 n 2 2 n 1 2 convergenta 8 1000(1.08) 811 4 n 2 1 n for all n, so convergent17131.11 1 n 3 1 n for all n, so convergent.1n 1 n 1 n for all n, so divergent.25 1 n 2 n for all n, so divergent19. a n 2n 2 21 120. The general term is (2 n) 2 or 4 n 2 .21. The general term is n 322. The general term is n n23. The general term is n 5convergent31a. No, MagicSoft let a 1 1,000,000 to arrive attheir figure. The first term of this series is1,000,000 0.70 or 700,000.31b. The series is geometric where a 1 700,000 and32. the harmonic series: 1 1 2 1 3 1 4 33a. Culture A: 1400 cells, Culture B; 713 cellsCulture A generates an arithmetic sequencewhere a 1 1000, d 200 and n 8a 8 1000 (8 1)200 2400Only considering cell growth, there are2400 1000 or 1400 new cells.Culture B generates a geometric sequence wherea 1 1000 and r 1.08A part of a cell cannot be generated.Only considering cell growth, there are1713 1000 or 713 new cells.33b. Culture A: a 31 1000 (31 1)200 70007000 1000 6000 cellsCulture B: a 31 1000(1.08) 311 10,06210,062 1000 9062Culture B; at the end of one month, culture Awill have produced 6000 cells while culture Bwill have produced 9062 cells.Chapter 12 416
34a. 2 2nn 11 34b. S lim n→S n lim 2 n 1n→2n1 lim 2(2n 1)n→2 n 2 lim 2 n→2 n 2 seconds35a. When the minute hand is at 4, 20 minutes havepassed. This is 20 160 or 1 3 hour.35b. 1 3 the distance between 4 and 5 is 1 3 (5) or 5 3 minutes. This is 5 3 1601or 3 6hour.35c. 1 3 1 3 6 (5) 6 536 minutes 6 536 5 3 5 3 6minute5 3 6 1 6 0 1 4 32 hour 1 3 1 1 3 6 4 32 (5) 7 85432minutes 7 85432 6 536 5 4 32 minute5 1 4 1 51hour32 601 4 32 , 511848435d. The sequence 1 3 1 1, 3, 6 41where r 1.2lim S n S a1n→ 1 r36. lim 1 3 —11 1 2 141 hour32 1, 51 84 is geometricThe hands will coincide at 4 141 o’clock,approximately 21 min 49 s after 4:00.4n2 5n→3 n2 2n lim 4 n2n2 n52 —n→ 3 nn 22 2 nn 2 4 3 237. r or 2a 9 22 91u41. AB 5 8, 1 (3) 3, 242. List all cubes from 1 to 200. Thereare five. The correct choice is E.Page 793 Mid-Chapter Quiz1. a 12 11 (19 1)(2)252. S 20 2 02[2(14) (20 1)6] 8603. 189 56r 41 2 78 r 3 3 2 r56 3 2 84, 84 3 2 12656, 84, 126, 1894. r 63or 2S 8 3 3(2) 81 ( 2)255 n2 2n 5n→n2 15. lim lim ——2n→ 16. a 1 250, r 0.55a 2 250(0.55) or 137.5a 3 137.5a 4 137.5(0.55) or 75.625a 5 75.625 137.5 137.S n 250 1 1 861 ft1 17. a 1 2,5r 1 0 2 42 16238. 19 11 (7 1)d30 6d5 d11 5 6, 6 5 1, 1 5 4,4 5 9, 9 5 1411, 6, 1, 4, 9, 14, 1939. 45.9 e 0.075tln 45.9 0.075t51.02 t40. 6 12r cos (v 30°) 1 2 r(cos v cos 30° sin v sin 30°) 1 2 32r cos v 1 2 r sin v0 32x 1 2 y 1 2 0 3x y 112 5S —11 1 05 n 0.55 n 2n2 2 nn2 n52 n n2 n12 5 0.558. a n 1 2 10, n a n1 1 2 … 10 n1 2 … (n 1)10 n1n→ 1 2 … n10n10 n [1 2 … (n 1)]10 n 10(1 2 … n)r lim lim n→ n 1n→10 10 lim(n 1)1n n 31 12 83 274 645 125As n → , n 1→ . Since r 1, the series isdivergent.9. The series is geometric where r 1 3 .Since r 1, it is convergent.417 Chapter 12
10. 5001 0. 124 515a 1 515, r 1.03 n 4S 4 $2154.5712-5Sigma Notation and the n th term.Pages 797–798 Check for Understanding1. The series 4 6 8 10 12 can berepresented by 42n 4 or by 52n 2.n02a. Sample answer: (1) n12b. Sample answer: (1) n3a. 9; 2, 3, 4, 5, 6, 7, 8, 9, 103b. t b a 13c. t b a 1 3 (2) 1 63d. 3 1 1 1 k 2 1k23 31 3 3n1 1 1 1 2 1 3 1 4 1 5 1 6 1 1 1 3 0 5 1 3 2 3 1 1 2 1 3 1 4 1 5 1 5 There are 6 terms.4. 6(n 3) (1 3) (2 3) (3 3) n1(4 3) (5 3) (6 3) (2) (1) 0 1 2 3 35. 54k 4 2 4 3 4 4 4 5k2 8 12 16 20 566. 4 1 1 1 1 1 1 2 a 2 0 2 1 2 2 2 3 2 4a07. p0 1 1 2 1 4 1 8 11 1 1 5165 3 4 p 5 3 4 0 5 3 4 1 5 3 4 2 5 3 4 3 5 3 4 10 5 1 546 4 516 1 35645S n 1 3 4 208. 55n 9. 3n110. 4(8 6n) 11. n112. (3) nn1515 515(1.03) 41 1.03 5 3 4 k0n2(3 k 1)3 1 2 n13a. 60n1389(0.63) n1S 60 1051 ft38913b. S 1 0.63 1051 ftPages 798–800 Exercises14. 4(2n 7) (2 1 7) (2 2 7) (2 3 7)n1 (2 4 7) (5) (3) (1) 1815. 55a 5(2) 5(3) 5(4) 5(5)a2 10 15 20 25 7016. 8(6 4b) (6 4 3) (6 4 4) (6 4 5)b3 (6 4 6) (6 4 7) (6 4 8) (6) (10) (14) (18) (22) (26)9617. 6(k k 2 ) (2 2 2 ) (3 3 2 ) (4 4 2 ) k2(5 5 2 ) (6 6 2 ) 6 12 20 30 42 11018. 8 n 5 6 7 8 n 5 6 7 8 n519. 8j420. 3m021. 3r14389 389(0.63) 601 0.6344 5 3 7 3 2 12 1 3 2 j 2 4 2 5 2 6 2 7 2 8 16 32 64 128 256 4963 m1 3 01 3 11 3 21 3 31 1 3 1 3 9 13 1 3 1 2 4 r 1 2 4 1 1 2 4 2 1 2 4 3 4 1 2 16 1 2 64 1 2 85 1 2 22. 5(0.5) i (0.5) 3 (0.5) 4 (0.5) 5i3 8 16 32 5623. 7k! 3! 4! 5! 6! 7!k3 6 24 120 720 5040 591024. 104(0.75) p 4(0.75) 0 4(0.75) 1 4(0.75) 2 p04(0.75) 3 4(0.75) 4 3 2.25 1.6875 4(0.75) 4S 1 0.75 1644Chapter 12 418
25. 4 2 5 n 4 2 5 1 4 2 5 2 4 2 5 3 4 2 5 47b. 0 1(1 x) 2(2 x) 3(3 x) 4(4 x)n1 8 5 1 625 32 1 25 4 2 5 5(5 x) 251 x 4 2x 9 3x 16 4x 25 5x 25 8 5 55 15x 25S —15x 301 2 5 x 248a. false; 73 k 3 3 3 4 3 726. 5n227. 4k129. 12k431. 4k133. 10k2 8 3 or 2 2 3 n i n (2 i 2 ) (3 i 3 ) (4 i 4 ) (5 i 5 ) (1) (3 i) (5) (5 i) 14(3k 3) 28. 44kk02k 30. 3k02 5 k 32. 315k 135. (1) k k 2k237. 39. 41. n0(1) n1 3 2k 2 k 3k1k112n 42. (a 2)! (a 2) !a!a(a 1)(a 2)!1 a(a 1)43. ( a 1)!( a 2)!( a b)!44. (a b 1)!45. 43.6446a. n12 k 2 3k! a(a 1)(a 1)500,000(0.35) nk034. 36. (2) 3k(13 4k)2k2k 1k1k038. k140. (a 1)(a)(a 1)(a 2)!(a 2)!(a b)(a b 1)!(a b 1)!5k ! (k 1)!k k 2 13kk!k2 a175,00046b. S 1 0.35 269,239 people46c. 2 69,230500,000 53.8%46d. The ad agency assumes that the people who buythe tennis shoes will be satisfied with theirpurchase.47a. (x 3) (x 6) (x 9) (x 12) (x 15) (x 18) 36x 63 36x 60x 60k3 93 b 3 7 3 8 3 9b7Since there are two 3 7 terms, 73 k 9k3 93 a .a348b. true; 8(2n 3) 1 3 13 49n23 bb7 9(2m 5) 1 3 13 49m3Since 49 49, 8(2n 3) 9(2m 5).n2n348c. true; 2 7n 2 18 32 98 270n3 72n 2 18 32 98 270n3Since 270 270, 2 7n 2 72n 2 .n3 n348d. false; 10(5 n) 6 7 15 105k1 9(4 p) 4 5 13 85p0Since 85 105, 10(5 n) 9(4 p).k1p049a. 6!49b. 5! or 12049c. 4! or 24, “LISTEN”50a. On an 8 8 chessboard, there is 1-8 8 square.On an 8 8 chessboard, there are 4-7 7squares, one in each of the four corners.50b. For the 6 6 squares, begin in one corner.For different configurations, you can move itover, up to 2 more spaces, or down, up to 2 morespaces. Thus, there are 3 3 or 9-6 6 squares.Continue this procedure for the other sizes ofsquares.9-6 6, 16-5 5, 25-4 4, 36-3 3, 49-2 2,and 64-1 150c. 8n 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2n1 1 4 9 16 25 36 49 64 204351. The general term is n .3 1 n 2 nfor all n, so divergent52. 0.42(21) 8.82 litersIf with each stroke 20% is removed, then 80%remains.a 1 21(0.80) 16.8a 2 16.8(0.80) 13.44a 3 13.44(0.80) 10.752a 4 10.752(0.80) 8.6016It will take 4 strokes for 42% of the air to remain.419 Chapter 122
53. 322 a 1 2 51322 4a 182 a 1822 16, 162 162,1622 3282, 16, 162, 3254. log 10 0.001 355. x 2 y 2 Dx Ey F 0(0, 9) → 81 9E F 0(7, 2) → 49 4 7D 2E F 0(0, 5) → 25 5E F 09E F 81 9(4) F 815E F 25F 4514E 56 53 7D 2(4) 45 0E 4 7D 0D 0x 2 y 2 4y 45 0x 2 (y 2) 2 4956. (2 i)(42 i) 8 52i i 2 7 5i257. v y 59 sin 63° 52.57 ft/s59 ft/sv x 59 cos 63° 26.79 ft/s63˚58. d 1 2x 1 3y1 9314y1 47 1 2x 1 3y1 9 x 1 4y1 41371, d 2 x 1 217x 1 317y 1 917 13x 1 413y 1 413217 13x 413 317y 917 413 059. 5 m9 m 2 3 3(5 m) 2(9 m)15 3m 18 2mm 3 The correct choice is D.12-6The Binomial TheoremPages 803–804 Check for Understanding1a. n 0: 1n 1: 1 1 or 2n 2: 1 2 1 or 4n 3: 1 3 3 1 or 8n 4: 1 4 6 4 1 or 16n 5: 1 5 10 10 5 1 or 321, 2, 4, 8, 16, 321b. 2 n2a. The second term of (x y) 3 is 3x 2 y.It is negative.The second term of (x y) 4 is 4x 3 y.It is negative.The second term of (x y) 5 is 5x 4 y.It is negative.2b. The third term of (x y) 3 is 3xy 2 . It is positive.The third term of (x y) 4 is 6x 2 y 2 . It is positive.The third term of (y y) 5 is 10x 3 y 2 . It is positive.2c. Even indexed terms are negative and oddindexed terms are positive.3. The sum of the exponents of each term is n.4. The exponents must add to 12, so the exponent ofy is 12 7 or 5.To find the coefficient of the term use the formula(x y) n n n! r!(nx nr y r .r0 r)!Evaluate the general term for n 12 and r 5.12! 5 ! 7!x 7 y 5 792x 7 y 5 .5. c 5 5c 4 d 10c 3 d 2 10c 2 d 3 5cd 4 d 5 6 56. (a 3) 6 a 6 3 6a 5 32 1 a24 33 6 5 41 2 3a 3 34 6 5 4 31 2 3 4a 235 6 5 4 3 21 2 3 4 5a36 6 5 4 3 2 11 2 3 4 5 6 a 6 18a 5 135a 4 540a 3 1215a 2 1458a 7297. (5 y) 3 5 3 (y) 0 3 5 2 (y) 3 2 2 125 75y 15y 2 y 38. (3p 2q) 4 (3p) 4 (2q) 0 4(3p) 3 (2q) 17!9. r!(7 r)!7! 5!(7 5)!5 (y)2 1 4 3(3 p) 2 (2q) 2 4 3 2(3p)(2q)2 133 2 1(a) 7r (b) r 81p 4 216p 3 q 216p 2 q 2 96pq 3 16q 4(a) 75 (b) 5a 2 b 521a 2 b 59!10. r!(9 r)!9! 3!(9 3)!7 6 5 4 3 2 15 4 3 2 1 2 1(x) 9r 3 r 2523 x 63 2 1 5 0 (y) 33 2 14 3 2 1(3p) 0 (2q) 44 3 2 1(x) 93 3 39 8 7 6 5 4 3 2 13 2 1 6 5 4 3 2 1x 6 3311. (H T) 5 H 5 5H 4 T 10H 3 T 2 10H 2 T 3 5HT 2 T 511a. 1 11b. 1011c. 1 5 or 6 11d. 10 10 5 1 or 26Pages 804–805 Exercises12. (a b) 8 a 8 8a 7 b 28a 6 b 2 56a 5 b 3 70a 4 b 4 56a 3 b 5 28a 2 b 6 8ab 7 b 813. (n 4) 6 n 6 24n 5 240n 4 1280n 3 3840n 2 6144n 409614. (3c d) 4 81c 4 108c 3 d 54c 2 d 2 12cd 3 d 415. (2 a) 9 512 2304a 4608a 2 5376a 3 4032a 4 2016a 5 672a 6 144a 7 18a 8 a 9Chapter 12 420
d 5 22 1 7 6 5 d 4 2 3 7 6 5 4 d3 2 13 244 3 2 116. (d 2) 7 d 7 2 0 7 d 6 2 1 7 6 2 d 7 14d 6 84d 5 280d 4 560d 3 672d 2 448d 128 33(x) 2 15 4 3 3 2 (x) 5 4 3 2 31 (x)343 2 14 3 2 15 4 3 2 1 3 0 (x)55 4 3 2 117. (3 x) 5 3 5 (x) 0 5 3 4 (x 1 ) 1 5 4 27 6 5 4 3 d 2 2 55 4 3 2 17 6 5 4 3 2 d 1 2 66 5 4 3 2 17 6 5 4 3 2 1 d 0 2 77 6 5 4 3 2 1 243 405x 270x 2 90x 3 15x 4 x 5(4a) 2 (b)2 1 4 3 2(4a) 1 (b) 3 4 3 2 1(4a)3 2 10 (b)44 3 2 118. (4a b) 4 (4a) 4 (b) 0 4(4a) 3 (b) 1 4 3 2 256a 4 256a 3 b 96a 2 b 2 16ab 3 b 419. (2x 3y) 3 (2x) 3 (3y) 0 3(2x) 2 (3y) 1 3 2(2 x)(3y) 22 13 2 1(2x) 0 (3y) 33 2 1 8x 3 36x 2 y 54xy 2 27y 320. 3m 2 4 (3m) 4 2 0 4(3m) 3 2 1 4 3(3 m) 2 (2) 22 1 4 3 2(3m)1 (2)33 2 14 3 2 1(3m) 0 (2) 44 3 2 1 81m 4 1082m 3 108m 2 242m 421. c 1 6 c 6 (1) 0 6c 5 (1) 1 6 5( c ) 4 (1) 2 6 5 4(c)3 (1) 32 13 2 1 6 5 4 3(c)2 (1) 44 3 2 1 6 5 4 3 2(c)1 (1)55 4 3 2 16 5 4 3 2 1(c) 0 (1)6 6 5 4 3 2 1 c 3 6c 2 c 15c 2 20cc 15c 6c 122. 1 2 n 2 1 2 n 5 (2) 0 5 1 2 n 4 (2) 1 5 4 3 1 2 n 2 (2) 33 2 15 4 3 2 1 1 2 n 0 (2) 55 4 3 2 15 4 1 2 n 3 (2) 2 2 15 4 3 2 1 2 n 1 (2) 44 3 2 11 3n25 5 8 n 4 5n 3 20n 2 40n 3223. 3a 2 3 b 4 (3a) 4 2 3 b 0 4(3a) 3 2 3 b 14 3(3a) 2 2 3 b 2 2 14 3 2(3a) 1 2 3 b 33 2 1 4 3 2 1(3a)0 2 3 b 44 3 2 1 81a 4 72a 3 b 24a 2 b 2 3 29 ab 3 1 681 b 4(p2) 6 (q)2 1 8 7 6(p2) 5 (q) 3 8 7 6 5(p 2 ) 4 (q) 43 2 1 4 3 2 18 7 6 5 4(p 2 ) 3 (q)5 5 4 3 2 18 7 6 5 4 3(p 2 ) 2 (q)6 6 5 4 3 2 18 7 6 5 4 3 2(p 2 ) 1 (q)7 7 6 5 4 3 2 18 7 6 5 4 3 2 1(p 2 ) 0 (q)8 8 7 6 5 4 3 2 124. (p 2 q) 8 (p 2 ) 8 (q) 0 8(p 2 ) 7 (q) 1 8 7 2 p 16 8p 14 q 28p 12 q 2 56p 10 q 3 70p 8 q 4 56p 6 q 5 28p 4 q 6 8p 2 q 7 q 825. (xy 2z 3 ) 6 (xy) 6 (2x 3 ) 0 6(xy) 5 (2z 3 ) 19!26. 4!(9 4)!827. 3!(8 3)!28. 3!(779!29. 6!(9 6)! x 6 y 6 12x 5 y 5 z 3 60x 4 y 4 z 6 160x 3 y 3 z 9 240x 2 y 2 z 12 192xyz 15 64z 18(x) 94 (y) 4 x 5 y 4 126x 5 y 4!(a) 83 2 38 7 6 5 4 3 2 13 2 1 5 4 3 2 1a 5 221122a 5!(2a)3)!73 (b) 3 35 16a 4 (b 3 )560a 4 b 3(3c) 96 (2d) 6 84 27c 3 64d 6 145,152c 3 d 610!30. ( 1 7!(1 0 7)! 2 x) 107 (y) 7 120 1 8 x 3 (y 7 )15x 3 y 711!31. 5!(1 1(2p) 5)!115 (3q) 5 462 64p 6 (243q 5 )7,185,024p 6 q 532. The middle term is the fifth term.8! 4!(8 x 84 y 4 70x 2 y 2 4)! 6 5(x y) 4 ( 2z 3 ) 22 16 5 4 3(xy) 2 (2z 3 )4 4 3 2 16 5 4 3 2(xy) 1 (2z 3 )55 4 3 2 16 5 4 3 2 1(xy) 0 (2z 3 )66 5 4 3 2 16 5 4(xy) 3 (2z 3 )33 2 19 8 7 6 5 4 3 2 14 3 2 1 5 4 3 2 133. (M W) 8 M 8 8M 7 W 28M 6 W 2 56M 5 W 3 70M 4 W 4 56M 3 W 5 28M 2 W 6 8MW 7 W 870 56 28 8 1 or 16334. Sample answer: Treat a b as a single term andexpand [a b) c] 12 using the BinomialTheorem. Then evaluate each (a b) n term in theexpansion using the Binomial Theorem.35. (T F) 12 T 12 12T 11 F 66 10 F 2 220T 9 F 3 495T 8 F 4 792T 7 F 5 924T 6 F 6 792T 5 F 7 495T 4 F 8 220T 3 F 9 66T 2 F 10 12TF 11 F 1235a. 49535b. 924 792 495 220 66 12 1 or 251036. Find the term for which both x’s have the sameexponent. This will occur for the middle term ofthe expansion, the 4th term when n 6. Use the421 Chapter 12
Binomial Theorem to find the 4th term for the1expansion of 3x 4 x 6 .6! 3 !3(3x)!3 1 3 4 x 1 351637a. Sample answer: 1 0.0137b. Sample answer: 1.04060401(1 0.01) 4 1 4 4 1 3 0.01 1 6 1 2 0.01 2 4 1 1 0.01 3 0.01 4 1.0406040137c. 1.04060401; the two values are equal.38. 75 2k (5 2 2) (5 2 3) (5 2 4)x2 (5 2 5) (5 2 6) (5 2 7) 1 (1) (3) (5) (7) (9)2439. a n 2 n2n1n!a n1 (n 1)!r lim n→ lim n→2 n lim 0convergent40. This is a geometric series where r 1 2 .S 2n1 (n 1)!—— 2 nn!2 n 2 n!(n 1) n!2n→ n 1 2 3 —1 1 2 1 1 3 41. P n P 1 (1 i) ni 1 1 150,000 P 0 .081212(30)———.08 150,000 P(136.283491)$1100.65 P42. MK [4 (2)]iu (8 6)ju (2 3)ku 6i u 2j u 5k u43. s rv0.25 r 4 r 0.3183098862 mir 0.3183098862(5280) 1681 feet44. Test all answer choices that are prime integers.You can eliminate answer choice C.A: 3(2) ? 10 ? 56 (2)6 ? 10 ? 53 ; falseB: 3(3) ? 10 ? 56 (3)9 ? 10 ? 52 ; falseD: 3(11) ? 10 ? 56 (11)33 ? ? 55 10 ; true The correct choice is D.6 0 1 212-7Special Sequences and SeriesPage 809 Graphing Calculator Exploration1. Sample answer (without zooming): 2.4 x 2.42. Sample answer for greatest difference: about 0.08;The least difference is 0.3. Sample answer: 3.4 x 3.4; sample answer:about 0.15; 04. Sample answer: 3.8 x 3.8; sample answer:about 0.05; 05. larger1x6. 1 1 1!Pages 811–812 Check for Understanding1. The approximation given in Example 1 only usedthe first five terms of the exponential series.Using more terms of the exponential series wouldgive an approximation closer to that given by thecalculator.2. Sample answer: 2 x 1.53. The problem seems to imply that siblings mate.Genetically, this can lead to problems. Anotherproblem is the assumption that each birthproduces only two offspring, one male and onefemale. Rabbits are more likely to give birth tomore than two offspring and the ratio of male tofemale births is not guaranteed to be 1 to 1.4. a n1 a n a n1 for n 25. ln (7) ln (1) ln (7) i 1.94596. ln (0.379) ln (1) ln (0.379) i 0.97027. e 0.8 1 0.8 (0 .82 ) 2! (0 .83 ) 3! (0 .8) 4 4! 1 0.8 0.32 0.08 0.017 2.228. e 1.36 1 1.36 (1. 32 6) 2! (1. 33 6) 3! (1. 36) 4 4! 1 1.36 0.925 0.419 0.143 3.859. sin x x x 33 ! x 55 ! x 77 ! x 99 !sin sin 3.1416 3.1416 (3.1 43 16) 3! (3.1 45 16) 5! (3.1 416) 7 7! (3.1 416) 99! 3.1416 5.1677 2.5502 0.5993 0.0821 0.006910. 2cos 3 4 i sin 3 4 2e i 3 4i11. 1 3i 2 1 2 32 2cos 2 3 i sin 2 3 2e i 2 3 Chapter 12 422
12a. A Pe rt2P Pe (0.06)5P e 0.3.3) 2 1 0.3 (0 2! 1.345approximately 1.345P12b. No, in five years she will have increased hersavings by about 34.5%, not 100%.12c. The approximation is accurate to two decimalplaces.Pages 812–814 Exercises13. ln (4) ln (1) ln 4 i 1.386314. ln (3.1) ln (1) ln (3.1) i 1.131415. ln (0.25) ln (1) ln (0.25) i 1.386316. ln (0.033) ln (1) ln (0.033) i 3.411217. ln (238) ln (1) ln (238) i 5.472318. ln (1207) ln (1) ln (1207) i 7.0959.1) 2.1) 3.1) 419. e 1.1 1 1.1 (1 2 ! (1 3 ! (1 4 ! 1 1.1 0.605 0.222 0.06 2.9920. e 0.2 1 (0.2) (0 . 2) 22 ! (0 . 2) 33 ! (0 . 2) 44 ! 1 0.2 0.02 0.0013 0.00007 0.8221. e 4.2 1 4.2 (4 .2) 2 (4 .2) 3 (4 .2) 42!3!4! 1 4.2 8.82 12.348 12.965 39.3322. e 0.55 1 0.55 (0. 55) 2 (0. 55) 3 (0. 55) 42!3!4! 1 0.55 0.151 0.028 0.004 1.7323. e 3.5 1 3.5 (3 .5) 22 ! (3 .5) 33 ! (3 .5) 44 ! 1 3.5 6.125 7.146 6.253 24.0224. e 2.73 1 2.73 (2. 73) 2 (2. 73) 3 (2. 73) 42!3!4! 1 2.73 3.726 3.391 2.314 13.1625. cos x 1 x 22 ! x 44 ! x 66 ! x 88 !cos cos 3.1416 1 (3.1 416) 2 (3.1 416) 4 (3.1 416) 6 (3.1 42!4!6!8! 1 4.9348 4.0588 1.3353 0.2353 0.9760actual value: cos 126. sin x x x 33 x 55 x 77 x 99 sin 4 sin 0.7854! 0.7854 (0.7 3! (0.7 854) 99!!!!854) 3 (0.7 854) 5 (0.7 854) 75!7!845 0.7854 0.4 6 0. 120 0.7071actual value: sin 4 22 0.707116) 829 89 0 .18430.11375 040 3 62,88027. cos x 1 x 2 ! x 4 !cos 6 cos 0.52362 4 x 66 ! x 88 !236) 2 (0.5 236) 4 (0.5 236) 6 (0.5 236) 84!6!8!742 0.7 51624 0. 0272 060 0 . 00564 0,320 1 (0.5 2! 1 0.2 2 0.8660actual value: cos 6 32 0.866028. sin x x x 33 ! x 55 ! x 77 ! x 99 !sin 2 sin 1.5708 1.5708 0.6460 0.0797 0.0047 0.0002 1.0000actual value: sin 2 129. 5cos 5 3 i sin 5 3 5e i 5 330. i cos 2 i sin 2 e i 2 31. r 1 2 1 2 or 2Q Arctan 1 1 or 4 1 i 2cos 4 i sin 4 2e i 4 32. r (3) 2 1 2 or 21v Arctan or 3 6 3 i 2cos 6 i sin 6 2e i 6 33. r 2 2 2 2or 2v Arctan 22 or 3 4 2 2i 2cos 3 4 i sin 3 4 2e i 3 434. r 4 3 2 (4) 2 or 8v Arctan 443 or 7 643 4i 8cos 7 6 i sin 7 6 8e i 7 635. r 3 2 3 2 or 32v Arctan 3 3 or 4 3 3i 32cos 4 i sin 4 32e i 4 36. Sample answer: A transcendental number is onethat cannot be the root of an algebraic equationwith rational coefficients. Examples are and e.37. eix e ix cos x i sin x (cos x i sin x)2i 2i 2i sin x2i eix e ix2 sin x 2c 2os x cos x38. See students’ work.cos x i sin x cos x i sin x2 423 Chapter 12
39. If you add the numbers on the diagonal lines asshown, the sums are the terms of the Fibonaccisequence.40a. 1 1 , 2 1 , 3 2 , 5 3 , 8 5 , 1 3840b. neither40c., 2 11, 3 43 21 , 5 534 , 8 95540d. yes; 1.61840e. The two ratios are equivalent to three decimalplaces.40f. See students’ work.41a. A Pe rt 5000e 0.05(13) 5000e 0.6565) 265) 365) 4 50001 0.65 (0. 2 ! (0. 3 ! (0. 4 ! $9572.2941b. No, the account will be short by more than$30,000.41c. about 42 years; 47 years old41d. 40,000 Pe 0.65$20,882 P42a. Every third Fibonacci number is an evennumber.42b. Every fourth Fibonacci number is a multiple of 3.(2x) 4 (y)2 1 6 5 4(2x) 3 (y) 3 6 5 4 3(2x)3 2 12 (y)44 3 2 16 5 4 3 2(2x)1 (y)55 4 3 2 16 5 4 3 2 1(2x)0 (y)66 5 4 3 2 12 kk143. (2x y) 6 (2x) 6 (y) 0 6(2x) 5 (y) 1 6 5 244. 61 5 10 10 5 11 6 15 20 15 6 1F nF n 121.751.51.2510.750.50.25O11 1 351 2 1 813.1 3 3 1 ..1 4 6 4 11 2 3 4 5 6 7 8 9 101 64x 6 192x 5 y 240x 4 y 2 160x 3 y 3 60x 2 y 4 12xy 5 y 612n0.0145a. a 1 0.005, r 0 . 005or 2a 3 0.005(2) 31 0.020 cma 4 0.020(2) 0.040 cm45b. 0.005(2) n145c. a 10 0.005(2) 101 2.56 cma 100 0.005(2) 1001 3.169 10 27 cm8 2 3 46. 8 1 3 or 28 1 3 47. y 2 Dx Ey F 0(0, 0): F 0(2, 1): 1 2D E F 0(4, 4): 16 4D 4E F 02D E 1 4D 2E 24D 4E 16→4D 4E 162E 142D E 1 E 72D 7 1 y 2 3x 7y 02D 6D 3448. r 1 6or 4v 8 4 8 or 15 84cos 15 8 i sin 15 849. u2 r 30 2 50 2 2(30)(50) cos 140°ru 75.5 N75.5 30 sin 140° si n vsin v 30 s in140°75.5v Arcsin 30 s 7 14°4850a. v t 5(2 )radians1 secondin140°5.5 10 radians per second50b. v r v t 1 2 ft(10 radians/s) 15.7 ft/s51. Let m multiple choice. Let e essay.m e 301m 12e 96f(m, e) 5m 20ef(24, 6) 5(24) 20(6) 240f(0, 8) 5(0) 20(8) 160f(30, 0) 5(30) 20(0) 150f(0, 0) 0To receive the highest score, answer 24 multiplechoice and 6 essay.e3020m e 30m 12e 9610 (0, 8)(24, 6)m 0(0, 0) 10e 020 m(30, 0)Chapter 12 424
52. (DC) 2 2 2 18 2(DC) 2 14DC 1414 2 5 2 (BC) 239 (BC) 239 BC The correct choice is C.12-8Sequences and IterationPage 819 Check for Understanding1. Iteration is the repeated composition of a functionupon itself.2. It is the sequence of iterates produced when acomplex number is iterated for a function f(z).3. If the prisoner set is connected, then the Julia setis the boundary between the prisoner set and theescape set. If the prisoner set is disconnected, thenthe Julia set is the prisoner set.4. f(1) (1) 2 1f(1) 1 2 1f(1) 1 2 1f(1) 1 2 11, 1, 1, 15. f(2) 2 2 5 1f(1) 2 (1) 5 7f(7) 2 (7) 5 19f(19) 2 (19) 5 431, 7, 19, 436. z 0 6iz 1 0.6(6i) 2i 5.6iz 2 0.6(5.6i) 2i 5.36iz 3 0.6(5.36i) 2i 5.216i7. z 0 25 40iz 1 0.6(25 40i) 2i 15 26iz 2 0.6(15 26i) 2i 9 17.6iz 3 0.6(9 17.6i) 2i 5.4 12.56i8. z 0 0, f(z) z 2 (1 2i)z 1 0 2 (1 2i) 1 2iz 2 (1 2i) 2 (1 2i) 1 4i 4i 2 1 2i 2 6iz 3 (2 6i) 2 (1 2i) 4 24i 35i 2 1 2i 31 22i9. z 0 1 2i, f(z) z 2 (2 3i)z 1 (1 2i) 2 (2 3i) 1 4i 4i 2 2 3i 1 iz 2 (1 i) 2 (2 3i) 1 2i i 2 2 3i 2 5iz 3 (2 5i) 2 (2 3i) 4 20i 25i 2 2 3i 19 23i10. p n 2 440 0.60p 1 0.60 1.75(0.60)(1 0.60) 1.02(1.02)(40) 41p 2 1.02 1.75(1.02)(1 1.02) 0.9843(0.9843)(40) 39p 3 0.9843 1.75(0.9843)(1 0.9843) 1.0113(1.0113)(40) 40p 4 1.0113 1.75(1.0113)(1 1.0113) 0.9913(0.9913)(40) 40p 5 0.9913 1.75(0.9913)(1 0.9913) 1.0064(1.0064)(40) 40p 6 1.0064 1.75(1.0064)(1 1.0064) 0.9951(0.9951)(40) 40p 7 0.9951 1.75(0.9951)(1 0.9951) 1.0036(1.0036)(40) 40p 8 1.0036 1.75(1.0036)(1 1.0036) 0.9973(0.9973)(40) 40p 9 0.9973 1.75(0.9973)(1 0.9973) 1.002(1.002)(40) 40p 10 1.002 1.75(1.002)(1 1.002) 0.9985(0.9985)(40) 4041, 39, 40, 40, 40, 40, 40, 40, 40, 40Pages 820–821 Exercises11. f(x 0 ) f(4) 3(4) 7 5f(x 1 ) f(5) 3(5) 7 8f(x 2 ) f(8) 3(8) 7 17f(x 3 ) f(17) 3(17) 7 4412. f(2) (2) 2 4f(4) 4 2 16f(16) 16 2 256f(256) 256 2 65.53613. f(4) (4 5) 2 1f(1) (1 5) 2 16f(16) (16 5) 2 121f(121) (121 5) 2 13,45614. f(1) (1) 2 1 0f(0) 0 2 1 1f(1) (1) 2 1 0f(0) 0 2 1 1425 Chapter 12
15. f(0.1) 2(0.1) 2 0.10.08f(0.08) 2(0.08) 2 (0.08) 0.09f(0.09) 2(0.09) 2 0.09 0.07f(0.07) 2(0.07) 2 (0.07) 0.0816a. t 1 2 1 2t 2 2 2 1t 3 2 1 2t 4 2 2 1t 10 2 2 116b. t 1 2 4 1 2 t 2 4t 3 2 4 2 1 2 1 2 2 1 2 t 4 42t 10 4 1 2 16c. t 1 2 7 2t 2 7 2 7 t 3 2 7 2t 4 7 2 7 2t 10 7 2 7 216d. The values of the iterates alternate between x 0and x 0 .17. z 0 5iz 1 2(5i) (3 2i) 3 8iz 2 2(3 8i) (3 2i) 6 16i 3 2i 9 14iz 3 2(9 14i) (3 2i) 18 28i 3 2i 21 26i18. z 0 4z 1 2(4) (3 2i) 11 2iz 2 2(11 2i) (3 2i) 22 4i 3 2i 25 6iz 3 2(25 6i) (3 2i) 50 12i 3 2i 53 14i19. z 0 1 2iz 1 2(1 2i) (3 2i) 2 4i 3 2i 5 2iz 2 2(5 2i) (3 2i) 10 4i 3 2i 13 2iz 3 2(13 2i) (3 2i) 26 4i 3 2i 29 2i20. z 0 1 2iz 1 2(1 2i) (3 2i)2 4i 3 2i 1 6iz 2 2(1 6i) (3 2i) 2 12i 3 2i 5 14iz 3 2(5 14i) (3 2i) 10 28i 3 2i 13 30i21. z 0 6 2iz 1 2(6 2i) (3 2i) 12 4i 3 2i 15 2iz 2 2(15 2i) (3 2i) 30 4i 3 2i 33 2iz 3 2(33 2i) (3 2i) 66 4i 3 2i 69 2i22. z 0 0.3 iz 1 2(0.3 i) (3 2i) 0.6 2i 3 2i 3.6 4iz 2 2(3.6 4i) (3 2i) 7.2 8i 3 2i 10.2 10iz 3 2(10.2 10i) (3 2i) 20.4 20i 3 2i 23.4 22i23. z 0 1 3 2 3 iz 1 3 1 3 2 3 i 2i 1 2i 2i 1z 2 3(1) 2i 3 2iz 3 3(3 2i) 2i 9 6i 2i 9 8i24. z 0 0 i, f(z) z 2 1z 1 (i) 2 12z 2 (2) 2 1 3z 3 3 2 1 8Chapter 12 426
25. z 0 i, f(z) z 2 1 3iz 1 i 2 1 3i3iz 2 (3i) 2 1 3i8 3iz 3 (8 3i) 2 1 3i 64 48i 9 1 3i 56 45i26. z 0 1, f(z) z 2 3 2iz 1 1 2 3 2i 4 2iz 2 (4 2i) 2 3 2i 16 16i 4 3 2i 15 18iz 3 (15 18i) 2 3 2i 225 540i 324 3 2i96 542i27. z 0 1 i, f(z) z 2 4iz 1 (1 i) 2 4i 1 2i 1 4i2iz 2 (2i) 2 4i4 4iz 3 (4 4i) 2 4i 16 32i 16 4i 28i28. z 0 22 22i, f(z) z 2z 1 22 22i 2 1 2 i 1 2 i 2 1z 2 (i) 2 1z 3 1 2 129. z 0 1 i, f(z) z 2 2 3iz 1 (1 i) 2 2 3i 1 2i 1 2 3i 2 iz 2 (2 i) 2 2 3i 4 4i 1 2 3i 5 7iz 3 (5 7i) 2 2 3i 25 70i 49 2 3i22 73i30. p 1 p 0 rp 0p 1 2000 (0.052)(2000) $2104p 2 2104 (0.052)(2104) $2213.41p 3 2213.41 (0.052)(2213.14) $2328.51p 4 2328.51 (0.052)(2328.51) $2449.59p 5 2449.59 (0.052)(2449.59) $2576.9731.2002 1984 18; After 18 years, about 54% ofthe maximum sustainable population is present.32. f(z) z 2 c1 15i (2 3i) 2 c1 15i 4 12i 9i 2 c4 3i c33. 22x 1 2 22x 1 2 234. See students’ work. Sample topics for discussionare judging soil quality and detection of heatstress in cows.35a. 1.414213562, 1.189207115, 1.090507733,1.04427378235b. f(z) z, z 0 235c. 135d. 136. 2cos 3 i sin 3 2e i 3 8!37. 4!(8 4)!Iteration x percent x 2.5x(1 x)1 0.10 0.3252 0.325 0.87343 0.8734 1.14984 1.1498 0.71925 0.7192 1.22416 1.2241 0.53837 0.5383 1.15968 1.1596 0.69699 0.6969 1.22510 1.225 0.535911 0.5359 1.157712 1.1577 0.701313 0.7013 1.22514 1.225 0.535915 0.5359 1.157716 1.1577 0.701317 0.7013 1.22518 1.225 0.5359(2a) 84 (3b) 4 70 16a 4 81b 4 90,720a 4 b 438. Convergent; the series is geometric with⏐r⏐ 1 4 1.39. The distance between the vertices is 130 ft.2a 130, so a 75.ce a 7 5 9 165 b 2 c 2 a 2b 2 91 2 65 2 4056ax 2 2 by 2 2 1 ⇒ 42x2 y225 40 56 1427 Chapter 12
40. n 1 sin I n 2 sin r1.00 sin 42° 2.42 sin rr Arcsin 1.00 2r 16°41a.sin42°.4241b. Let h height of the building.Let x distance from the point of elevation tothe center of the base of the building.tan 42° h x tan 56° h 40xhx tan 42°tan 56° h 40htan 56° (h 40 )tan42°h1.6466 1 4 0h0.6466 4 0 hh 62 feetNo, the height of the building is about 62 feet fora total of about 102 feet with the tower.42. x 2 for all x, so infinite discontinuity43. yx y 14O42˚(3, 9)y 9y 5(3, 5)56˚(5, 9)(8, 5)40 ft(8, 6)x 3 x 8f(x, y) 2x 8y 10f(3, 9) 2(3) 8(9) 10 88f(5, 9) 2(5) 8(9) 10 92f(8, 6) 2(8) 8(6) 10 74f(8, 5) 2(8) 8(5) 10 66f(3, 5) 2(3) 8(5) 10 56max: 92, min: 5644. H L H L22H 2L H LH 3L H L 3 The correct choice is D.x12-9Mathematical InductionPage 826 Check for Understanding1. The n 1 case shows that the premise is true foran infinite number of cases.2. Provide a counterexample.3a. n(n 2)3b. Since 3 is the first term in the sequence ofpartial sums and 1(1 2) 3, the formula isvalid for n 1.Since 8 is the second term in the sequence ofpartial sums and 2(2 2) 8, the formula isvalid for n 2.Since 15 is the third term in the sequence ofpartial sums and 3(3 2) 15, the formula isvalid for n 3.3c. S k ⇒ k(k 2); S k1 ⇒ (k 1)(k 3)4. 8 n 1 7r for some integer r.5. Sample answer: If we wish to prove that we canclimb a ladder with an indefinite number of steps,we must prove the following. First, we must showthat we can climb off the ground to rung 1. Next,we must show that if we can climb to rung k, thenwe can climb to rung k 1.6. Step 1: Verify that the formula is valid for n 1.Since 3 is the first term in the sequence and1(1 2) 3, the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.S k ⇒ 3 5 7 … (2k 1) k(k 2)S k1 ⇒ 3 5 7 … (2k 1) (2k 3) k(k 2) (2k 3) k 2 4k 3 (k 1)(k 3)Apply the original formula for n k 1.(k 1)[(k 1) 2] (k 1)(k 3)The formula gives the same result as adding the(k 1) term directly. Thus if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2,n 3, and so on indefinitely. Thus, the formula isvalid for all positive integral values of n.7. Step 1: Verify that the formula is valid for n 1.Since 2 is the first term in the sequence and2(2 1 1) 2, the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.S k ⇒ 2 2 2 2 3 … 2 k 2(2 k 1)S k1 ⇒ 2 2 2 2 3 … 2 k 2 k1 2(2 k 1) 2 k1 2 2 k1 2 2(2 k1 1)When the original formula is applied for n k 1,the same result is obtained. Thus if the formula isvalid for n k, it is also valid for n k 1. Sincethe formula is valid for n 1, it is also valid forn 2, n 3, and so on indefinitely. Thus, theformula is valid for all positive integral valuesof n.Chapter 12 428
8. Step 1: Verify that the formula is valid for n 1.Since 1 2 is the first term in the sequence and11 2 1 1 2 , the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.S k ⇒ 1 2 1 1 11 2 2 2 3 … 2 k 1 2 kS k1 ⇒ 1 2 212 213 … 211k 2 k1 1 21 1 2 22 k 1 2 k1k 2 k11 2 k12 11 2 k111 1 2 kWhen the original formula is applied for n k 1,the same result is obtained. Thus if the formula isvalid for n k, it is also valid for n k 1. Sincethe formula is valid for n 1, it is also valid forn 2, n 3, and so on indefinitely. Thus, theformula is valid for all positive integral valuesof n.9. S n : 3 n 1 2r for some integer rStep 1: Verify that S n is valid for n 1.S 1 ⇒ 3 1 1 or 2. Since 2 2 1, S n is valid forn 1.Step 2: Assume that S n is valid for n k and showthat it is also valid for n k 1.S k ⇒ 3 k 1 2r for some integer rS k1 ⇒ 3 k1 1 2t for some integer t3 k 1 2r3(3 k 1) 3 2r3 k1 3 6r3 k1 1 6r 23 k1 1 2(3r 1)Thus, 3 k1 1 2t, where t 3r 1 is aninteger, and we have shown that if S n is valid,then S k1 is also valid. Since S n is valid for n 1,it is also valid for n 2, n 3, and so onindefinitely. Hence, 3 n 1 is divisible by 2 for allintegral values of n.10a. 6 4 10 10b. a n n(n 1)210 5 1515 6 2121 7 2828 8 3610, 15, 21, 28, 3610c. Step 1: Verify that the formula is valid for n 1.Since 1 is the first term in the sequence and 1(1 1 )(1 2)6 1, the formula is valid for n 1.Step 2: Assume that the formula is valid forn k and derive a formula for n k 1.S k ⇒ 1 3 6 … k(k 1)2 (k 1) (k 2)6S k ⇒ 1 3 6 … k(k 1)2 (k 1) (k 2)2 k(k 1 )(k 2)6 (k 1) (k 2)2k(k 1)(k 2) 3(k 1)(k 2) 6(k 1)(k 2)(k 3) 6Apply the original formula for n k 1.(k 1)[(k 1) 1][(k 1) 2]6(k 1)(k 2)(k 3)6The formula gives the same result as adding the(k 1) term directly. Thus if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2,n 3, and so on indefinitely. Thus, the formulais valid for all positive integral values of n.Pages 826–828 Exercises11. Step 1: Verify that the formula is valid for n 1.Since 1 is the first term in the sequence and(1)[2(1) 1] 1, the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.S k ⇒ 1 5 9 … (4k 3) k(2k 1)S k1 ⇒ 1 5 9 … (4k 3) (4k 1) k(2k 1) (4k 1) 2k 2 3k 7 (k 1)(2k 1)Apply the original formula for n k 1.(k 1)[2(k 1) 1] (k 1)(2k 1)The formula gives the same result as adding the(k 1) term directly. Thus if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2.Since it is valid for n 2, it is also valid for n 3,and so on indefinitely. Thus, the formula is validfor all positive integral values of n.12. Step 1: Verify that the formula is valid for n 1.Since 1 is the first term in the sequence and (1)[3(1 ) 1]2 1, the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.S k ⇒ 1 4 7 … (3k 2) k(3k 1)2S k1 ⇒ 1 4 7 … (3k 2) (3k 1) k(3k 1)2 (3k 1) k(3k 1)2 2(3k 1)2 3k2 5k 22 (k 1)( 3k 2)2Apply the original formula for n k 1.(k 1)[3(k 1) 1]2 (k 1)( 3k 2)2The formula gives the same result as adding the(k 1) term directly. Thus if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2,n 3, and so on indefinitely. Thus, the formula isvalid for all positive integral values of n.429 Chapter 12
13. Step 1: Verify that the formula is valid for n 1.Since 1 2 is the first term in the sequence and1 2 1 1 1 2 , the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.S k ⇒ 1 2 1 4 1 8 1 1… 2 k 2 k 1S k1 ⇒ 1 2 1 4 1 8 1 1… 215. Step 1: Verify that the formula is valid for n 1.Since 1 is the first term in the sequence and1[2(1) 1][2(1) 1]3 1, the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.S k ⇒ 1 2 3 2 5 2 … (2k 1) 2 k(2k 1)(2k 1) 3k 2 k1S k1 ⇒ 1 2 3 2 5 2 … (2k 1) 2 (2k 1) 211k(2k 1)(2k 1) 2 k 1 2 k1 (2k 1) 231k(2k 1)(2k 1) 3(2k 1) 2 22 k 1 2 k1 2321 2 k1 1 2 k1[k(2k 1) 3(2k 1)](2k 1) 31 2 k1 (2k 12 5k 3)(2k 1)3(2k 3)(k 1)(2k 1) 3Apply the original formula for n k 1.(k 1)[2(k 1) 1][2(k 1) 1] (k 1)(2k 1)(2k 3) 33The formula gives the same result as adding the(k 1) term directly. Thus if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2,n 3, and so on indefinitely. Thus, the formula isvalid for all positive integral values of n.16. Step 1: Verify that the formula is valid for n 1. 1)2Since S41 ⇒ 1 and 2 1 1 1, the formula is validfor n 1.Step 2: Assume that the formula is valid for n k k2 (k 1) 24 (k 1) 3and derive a formula for n k 1.S k2 (k 1) 2 4(k 1)3k ⇒ 1 2 4 … 2 k1 2 k 14S k1 ⇒ 1 2 4 … 2 k1 2 k 2 k 1 2 kWhen the original formula is applied for n k 1,the same result is obtained. Thus if the formula isvalid for n k, it is also valid for n k 1. Sincethe formula is valid for n 1, it is also valid forn 2, n 3, and so on indefinitely. Thus, theformula is valid for all positive integral values of n.14. Step 1: Verify that the formula is valid for n 1.Since 1 is the first term in the sequence and 12 (1 1) 24 1, the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.S k ⇒ 1 8 27 … k 3 k2 (kS k1 ⇒ 1 8 27 … k 3 (k 1) 3 (k 1)2 [k 2 4(k 1)]4(k 1)2 (k 2 4k 4)4 (k 1)2 (k 2)24(k 1) 2 [(k 1) 1] 2 (k 1)2 (k 2) 24 4Apply the original formula for n k 1.The formula gives the same result as adding the(k 1) term directly. Thus if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2,n 3, and so on indefinitely. Thus, the formula isvalid for all positive integral values of n. 2(2 k ) 1 2 k1 1When the original formula is applied for n k 1,the same result is obtained. Thus if the formula isvalid for n k, it is also valid for n k 1. Sincethe formula is valid for n 1, it is also valid forn 2, n 3, and so indefinitely. Thus, theformula is valid for all positive integral values of n.17. S n ⇒ 7 n 5 6r for some integer rStep 1: Verify that S n is valid for n 1.S 1 ⇒ 7 1 5 or 12. Since 12 6 2, S n is valid forn 1.Step 2: Assume that S n is valid for n k and showthat it is also valid for n k 1.S k ⇒ 7 k 5 6r for some integer rS k1 ⇒ 7 k1 5 6t for some integer t7 k 5 6r7(7 k 5) 7 6r7 k1 35 42r7 k1 5 42r 307 k1 5 6(7r 5)Thus, 7 k1 5 6t, where t 7r 5 is aninteger, and we have shown that if S n is valid,then S k1 is also valid. Since S n is valid for n 1,it is also valid for n 2, n 3, and so onindefinitely. Hence, 7 n 5 is divisible by 6 for allintegral values of n.Chapter 12 430
18. S n ⇒ 8 n 1 7r for some integer rStep 1: Verify that S n is valid for n 1.S 1 ⇒ 8 1 1 or 7. Since 7 7 1, S n is valid forn 1.Step 2: Assume that S n is valid for n k and showthat it is also valid for n k 1.S k ⇒ 8 k 1 7r for some integer rS k1 ⇒ 8 k1 1 7t for some integer t8 k 1 7r8(8 k 1) 8 7r8 k1 8 56r8 k1 1 56r 78 k1 1 7(8r 1)Thus, 8 k1 1 7t, where t 8r 1 is aninteger, and we have shown that if S n is valid,then S k1 is also valid. Since S n is valid for n 1,it is also valid for n 2, n 3, and so onindefinitely. hence, 8 n 1 is divisible by 7 for allintegral values of n.19. S n ⇒ 5 n 2 n 3r for some integer rStep 1: Verify that S n is valid for n 1.S 1 ⇒ 5 1 2 1 or 3. Since 3 3 1, S n is valid forn 1.Step 2: Assume that S n is valid for n k and showthat it is also valid for n k 1.S k ⇒ 5 k 2 k 3r for some integer rS k1 ⇒ 5 k1 2 k1 3t for some integer t5 k 2 k 3r5 k 2 k 3r5 k 5 (2 k 3r)(2 3)5 k1 2 k1 3(2 k ) 6r 9r5 k1 2 k1 2 k1 3(2 k ) 6r 9r 2 k1 3(2 k ) 15r 3(2 k 5r)Thus, 5 k1 2 k1 3t, where t 2 k 5r is aninteger, and we have shown that if S n is valid,then S k1 is also valid. Since S n is valid for n 1,it is also valid for n 2, n 3, and so onindefinitely. Hence, 5 n 2 n is divisible by 3 for allintegral values of n.20. Step 1: Verify that the formula is valid for n 1.Since a is the first term in the sequence and 1 2 [2a (1 1)d] a, the formula is valid forn 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.S k ⇒ a (a d) (a 2d) … [a (k 1)d] k 2 [2a (k 1)d]S k1 ⇒ a (a d) (a 2d) … [a (k 1)d] (a kd) k 2 [2a (k 1)d] (a kd)k[2a (k 1)d] 2(a kd)22ak k(k 1)d 2a 2kd 2(k 1)2a [k(k 1) 2k]d 2 (k 1)2a (k2 k)d2(k 1)2a k(k 1)d 2 (k 1)2(2a kd)Apply the original formula for n k 1. (k 1)2{2a [(k 1) 1]d} (k 2(2a kd)The formula gives the same result as adding the(k 1) term directly. Thus if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2,n 3, and so on indefinitely. Thus , the formula isvalid for all positive integral values of n.21. Step 1: Verify that the formula is valid for n 1.Since 1 2 is the first term in the sequence and1 1 1 1 2 , the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.1S k ⇒ 2 13 3 1 k4… k(k 1) k 11S k1 ⇒ 1 12 2 13 3 114… k(k 1) (k 1) (k 2)k1 k 1 (k 1) (k 2)k(k 2) 1 ( k 1) (k 2)k 2 2k 1(k 1)(k 2)(k 1) 2 (k 1)(k 2) k 1k 2Apply the original formula for n k 1.( k 1) (k 1) 1 k 1k 2The formula gives the same result as adding the(k 1) term directly. Thus if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2,n 3, and so on indefinitely. Thus, the formula isvalid for all positive integral values of n.22. S n ⇒ 2 2n1 3 2n1 5r for some integer rStep 1: Verify that S n is valid for n 1.S 1 ⇒ 2 2(1)1 3 2(1)1 or 35. Since 35 5 7, S n isvalid for n 1.Step 2: Assume that S n is valid for n k and showthat it is also valid for n k 1.S k ⇒ 2 2k1 3 k1 5r for some integer rS k1 ⇒ 2 k3 3 2k3 5t for some integer t2 2k1 3 2k1 5r2 2k1 5r 3 2k12 2k1 2 2 (5r 3 2k1 )(3 2 5)2 2k3 45r 25r 3 2k3 5(3 2k1 )2 2k3 3 2k3 45r 25r 3 2k3 5(3 2k1 ) 3 2k3 20r 5(3 2k1 ) 5(4 3 2k1 )Thus, 2 2k3 3 2k3 5t, where t 4 3 2k1 isan integer; and we have shown that if S n is valid,then S k1 is also valid. Since S n is valid for n 1,it is also valid for n 2, n 3, and so onindefinitely. Hence, 2 2n1 3 2k1 is divisible by 5for all integral values of n.1)431 Chapter 12
23. Step 1: Verify that the formula is valid for n 1.Since S 1 ⇒ [r(cos v i sin v] 1 or r(cos v i sin v)and r 1 [cos (1)v i sin (1)v] r(cos v i sin v), theformula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.That is, assume that [r(cos v i sin v)] k r k (cos kv i sin kv).Multiply each side of the equation byr(cos v i sin v).[r(cos v i sin v)] k1 [r k (cos kv i sin kv] [r(cos v i sin v)] r k1 [cos kv cos v (cos kv(i sin v) i sin kv cos v i 2 sin kv sin v] r k1 [(cos kv cos v sin kv sin v) i(sin kv cos v cos kv sin v)] r k1 [cos (k 1)v i sin (k 1)v]When the original formula is applied forn k 1, the same result is obtained. Thus ifthe formula is valid for n k, it is also valid forn k 1. Since the formula is valid for n 1, itis also valid for n 2, n 3, and so onindefinitely. Thus, the formula is valid for allpositive integral values of n.24a.24b. 4 1 39 4 516 9 71, 3, 5, 7, …24c. 2n 124d. n 224e. Step 1: Verify that the formula is valid for n 1.Since 1 is the first term in the sequence and1 2 1, the formula is valid for n 1.Step 2: Assume that the formula is valid forn k and derive a formula for n k 1.S k ⇒ 1 3 5 7 … (2k 1) k 2S k1 ⇒ 1 3 5 7 … (2k 1) (2k 1) k 2 (2k 1) k 2 2k 1 (k 1) 2When the original formula is applied forn k 1, the same result is obtained. Thus ifthe formula is valid for n k, it is also valid forn k 1. Since the formula is valid for n 1,it is also valid for n 2, n 3, and so onindefinitely. Thus, the formula is valid for allpositive integral values of n.25. S 1 ⇒ n 2 5n 2r for some positive integer rStep 1: Verify that S 1 is valid for n 1.S 1 ⇒ 1 2 5 1 or 6. Since 6 2 3, S 1 is valid forn 1.Step 2: Assume that S n is valid for n k and showthat it is valid for n k 1.S k ⇒ k 2 5k 2r for some positive integer rS k1 ⇒ (k 1) 2 5(k 1) 2t for some positiveinteger t(k 1) 2 5(k 1) k 2 2k 1 5k 5 (k 2 5k) (2k 6) 2r 2(k 3) 2(r k 3)Thus, k 2 5k 2t, where t r k 3 is aninteger, and we have shown that if S n is valid,then S k1 is also valid. Since S n is valid for n 1,it is also valid for n 2, n 3, and so onindefinitely. Hence, n 2 5n is divisible by 2 for allpositive integral values of n.26a. Number of people Number of Interactions2 1 2 13 1 2 3 34 1 2 3 4 6n n(n 1)226b. Step 1: Verify that S n ⇒ 0 1 2 3 … (n 1) n(n 1)2is valid for n 1.Since 0 is the first term in the sequence and 1(1 1)2 0, the formula is valid for n 1.Step 2: Assume that the formula is valid forn k and derive a formula for n k 1.S k ⇒ 0 1 2 … (k 1) k(k 1)2S k1 ⇒ 0 1 2 … (k 1) k k(k 1)2 kApply the original formula for n k 1.(k 1)[(k 1) 1] 2 k(k 1)2 k(k 1 ) 2k2 k2 k 2k2 k(k 1)2The formula gives the same result as adding the(k 1) term directly. Thus if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid forn 2, n 3, and so on indefinitely. Thus, theformula is valid for all positive integral valuesof n.26c. Yes; 15 people will require 15( 14)2or 105interactions and last approximately 105(0.5) or52.5 minutes.Chapter 12 432
27. Step 1: Verify that S n ⇒ (x y) n x n nx n1 y n(n 1)2!)(n 2)xn2y2 n(n 13 !x n3 y 3 … y n isvalid for n 1. Since S 1 ⇒ (x y) 1 x 1 y 1 orx y, S n is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1. 1)S k ⇒ (x y) k x k kx k1 y k(k 2!x k2 y 2 k(k 1 )(k 2)3! x k3 y 3 … y kS k1 ⇒ (x y) k (x y) (x y)(x k kx k1 y k(k 1)2!)(k 2)xk2y2 k(k 13 !x k3 y 3 … y k )(x y) k1 x(x k kx k1 y k(k 1)2!x k2 y 2 k(k 1 )(k 2)3! k(k 1)2!x k3 y 3 … y k ) y(x k kx k1 y )(k 2) xk2y2 k(k 1 x 3 !k3 y 3 … y k ) x k1 kx k y k(k 1)2!x k1 y 2 …xy k x k y kx k1 y 2 k(k 1)2!x k2 y 3… y k1 xk1 (k 1)xky kxk1y2 k(k 2 !x k1 y 2 … y k1 1) x k1 (k 1)x k y k(k 2!x k1 y 2… y k11)When the original formula is applied for n k 1,the same result is obtained. Thus if the formula isvalid for n k, it is also valid for n k 1. Sincethe formula is valid for n 1, it is also valid forn 2, n 3, and so on indefinitely. Thus, theformula is valid for all positive integral values of n.28a. 0.9 0.09 0.009 …28b. 190 n28c. S a 1 a 1 r n1 r9 9 1 0 1 0 1 n 1 0 11 1 09 1 n 1 0 1 1 0 91 0 1 110 1 10 n 110n28d. Step 1: Verify that S n : 0.9 0.9 0.009 …9 1 0 n 10 n 110nis valid for n 1.Since 0.9 is the first term in the sequence and 10 11101 0.9, the formula is valid for n 1.Step 2: Assume that the formula is valid forn k and derive a formula for n k 1.9 k 1S k ⇒ 0.9 0.09 0.009 … 1S k1 ⇒ 0.9 0.09 0.009 … 19 10 k 1 910k 10 k1 10(10 k 1) 910k1 10k1 10 910k1 10 k1 110k10 k 10 10k0 k 9 10 k1 When the original formula is applied forn k 1, the same result is obtained. Thus ifthe formula is valid for n k, it is also valid forn k 1. Since the formula is valid for n 1,it is also valid for n 2, n 3, and so on inindefinitely. Thus, the formula is valid for allpositive integral values of n.28e. lim 10 k 11n→10n lim 1 n→1 0 n 1 0 or 1Thus, 0.999… 1.29. z 1 2(4 i) i 8 iz 2 2(8 i) i 16 iz 3 2(16 i) i 32 i30. 64 6 (29 1)d70 28d 5 2 d31. 25x 2 4y 2 100x 40y 100 025(x 2 4x) 4(y 2 10y) 10025(x 2) 2 4(y 5) 2 100 100 10025(x 2) 2 4(y 5) 2 1002) 2 5) 2 1 (x 4 (y 25ellipse32. 250, k 4A (25 0) 2 4(4) 12,271.85 m 233. 2 2; y 3 4 sin 2x34. cos 1 32 30° Since 3 is2negative, x isin the secondor third quadrants.x 180° 30° or 150°x 180° 30° or 210°35. Since the areaof the square Eis 25, each sidemeasures 5.AC60˚1Since AE and BE measure 1 unit, AH and BFeach measure 4 units. AB is the hypotenuse ofan isosceles right triangle with legs that measure1 unit. So AB measures 2 units. CA is thehypotenuse of an isosceles right triangle withlegs that measure 4 units. So CA measures42 units. Since ABCD is a rectangle, theperimeter is 2(2 42) 2 (52) 102 units.The correct choice is B.2FH B GD30˚ 3433 Chapter 12
Chapter 12 Study Guide and AssessmentPage 829 Understanding the Vocabulary1. d 2. i 3. m 4. j5. k 6. f 7. c 8. e9. b 10. hPages 830–831 Skills and Concepts11. d 4.3 3 or 1.35.6 1.3 6.9, 6.9 1.3 8.2,8.2 1.3 9.5, 9.5 1.3 10.86.9, 8.2, 9.5, 10.812. a 20 5 (20 1)(3)5213. 4 6 (5 1)d10 4d2.5 d6 (2.5) 3.5, 3.5 (2.5) 1,1 (2.5) 1.56, 3.5, 1, 1.5, 414. d 23 (30) or 7a 14 30 (14 1) 7 61S 14 1 42(30 61) 21715. S n n 2 [2a 1 (n 1)d]250.2 n 2 [2(2) (n 1)(1.4)]250.2 2n 0.7n(n 1)0 0.7n 2 1.3n 250.21.3 (1.3) 3 .7)(2 4(050.2)2(0.7) 1.3 26.51.4n n 18 or n 19.86Since n is a positive whole number, n 18.716. r 4 9or 1 7 1 1 7 1 7 , 1 7 1 7 1 1 4, 9 4 9 1 7 1 3 43 1 7 1 1, 4, 9 3 43 17. a 15 2.2(2) 151 36,044.818. 8 a 1 (0.2) 718 0.000064a 1125,000 a 119. 125 0.2r 51625 r 45 r0.2(5) 1, 1(5) 5, 5(5) 250.2, 1, 5, 25, 12520. r 2.41.2or 2S 9 1.2 1.2( 2) 91 ( 2) 1.2 614.43 205.221. r 4 24 or 2S 8 4 4( 2) 81 222. lim23. lim60 1 21 21 2 60( 1 2)1 2 601 23nn→4n 1 6n 3n→n 3 4 limn→ 6 n 6 0 6n limn→ n3 nn324. Does not exist; lim 2 n→3n3 lim 2n→3; since limbecomes increasingly large as n approachesinfinity, the sequence has no limit. 4 n3 3nn→n4 4n325. lim lim n→ lim n→ 0 1 01226. 5.123 5 11231a 1 1, r 10S n 5 27. r 1500000 1231 00011 10 00 5 1 2399941 53 3304260or 0.4260 0.4S n 11 210028. a n n 5 n 2, an1 (n 5 (n 1) 5n21n→n25 nr lim lim 5n (n 2nn→5 n2 5 n22 lim limnn→ 5 n2 1 5 0 0 1 5 3 n n 4 n 1 n n4nn 34 3 nn 4 n 4n4 4 n3n43 123000 1,0 00,000 1) n211) 2 nn→ n2 lim …1n→ n 2convergent29. The general term is n 5 5nor 1 . n5 11 n n for all n, so divergent230. The general term is . n n2 n1 for all n, so divergentn 2 nn→3Chapter 12 434
31. 9(3a 3) (3 5 3) (3 6 3) (3 7 3)a 5 (3.8 3) (3.9 3) 12 15 18 21 24 9032. (0.4) k (0.4) 1 (0.4) 2 (0.4) 3 … (0.4) k10.4S 1 0.4 2 3 33. (2n 1) 34. 91(n 2 1)a0a135. (a 4) 6 a 6 6! 1!(6 a 5 (4) 1 6! 2!(6 a 4 1)! 2)!! 4)! (4) 2 6! 3!(6 a 3 (4) 3 6 4!(6 3)!! 5)! a 2 (4) 4 6 5!(6 a 1 (4) 5 6! 6!(6 6)! a 0 (4) 6 a 6 24a 5 240a 4 1280a 3 3840a 2 6144a 409636. (2r 3s) 4 (2r) 4 4! 1!(4(2r) 3 (3s) 110!37. 4!(1 0 4)!8!38. 2!(8 2)!10!39. 7!(1 0 7)!12!40. 5!(1 2 5)!4! 2!(4 2)!4! 3!(4 3)!4! 4 !0! 1)!(2r) 2 (3s) 2 (2r)(3s) 3 (2r) 0 (3s) 4 16r 4 96r 3 s 216r 2 s 2 216rs 3 81s 4 x 104 (2) 4 210 x 6 16 3360x 6 4m 82 1 2 28 4096m 6 1 114,688m 6 x 107 (3y) 7 120 x 3 2187y 7 262,440x 3 y 7 (2c) 125 (d) 4 792 128c 7 (d) 541. 2cos 3 4 i sin 3 4 2e i 3 442. 4i 4cos 2 i sin 2 4e i 2 43. r 22) 2 ( 2 or 22v Arctan 22 or 7 42 2i 22cos 7 4 i sin 7 4 22 e i 7 444. r (33) 2 3 2 or 6v Arctan 333 or 6 33 3i 6cos 6 i sin 6 6e i 6 45. f(2) 6 3 2 0f(0) 6 3 0 6f(6) 6 3 6 12f(12) 6 3 (12) 420, 6, 12, 42 101,376c 7 d 5 46. f(3) (3) 2 4 13f(13) 13 2 4 173f(173) 173 2 4 29,933f(29,933) 29,933 2 4 895,984,49313; 173; 29.933; 895, 984, 49347. z 0 4iz 1 0.5(4i) (4 2i) 2i 4 2i 4z 2 0.5(4) (4 2i) 2 4 2i 6 2iz 3 0.5(6 2i) (4 2i) 3 i 4 2i 7 3i48. z 0 8z 1 0.5(8) (4 2i) 4 4 2i 2iz 2 0.5(2i) (4 2i) i 4 2i 4 3iz 3 0.5(4 3i) (4 2i) 2 1.5i 4 2i 6 3.5i49. z 0 4 6iz 1 0.5(4 6i) (4 2i)2 3i 4 2i 2 iz 2 0.5(2 i) (4 2i) 1 0.5i 4 2i 5 1.5iz 3 0.5(5 1.5i) (4 2i) 2.5 0.5i 4 2i 6.5 2.75i50. z 0 12 8iz 0.5(12 8i) (4 2i) 6 4i 4 2i 10 6iz 2 0.5(10 6i) (4 2i) 5 3i 4 2i 9 5iz 3 0.5(9 5i) (4 2i) 4.5 2.5i 4 2i 8.5 4.5i51. Step 1: Verify that the formula is valid for n 1.Since the first term in the sequence is 1 and 1(1 1)2 1, the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive a formula for n k 1.S k ⇒ 1 2 3 … k k(k 1)2S k1 ⇒ 1 2 3 … k (k 1) k(k 1)2 2(k 1)2 k2 k2 2k 22 k2 k 2k 22 k2 3k 22 (k 1) (k 2)2Apply the original formula for n k 1.(k 1)[(k 1) 1] 2 (k 1) (k 2)2The formula gives the same result as adding the(k 1) term directly. Thus, if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2,n 3, and so on indefinitely. Thus, the formula isvalid for all positive integral values of n.435 Chapter 12
52. Step 1: Verify that the formula is valid for n 1. 54b. S 12 1 2Since the first term in the sequence is 3 and1(1 1)(2 1 7) 2304 ft 6 3, the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand derive formula for n k 1.S k ⇒ 3 8 15 … k(k 2) k(k 1) (2k 7)6S k1 ⇒ 3 8 15 … k(k 2) (k 1)(k 3) k(k 1) (2k 7)6 k(k 1) (2k 7)6 6(k 1 )(k 3)6(k 1)[(k 1) 1][(2(k 1) 7]6 (k 1)(k 3)(k 1)(k 2)(2k 9)6The formula gives the same result as adding the(k 1) term directly. Thus, if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2,n 3, and so on indefinitely. Thus, the formula isvalid for all positive integral values of n.53. S n ⇒ 9 n 4 n 5r for some integer rStep 1: Verify that S n is valid for n 1.S 1 ⇒ 9 1 4 1 or 5. Since 5 5 1, S n is valid forn 1.Step 2: Assume that S n is valid for n k and showthat it is also valid for n k 1.S k ⇒ 9 k 4 k 5r for some integer rS k1 ⇒ 9 k1 4 k1 5t for some integer t9 k 4 k 5r9 k 4 k 5r9(9 k ) (4 k 5r)(4 5)9 k1 4 k1 5(4 k ) 20r 25r9 k1 4 k1 4 k1 5(4 k ) 20r 25r 4 k1 5(4 k ) 45r 5(4 k 9r)Thus, 9 k1 4 k1 5t, where t 4 k 9r is aninteger, and we have shown that if S n is valid,then S k1 is also valid. Since S n is valid for n 1,it is also valid for n 2, n 3, and so onindefinitely. Hence, 9 n 4 n is divisible by 5 for allintegral values of n.Page 833 Applications and Problem Solving54a. a 1 16, d 48 16 or 32A 12 16 (12 1)32 368 ft2(16 368)55. If the budget is cut 3% each year, 97% remainsafter each year.a 1 160,000,000, r 0.97a 11 160,000,000(0.97) 111 $117,987,860.3056a. One side of the original triangle measures 6 3 or2 units. Half of 2 units is 1 unit. Each side of thenew triangle measures 1 unit, so its perimeter is1 1 1 or 3 units.56b. ak(k 1)(2k 7) 6(k 1)(k 3)1 6, r 3 6 or 1 2 66S (k 1)[k(2k 7) 6(k 3)] 1 162 (k 11)(2k2 7k 6k 18)6(k 1)(2k2 13k 18)6Page 833 Open-Ended Assessment(k 1)(k 2)(2k 9)6Apply the original formula for n k 1.1a. Arithmetic; arithmetic sequences have commondifferences, while geometric sequences havecommon ratios.1b. Sample answer: 1, 4, 7, 10, …; a n 1 3(n 1)2. Sample answer: 6 5n 23n; lim 6 5n 2n→3n limn→ 2 n 5 32lim n→n 0, but since lim 5 n n→3 becomes increasinglylarge as n approaches infinity, the sequence hasno limit.Chapter 12 SAT & ACT PreparationPage 835 SAT and ACT Practice1. If 40% of the tapes are jazz then 60% of the tapesmust be blues. There are 80 tapes. Find 60% of 80.0.60(80) 48The correct choice is D.2. Because of alternate interior angles,150 130 unmarked angle of right 20 unmarked angle of right In the right triangle, x 20 90 or x 70.The correct choice is C.3. fraction pumped d gallonsk gallons d k Change this fraction into a percent by multiplyingby 100.percent pumped d k 100 or 10 0d %kThe correct choice is A.4. First calculate the number of caps Andrei has now.He starts with 48 and gives away 13, so he has48 13 35 left. Then he buys 17, so he has35 17 52. Then he trades 6 caps for 8 caps.this leaves him with 52 6 8 or 54. His total isnow 54.n Chapter 12 436
48Percent increase 54 48 100 12.5%The correct choice is B.5. Since the figure is not drawn to scale, do notassume that the two lines are parallel, eventhough they may appear parallel. Since AB AC,ABC is isosceles. So m∠B m∠ACB.m∠B m∠ACB 80 1802 m∠ACB 100m∠ACB 50Since AD is a line segment, x 70 50 180.So, x 60.Consider right triangle CDE.x y 9060 y 90y 30x y 60 30 or 30The correct choice is C.6. The population of Rockville is now 20,000 and willdouble every 8 years. So in 8 years the populationwill be 40,000 and in 16 years will be 80,000. Sof(8) 40,000 and f(16) 80,000. Choice A isincorrect since f(8) 20,000. Choice B is incorrectsince f(16) 2(20,000) 2 or 800,000,000. Choice Cis incorrect since f(8) . Choice E is20,000 64incorrect since f(8) 20,000. For Choice D,f(8) = 40,000 and f(16) = 80,000.The correct choice is D.7. Notice that the figure is not drawn to scale. ∠Acould be a right angle. To be sure it is, find theslope of AB and compare it to the slope of AC.Slope of AB 1 0 46 5 6 1 Slope of AC 1 6 Since the slopes are negative reciprocals, the linesegments are perpendicular, so m∠A 90.Therefore, ABC is a 30°-60°-90° right triangle.The hypotenuse, AC, is twice the length of the legopposite the 30° angle, AB.AB (6 ) 50 2 (1) 4 2 1 6 3 or 37AC 237The correct choice is D.8. Method 1: Substitute each answer choice for x totest both inequalities.A: (6) 6 0 and 1 2(6) 1.0 0 and 13 1; falseMethod 2: Solve each inequality for x.x 6 0 and 1 2x 1x 62x 2x 1The solution is 6 x 1. All of the answerchoices except A are in this range.The correct choice is A.9. The increase from 99 to 100 is 1. So the percent1increase from 99 to 100 is .9 911 is greater than , or 1%.9 91 00The correct choice is A.10. Choose a number for the total number of cars inthe parking lot. Since the fractions havedenominators of 2, 4, and 5, choose a number thatis divisible by 2, 4, and 5. Let the number of carsin the parking lot equal 40. 1 5 40 8 blue cars 1 2 8 blue cars 4 blue convertibles 1 4 the number of convertibles 4the number of convertibles 16Not Blue and Not Convertible8 Blue 4 Convertible16The number of cars that are neither blue norconvertible is the total number minus the bluecars minus the convertibles plus the number thatare both blue and convertible.Neither blue nor convertible 40 8 16 4 20percent that are neither blue nor convertible 2 040 100 50%The answer is 50.437 Chapter 12
Chapter 13 Combinatorics and Probability13-1Permutations and CombinationsPages 842–843 Check for Understanding1. Sample answer: Both are used to determine thenumber of arrangements of a group of objects.However, order of the objects is important inpermutations. When order of the objects is notimportant, combinations are computed.2. Select 2 jacks out of 4—C(4, 2)Select 3 queens out of 4—C(4, 3)number of hands—C(4, 2) C(4, 3)3. Sam is correct. The room assignments are anordered selection of 5 rooms from the 7 rooms. Apermutation should be used.4.blue ————— S-blueS green ———— S-greengray ————— S-grayblue ———— M-blueM green———— M-greengray ———— M-grayblue ————— L-blueL green ———— L-greengray————— L-grayblue ———— XL-blueXL green ——— XL-greengray ———— XL-gray5. Using the Basic Counting Principle,4 3 5 5 300.6. independent6!7. P(6, 6) (6 7205!8. P(5, 3) (5 6)!6 5 4 3 2 1 13)! 5 4 3 2 12 1 6012!9. P ( 12,8)(12 8)!P(6,4) ——6! (6 4)! 1 2!2!6!4!12 11 10 9 8 7 6 5 4 3 2 1 2 1 6 5 4 3 2 1 4 3 2 1 55,4407!10. C(7, 4) (7 11. C(20, 15) 3520! (20 15!) 15!20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 5 4 3 2 1 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 15,5044)! 4!7 6 5 4 3 2 13 2 1 4 3 2 14!12. C(4, 3) C(5, 2) (4 5!3)!3! (5 2)! 2! 4 3 2 11 3 2 1 5 4 3 2 13 2 1 2 1 4 10 or 4013. Using the Basic Counting Principle,10 9 8 7 6 5 4 3 2 1 3,628,800.15!14. C(15, 9) (15 500515a. Using the Basic Counting Principle,10 10 10 10 10 100,000.15b. Using the Basic Counting Principle,1 9 9 9 10 729015c. Using the Basic Counting Principle,(10 10 10 10 10 10 10 10 10) (10 10 10 10 10) 999,900,000Pages 543–545 Exercises16. Using the Basic Counting Principle, 2 6 4 48.7!17. P(7, 7) (7 504018a. Using the Basic Counting Principle,9 10 10 10 10 10 10 9,000,000.18b. Using the Basic Counting Principle,5 5 5 5 5 5 5 78,125.18c. Using the Basic Counting Principle,10 10 10 10 10 10 1 1,000,000.18d. Using the Basic Counting Principle,1 1 1 10 10 10 10 10,000.19. dependent 20. independent21. dependent8!22. P(8, 8) (8 40,3206!23. P(6, 4) (6 3605!24. P(5, 3) (5 3)! 5 4 3 2 12 1 607!25. P(7, 4) (7 7)!7 6 5 4 3 2 118)!8 7 6 5 4 3 2 114)!6 5 4 3 2 12 14)!7 6 5 4 3 2 13 2 1 8409)! 9!15 14 13 12 11 10 9 8 7 6 5 4 3 2 16 5 4 3 2 1 9 8 7 6 5 4 3 2 1Chapter 13 438
26. P(9, 5) (9 9! 15,12010!27. P(10, 7) (105)!9 8 7 6 5 4 3 2 14 3 2 1 604,8006!28. P ( 6,3)(6 3)!P(4,2)—4! (4 2)! 6 ! 2!4!3!6 5 4 3 2 1 2 14 3 2 1 3 2 1 106!29. P ( 6,4)(6 4)!P(5,3)—5! (5 3)! 6 ! 2!5!2!6 5 4 3 2 1 5 4 3 2 1 66! 7!30. P(6, 3) P(7, 5) (6 3)! (7 5)!P(9,6) ———9! (9 6)! 6 ! 7!3!9!3!2!6 5 4 3 2 1 7 6 5 4 3 2 19 8 7 6 5 4 3 2 1 2 1 55!31. C(5, 3) (5 3)! 3! 5 4 3 2 12 1 3 2 1 1010!32. C(10, 5) (10 2524!33. C(4, 2) (4 2)! 2! 4 3 2 12 1 2 1 612!34. C(12, 4) (12 4959!35. C(9, 9) (9 114!36. C(14, 7) (14 7)!10 9 8 7 6 5 4 3 2 13 2 15)! 5!10 9 8 7 6 5 4 3 2 15 4 3 2 1 5 4 3 2 19)! 9! 34323!37. C(3, 2) C(8, 3) (3 4)! 4!12 11 10 9 8 7 6 5 4 3 2 18 7 6 5 4 3 2 1 4 3 2 17)! 7!14 13 12 11 10 9 8 7 6 5 4 3 2 17 6 5 4 3 2 1 7 6 5 4 3 2 18!2)! 2! (8 3)! 3! 3 2 11 2 1 3 56 or 1688 7 6 5 4 3 2 15 4 3 2 1 3 2 138. C(7, 3) C(8, 5)7!8! (7 3)! 3! (8 5)! 5!7 6 5 4 3 2 14 3 2 1 3 2 1 35 56 or 196039. C(5, 1) C(4, 2) C(8, 2)5!4!8! (5 1)! 1! (4 2)! 2! (8 2)! 2! 5 4 3 2 14 3 2 1 1 4 3 2 12 1 2 1 5 6 28 or 84014!40. C(14, 4) (14 100114!41. C(14, 5) (14 200242. C(18, 12) 18,56443. C(3, 2) C(5, 1) C(8, 2)3!5!8! (3 2)! 2! (5 1)! 1! (8 2)! 2! 3 2 11 2 1 5 4 3 2 14 3 2 1 1 3 5 28 or 42011!44. P(11, 11) (11 39,916,80013!45a. C(13, 3) C(13, 2) (13 13!3)! 3! (13 2)! 2! 13 123 2 11 1 13 122 1 286 78 or 22,30845b. C(4, 1) C(4, 2) C(4, 2)4!4!4! (4 1)! 1! (4 2)! 2! (4 2)! 2! 4 3 2 13 2 1 1 4 3 2 1 2 12 1 4 3 2 12 1 2 1 4 6 6 or 14412!45c. C(12, 5) (12 8 7 6 5 4 3 2 13 2 1 5 4 3 2 14)! 4!14 13 12 11 10 9 8 7 6 5 4 3 2 110 9 8 7 6 5 4 3 2 1 4 3 2 15)! 5!14 13 12 11 10 9 8 7 6 5 4 3 2 19 8 7 6 5 4 3 2 1 5 4 3 2 118! (18 12)! 12!18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 16 5 4 3 2 1 12 11 10 9 8 7 6 5 4 3 2 1 1)!11 10 9 8 7 6 5 4 3 2 115)! 5!12 11 10 9 8 7 6 5 4 3 2 17 6 5 4 3 2 1 5 4 3 2 1 79246a. Using the Basic Counting Principle,10 10 10 10 10 100,00046b. Using the Basic Counting Principle,10 9 8 7 6 30,24046c. Using the Basic Counting Principle,5 5 4 4 4 1600.5! 4!P(5, 2) P(4, 3) (5 (4 2)! 3)! 5 4 3 3 2 2 1 1 4 3 2 11 20 24 or 4808 7 6 5 4 3 2 16 5 4 3 2 1 2 18 7 6 5 4 3 2 16 5 4 3 2 1 2 1439 Chapter 13
47. C(3, 1) C(4, 1) C(6, 1) C(14, 6)3!4!6!14! (3 1)! 1! (4 1)! 1! (6 1)! 1! (14 6)! 6! 3 2 12 1 1 4 3 23 2 11 1 6 5 4 3 2 15 4 3 2 1 114 13 12 11 10 9 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 6 5 4 3 2 1 3 4 6 3003 or 216,21642!48a. P(42, 42) (42 42)! 42! 1.4 10 512!30)! 30!42!12 ! 30!48b. C(42, 30) (42 4 1.1 10 1048c. C(5, 3) C(12, 6) C(10, 6) C(15, 5)5!12!10!15! (5 3)! 3! (12 6)! 6! (10 6)! 6! (15 5)! 5! 10 924 210 3003 5.8 10 949. P(n, n 1) P(n, n)n!n! [n ( n 1)]! (n n)! n !1! n !0!n! n!6!50a. P(6, 6) (6 0! 1 72050b. There are 3 ways to arrange the 3 couples, and 2ways to arrange each of the two members withina couple.P(3, 3) P(2, 2) P(2, 2) P(2, 2)3! 2! 2! 2! (3 3)! (2 2)! (2 2)! (2 2)! 3 2 11 2 1 1 2 1 1 2 1 1 6 2 2 2 or 4811!51a. C(11, 4) (11 3306!51b. C(6, 2) C(5, 2) (6 10!52. C(10, 2) (10 6)!6 5 4 3 2 1 1 454)! 4!11 10 9 8 7 6 5 4 3 2 17 6 5 4 3 2 1 4 3 2 15!2)! 2! (5 2)! 2!6 5 4 3 2 1 5 4 3 2 1 4 3 2 1 2 1 3 2 1 2 1 15 10 or 1502)! 2!10 9 8 7 6 5 4 3 2 18 7 6 5 4 3 2 1 2 153a. 35 52 6 or 37(2 5 9) 59253b. Yes; let h, t, and u be the digits.100h 10t u100h 10u t100t 10h u100t 10u h100u 10t h 100u 10h t200(h t u) 20(h t u) 2(h t u) 222(h t u) 222(h t u)6 37(h t u)54. 2140 (1.058) $2264.122264.12(1.058) $2395.442395.44(1.058) $2534.3855. 10n 3 1 3 2 3 . . . 10 3 3025n156. 7.1 x 83.1x ln 7.1 ln 83.1x l n 83.1ln7.1x 2.2657. x e 0.346 Use a calculator. 1.458. y 4x 2x sin 45° y cos 45° 4(x cos 45° y sin 45°) 2 2x 2y 2 24 2x 22 2y 259.150˚180˚120˚2x 2y 8 1 2 x 2 xy 1 2 y 2 0 4(x) 2 8xy 4(y) 290˚60˚1 2 3 430˚ 2x 2y210˚330˚240˚ 270˚300˚r 2, r 2 cos 2v2 2 cos 2v1 cos 2v2v 0° or 2v 360°v 0° v 180°(2, 180°), (2, 0°)60. v x 28 cos 45°, v y 28 sin 45° 28 22 28 22 19.80 ft/s 19.80 ft/s61. sin 2x 2 sin x 02 sin x cos x 2 sin x 02 sin x (cos x 1) 02 sin x 0 or cos x 1 0sin x 0 cos x 1x 0°, x 180°, x 360°, or x 180°So, x 0°, 180°, 360°62. y 8 cos (v 30°)amplitude 8period 36 0°1or 360°phase shift 30°0˚Chapter 13 440
63. Find B.B 180° 90° 27° or 63°Find a.atan 27° 15 .215.2 tan 27° a7.7 aFind c.cos 27° 15 .2c15.2c co s 27°c 17.164. Each hour, an hour hand moves through 3 6012or30°. Since 12 minutes is 1 5 of an hour, the hourhand moves through an additional 1 5 (30) or 6°.2(30°) 6° 66°The correct choice is A.13-2Permutations with Repetitionsand Circular PermutationsPages 848–849 Check for Understanding1. The circular permutation has no beginning or end.Therefore, the number of different arrangements1is always n of a revolution.2. Sample answer: house or phone numbers wheresome of the digits repeat3. Sample answer: The number of permutations of ncharms on a bracelet with a clasp is n !.28!4. 2! 2! 10,0809!5. 2! 2! 22,6802! 2!14!6. 2! 4 ! 5!2! 7,567,5607. circular; (11 1)! or 3,628,8008. circular; (8 1)! or 50409. circular; (12 1)! or 39,916,80010. linear; 5 ! or 6029!11. 5! 7562!2!Pages 849–851 10,08012. 2!8!2!10!13. 2 ! 2!14. 8 !2!10!15. 3 ! 2!12!16. 2! 2!2!9!17. 4! 2!2!7!18. 2! 2!2! 907,200 20,160 302,400 59,875,200 3780 630Exercises9!19. 5! 4! 12650! 20. 2.35 10 484! 3! 4! 2! 8! 8! 3! 2! 2! 3! 2! 4!21. circular; (12 1)! 39,916,80026!22. linear; 6! 3! 5.1 10 127!10!23. circular; (9 1)! 40,32024. circular; (5 1)! 2425. circular; (8 1)! 504026. linear; 6! 72027. linear; 10! 3,628,80028. circular; (9 1)! 40,32029. circular; (14 1)! 3,113,510,400230. circular; (20 1)! 1.22 10 1731. circular; (32 1)! 8.22 10 3332. linear; 25! 1.55 10 258!33. 2!2 !2!2! 252034a. (7 1)! 72034b. 7! 50401! 46,200135. 3! 4!3!11!36a. 2! 2!2! 4,989,60036b. integral calculus3! 7.85 10 17437a. 20! 14!9!37b. (43 1)! 1.41 10 5137c. 43! 6.04 10 5238. Let n total number of symbols.Let n 3 number of dashes.n! 3! (n 35 3)!n(n 1)(n 2)(n 3)!(n 3)! 210n 3 3n 2 2n 210 0Use a graphing calculator to find the solution atn 7.6!39. C(6, 3) (6 2040. (5x 1) 3 C(3, 0) (5x) 3 (1) 0 C(3, 1) (5x) 2 (1) 1 C(3, 2) (5x) 1 (1) 2 C(3, 3) (5x) 0 (1) 3 125x 3 75x 2 15x 141. x log 2 413x lo lg10413og1023)! 3!6 5 4 3 2 13 2 1 3 2 1x 8.6942. h 6, k 13 h p3 6 p3 p(y k) 2 4p(x h)(y 1) 2 12(x 6)43. 2(4 3i)(7 2i) 2(28 29i 6i 2 ) 56 58i 12(1) 44 58i441 Chapter 13
uiu u j k44. u v u w 2 0 3 2 5 0 0 3 u i 2 3 u j 2 0 u k5 0 2 0 2 515i u 6j u 10k u 15, 6, 10since 2, 0, 3 15, 6, 10 30 0 30 or0 and 2, 5, 0 15, 6, 10 30 30 0 or 0,then the resulting vector is perpendicular to u vand u w .45. x cos 45° y sin 45° 8 0 2x 2y 8 02 22x 2y 16 0y 45˚8410. P(s) C(4, C1) C(3, 2)( 7,3) 4 335 1 235 2 335 1 2352—3 2 3351) C(4, 2)( 7,3) 3 635 1 835 1 735 1 835— 87 1 735000or 4 5 odds or 1 211. P(s) C(3, Codds or 1 112. P(rain) 18P(f) 1 P(s) 1 1325 P(f) 1 P(s) 1 13P(not rain) 1 4 5 1 5 85 84 O448xodds 1 5 — 4 5 or 1 4 8The sparks will be highest at the y-intercept, 82inches above the center of the wheel. This is82 8 or about 3.31 inches above the wheel.46. If x 2 36, then x 6 or x 6.2 x1 2 61 or 322 x1 2 61 or 1128 The correct choice is E.13-3Probability and Odds855–856 Check for Understanding1. The probability of the event happening is 5050.2. Answers will vary; see students’ work.3. Sample answer: The probability of the successfuloutcome of an event is the ratio of the number ofsuccessful outcomes to the total number ofoutcomes possible. The odds of the successfuloutcome of an event is the ratio of the probabilityof its success to the probability of its failure.34. Geraldo is correct. P(win) 2 3 3 5 or 60%.7 75. P(softball) 3 7 11 2 1or 1 3 3 7 06. P(not a baseball) 3 07. P(golf ball) 3 7 11or 078. P(woman) 7 or 171 ( 4,3)47 11or 1 21 9. P(s) C C(7,3)P(f) 1 P(s)4 1 34 3 54 35 3 13 5odds —4or 315or 3 135 Pages 856–85813. P(face card) 4 5Exercises4 42 1 252 3or 1 3 6 62 2 452 6or 1 3 102 2 0 552or 1 310 10 10 10 52 4 052 or 1 0135 1 0or 1 2 22 3 1 0or 1 5 2 72 3or 1 03 82 3 1 0or 4 5 ( 4,2)( 40,2)14. P(a card of 6 or less) 6 6 515. P(a black, non-face card) 10 516. P(not a face card) 517. P(red) 5 2 3218. P(white) 5 19. P(not pink) 5 5 20. P(red or pink) 5 5 21. P(2 pop) CC6 7 80 1or 1 30 C(8,2)22. P(2 country) C23. P(1 rap and 1 rock) 24. P(not rock) C C( 40,2)287 7 80or 1 95 C(10, 1) C(18, 1)C(40, 2) 10 18735 1 803780or 1 3( 22,2)( 40,2) 2 3177780or 2 6025. Using the Basic Counting Principle, there are 1 1or 1 way to roll both fives. Using the Basic1Counting Principle, P(both fives) 3.6Chapter 13 442
26. P(s) C C(odds 3 1 5 1 5 27. P(s) C Codds ( 3,2)6,2)or 1 4 ( 4,2)( 6,2)6 1 5 2 5 28. P(s) C(1, Codds 1 5 — 4 5 2 5 — 3 5 or 2 3 1) C(3, 1)( 6,2) 1 3153 1 5or 1 5 1 5 — 4 5 29. P(s) C(1, Cor 1 4 P(f) 1 P(s) 1 1 5 4 5 P(f) 1 P(s) 1 2 5 3 5 P(f) 1 P(s) 1 1 5 4 5 1)( 6 C(2, 1), 2) C(1, 1)C ( 6 C(3, 1), 2) C(2, 1) C(3, 1)C ( 6,2)2 3 6 11 1 5 1 5 1 5or 1 5P(f) 1 P(s)1 1 11 45 or 1 511 1 5—41 5( 11,3)( 27,3)165 2 92511 1 95 1 84195111 95—1 1 8 4 841 95C(13, 2) C(11, 1)C(27, 3) 7 8 112925858 2 925or 2 275 5 375 2 275—23 5 37 5( 14,3)( 27,3)364 2 92528 2 25 1 97225282 25—8 1 9 7 972 25odds or 1 41 C30. P(s) CP(f) 1 P(s)11 1 1odds or 119531. P(s) P(f) 1 P(s)odds or 2 5 1 2 732. P(s) C CP(f) 1 P(s)28 1 2odds or 1225 2533. P(s) C(3, Codds or 291 134. P(s) 1 249or 2 50 P(f) 1 P(s) 1 3922535. P(s) 4 5 P(f) 1 P(s)odds 1) C(24, 2)( 27,3) 3 276292582892 2 925or 3 25 2 33325923 25—2 2 3 333 325 4 5 — or 4 1 1 5 C(13, 3) C(13, 2)C(52, 5)286 78 2 ,598,960 12267,696 2 ,598,960429 4 16536. P(s) P(4, 2)P(f) 1 P(s)429 1 4165or 3 73641654294 165—29736 3 73641654or 1 5 odds or 34 1 4 5 or 1 5 137. P(s) 1 38. P(s) 0.325 P(f) 1 P(s) 1 0.325 or 0.675odds 0 . 3250.675or 1 327139a. P(s) 1 0 1 9 1 8 1or 7 20 1 139b. P(f) 1 0 1 0P(s) 1 P(f)1 1 101 1 0or 10100 odds — or 99 140a. P(both males) 1 2 1 2 or 1 4 40b. P(s) 1 1 2 P(f) 1 P(s)odds 41a. P(s) 41b. P(s) C(15 C99900or 1 0009991 0009 10100 1 _ 2 or 1 1 1 2 1 2 1 1 2 or 1 2 C(15, 10) C(5, 0)C(20, 10) 3 003 1 21184,756or 1 292 , 8) C(5, 2)( 20,10) 6 435 10184,75664,350 1 84,756 4 21646 2 25646 2 25646—25 4 2 1 216 46odds or 2 4P(f) 1 P(s) 1 262546443 Chapter 13
42a. P 179,820 151,32284,475 3273 179,820 151,322 3 31,142418,890 0.79142b. P(f) 1 4odds 51,32218,89043. P A Q 2 67,568418,890—— 1 51,3224 18,890 2 67,568151,322or 1 33,78475,661P(s) 1 P(f) 1 1 51,322418,890 2 67,568418,890x 8xuGiven a pipe PQ and a random cut point, A,APAQ 18. If AP is x inches long, then AQ is 8xuinches long. Now, the cut must be made along APso that the longer piece will be 8 or more times aslong as the shorter piece. Thus, the probabilityu xthat the cut is on AP is x 8x 1 9 . Since the cutcan be made on either end of the pipe, the actualprobability is 2 9 .44. This is a circular permutation.(6 1)! 5! or 12010!45. C(10, 4) (10 4)! 4!10 9 8 7 6 5 4 3 2 16 5 4 3 2 1 4 3 2 1 21046. S n n 2 [2a 1 (n 1)d]S 14 1 42[2(3.2) (14 1)1.5] 7(25.9) 181.347. Let y 7 log 7 2xlog 7 y log 7 2xy 2xSo, 7 log 7 2x 2x.48. Center: (7, 2)r 2 (10 7) 2 (8 2) 2 3 2 (10) 2 or 109(x 7) 2 (y 2) 2 10949. r 3 2 or 6v ( 4 ) or 5 46(cos 5 4 i sin 5 )4 6 22 232 32i50. u 3, 5 4, 2 3 (4), 5 2 1, 32 i51. Drawing the altitude from one vertex to theopposite side forms a 30°-60°-90° right trianglewith hypotenuse 2s. The short side of this righttriangle measures s. So the altitude drawn haslength s3. This is the height of the equilateraltriangle. The base measures 2s. So the area of theequilateral triangle is 1 2 2s s3 3s2 .The correct choice is B.2sPage 858 Mid-Chapter Quiz15!1. P(15, 5) (15 360,36020!2. C(20, 9) (20 9)! 9! 167,9603. Using the Basic Counting Principle,26 26 26 10 10 10 10 175,760,000.12!4. P(12, 5) (12 95,0405. Using the Basic Counting Principle,18 18 3 6 5832.! 181,4406. 9 2!10!7. 3! 4!3! 42008. This is a circular permutation.(8 1)! 50409. P(both hearts) C ( 13,2)C(52,2)10. P(s) C(3, C78 1 326 or 1171) C(3, 1)( 12,2) 3 3669 3 6 6or 2 2 1 922 32 2— 1 9 92 2odds or 1313-4ss 3s2s 5)!15 14 13 12 11 10 9 8 7 6 5 4 3 2 110 9 8 7 6 5 4 3 2 1 5)!12 11 10 9 8 7 6 5 4 3 2 17 6 5 4 3 2 1P(f) 1 P(s) 1 232 Probabilities of CompoundEventsPages 863–864 Check for Understanding1. The occurrence of one event does not affectanother for independent events. The occurrence ofthe first event affects the occurrence of a secondfor dependent events.Chapter 13 444
2a.4 4 720. dependent, 1 5 1 4 1 3 8diamonds spades, 1 95 clubs,21. dependent, 2 652 2 551 2 450 2 349 2 248 2 147 2 046 1 945 1 844 aceshearts 1 743 1 642 1 541 1 4 1940 1,16 0,05422. dependent, 1 228 8 287 2 6 32 8 195 4 7 3523. independent, 2b. No, one of the aces can be an ace of diamonds.1 6 1 6 1 6 1 024 24. independent2c. P(ace or diamond) P(ace) P(diamond) P(ace and diamond)P(winning) 4 7 3. Answers will vary; see students’ work.P(winning next four games) 4 7 4 7 4 7 4 7 6 3 14. independent, 3 6 3 6 2567 2 245. dependent, 1 0 3 4019 2 1 525. inclusive, 16. dependent, 4 6 3 5 2 6 1 6 1 11 3 6 3 65 426. inclusive, 54 67. exclusive, 1 3 1 3 1 02 2 652 522 173 1327. inclusive, 2 68. inclusive, 1 527 1 6 217 2 7 2 52 1 252 562 183 027 28. exclusive4 4 8 2P(at least 3 males)9. exclusive, 5 2 5 2 5 2or 1 3 P(3 males) P(4 males) P(5 males)10. P(selecting 5 even numbers) 3 775 3 674 3 573 3 472 3 371 C(5, 3)C(9 C(4, 2), 5) C(5, 4) 0.025C ( 9 C(4, 1), 5) C(5, 5) C(5, 0)C ( 9,5)6020111. P(selecting 5 two digit numbers) 6 675 6 574 6 473 6 372 6 2171 26 1 26 1 26 819 0.5181 26or 1 429. exclusive12. P(5 odd numbers or 5 multiples of 4)P(sum of 6 or sum of 9) P(sum of 6) P(sum of 9) P(5 odd numbers) P(5 multiples of 4) 3 875 3 774 3 673 3 572 3 471 1 775 1 674 1 573 1 472 1 5371 3 6 346 9 0.029 0.0004 3 6or 1 4 0.02930. exclusive13. P(5 even numbers or 5 numbers less than 30) P(at least three women)P(5 even numbers) P(5 numbers less than 30) P(3 women) P(4 women) P(5 women) P(5 even numbers and 5 numbers less than 30)P(6 women) 3 775 3 674 3 573 3 472 3 371 2 975 2 874 2 773 2 672 2 571 C(7, 3)C(14 C(7, 3), 6) C(7, 4) C ( 14 C(7, 2), 6) C(7, 5) C(7, 1)C ( 14,6)1 475 1 374 1 273 1 712 1 701 C(7, 6) C(7, 0)C(14,6) 0.025 0.007 0.001 1 22573514773003 3 003 3 003 30 03 0.0322114 14. P(none if 6 clocks are damaged)3 003or 3 0242996 1 00 9 599 9 498 9 397 9 296 9 131. exclusive95P(at least 4 tails) 4 35,643 P(4 tails) P(5 tails) P(6 tails)560,17515. P(at least 1 right handed pitcher C(6, 4) 1 2 6 C(6, 5) 1 2 6 C(6, 6) 1 2 6P(1 right-handed pitcher) P(2 right-handed 1 5pitchers)64 1 664 6 4 C(8, 1) C(5, 1)C(13 , 2) C(8, 2) C(5, 0)C 2 264 1or 312( 13,2) 4 078 2 832. inclusive78 6 878 or 3 4 4 352 5 1 2 652 2 551 2 152 5 1 3912 2 652 50 266 52 26252 66055 2 652or 2 21Pages 864–867 Exercises33. exclusive16. dependent, 5 9 4 8 5 1P(at least 2 rock) P(2 rock) P(3 rock)817. independent, 5 9 5 9 2 5 C(6, 2)81C(11 C(5, 1), 3) C(6, 3) C(5, 0)C ( 11,3)18. independent, 1 6 1 6 17520 3 61 65 1 6519. dependent, 4 7 3 6 2 7 959 165or 1 33 445 Chapter 13
34. P(all red cards) 2 652 2 551 2 450 2 349 2 248 253 9 99635. P(both kings or both aces) P(both kings) P(both aces) 4 352 5 112 2 652 12 2 652 24 2 652 2or 2 21 54 32 5 1 36. P(all diamonds) 1 352 1037. P(both red or both queens) P(both red) P(both queens) P(both red queens) 2 652 2 551 4 352 5 1 2 152 5 1 65012 2 652 2 652 2 26 52 66055 2 652or 2 215 438. P(2 pennies) 2 1 2 01 2 139. P(2 nickels or 2 silver-colored coins) 7 621 2 0 2 40420or 4 7 1 261 1 520 27 61 2 0 40. P(at least 1 nickel) 1 P(no nickels) 1 1 241 1 320 1 730 41. P(2 dimes or 1 penny and 1 nickel) P(2 dimes) P(1 penny and 1 nickel)C(9,2) C ( 21,2) C(5, 1) C(7, 1)C(21,2)3635 2 10 2 1071 2 10542. P(all female) 1 0 4 9 3 8 2 7 1 4 243. P(all female or all male) P(all female) P(all male)5 1 01 1 4 2 4 22 1 4 2or 2 1 4 9 3 8 2 7 5 1 4 9 3 8 2 7 44. P(at least 3 females) P(3 females) P(4 female) C(5, 3)C(10 C(5, 1), 4) C(5, 4) C(5, 0)C ( 10,4)505 2 10 2 10 5511 2 10or 4 245. P(at least 2 females and at least 1 male) P(2 females and 2 males) P(3 females and 1 male) C(5, 2)C(10 C(5, 2), 4) C(5, 3) C(5, 1)C ( 10,4) 1 0050210 2 10 1 50210or 5 7 046. P(word processing or playing games) P(word processing) P(playing games) P(both) 2 5 1 3 1 4 2 960 47. P(rain or lightning) P(rain) P(lightning) P(both) 3 5 2 5 1 5 4 5 48. P(even sum) P(3 even cards) P(2 odd cards and 1 even card) C(5, 3)C(9 C(4, 0), 3) C(4, 2) C(5, 1)C ( 9,3) 1 084 3 084 4 084 or 1 02149. P(at least 3 women) P(3 women) P(4 women) P(5 women) C(6, 3)C(13 C(7, 2), 5) C(6, 4)C ( 13 C(7, 1), 5) C(6, 5) C(7, 0)C ( 13,5)4201056 1 287 1 287 12 8753159 1 287or 1 4350. P(at least 1 doctor) P(1 doctor) P(both doctors) 933100 1 00 7 7 190 0 19589021 10 , 000 1 0,0009979 1 0,00000 1939700 1 0051. P(supplies or money) P(supplies) P(money) P(both)52a.812625375 2 500 2 500 2 500 1 0625312500or 1 250P(A)P(C)P(B)52b. P(A or B or C) P(A) P(B) P(C) P(A andB) P(A and C) P(B and C) P(A and B andC). You must add the intersection of all threesets which have not been accounted for.53. P(action video or pop/rock CD or romance DVD) 4 7 1 2 5 1 1 2 9 1 7 1 4 414 0.93Chapter 13 446
54a. First consider the probability that no 2 studentshave the same birthday. The first person in theclass can have any birthday; there are 366choices out of 366 days. The second person hasonly 365 choices out of 366 days, and so on.So, P(2 students with the same birthday) 1 P(no 2 students have the same birthday). 1 3 3 1 1 6666 3 65366 3 64366 3 66!348! 366 1866,18)6618 . . . 3 34966 1 P(3 3 0.34654b. 1 P(3 66,n)366n 1 2 54c. In part a, there is only a 0.346 probability that 2students have the same birthday. This is toosmall. Substitute numbers greater than 18 for nin the inequality of part b. When n is 23, P isabout 0.51. So, 23 18 or 5 more students areneeded in the class.55a. inclusive55b.55c. P(flooded engine or dead battery) P(flooded engine) P(dead battery) P(both) 1 2 2 5 110 4 5 56. P(two threes given a sum of six) 1 5 57. (7 1)! 72058. Let b basket. Let m miss.Expand (b m) 20 20 20! r!(2b 20 r m rr00 r)!Find the coefficient of the b 15 m 5 term where r 5.20! 5!(2b 205 m 5 15,504b 15 m 50 5)!15,50459. No, the spill will spread no more than 2000 metersaway.480a 1 1200; r 1 200or 0.4a1s 1 r1200 1 0.04 2000366! (366 348)! 366 18DeadBatteryFloodedEngine60. 12 x 2 3 x 4(x 2) log 12 (x 4) log 3x log 12 2 log 12 x log 3 4 log 3x log 12 x log 3 4 log 3 2 log 12x (log 12 log 3) 4 log 3 2 log 12x x 6.754961. Let y income and x number of $1.00 increases.income (number of customers) (costof a ticket)y (400 20x)(3 x)y 1200 340x 20x 2y 1200 20(x 2 17x)y 1200 1445 20(x 2 17x 72.25)(y 2645) 20(x 8.5) 2The vertex of the parabola is (8.5, 2645). Anincrease of $8.50 will give a maximum profit of$2645. The price of each ticket should be 3 8.5or $11.50.62. A 248270 4k2( 7),560 2 231 271.5 yards63. x x 1 ta 1 y y 1 ta 2x 1 t(2)y 5 t(4)x 1 2ty 5 4tx 1, y 5 t2, 464. 2 tan x 4 0tan x 2tan 1 (tan x) tan 1 2x 63° 2665. Since ADB CBD and they are alternateinterior angles, AD L BC. Simply because a 45does not mean b 45, so you cannot conclude that 3 bisects ABC.The correct choice is B.13-5Conditional ProbabilityPages 870–871 Check for Understanding1. Sample answer: If A and B are independentevents, then P(A⏐B) P(A). Thus, the formulafor conditional probability becomes P(A) P(A andB)P(B)4 log 3 2 log 12log 12 log 3 or P(A) P(B) P(A and B). This is theformula for the probability of independent events.2. S {J spades, Q spades, K spades, J clubs,Q clubs, K clubs}3. Answers will vary; see students’ work.4. P(cubes match ⏐ sum greater than 5) 4 3 6 — 2 636 2 1 3447 Chapter 13
5. P(queen ⏐ face card) 45 2— 1 25 1 3 6. P(all heads ⏐ first coin is a head) 7. P(all heads ⏐ at least 1 head) 1 7 8. P(all heads ⏐ at least 2 heads) 1 8 — 4 8 1 4 1 8 — 7 8 1 8 — 4 8 1 4 23 6 — 1 036 1 5 9. P(numbers match ⏐ sum greater than or equal to 9)10. P(sum is even ⏐ sum greater than or equal to 9)43 6 — 1 03 6 2 5 11. P(numbers match or sum is even ⏐ sum greaterthan or equal to 9) 12a. P(disease prevented) 1643 6— 1 03 6 2 5 8 6200 100 1 320 682 00— 1 0 02 00 1 725 12b. P(disease prevented ⏐ vaccine) 12c. P(disease prevented ⏐ conventional treatment)622 00 — 1 02 000 3 150 13a. P(legal ⏐ accepted) 13b. P(rejected ⏐ legal) 691 00—701 00 6 970 6 1 00 —751 002 2 52 13c. P(not rejected ⏐ counterfeit) Pages 872–874Exercises14. P(1 head ⏐ at least 1 tail) 15. P(Democrat ⏐ man) 16. P(first bag ⏐ first chip is blue) 1 1 00 —251 001 2 5 2 4 — 3 4 2 3 41 2—81 2 1 2 4 1 6 — 1 016 2 5 1 224 — 2 024 3 5 3 2 15!4 3 2 1 4 3 2 15!17. P(girls are separated ⏐ girl at an end) 18. P(number end in 52 ⏐ number is even) 1 8 2 072— 3 27 2 5 8 25 2— 2 652 1 1 325 2— 2 65 21 1 365 2— 2 65 2 3 1 305—2 2 65 2 015 2— 2 652 1 2 619. P(2 odd numbers ⏐ sum is even) 20. P(ace ⏐ black) 21. P(4 ⏐ black) 22. P(face card ⏐ black) 23. P(queen of hearts ⏐ black) 24. P(6 of clubs ⏐ black) Chapter 13 448
36. A the sum of the cards is 7 or less25. P(jack or ten ⏐ black) —2B at least one card is an ace 2 652 B both cards not an ace2 1P(B) C ( 48,2)3C(52,2) 1 8822126. P(second marble is green ⏐ first marble wasP(B) 1 P(B) 1 1 8833221 2 21green) 2 7 27. P(second marble is yellow ⏐ first marble wasgreen) 3 8 5 7 — 3 8 5 7 5 8 4 7 — 5 8 28. P(second marble is yellow ⏐ first marble is yellow) 4 7 C(1, 1) C(1, 1) C(4, 1)C(6, 3) C(1, 1) C(5, 2)C(6,3)4 1 0or 2 5 C(1, 1) C(1, 1) C(3, 1)C(6, 3)29. P(salmon ⏐ bass) 30. P(not walleye ⏐ trout and perch) 3 4 31. P(bass and perch ⏐ not catfish)3 1 0 C ( 5,3)C(6,3)32. P(perch and trout ⏐ neither bass nor walleye)33. 3 8 2 7 —C(1, 1) C(1, 1) (4, 1)C(6, 3)C(1, 1) C(1, 1) C(3, 1)C(6, 3)C(1, 1) C(1, 1) (2, 1)C(6, 3) C ( 4,3)C(6,3) 2 4 or 1 2 Brown Hair 3 8 54Brown EyesP(A and B) P(A ⏐ B) P(A PC(4, 2) C(4, 1) C(20, 1)C(52, 2)43 6 63andB)(B)436 63—332 21 4 399 37. P(sum greater than 18 ⏐ queen of hearts)38a.1 5 2 1 951 1 5 2 5 151 1 951 —C38b. P(cancer ⏐ smokes) 252 00—302 00 5 6 39. A person buys somethingB person asks questionsP(A ⏐ B) 15 25 5 1 20500—or 4 5 1 55 000Four out of five people who ask questions willmake a purchase. Therefore, they are more likelyto buy something if they ask questions.40. Sample answers: The rolls are independent. Thenumber cubes do not have a memory, whetherthey are fair or biased. Probability does notguarantee an outcome.41. P(passes ⏐ studied) S U155P(passes and studied)P(studied)50% 10% 20%20%Exercises 33–35P(brown eyes ⏐ brown hair) 0 0. 1 6 . 106034. P(no brown hair ⏐ brown eyes) 0 0. 2 3 . 203035. P(no brown eyes ⏐ no brown hair) 0 0. 1 2 . 2040 4 25 3 P(studied)P(studied) 2 3 5 4 or 5 6 6000 3or 530 021 1 000 —66 1 000 7 2 234000or 4 67500642a. P(defective) 142b. P(chip from 3-D Images ⏐ defective)42c. P(functioning) 19449 Chapter 13
42d. Sample answer: A chip from CyberChip Corp.has the least probability of being defective.25P(defective from CyberChip) 5 00 0.0521P(defective from 3-D Images) 3 00 0.0720P(defective from MegaView Designs) 2 00 0.1043. P(A ⏐ B) P(A and BP(B)by definition.So, if P(A) P(A ⏐ B) then by substitution P(A) P(A andB)P(B) or P(A and B) P(A) P(B). Therefore,the events are independent.44. P(at least 3 women) P(3 women) P(4 women) P(5 women) C(6, 3)C(13 C(7, 2), 5) C(6, 4)C ( 13 C(7, 1), 5) C(6, 5) C(7, 0)C ( 13,5)4201056 1 287 1 287 12 8753159 1 287or 1 4345. C(9, 4) 12646. 3(0.5) b 3(0.5) 1 3(0.5) 2 3(0.5) 3 . . .b1 1.5 0.75 0.375 . . .a1r 1 . 50.5S a 1 1.5, r 0.5 or 347. They are reflections of each other over the x-axis.48.y53. xx (4 2x)(3 2x) 612 14x 4x 2 6xx4x 2 14x 6 02(2x 1)(x 3) 0xxx 1 2 or x 3xx4 ftIf x 3 ft then the photo would have a negativelength and width. So, x 1 2 ft or 6 in.54. f(x) x 25 46 ft 2 A E D5 (x 2) (x 2)f(x) is discontinuous when x 2;f(x) is undefined when x 2.55. Drop an altitude Bfrom B to AE, fromuC to ED , and fromE to BC. Labelhhthe diagram asshown.area of unshaded region 1 2 (BC)harea of shaded region 1 2 (AE)h 1 2 (ED)h 1 2 h(AE ED)BC AE ED since opposite sides of aparallelogram are equal. So, the ratio of the areasis 1 2 h(BC) 1 2 h(BC) or 11.The correct choice is B.Ch64264 O 2 4 646x13-6The Binomial Theoremand ProbabilityPage 877 Graphing Calculator Exploration1. S {0, 1, 2}49. r cos v 2 5 0rcos v cos 2 sin v sin 2 50 r sin v 5y 550. x 4t y 3 2tx 4 t y 2 tx 4 y 323y(0, 3)(4, 5)2. P (Bobby wins) 2 3 3. Answers will vary. In 40 repetitions, it may bearound 0.22. This means that there were exactly5 wins for 8 or 9 of the 40 repetitions.4. P(winning 5 games) C(6, 5) 2 3 5 1 3 1 0.265. There is not a large enough sample of trials.6. Increase the number of repetitions.y 2x 351. A 1 2 r 2 v 1 2 (8) 2 9 81 54.7 ft 271 1 80(4, 1)OxPage 878 Check for Understanding1a. Yes, it meets all three conditions.1b. No, there are more than 2 possible outcomes.1c. No, the events are not independent.2. Sample answer: the probabilities derived from asimulation rather than an actual event52. tan 27° 2x5 12.7 x; 12.7 mChapter 13 450
3. First, determine P(right) and P(wrong). Second,set up the binomial expansion (p r p w ) 5 . Third,determine the term of the expansion. Fourth,substitute the probability values for p r and p w .Last, compute the probability of getting exactly 2correct answers.4. P(only one 4) C(5, 1) 1 6 1 5 6 415. P(no more than 3 times correct) 1 P(correct 4 times) 1 C(4, 4) 2 3 4 1 3 0 1 1 1 681 1 1 1 681 3 125 6 587776 6 4 9 1 9 287 C(5, 2) 1 6 2 5 6 317. P(7 correct) C(10, 7) 1 3 1257776 3 1257776 1 2502 7 1 2 317776 120 1 6 2528 1 8 15648 1 28 C(5, 5) 1 6 5 5 6 025025 7 776 7 776 1 C(10, 10) 1 2 10 1 2 077 761 1123 210 6 4 1 6 120 1 6 481 10 5 12 1 2 1 10124 13861 1 02477 7670 5 3 010 1 9351216,80719. P(all incorrect) C(10, 0) 1 2 0 1 2 101 00,000 1 1 10124 C(5, 1) 3101 7 4110 10 24 36,015720320. P(at least half correct) 1 00,000or 2 0,0001 1 386 1 [P(rain on 4 days) P(rain on 5 days) 252 3 2 3 2 1 024 1 C(5, 4) 3104 7 110 C(5, 5) 3105 7102523860 1 024 1 024 1 2835243100,000 10 0,000 3 1951296,922 1 00,000or 4 8,46121. P(4 heads) C(4, 4) 1 50,0003 4 2 3 01 1 8 1 138407681 1 5,625or 3 1258 122. P(3 heads) C(4, 3) 1 3 3 2 3 11 0.1372 4 2 7 2 3 8 8 1Exercises23. P(at least 2 heads)1 1 1 6 18 19 4 9 8 1 8 1 8 11 2 48 181 89111 2 78 4 2 7 1 3 1 1 681 11 81 210 40 1 96 2 56627 0.0165. P(no more than two 4s) P(no 4s) P(one 4) P(two 4s) C(5, 0) 1 6 0 5 6 5 C(5, 1) 1 6 1 5 6 46. P(at least three 4s) P(three 4s) P(four 4s) P(5 fours) C(5, 3) 1 6 3 5 6 2 C(5, 4) 1 6 4 5 6 17. P(exactly five 4s) C(5, 5) 1 6 5 5 6 08. P(not having rain on any day) C(5, 5) 19. P(having rain on exactly one day)10. P(having rain no more than three days)11. P(4 do not collapse) C(6, 4) 4 5 4 1 5 212. P(10 stocks make money) C(13, 10) 5 8 10 3 8 3Pages 878–88013. P(never the correct color) C(4, 0) 2 3 0 1 3 414. P(correct at least 3 times) C(4, 3) 2 3 3 1 3 1 C(4, 4) 2 3 4 1 3 01 16. P(correct exactly 2 times) C(4, 2) 2 3 2 1 3 218. P(at least 6 correct) C(10, 6) 1 2 6 1 2 4 C(10, 7) 1 2 7 1 2 3 C(10, 8) 1 2 8 1 2 2 C(10, 9) 1 2 9 1 2 128 1 8 45 21 C(10, 5) 1 2 5 1 2 5 answer to 1856 1 4 C(4, 2) 1 3 2 2 3 2 answer to 22 answer to 2124. P(6 correct answers) C(10, 6) 1 4 6 3 4 4451 Chapter 13
25. P(half answers correct) C(10, 5) 1 4 5 3 4 5 252 10124243 1 024 0.05826. P(from 3 to 5 correct answers) C(10, 3) 1 4 3 3 4 7 C(10, 4) 1 4 4 3 4 6 C(10, 5) 1 4 5 3 4 51 21871 120 6 4 1 6,384 210 21 243 252 10 24 1 02456 729 4 096 0.25 0.15 0.06 0.4627. P(all point up) C(10, 10) 2 5 10 3 5 01024 1 9,7 65,625 1 1.049 10 428. P(exactly 3 point up) C(10, 3) 2 5 3 3 5 7 120 1825 2187 7 8,125 0.21529. P(exactly 5 point up) C(10, 5) 2 5 5 3 5 5 252 332125 243 3 125 0.20130. P(at least 6 point up) C(10, 6) 2 5 6 3 5 4 C(10, 7) 2 5 7 3 5 3 C(10, 8) 2 5 8 3 5 2 C(10, 9) 2 5 9 3 5 1 C(10, 10) 2 5 10 3 5 0 210 156 45 392 1 9,714 81,625 6 255690,625 2 502465,625 1 120 781 10 1,952827, 125 1 251253,125 3 5 0.16631. P(3 heads or 3 tails) C(3, 3) 1 2 3 1 2 0 C(3, 3) 1 2 3 1 2 0 1 1 8 1 1 1 8 1 1 4 32. P(at least 2 heads) C(3, 2) 1 2 2 1 2 1 C(3, 3) 1 2 3 1 2 0 3 1 4 1 2 1 1 8 1 4 8 1 2 33. P(exactly 2 tails) C(3, 2) 1 2 2 1 2 1 3 1 4 3 8 34a. The values of the function for 0 x 6 are thecoefficients of the binomial expansion.34b. Change 6 nCr X to 8 nCr X on the Y-menu. 1 2 35. Enter 16 nCr X on the Y-menu of your calculator.16 nCr X represents the coefficients of thebinomial expansion where X is the number ofgames won.P(winning at least 12 games) 1820 71012 3 410 560 71013 3 3 1 0 120 71014 3 210 16 71015 3 110 1 71016 310 1 0.45 or 45%36a. Asuccess means that a missile hits its target.There are 6 trials and the probability of successon each trial is 20% or 1 5 .36b. P(between 2 and 6 missiles hit the target) 1 [P(0 missiles hit the garget) P(1 missilehits the target)] 1 C(6, 0) 1 5 0 4 5 6 C(6, 1) 1 5 1 4 5 5 4096 1 1 1 1 5,625 6 1 5 1 3 1 077312502412537. P(all men or all women) C(10, 10) 41010 6 010 C(10, 10) 61010 140 0 0.006238a. P(all carry the disease) C(20, 20) 11020 9 010 1 10 2038b. P(exactly half have the disease) C(20, 10) 11010 9 1010 6.4 10 639. P(at least 2 people do not show up) 1 [P(0 people do not show up) P(1 person does not show up)] 1 C(75, 0) 4100 0 9675100 C(75, 1) 4100 1 9610074 0.80740. P(less than or equal to 3 policies) 1 P(4 policies) 1 C(4, 4) 1 2 4 1 2 0 1 1 1 1 1 1 5166 or about 0.9441a. If Trina walks 100 meters, then she has flippedthe coin 10 times. To end up where she began,she walked north and south 5 times each.P(back at her starting point) C(10, 5) 1 2 5 1 2 5 252 312 312 0.24641b. The closest Trina can come to her starting pointis if she flips 6 heads and 4 tails or 4 heads and6 tails. However, this places her 20 meters fromher starting point. The answer for part b is thesame as that for part a, 0.246.Chapter 13 452
41c. P(exactly 20 meters from the starting point) C(10, 6) 1 2 6 1 2 4 C(10, 4) 1 2 4 1 2 61 11 210 6 4 1 6 210 1 6 1 6 4 0.4142. P(sum less than 9 ⏐ both cubes are the same)43 6 —63 6 2 3 43. P(letter is contained in house or phone)5 5 3 2 6 2 6 2 67 2 644a. 80, 75, 70, . . .44b. T 80 5n44c. T 125n 4 0,0001 000or 40125 g 5(40)75 g; 75° F45. 3 x 1 6 x(x 1)log 3 x log 6x log 3 log 3 x log 6x log 3 x log 6 log 3x(log 3 log 6) log 3log3x log 3 log 6x 0.3846. h 0, k 3, a 7, b 5, c 26Center: (0, 3)Foci: (26, 3)Vertices: major axis → (7, 3) and (7, 3)minor axis → (0, 2) and (0, 8)47. 2cos 2 i sin 2 2 0 2 i 0 2iu48. WX 6 8, 5 (3) or 2, 8u⏐WX ⏐ (6 ) 8 2 (5 (3)) 2 (2) 2 8 249.71 cm 68 or 217106˚55 cmdd 2 71 2 55 2 2(71)(55) cos 106°d 101.1 cm71 cm74˚106˚55 cmdd 2 71 2 55 2 2(71)(55) cos 74°d 76.9 cm50. 4⏐ 1 12 21 62 724 32 44 721 8 11 18 ⏐ 0Since the remainder is 0, x 4 is a factor of thepolynomial.Area of page for textArea of entire page51. The answer is 7/12.63 1 08 172 Chapter 13 Study Guide and AssessmentPage 881 Understanding the Vocabulary1. independent 2. failure3. 1 4. probability5. permutation 6. permutation withrepetitions7. mutually exclusive 8. sample space9. conditional 10. combinatoricsPages 882–884 Skills and Concepts11. Using the Basic Counting Principle, 3 2 1 or 6.12. Using the Basic Counting Principle, 5 4 3 2 1or 120.13. Using the Basic Counting Principle,6 5 4 3 2 1 or 720.6!14. P(6, 3) (6 1208!15. P(8, 6) (8 3)!6 5 4 3 2 13 2 1 20,1605!16. C(5, 3) (5 3)! 3! 5 4 3 2 12 1 3 2 1 1011!17. C(11, 8) (11 6)!8 7 6 5 4 3 2 12 18)! 8!11 10 9 8 7 6 5 4 3 2 13 2 1 8 7 6 5 4 3 2 1 1656!18. P ( 6,3)(6 3)!P(5,3) 5! (5 3)!6 5 4 3 2 1 2 1 5 4 3 2 1 3 2 1 25!19. C(5, 5) C(3, 2) (5 (9 1 1)(12 1.5 1.5)9 12 1 1 3 1 33!5)! 5! (3 2)! 2! 2 1 2 1453 Chapter 13
20. There are P(5, 5) ways to arrange the other booksif the dictionary is on the left end. The same istrue if the dictionary is on the right end.5!2 P(5, 5) 2 (5 5)! 2 1 2 5 4 3 1 2403!21. C(3, 2) C(7, 2) (3 5!22. 2! 2! 5 4 3 2 12 1 2 1 3010!23. 2! 3!3!24. 8 !2!6!25. 3! 2!9!26. 3! 2! 50,400 20,160 60 30,24027. P(3 pennies) 7!2)! 2! (7 2)! 2! 3 2 11 2 1 3 21 or 6328. P(2 pennies and 1 nickel) 29. P(3 nickels) 30. P(1 nickel and 2 dimes) 31. P(s) 116odds 11 6— 11 561 1 5; P(f) 1 1332. P(s) 2 0;P(f) 1 2odds 70 3 1 7133. P(s) 1 40 ; P(f) 1 1odds 8 7 6 5 4 3 2 12 16 5 4 3 2 13 2 1 2 19 8 7 6 5 4 3 2 13 2 1 2 132 0— 1 72 01 1 40 — 1 391401 1 39 10 9 8 7 6 5 4 3 2 12 1 3 2 1 3 2 1C(7, 3) C(4, 0) C(5, 0)C(16, 3) 35 1 15601 1 6C(7, 2) C(4, 1) C(5, 0)C(16, 3) 21 4 15603 2 0C(7, 0) C(4, 3) C(5, 0)C(16, 3) 1 4 15601 1 40 C(7, 0) C(4, 1) C(5, 2)C(16, 3) 1 4 105601 1 45639407 6 5 4 3 2 15 4 3 2 1 2 1134. P(s) 1;4P(f) 1 1odds 114— 11 341 1 335. independent, P(sum of 2) P(sum of 6)341 5 3 6 3 65 12 9636. dependent, P(two yellow markets)4 1 0 3 9 2 1 537. P(selecting a prime number or a multiple of 4) P(prime number) P(a multiple of 4)6 3 1 4 1 49 1 438. P(selecting a multiple of 2 or a multiple of 3) P(multiple of 2) P(multiple of 3) P(multiple of 2 and 3)7 4 2 1 4 1 4 1 49 1 439. P(selecting a 3 or a 4) P(3) or P(4)1 1 1 4 1 4 1 7 40. P(selecting an 8 or a number less than 8) P(8) P(less than 8)1 7 1 4 1 4 4 7 41. P(sum less than 5 ⏐ exactly one cube shows 1)43 6— 1 03 2 5 43—653 6 4 5 43 6— 3 036 42. P(different numbers ⏐ sum is 8) 43. P(numbers match ⏐ sum is greater than or equalto 5)6 2 1 544. P(exactly 1 head) C(4, 1) 1 2 1 1 2 3 4 1 2 1 4 1 8 45. P(no heads) C(4, 0) 1 2 0 1 2 41 1 1 1 61 1 6Chapter 13 454
46. P(2 heads and 2 tails) C(4, 2) 1 2 2 1 2 2 6 1 4 3 8 1 4 47. P(at least 3 tails) P(3 tails) P(4 tails) C(4, 3) 1 2 3 1 2 1 C(4, 4) 1 2 4 1 2 0 4 1 8 1 2 1 1 1 6 1 1 4 1 5 1 6or 1 6Page 885 Applications and Problem Solving48. C(1, 1) C(6, 4) 1 15 or 1549. 7 ! 2520250. P(at least 1 good chip) 1 P(both defective chips)C(3,2) 1 C ( 15,2)1 1 3 5 3 435 751a. P(female name excluding Reba) 1 5151b. P(Reba’s name, then a male name) 1 5 7 1 41 3 0Page 885 Open-Ended Assessment1. Yes; sample answer: x 1 2 1 1 2 , so x 1 6 ; Twomarbles are chosen from a box containing 6 red, 4blue, and 2 green marbles. What is the probabilityof choosing a red and a green marble?2. Sample answer: In a permutation, the order ofobjects is very important. In a combination, theorder of objects is not important.Chapter 13 SAT & ACT PreparationPage 887 SAT and ACT Practice1. You might want to draw a diagram of the 20 coins.The first and last coins are heads. The totalnumber of heads is 10. There could be 9consecutive heads followed by 10 tails and thenthe final head. The correct choice is D.H H H HH H H HHH2. 4x – 3 8 324x – 3 24x – 3 6(x – 3) 2 6 2x – 3 36x – 3 39The correct choice is E.3. Find the probability of selecting a green marblefrom the jar now.number of green marblestotal marbles3 1 5or 1 5 Let x represent the number of green marblesadded so the probability equals 2 1 5 or 2 5 .number of green marblestotal marbles3 x 1 5 x 2 5 5(3 x) 2(15 x)15 5x 30 2x3x 15x 5The correct choice is C.sumofterms4. Average nu20 mberoftermssum of five terms 5sum of five terms 100Since one of the numbers is 18, the sum of theother four is 100 18 or 82. The correct choiceis C.5. Select specific numbers for the problem. Letx y 8. Let x z 12. Let z 7.x 7 12, so x 5, an odd number.5 y 8, so y 3, an odd number.Statement I is false, and statement III is true.Since y z or 3 7 10, an even number,statement II is true also. The correct choice is E.6. Since the probability of selecting a blue marble is 1 5 and the total number of blue and white marblesis 200, the number of blue marbles must be 40. Sothe number of white marbles must be 160. After100 white marbles are added, the total number ofwhite marbles is 260 and the total of all marblesis 300. The probability of selecting a white marbleis 2 60300or 1 31.5The correct choice is E.7. ∠A and ∠C must be equal because they arecorresponding angles. The correct choice is A.8. The sample space, or total possible outcomes, isthe 52 cards in a deck. The outcome “drawing adiamond” consists of the 13 cards that arediamonds.P(diamond) 1 352 or 1 4 The correct choice is C.9. 7 different entrees are offered. 3 are chosen.The number of combinations that can be chosen7!7!is C(7, 3) 35.3!(7 3)! 3 !4!The correct choice is B.455 Chapter 13
10.12The probability that a dart thrown randomly atthe target will land in the shaded region is equalto the ratio of the area of the shaded region to thearea of the entire target.P P P area of shaded regionarea of target (1) 2 (2) 2 4or1 41The answer is 4.Chapter 13 456
Chapter 14 Statistics and Data Analysis14-1The Frequency Distribution6f. Sample answer:Ages of PresidentsPages 892–893 Check for Understanding1. A line plot, a bar graph, a histogram, and afrequency polygon all show data visually. A lineplot shows the frequency of specific quantities byusing symbols and a bar graph shows thefrequency of specific quantities by using bars. Ahistogram is a special bar graph in which thewidth of each bar represents a class interval. Afrequency polygon shows the frequency of a classinterval using a broken line graph.2. Choose an appropriate class interval. Use tallymarks to determine the number of elements ineach class interval.3a. No; there would be too many classes.3b. Yes; there would be 9 classes.3c. Yes; there would be 5 classes.3d. No; there would only be 3 classes.3e. No; there would only be 2 classes.4. See students’ work.5a.Age70+1900 60-69200050-5940-4930-3920-2910-190-916 12 8 4 0 0 4 8 12 16Percent5b. In 1999, there are larger percents of oldercitizens than in 1990.6a. range 69 42 or 276b. Sample answer: 56c. Sample answer: 40, 45, 50, 55, 60, 65, 706d. Sample answer: 42.5, 47.5, 52.5, 57.5, 62.5, 67.56e. Sample answer:Ages Frequency40–45 245–50 650–55 1255–60 1260–65 765–70 36g. Sample answer: 50–60Pages 893–8967a.Exercises7b. 432207c. Sample answer: to determine where most oftheir customers live so they can target theiradvertising accordingly8a.Frequency128400 40 50 60 70Age 43026 43214 4322143212 43220 43229 45414ZIP CodesMenAge65+50-6435-4920-3416-19Women80 60 40 20 0 0 20 40 60 80Minutes Behind the Wheel8b. Sample answer: Men spend more hours drivingthan women.9a. Rental Revenue Year Sales Revenue2003200019971990198510 8 6 4 2 0 0 2 4 6 8 10 12 14Dollars (in billions)9b. Sales; the sales revenue is increasing, and therental revenue has started to decrease.10a. range 53 4 or 4910b. Sample answer: 1010c. Sample answer: 0, 10, 20, 30, 40, 50, 6010d. Sample answer: 5, 15, 25, 35, 45, 55457 Chapter 14
10e. Sample answer:Grams of Fat Frequency0–10 710–20 1120–30 1030–40 740–50 250–60 110f. Sample answer:Grams of Fat inFast-Food SandwichesFrequency1010g. Sample answer: 10–2011a. range 72 16 or 5611b. Sample answer: 1011c. Sample answer: 10, 20, 30, 40, 50, 60, 70, 8011d. Sample answer: 15, 25, 35, 45, 55, 65, 7611e. Sample answer:Number of Nations Frequency10–20 220–30 330–40 840–50 150–60 160–70 270–80 111f. Sample answer:864200 10 20 30 40 50 60Grams of Fat8Olympic WinterGames11g. Sample answer:Frequency 412a. range 1023 404 or 61912b. Sample answer: 10012c. Sample answer:Height (feet) Frequency400–500 5500–600 4600–700 2700–800 3800–900 1900–1000 21000–1100 312d. Sample answer:862Frequency 32Olympic WinterGames00 10 20 30 40 50 60 70 80Number of Nations5410 0400 600 800 1000Height (feet)12e. Sample answer: 400–50013a. American League Year National League200320022001200019991998Frequency64200 10 20 30 40 50 60 70 80Number of Nations199719961995199480 60 40 20 0 0 20 40 60 80Greatest Number of Stolen Basesfor a Single PlayerChapter 14 458
13b. Sample answer:13c. Sample answer:13d. 3 players13e. 7 players14. Sample answer: 0.1, 0.2, 0.3, 0.4, 0.6, 0.7, 0.8, 0.9,1.1, 1.2, 1.3, 1.4, 1.6, 1.7, 1.8, 1.9, 2.1, 2.2, 2.3, 2.415.300250200Stolen Bases Frequency30–40 140–50 650–60 660–70 470–80 36Frequency4200MillionsofTons30 40 50 60 70 80Stolen Bases16b.Sales16c. See students’ work.17. See students’ work.18. This is a biannual experiment, where 8 mums areinvolved and there are only two possible outcomes,survival S and not survival N.P(exactly 6 surviving) C(8, 6)(S) 6 (N) 2 28(0.8) 6 (0.2) 2 0.29 or 29%19. (c 2d) 7 7 7! r!(7(c) 7r (2d) rr0 r)!To find the second term, evaluate the general termfor r 1.7! r!(7(c) 7r (2d) r 7! 1!(7(c) 71 (2d) 1 r)!$1,200,000$800,000$400,00020. 3.6 x 58.9x ln 3.6 ln 58.9x 3.1821. yO$09xy 361998 1999 2000Year 1)! 7c 6 2d14c 6 dx150100500Wheat Rice CornChinaIndiaUnited States22. (x y) 2 x 2 2xy y 2 (x 2 y 2 ) 2(xy) 16 2(8) 3216a. The first interval on the vertical axis represents$800,000, but the other intervals represent only$200,000. Therefore, the sales for 1999 appear tobe twice the sales of 1998, but in reality they arenot.14-2Measures of Central TendencyPage 903 Check for Understanding1. Mean, median, mode; to find the mean, add thevalues in a set of data and divide the sum by thenumber of values in the set. To find the median,arrange the values in a set of data from least togreatest. If there is an odd number of values inthe set, the median is the middle value. If there isan even number of values in the set, the median isthe mean of the two middle values. To find themode, find the item of data that appears morefrequently than any other in the set.2. Sample answer: {1, 2, 2, 3, 4, 4, 5}The modes are 2 and 4.459 Chapter 14
3. Write the stems 9, 10, 11, 12, 13, and 14 on theleft. Write the tens digits as leaves to the right ofthe appropriate stems. Be sure to order the leaves.4. Tia; the median 2.5 and the mode 2 do notrepresent the greater numbers. The mean 8.5 ismore representative of all 8 items in the data.5. X 1 4 (10 10 45 58) or 30.7545M d 10 2or 27.5Mode 1016. X 1(210 22 23 24 28 29 31 31 34 37) 28M d 28 292or 28.5Mode 317. X 1 3 (91 94 95 98 99 105 105 107 107 107 111 111 112) 100 103.23 100 10,323M d 10,500Mode 10,7008a. 2 8 15 6 38 31 13 7 120120 members8b. 8f i 1208c.i1 8(f i X i ) 2(3) 8(7) 15(11) 6(15) i138(19) 31(23) 13(27) 7(31) 2320X 2 320120or about 19.3Visits Members Cumulative Members1–5 2 25–9 8 109–13 15 2513–27 6 3117–21 38 6921–25 31 10025–29 13 11329–33 7 120Half of the data has been gathered in the 17–21class. This is the median class.8d. 69 31 38 21 17 460 31 29 M d 17 x 3 84 2 9xx 3.052631579M d 17 xM d 17 3.1M d 20.19a. stem leaf0 6 7 7 7 91 3 3 4 4 5 6 7 8 9 92 0 0 0 1 1 1 1 1 3 3 83 0 1 1 1 2 4 4 6 84 1 1 1 2 71|3 1319b. X 4(60 3(7) 9 2(13) 2(14) 15 16 17 18 2(19) 3(20) 5(21) 2(23) 28 30 3(31) 32 2(34) 36 38 3(41) 42 47) 23.559c. M d 219d. Mode 219e. Since the mean 23.55, the median 21, and themode 21 are all representative values, any ofthem could be used as an average.Pages 904–907Exercises10. X 1 4 (140 150 160 170) or 155160M d 150 2or 155Mode: none11. X 1 5 (3 3 3 6 12) or 5.4M d 3Mode 312. X 1 4 (17 19 19 21) or 19M d 19Mode 1913. X 1 8 (3 5 5 8 14 15 18 18) 10.75M d 8 142or 11Mode 5 and 18114. X 1(542 58 62 63 64 70 76 76 87 87 98) 73.5M d 70 762or 73Mode 87115. X 1(52 6 6 6 7 8 9 10 11 11 11 12) 8.5M d 8 92or 8.5Mode 6 and 1116a. X 1 9 (117 124 139 142 145 151 155 160 172) 145 lb16b. M d 145 lb16c. X 1 9 (122 129 144 147 150 156160 165 177) 150 lbM d 150 lbEach will increase by 5 lb.17. X 1 2 (35 2(38) 39 44 3(45) 48 2(57) 59) 45.8M d 45Mode 45Chapter 14 460
118. X 1(5.24 5.4 5.6 6.0 6.1 6.7 6.8 6.9 7.1 7.6 8.0 8.2 8.6 9.1) 6.95M d 6.8 6.92or 6.85Mode: none119. X 1(905 91 97 98 99 105 106 109 113 3(118) 120 2(125)) 10 1088M d 1090Mode 118020. stem leaf1 0 5 5 5 5 7 72 0 0 0 5 5 5 5 7 83 0 0 5 5 54 65 51|0 1021a. 135(11) $1485; 145(24) $3480;155(30) $4650; 165(10) $1650;175(13) $2275; 185(8) $1480;195(4) $78021b. 1485 3480 4650 1650 2275 1480 780 $15,80021c. 11 24 30 10 13 8 4 100100 employees21d. X 15 , 800100or about $15821e.Weekly Frequency CumulativeWagesFrequency$130–$140 11 11$140–$150 24 35$150–$160 30 65$160–$170 10 75$170–$180 13 88$180–$190 8 96$190–$200 4 100Half of the data has been gathered in the$150–$160 class. This is the median class.21f. 65 35 30 160 150 1050 35 15 M d 150 x 3 010 1 5xx 5M d 150 xM d 150 5M d 15521g. Both values represent central values of the data.22. 7.5 1 5 (2 4 5 8 x)37.5 19 x18.5 x23. 6 1 4 (x 2x 1 2x 3x 1)24 8x3 x24. Order the values from least to greatest. Themedian lies between the fourth and fifth terms.2, 3, 3.2, 8, x, 11, 13, 14x 8, since M d 8 82or 8.125a. X 1 79 [9(245) 14(275) 23(325) 30(375) 33(425) 28(475) 18(525) 12(575) 7(625) 3(675) 1(725) 1(775)] 425.625b.Scores 17 92 89.5; Half of the data has been gathered inthe 400–450 class. This is the median class.25c. 109 76 33 450–400 5089.5 76 13.5 M d 400 x 3 350 13 .5xNumberof StudentsCumulativeNumber ofStudents200–250 9 9250–300 14 23300–350 23 46350–400 30 76400–450 33 109450–500 28 137500–550 18 155550–600 12 167600–650 7 174650–700 3 177700–750 1 178750–800 1 179x 20.45M d 400 xM d 400 20.5M d 420.526a. X 1 5 (3.6 3.6 3.7 3.9 4.8) 3.92M d 3.7Mode 3.626b. Only the mean would change. It would increaseto 4.6.X 1 5 (3.6 3.6 3.7 3.9 8.2) 4.626c. The mean increases slightly; the medianincreases slightly; the mode stays the same.X 1 4 (3.6 3.6 3.7 3.9) 3.7M d 3.6 3.72or 3.65Mode 3.627a. Sample answer: {1, 2, 2, 2, 3}27b. Sample answer: {4, 5, 9}27c. Sample answer: {2, 10, 10, 12}27d. Sample answer: {3, 4, 5, 6, 9, 9}461 Chapter 14
28a.stem leaf0 1 1 1 1 1 1 1 2 2 2 2 2 3 3 33 3 4 4 4 5 5 5 5 6 6 7 7 7 88 8 8 9 9 9 91 0 1 3 3 3 5 8 9 92 5 93 245 35|3 53128b. X 5[7(1)0 5(2) 5(3) 3(4) 4(5) 2(6) 3(7) 4(8) 4(9) 10 11 3(13) 15 18 2(19) 25 29 32 53] 8.728c. M d 628d. Mode 128e. The mean 8.7 and the median 6 arerepresentative of the data, but the mode 1 is notrepresentative of the data.129a. X 2[1(170)7 6(190) 10(210) 6(230) 3(250) 1(270)] 215.229b.Goals Number of CumulativeTeams Number of Teams160–180 1 1180–200 6 7200–220 10 17220–240 6 23240–260 3 26260–280 1 27 2 72 13.5; Half of the data has been gathered inthe 200–220 class. This is the median class.29c. 17 7 10 220 200 2013.5 7 6.5 M d 200 x 1 020 6 .5 xx 13M d 200 xM d 200 13M d 213129d. X 2(2687 248 245 242 239 239 237 236 231 230 217 215 214 211 210 210 207 205 202 200 196 194 192 190 189 184 179) 215.9M d 21129e. The mean calculated using the frequencydistribution is very close to the one calculatedwith the actual data. The median calculatedwith the actual data is less than the onecalculated with the frequency distribution.30a. Let X 50 9(X f i ) 0 (50 5) (50 20) (50i1 37) (50 44) (50 52) (50 68) (50 71) (50 85) (50 x)0 68 xx 68The weight should be hung 68 cm from the end.30b. Let X 50 9(X f i ) 0 (50 5) (50 20) (50i1 37) (50 44) (50 52) (50 68) (50 71) (50 85) (50 x) (50 x)0 118 2x118 2x59 xThe weight should be hung 59 cm from the end.131a. X 1(540 55 59 59 61 62 65 75 162 226) 1000 $87,80031b. M d 61,000 62,0002or $61,50031c. Mode $59,00031d. The mean, since it is the greatest measure ofcentral tendency.31e. The mode, since it is the least measure of centraltendency.31f. Median; the mean is affected by the extremevalues of $162,000 and $226,000, and only twopeople make less than the mode.31g. Sample answer: I have been with the companyfor many years, and I am still making less thanthe mean salary.132a. X 1 00 [12(2.00) 15(2.50) 31(3.00) 37(3.50) 5(4.00) 30.432b.Grade Point Frequency CumulativeAveragesFrequency1.75–2.25 12 122.25–2.75 15 272.75–3.25 31 583.25–3.75 37 953.75–4.25 5 100The median class is 2.75 3.25.58 27 31 3.25 2.75 0.5050 27 13 M d 2.75 x31 0 .50 1 x3x 0.2096774194M d 2.75 xM d 2.75 0.21M d 2.96Chapter 14 462
33. He is shorter than the mean (5 11.6) and themedian (5 11.5).1X 1(670 68 69 69 71 72 73 74 75 78) 71.6 or 5 11.6M d 71 72234. 71.5 or 5 11.535. dependent; 131 2 3 1 0 5 51 1n36. a n 3 n; a n1 3nnr lim n→ 3 n(n 1)n→3n1(n) 3n ( n 1)n→3n 3n n 1n→3n lim lim limn lim n→3 n 1 3 n 1 3 0 or 1 3 Since r 1, the series is convergent.37. F n P (1 i38.222018161412Frequency1086420 0 n 13 n 1 —n 3 ni) n 1 1500 (1 0 060 65 70 75Speed Limit.03) 20 1.03 $40,305.56y8642O864 2 2 4 6 8x46839. To find the area of the triangle use Hero’s formula:Area s(s a)(s b)(s , c) wheres 1 (a b c) and a 10, b 7, and c 5.2So, s 1 2 (10 7 5) 1 (22) 11.2Area 11(11 (11 10) 7)(11 5) 11 1 4 6 or 264 16.25.The correct choice is A.14-3Measures of VariabilityPage 914 Check for Understanding1. The median of the data is 70, Q 1 is 60, and Q 3 is100. The interquartile range is 40 and the semiinterquartilerange is 20. The outliers are 170 and180. The data in the first two quartiles are closetogether in range. The last two quartiles are morediverse.2. square the standard deviation3. Both the mean deviation and the standarddeviation are measures of the average amount bywhich individual items of data deviate from themean of all the data. The mean deviation uses theabsolute values of the deviations. Standarddeviation uses the squares of the deviations.4. See students’ work.5. interquartile range Q 3 Q 1 41 25 16Semi-interquartile range 1 621520 86. X 1 8 (4.45 5.50 5.50 6.30 7.80 11.00 12.20 17.20) 8.74MD 1 8 (⏐4.29⏐ ⏐3.24⏐ … ⏐8.46⏐) $3.54 1 8 (4. 292 (3.2 4)2 … 8.462 $4.1117. X 2 00 [15(5000) 30(15,000) 50(25,000) 60(35,000) 30(45,000) 15(55,000)] 30,250 (25,250)2 15 (15,250) 2 30 … 24,750 2 15200 13,226.3918a. X 1(65.72 65.9 … 65.9) 70.37570.3M d 69.0 2or 69.65 (4.875)2 (4.475) 2 … 6.225212 4.2518b. X 1(57.32 63.3 … 57.5) 80.4882.1M d 77.5 2or 79.8 (23.18) 2 (22.98) 2 … 25.32 212 17.062530354045463 Chapter 14
8c. Los AngelesLas Vegas8d. Los Angeles8e. Los Angeles is near an ocean; Las Vegas is in adesert.Pages 915–917 Exercises9. interquartile range Q 3 Q 1 24 17 7semi-interquartile range 7 2 or 3.510. interquartile range Q 3 Q 1 21.5 12 9.5semi-interquartile range 9 .5 2 or 4.7511. interquartile range Q 3 Q 1 10.5 7.6 2.9semi-interquartile range 2 .9 2 or 1.4512.55 60 65 70 75 80 85 90 95100 10555 60 65 70 75 80 85 90 95100 10514 16 18 20 22 24 26 28 300 10 20 30 40 50 60 70 805 6 7 8 9 10 11 12 13 14 15 1613. X 1 6 (152 158 … 721) 381MD 1 6 (⏐229⏐ ⏐223⏐ … ⏐340⏐) 211(229) 2 (223) 2 … 34026 223.14114. X 1(5.70 5.7 … 3.8) 4.891MD 1(⏐0.81⏐0 ⏐0.81⏐ … ⏐1.09⏐) 0.6720.81 2 0.81 2 … (1.09)210 0.73115. X 1(3692 376 … 454) 403.51MD 1(⏐34.5⏐2 ⏐27.5⏐ … ⏐50.5⏐) 20.25(34.5) 2 (27.5) 2 … 50.5212 25.3116. X 1 5 (13 22 34 55 91) 43Variation 774117. X 1 20 [2(3) 8(7) … 7(31)] 19.33(16.33) 2 2 (12.33) 2 8 … 11.67 2 7120 6.48118. X 9[3(57)0 7(65) … 12(97)] 81.8(24.8) 2 3 (16.8) 2 7 … 15.2 2 1290 9.69119. X 8[2(80)5 11(100) … 7(180)] 129.65(49.65) 2 2 (29.65) 2 11 … 50.35 2 785 23.2920a. M d 259 mi20b. Q 1 129 mi; Q 3 360 mi20c. interquartile range Q 3 Q 1 360 129 231 mi20d. semi-interquartile range 23 12 115.5 mi20e. An outlier would lie 231 115.5 or 346.5 mioutside of Q 1 or Q 3 . There are no such points.20f.(30) 2 (21) 2 (9) 2 12 2 48 250 100 200 300 400 500 600 70020g. The data in the upper quartile is more diversethan the other quartiles.21. Sample answer: {15, 15, 15, 16, 17, 20, 24, 26, 30,35, 45}22a. M d 28222b. Q 1 42; Q 3 770Chapter 14 464
22c. interquartile range Q 3 Q 1 770 42 728semi-interquartile range 72 82 36422d. An outlier would lie 728 364 or 1092 pointsoutside of Q 1 or Q 3 . There are no such points.22e.122f. X 1(229 23 … 966) 404.42122g. MD 1(⏐382.4⏐9 ⏐381.4⏐ … ⏐561.58⏐) 316.9722h. Variance 118,712.5622i. 118,71 2.56 344.5522j. There is a great variability among the number ofteams in women’s sports.23a. Q 1 $3616, M d $4125, Q 3 $566423b. interquartile range Q 3 Q 1 5664 3616 204823c. An outlier would lie 2048 1024 or $3072outside of Q 1 or Q 3 . There are two such values,$26,954 and $27,394.23d.0 200 400 600 800 1000382.4 2 381.4 2 … 561.58 2190 5000 10,000 15,000 20,000 25,000 30,000123e. X 1(26849 2929 … 27,394) 6775.951MD 1(⏐4091.95⏐9 ⏐3846.95⏐ … ⏐20,618.05⏐) 4463.3923f. (4091.95)2 (3846.95) 2 … 20,618.05219 7103.4523g. The data in the upper quartile is diverse.24a.0 100 200 300 400 500 600124b. X 4(02 0 … 635) 60.401MD 4(⏐60.40⏐2 ⏐60.40⏐ … ⏐574.60⏐) 67.87(60.40) 2 (60.40) 2 … (514.60) 24224c. Variance 14,065.4824d. 14,06 5.48 118.6024e. The data in the upper quartile is diverse.125a. X 5[26(9)0 12(11) … 2(21)] 11(9 11)25b. 2 26 (11 11) 2 12 … (21 11) 2 250 2.9426. yes; when the standard deviation is less than 1;when both equal 0 or 127. See students’ work.128a. X 3[2(4.4)5 4.9 5.4 5.5 2(6.2) 6.4 6.5 6.9 7.1 7.4 7.5 7.6 7.7 7.8 7.9 8.0 8.2 8.4 8.5 8.6 8.7 8.8 3(8.9) 9.0 9.2 2(9.3) 9.5 9.6 9.8 9.9] 7.7528b. M d 8.028c. Mode 8.929a. range 68 23 or 4529b. Sample answer: 1029c. Sample answer: 20, 30, 40, 50, 60, 7029d. Sample answer:29e. Sample answer:30. 1 9 (9!) 40,302 ways31. x 1 0.5(8) 1 or 3x 2 0.5(3) 1 or 0.5x 3 0.5(0.5) 1 or 0.7532. 7 ft 7(12) or 84 in.84 9 93 in.3 31 in. 9 3Programs Sold Frequency20–30 230–40 140–50 250–60 560–70 214Frequency200 10 20 30 40 50 60 70ProgramsSold31 in. 3 12or 2 ft 7 in.The correct choice is C.465 Chapter 14
Page 917 Mid-Chapter Quiz1. Sample answer: 102. Sample answer:3. Sample answer:Frequency4. stem leaf5 4 56 2 2 4 57 1 5 6 7 8 98 0 2 4 5 6 7 8 9 9 99 0 2 3 3 5 6 8 95⏐4 5415. X 3(540 55 … 99) 81.16. M d 84 852or 84.57. Mode 898.19. MD 3(⏐27.1⏐0 ⏐26.1⏐ … ⏐17.9⏐) 10.4210. Sample answer: The data that are less than themedian are more spread out than the data greaterthan the median.14-4Exam Scores Frequency50–60 260–70 470–80 680–90 1090–100 81086420 0Physics Exam50 60 70 80 90 100Exam Score50 60 70 80 90 100The Normal Distribution3.45 55 65 75 85 95 105 45 55 65 75 85 95 105The second curve is less variable.4. Sample answer:5. 50th percentile; it contains half of the data.6a.445 480 515 550 585 620 6556b. Since 515 and 585 are within are standarddeviation of the mean, it contains 68.3% of thedata.6c. 99.7% of the data lie within 3 standarddeviations of the mean.550 3(35) 445 6556d. 550 480 70, 620 550 70tj 70t(35) 70t 2 → 95.5%0.995(200) 191 values7a. Since 22 and 26 are within one standarddeviation of the mean, it contains 68.3% of thedata.7b. 24 20.5 3.5, 27.5 24 3.5tj 3.5t(2) 3.5t 1.75 → 92.9%7c. 24 0.7(2) 22.6 and 24 0.7(2) 25.422.6 25.47d. 24 1.96(2) 20.08 and 24 1.96(2) 27.9220.08 27.928a.Pages 922–923 Check for Understanding1. The median, mean, and mode are the same.2. X 1.58b.64 67 70 73 76 79 8265 70 75 80 85 90 95Chapter 14 466
8c. Chemistry; the chemistry grade is 3 standarddeviations above the class mean, while thespeech grade is only 2 standard deviations abovethe class mean.Pages 923–9259a.Exercises13b. X tj 180 X tj 150140 t(20) 180 140 t(20) 15020t 40 20t 10t 2 t 0.5 95 .5% 2 47.75% 38 .3% 2 19.15%47.75 19.15 28.6%13c.50%25% 25%9b. 12 1(1.5) 10.513.59c. 12 7.5 4.5, 16.5 12 4.5tj 4.5t(1.5) 4.5t 3 → 88.7%9d. 12 9 3, 15 12 3tj 3t(1.5) 3t 2 → 95.5%10a. 0.683(200) 136.6; about 13710b. 0.955(200) 19110c. 0.6 8327.5 9 10.5 12 13.5 15 16.5 (200) 68.3; about 6811a. 45% corresponds to t 0.6.82 0.6(4) 79.6 84.411b. 80% corresponds to t 1.382 1.3(4) 76.8 87.211c. 82 76 6, 88 82 6tj 6t(4) 6t 1.5 → 86.6%11d. 82 80.5 1.5, 83.5 82 1.5tj 1.5t(4) 1.5t 0.375 → 31.1%12a. 25% corresponds to t 0.3.402 0.3(36) 391.2412.812b. 402 387 15, 417 402 15tj 15t(36) 15t 0.416 → 31.1%12c. 402 362 40, 442 402 40tj 40t(36) 40t 1.1 → 72.9%12d. 45% corresponds to t 0.6.402 0.6(36) 380.4 423.613a. X tj 150 X tj 100140 t(20) 150 140 t(20) 10020t 1020t 40t 0.5 t 2.5% 47.75% 38 .3% 2 19.15% 95 219.15 47.75 66.9%t 0.7 corresponds with 50% of the datacentered about the mean.The upper limit results in 75% of the data.140 0.7(20) 15414a. X tj 7 X tj 6.56 t(35) 7 6 t(3.5) 6.53.5t 1 3.5t 0.5t 0.29 t 0.14 23 .6% 2 11.8% 8 % 2 4%11.8 4 7.8%14b. X tj 6.2 X tj 5.56 t(0.35) 6.2 6 t(0.35) 5.50.35t 0.2 0.35t 0.514c.80t 0.57 t 1.43.8% 41.9% 45 .1% 2 22.55% 83 222.55 41.9 64.45%4.951005.30120 140 160 180 20080%10% 10%5.65 6.00 6.35 6.70 7.05t 1.3 corresponds with 80% of the datacentered about the mean. The lower limit resultsin the value above which 90% of the data lies.6 1.3(0.35) 5.54515a. P(no tails) 1 2 6 or 614 P(one tail) 6 1 2 1 2 5 3or 3 2P(two tails) 15 1 2 2 1 2 4 or 1 564 P(three tails) 20 1 2 3 1 2 3 5or 1 6P(four tails) 15 1 2 4 1 2 2 5or 6 4 P(five tails) 6 1 2 5 1 2 3 or 3 2P(six tails) 1 2 6 1or 6 4467 Chapter 14
15b.2015Frequency10500 1 2 3 4 5 6Number of Tails115c. X 6[0(1)4 1(6) 2(15) 3(20) 4(15) 5(6) 6(1)] 3(0 3)15d. j 2 (1 3) 2 … (6 3)264 1.215e. They are similar.16a.84%18.19a.465196%2% 2%56 61 66 71 7696% corresponds to t 2.1.61 2.1(5) 50.5 months30%20% 20%15% 15%16b.38% 8%21 0 1 2 3 tThe 92nd percentile is the upper limit to 84% ofthe data that is centered about the mean. 84%corresponds to t 1.4. The 92nd percentile is1.4 standard deviations above the mean.3257.6%21.2% 21.2%1 0 1 2 3 tt 0.8 corresponds to 57.6% of the data centeredabout the mean. 100 57.62 21.221.2 57.6 78.8 percentile17a. X tj 22.320.4 t(0.8) 22.30.8t 1.9t 2.38 → 98.4% 100 98.42 0.8%17b. 100 0.8 99.2%445158 65 72 79 8670% corresponds to t 1.0.65 1.0(7) 7219b. 65 1.0(7) 5819c. 30% corresponds to t 0.465 0.4(7) 67.8 68The lowest score for an A is 72, so the highestscore for a B is 71. The interval for B’s is 6871.20a. a normal distribution with a small standarddeviation20b. a normal distribution with a large standarddeviation20c. a distribution where values greater than themean are more spread out than values less thanthe mean20d. a distribution where all values occur with thesame frequency21a. 95% corresponds to t 1.96X tj 260 X tj 250255 1.96j 260 255 1.96j 2501.96j 5 1.96j 5j 2.55 j 2.55about 2.55 mL21b. X tj 357 X tj 353355 t(2.55) 357 355 t(2.55) 3532.55t 2 2.55t 2t 0.78 t 0.7857.6%22a. M d 147 1502or 148.522b. Q 1 110, Q 3 20022c. interquartile range Q 3 Q 1 200 110 9022d. semi-interquartile range 9 02or 4522e.100 200 300 400 500Chapter 14 468
23. X 1 9 (19 33 42 42 45 48 55 71 79) 48.2M d 45Mode 4224. y sec(k c) hk: 2 k 2 k 4c c: k c4c 4h 3y sec(4 4) 325a. Sample answer: Use a graphing calculator toenter the year data as L1 and the Enrollmentdata as L2. Then make a scatter plot. Thescatter plot indicates that a cubic function wouldbest fit the data. Perform a cubic regression tofind the equationy 0.05x 3 2.22x 2 29.72x 366.92.25b. Sample answer:2015 1965 50f(50) 0.05 50 3 2.22 50 2 29.72 50 366.92 2553 students26.60˚ 43. About 68.3%; the answer for Exercise 2 can berounded to 0.683, which equals 68.3%.4.5.0.954499740.99730026. The answer for Exercise 3 can be rounded to0.954, which is about 95.5%. The answer forExercise 4 can be rounded to 0.997, which equals99.7%.7. 0.9999; t 4 corresponds to P 0.999.8.x˚360˚125Since vertical angles are equal, m2 60.m1 60 30 180m1 90Since an exterior angle of a triangle is equal to themeasure of the sum of the two remote interiorangles, x 60 m1. So, x 60 90 and x 30.The correct choice is E.14-4B30˚Graphing Calculator Exploration:The Standard Normal CurvePage 9261. The domain of the function is the set of all realnumbers. The range is the set of all real numbers1y such that 0 y 2 . The graph is symmetricwith respect to the y-axis and has the x-axis as ahorizontal asymptote. the value of f(x) approaches0 as x approaches .2.1; no; since the curve is approaching the x-axisasymptotically, the area is probably not exactlyequal to 1.14-5Sample Sets of DataPage 930 Check for Understanding1. A sample is a subset of a population. However, asample must be similar in every way to thepopulation.2. Divide the standard deviation of the sample by thesquare root of the number of values in the sample.3. Use a larger sample.4. Tyler; every twentieth student to enter the schoolshould produce a representative sample. Thesenior English class will not representunderclassmen. The track team will not representstudents who prefer other types of activities.735. j X 1 or 7.3003.46. j X 2 50 0.220.68268949469 Chapter 14
7. A1% confidence level is given when P 99% andt 2.58.5j X 5 36 or 0.83internal: X tj X 45 2.58(0.83) 42.8547.158. A5% confidence interval is given when P 95%and t 1.96.5.6j X 3 00 0.323316156internal: X tj X 55 1.96j X 54.3755.633.59a. j X 1 50 0.299b. P 50% corresponds to t 0.7.interval: X tj X 27.5 0.7j X 27.3027.70 min9c. A1% confidence level is given when P 99%and t 2.58.interval: X tj X 27.5 2.58j X 27.7628.24 minPages 930–932 Exercises1. 810. j X 8 or 0.215.811. j X 2 50 0.377.812. j X 1 40 0.661413. j X 7 00 0.532.714. j X 1 30 0.2413.515. j X 3 75 0.705. 616. 0.056 NN 100N 100,0005. 317. j X 5 0 0.7495331881interval: X tj X 335 2.58j X 333.07336.934018. j X 6 or 54interval: X tj X 200 2.58(5) 187.1212.91219. j X 2 00 0.8485281374interval: X tj X 80 2.58j X 77.8182.9111.20. j X 121 0 00 0.3516452758interval: X tj X 110 2.58j X 109.09110.9121. P 90% corresponds to t 1.65.4j X 1 or 0.400interval: X tj X 68 1.65(0.4) 67.3468.66 in.2.422. j X 1 or 0.2400interval: X tj X 24 1.96(0.24) 23.5324.4717.123. j X 3 50 0.9140334473interval: X tj X 4526 1.96j X 4524.214527.792824. j X 3 70 1.455650686interval: X tj X 678 1.96j X 675.15680.850.6725. j X 8 0 0.0749082772interval: X tj X 5.38 1.96j X 5.235.53126a. X 6 4[1(4) 3(6) … 2(20)] 12.375(4 12.375)26b. j 2 (6 12.375) 2 … (20 12.375)264 3.373.3726c. j X 64 0.4226d. P 0.95 corresponds to t 1.96.interval: X tj X 12.375 1.96(0.42) 11.5513.20 min26e. tj X 1t(0.42) 1t 2.38 → about 98.4%1227. j X 4 5 1.788854382tj X 3t 1.68 → 91.1%100 91.1 8.9%1. 428a. j X 50 0.1979898987 or about 0.2028b. A5% confidence interval is given when P 95%and t 1.96.interval: X tj X 16.2 1.96j X 15.8116.59 mm28c. P 99% corresponds to t 2.58.interval: X tj X 16.2 2.58j X 15.6916.71 mm28d. P 0.80 corresponds to t 1.3interval: X tj X 16.2 1.3j X 15.9416.46 mm4529a. j X or 4.5100Chapter 14 470
29b. A1% confidence level is given when P 99%and t 2.58.interval: X tj X 350 2.58(4.5) 338.39361.61 hours29c. Sample answer: 338 hours, there is only 0.5%chance the mean is less than this number.30. 10.2064 9.7936 0.4128 0.4 1282 0.2064tj X 0.20642.58tj X 0.2064j X 0.08j X jN8N0.0.08 N 10N 100 packages1. 831a. j X 1 0 0.5731b. interval: X tj X 4.1 1.96(0.57) 2.985.22 hoursWith a 5% level of confidence, the average familyin the town will have their televisions on from2.98 to 5.22 hours.31c. Sample answer: None; the sample is too small togeneralize to the population of the city.3. 232a. j X 5 0 0.4532b. P 50% corresponds to t 0.7.interval: X tj X 42.7 0.7j X 42.3843.02 crackers32c. Sample answer: No; there is a 50% chance thatthe true mean is in the interval. However, since43 is near one end of the interval, they maywant to take another sample in the near future.33a. A5% confidence interval gives a P 95% andt 1.96.753.136 746.864 3.136tj X1.960j X 3.136j X 1.6X tj X 753.136X 1.96(1.6) 753.136X 750 hj33b. j X Nj1.6 1 60064 j; 64 h34a. 40,000 35,000 5000, 45,000 40,000 5000tj 5000t(500) 5000t 1 → 68.3%0.683(10,000) 6830 tires34b. X tj 30,00040,000 t(5000) 30,0005000t 10,000t 2 → 95.5%.5% 47.75% 95 20.4775(10,000) 4775 tires34c. 50% of 10,000 0.50(10,000) 5000 tires34d. X tj 50,00040,000 t(5000) 50,0005000t 10,000t 2 → 95.5% 100% 95.5%2 2.25%0.0225(10,000) 225 tires34e. X tj 25,00040,000 t(5000) 25,0005000t 15,000t 3 → 99.7% 100% 99.7%2 0.15%0.0015(10,000) 15 tires35. X 1 8 (44 49 55 58 61 68 71 72) 59.75MD 1 8 (⏐15.75⏐ ⏐10.75⏐ … ⏐12.25⏐) 8.25(15.75)j 2 (10.75) 2 … (12.25)28 9.5936. S n a 1 a1 rn1 r1n 10, a 1 1,6r 41 1 1 6 1(4)610S 10 1 4 349 , 52516or 21,845.312537. x yr cos v r sin vsin1 vc osv1 tan vtan 1 (1) v45° v38. tan x cot x 21tan x ta 2n x tan 2 x 1tanx 2tan 2 x 1 2 tan xtan 2 x 2tan x 1 0(tan x 1) 2 0tan x 1 0tan x 1x 45°39. *2 2 2 2(2) or 0*1 1 2 2(1) or 1*2 *1 0 (1) or 1The correct choice is C.Chapter 14 Study Guide and AssessmentPage 933 Understanding the Vocabulary1. box-and-whisker plot2. median3. standard error of the mean471 Chapter 14
4. range5. measure of central tendency6. population7. bimodal8. inferential statistics9. histogram10. standard deviationPages 934–936 Skills and Concepts11. range 14.0 9.0 or 512. 9.5, 10.5, 11.5, 12.5, 13.513.Women's Tennis ShoesFrequency181614121086420 09 10 11 12 13 14Weight (ounces)14. X 1 9 (2 4 4 4 5 5 6 7 8) 5M d 5Mode 415. X 1 5 (160 200 200 240 250) 210M d 200Mode 20016. X 1 5 (11 13 15 16 19) 14.8M d 15Mode: none17. X 1 8 (5.9 6.3 6.3 6.4 6.6 6.6 6.7 6.8) 6.45M d 6.4 6.62or 6.5Mode 6.3 and 6.618. X 1 8 (122 128 130 131 133 135 141 146) 133.25M d 131 1332or 132Mode: none19. interquartile range Q 3 Q 1 5 2 320. semi-interquartile range 3 2 or 1.5121. X 1(10 1 … 6) 3.41MD 1(⏐2.4⏐0 ⏐2.4⏐ … ⏐2.6⏐) 1.622. j 2.4) 2 2.4) (… 2 .6 22 1.7423. 88 78 10 98 88 10tj 10t(5) 10t 2 → 95.5%24. 88 86 2 90 88 2tj 2t(5) 2t 0.4 → 0.31125. 90% corresponds to t 1.65.interval: X tj 88 1.65(5) 79.7596.2526. 0.683(150) 102.4527. 0.955(150) 143.2528. 0.6 832(150) 51.2251. 529. j X 9 0 0.164.930. j X 1 20 0.452531. j X 4 or 1.25001832. j X 2 or 3.651533. j X 50 2.121320344interval: X tj X 100 2.58j X 94.53105.473034. j X 1 5 7.745966692interval: X tj X 90 2.58j X 70.02109.982435. j X 2 00 1.697056275interval: X tj X 40 2.58j X 35.6244.380.536. j X 2 00 0.03537. P 0.90 corresponds to t 1.65.range: X tj X 1.8 1.65(0.035) 1.741.86 h38. A5% confidence level is given when P 95% andt 1.96.range: X tj X 1.8 1.96(0.035) 1.731.87 h39. A1% confidence level is given when P 99% andt 2.58.range: X tj X 1.8 2.58(0.035) 1.711.89 h40. P 0.90 corresponds to t 1.65.1.4j X 1 or 0.1400range: X tj X 4.6 1.65(0.14) 4.374.83 hChapter 14 472
Page 937 Applications and Problem Solving41a. stem leaf1 0 3 5 6 7 92 1 3 4 53 9 91⏐0 1041b. X 1 2 (10 13 15 16 17 19 21 23 24 25 39 39) 21.7541c. M d 19 212or 2041d. Mode 3942. X tj 8075 t(2) 802t 5t 2.5 → 98.8% 100% 98.8%2 0.6%Page 937 Open-Ended Assessment1a. Sample answer: {2, 3, 10, 20, 40}1b. Sample answer: 152. See students’ work.Chapter 14 SAT & ACT PreparationPage 939 SAT and ACT Practice1. The percent increase is the ratio of the numberincrease to the original amount.Amy 1 080 1 8 Brad 3 070 3 7 Cara 2 080 1 4 Dan 3 060 1 2 0Elsa 2 90 2 9 The largest fraction is 1 2 . Dan has the greatestpercent increase. The correct choice is D.2. a b bc ab b (b bc)b b(1 c)1 1 cThe correct choice is B.3. Method 1:0.1%(m) 10%(n)m x%(10n)x0.001m 0.1n m 10 0 10nm 100nm 0.1xn100n 0.1xn100 xMethod 2Let m represent a large number such as 2000.0.1% of m 0.001(2000) or 210% of n 2, so 0.1n 2 or n 20.m x%(10n)x2000 10(100 20)1000 xThe correct choice is D.4. The numbers in S are positive numbers that areless than 100 and the square root of each numberis an integer. So the set S contains perfect squaresbetween 0 and 100.Make a list of thenumbers, n, in set S.From your list, you cansee that the median, ormiddle value for n, is 25.The correct choice is C.n n1 14 29 316 425 536 649 764 881 95. m∠DBA 90 30 60m∠EBC 90 40 50m∠ABC 180 m∠DBA m∠EBC 180 (60) (50) or 70The correct choice is E.6. X 1 6 (10 20 30 35 35 50) 30There are 3 numbers larger than 30: 35, 35, and50. The correct choice is D.7. yOxThe line of best fit has a rise of 2 and a run of 5. Sothe slope of the line of best fit is 2 5 . The closestanswer choice to 2 5 is 1 2 .The correct choice is D.8. Each year the number of employees increases by300. The last year of data is 2005. The expectedemployment in 2007, two years later, will be2 300 more employees than in 2005.3100 600 3700The correct choice is D.473 Chapter 14
9.To find the median of Set A, first rewrite theelements of Set A in order: 4, 1, 2, 3, 7, 11.Since the number of elements is even, the medianis the average of the middle two elements: 2 and 3.So the median of Set A is 2.5. To find the mean ofSet B, add all of the elements together and divideby the number of elements in the set. The sum ofthe elements is 15, and there are 6 terms. So themean is 1 5 or 2.5.6The difference between the median of Set A andthe mean of Set B is 2.5 2.5 or 0.The correct choice is C.110. X 1[8200 (65) (32) 0 1 2 3 32 64 820]1 1(10 2 3)6 1 0The answer is 0.6, 6/10, or 3/5.Chapter 14 474
Chapter 15 Introduction to Calculus15-1Page 9451.2.lim ex 1x→0xLimits 1x 2 43Graphing Calculator Explorationlim x→2 x 2 4 x 23. y is undefined when x 1.4. lim x→2x 2 x 2 43x 2 ( x 2)( x 2)x→2( x 2)( x 1) lim x 2x→2x 1 2 22 1or 4 limYes, the limit is the same.5. No; if the exact answer is a complicated fraction oran irrational number, you may not be able to tellwhat it is from the decimals displayed by acalculator.Page 946 Check for Understanding1. Sample answer: The limit of f(x) as x approaches ais the number that the values of f(x) get closer toas the values of x gets closer and closer to a.2. Sample answer: lim f(x) is the number that thex→1values of f(x) approach as x approaches 1. f(1) isthe number that you get if you actually plug 1 intothe function. They are the same if f(x) iscontinuous at x 1.3. Sample answer: If f(x) is continuous at x a youcan plug a into the function. If the function is notcontinuous, you may be able to simplify it andthen plug in a. If neither of these methods work,you can use a calculator. Examples will vary.4. The closer x is to 0, the closer y is to 3. So, lim f(x) x→0 3. However, there is a point at (0, 1), so f(0) 1.5. lim (4x 2 2x 5) 4(2) 2 2(2) 5x→216 4 5 176. lim (1 x 2 x cos x) 1 0 2 0 cos 0x→0 1 1 1 1x 2x 27. lim x→2 x 2 lim 4 x→2 (x 2)(x 2)1 lim x→2 x 21 2 2or 1 4 x 2 3xx(x 3)x→0 x3 lim 4xx→0 x ( x2 4)x 3 lim x→0 x 2 40 3 0 2 or 3 4 4 x 2 3x 10x→3 x2 lim ( x 5)( x 2) 5x 6 x→3( x 3)( x 2) ( 3 5)( 3 2)( 3 3)( 3 2) 8 ( 1)6(5)or 145 2x 2 5x 2x2 x 2 lim ( 2x 1)( x 2)( x 1)( x 2)x→2 lim 2 x 1x 1x→2 2( 2) 12 1or 18. lim9. lim10. limx→211a. v (r)0.160.140.120.10 v (r) 0.65(0.5 2 r 2 )0.080.060.040.02O 0.2 0.4 0.6 0.8 r11b. As the molecules get farther from the center andcloser to the pipe, r is increasing. As r increases,v(r) gets closer and closer to 0 in./s.Pages 946–948 Exercises12. The closer x is to 2, the closer y is to 1. So,f(x) 1. Also f(2) 1.limx→213. The closer x is to 0, the closer y is to 0. So,lim f(x) 0. However, there is point discontinuityx→0when x 0. So f(0) is undefined.14. The closer x is to 3, the closer y is to 4. So,lim f(x) 4. However, there is a point at (3, 2).x→3So f(3) 2.15. lim (4x 2 3x 6) 4(2) 3 3(2) 6x→216 6 61616. limx→1 (x3 3x 2 4) (1) 3 3(1) 2 4 1 3 4 0x 17. lim sin x→x sin 0 or 018. lim x→0(x cos x) 0 cos 0 0 1 or 1475 Chapter 15
x2 25x→5x 519. lim lim (x 5)(x 5)x→5x 5 lim x→5(x 5) 5 5 or 1020. lim 2 n 2x→0n lim 2n x→0 2(0) or 021. lim x→3x 2 22. lim x→123. limh→224. limx 2 3x2x 15x(x 3) x→3(x 5) (x 3)x lim x→3 x 53 3 5or 3 8 lim h2 4h 4h 2 lim 1 3 4 87or 8 7 (h 2)(h 2)h 2h→2 lim h 2h→22 2 or 02x2 3x2(3) 2 3(3)x→3 x 3 2x2 x 63 3 2(3) 2 3 618 9 27 18 9or 1 2 3 x2 2xx(x2 x 2)3 4x2 lim 2xx→0 x ( x2 4x 2)x2 x 2 lim x→0 x 2 4x 20 2 0 20 2 or 14( 0) 2 x cosx x cosxx→0x 2 x lim x→0 x (x 1)co lim s xx→0x 1cos0 0 1or 1 (x 2 ) 2 4x→04 lim x2 4x 4 4x→0x lim x(x 4)x→0xx25. lim x→0 x26. lim27. lim28. limx→229. limx→2230. lim x→4 xx 3 3x 2 4x 8x 6 (x 1) 2 1x 2 x 3 8x2 4 lim x→0(x 4) 0 4 or 4 limx→2 limx→2 limx→2 limx→2 lim x x→22 x2 2x 1 1x 2 x( x 2)x 2(x 2)(x 2 2x 4)(x 2)(x 2) x2 2x 4x 2(2) 2 2(2) 42 2 4 4 44or 3x 83 lim 64x→42 lim x→4 x 2 4x 1624 2 4( 4) 162 1 4 8or 2 41 3 3(1) 2 4(1) 81 62(x 4)(x 4)(x 2 4x 16)31. lim x→132. lim x→433. lim 1 x 1 x 1x 4x 2 lim x→1 1 x 1 x→1 x (1 x) lim lim x→1 1 x 11 lim x→4 lim x→4 lim 2h3 h 2 5hh→0h34. lim35.36. 1 xx x 1 or 1x 4x 2x→4x 2 4 2 or 4 limx 0 x→0 cos ( x ) cos ( 0 )limx→0 tan 2xx 2ln xlim x→1 ln (2 x 1)37.limx→1 1 8x 1x 2x 2 h(2h2 h 5)h→0h lim h→0(2h 2 h 5) 2(0) 2 0 5 or 5 cos ()or 0.5 0.5(x 4)(x 2) x 4Chapter 15 476
38.45. When P 0.99, t 2.58.Find X .1. 4 X 5 0 0.20Find the range.X tj X 16.2 2.58(0.20) 15.68416.716 mmlim 3x 9 sin3xx→0 x 2 4.546. P(not getting a 7) 1 0sinx39. lim aa 2 c 2 aa 2 02 P(never getting a 7 in five spins) 9 510c→0 a 259,049 1 00,000Letting c approach 0 moves the foci together, sothe ellipse becomes a circle. a 2 47. (x 3y)is the area of a5 5 5! r!(5 r)!(x) 5r (3y) rr0circle of radius a.To find the third term, evaluate the general term40.for r 2.2 x10 15! r!(5 r)!(x) 5r (3y) r 5!f(x) x 2!(5 2)!(x) 52 (3y) 2x y5! 2! 3!(x 3 )(9y 2 )1 0.06697 10(x 3 )(9y 2 )0.1 0.06908 90x 3 y 248. (16y 8 ) 3 4 4 16 3 y 83 40.1 0.06956 1 0.07177 2 3 y 6 8y 62 x10 1lim t 0.07 or 7%49. center 5 52, 5 12 t→0 (5, 2)41. No; the graph of f(x) sin 1 x oscillates infinitelyThe foci are on the x-axis.many times between 1 and 1 as x approaches 0,so the values of the function do not approach aa 9 12or 4unique number.b 1 (5)2or 342a.x 1 0.5 0.1 0.01 0.001cos x 0.540302 0.877583 0.995004 0.999950 1.0000001 x 220.5 0.875 0.995 0.99995 1.00000050. (x 4 2 5) 2 (y 3 2 2) 2150˚120˚ 1 → (x 1690˚60˚30˚5) 2 (y 2) 29 142b. yes; in the last three columns, all the decimalplaces of 1 x 22agree with those of cos x.) d(2)t43. lim lim2 64 d(tt→2t 2 16t→2t 2 16(t 2)( t 2)t→2t 2 lim lim 16(t 2)t→2 16(2 2) or 64 ft/s44a. As x approaches 0, the decimals for the values off(x) (1 x) 1 x approach 2.71828 ... , which is thedecimal expansion of e.44b. He ignored the exponent. As x approaches 0 fromthe positive side, 1 x approaches infinity. Anumber close to 1 raised to a large power neednot be close to 1. If x approaches 0 from thenegative side, 1 x approaches negative infinity. Anumber close to 1 raised to a large negativepower need not be close to 1, either.180˚210˚240˚0˚1 2 3 4270˚300˚330˚u51. WX 3 4, 6 0 7, 6u⏐WX ⏐ (3 4) 2 (6 0) 2 49 36 8552. C 25 in. 1 yd1 mi1 3 6 in. 17 60yd 0.0012395804 miv 6 5 C 52437.09741w v t v 360 0s 14.6 rps477 Chapter 15
153. csc 270° sin 270° 1 y 1 1or 154. Use a graphing calculator tofind the rational roots at 1 4 and 4 3 .y270˚(0, 1)x3. 0.54. 155.y 4x 5 2x 2 4x y100 4 10 010 4 10 51 610 3998045. 1y → as x → , y → as x → 100 4 10 1056. 1 2 (1)(6) (2)(3)3 6 6 (6) 1257. yes; opposite sides have the same slopem AB 1 8 y(8, 4)m DC 1 8 Am BC 1 (0, 3)4 B6. 3m AD 1 4 58. If 2 n 8, then n 3.3 n2 3 3215-2A 3 5 or 243D (2, 5)Graphing Calculator Exploration:The Slope of a CurvePage 9501-6. Exact answers are given. Accept all reasonableapproximations.1. 4xC(10, 4)7. The graph of a linear function is a line and themethods in this function will result in thecalculation of the slope of that line.8. If you zoom in on a maximum or minimum point,the graph appears flat. The slope is 0.9. At (0, 1), d ydx 1. For other points on the curve, thevalues for y and d ydxare approximately the same.2. 115-2Derivatives and AntiderivativesPages 957–958 Check for Understanding1. 4x 3 is the derivative of x 4 . x 4 is an antiderivativeof 4x 3 .12. Letting n 1 in the expression n 1x n1results in x 00, which is undefined.Chapter 15 478
3. f(x h) means substitute the quantity x h into 13. C(x) 1000 10x 0.001x 24. f(x) lim f(x h ) f(x) lim 2(x h ) 2xh→0hh→0h3(x h) 2 (3x 2) lim lim h 3xh h→0 lim 2 h 3x 3h 2 3x 2h→0h lim hh→0 2the function. On the other hand, f(x) h meanssubstitute x into the function, then add h to theresult. Using f(x) h instead of f(x h) in thedefinition of the derivative results in:C(x) 0 10 1x 11 0.001(2)x 21C(x) 10 0.002xC(1000) 10 0.002(1000) 8lim f(x) h f(x)h lim hh→0h→0h The marginal cost is $8. lim 1 h→0Pages 958–960 Exercises 114. f(x) lim f(x h ) f(x)You would always get 1.h→0h lim 3 h h→0h15. f(x) lim) f(x) 35. f(x) lim f(x h ) f(x)h→0h lim lim h→0 2xh h 2 hh→0h h(2x h 1)h→0h lim lim lim h→02x h 1 2x 0 1 or 2x 16. f(x) 2x 2 3x 5f(x) 2 2x 21 3 1x 11 0 4x 37. f(x) x 3 2x 2 3x 6f(x) 1 3x 31 2 2x 21 3 1x 11 03x 2 4x 38. f(x) 3x 4 2x 3 3x 2f(x) 3 4x 41 2 3x 31 3 1x 11 0 12x 3 6x 2 39. y x 2 2x 3y 2 1x 21 2 lx 11 0 d dx d ydx(x h) 2 x h (x 2 x)hh→0x 2 2xh h 2 x h x 2 xh 2x 2f(1) 2(1) 2 410. f(x) x 21F(x) 2 x 21 C1 1 3 x 3 C11. f(x) x 3 4x 2 x 311F(x) 3 x 21 1 1 x 111x 31 4 2 1 3x C 1 4 x 4 4 3 x 3 1 2 x 2 3x C12. f(x) 5x 5 2x 3 x 2 4F(x) 5 15 1 x 51 2 13 1 x 31 1 2 1x 21 4x C 5 6 x 6 1 2 x 4 1 3 x 3 4x C1h→0 f(x h h lim lim h→0h lim 7h→0h 716. f(x) lim f(x h ) f(x)h→0h lim h→0 3x 3h 3xh→0h3h lim lim h→0h317. f(x) lim f(x h ) f(x)h→0h lim h→0 lim h→0 lim 4h h→0h418. f(x) lim7(x h) 4 (7x 4)hh→07x 7h 4 7x 4h f(x h ) f(x)h→0h lim h→0 lim h→0 lim 4xh 2 h 2 5hh→0h h(4x 2h 5)h→0h lim 4x 2(0) 5 or 4x 5) f(x)19. f(x) lim h→0 f(x h h lim 3(x h) (3x) h4(x h) 9 (4x 9)h4x 4h 9 4x 9h2(x h) 2 5(x h) (2x 2 5x)h2x 2 4xh 2h 2 5x 5h 2x 2 5xh(x h) 3 5(x h) 2 6 (x 3 5x 2 6)hh→0x lim3 3x 2 h 3xh 2 h 3 5x 2 10xh 5h 2 6 x 3 5x 2 6–––––––––––––––hh→0 lim h→0h(3x 2 3xh h 2 10x 5h)h 3x 2 3x(0) 0 2 10x 5(0) 3x 2 10x479 Chapter 15
20. f(x) 8xf(x) 8 1x 11 821. f(x) 2x 6f(x) 2 1x 11 0 222. f(x) 1 3 x 4 5 f(x) 1 3 1x 11 0 1 3 23. f(x) 3x 2 2x 9f(x) 3 2x 21 2 1x 11 06x 224. f(x) 1 2 x 2 x 2f(x) 1 2 2x 21 1 1x 11 0 x 125. f(x) x 3 2x 2 5x 6f(x) 3x 31 2 2x 21 5 1x 11 0 3x 2 4x 526. f(x) 3x 4 7x 3 2x 2 7x 12f(x) 3 4x 41 7 3x 31 2 2x 21 7 1x 11 012x 3 21x 2 4x 727. f(x) (x 2 3)(2x 7) 2x 3 7x 2 6x 21f(x) 2 3x 31 7 2x 21 6 1x 11 0 6x 2 14x 628. f(x) (2x 4) 2 4x 2 16x 16f(x) 4 2x 21 16 1x 11 0 8x 1629. f(x) (3x 4) 3 27x 3 108x 2 144x 64f(x) 27 3x 31 108 2x 21 144 1x 11 0 81x 2 216x 14430. f(x) 2 3 x 3 1 3 x 2 x 9f(x) 2 3 3x 31 1 3 2x 21 1 1x 11 0 2x 2 2 3 x 131. y x 3 d ydx 3x 2f(1) 3(1) 2 332. y x 3 7x 2 4x 9 d ydx 3x 2 14x 4f(1) 3(1) 2 14(1) 4733. y (x 1)(x 2) x 2 x 2 d ydx 2x 1f(1) 2(1) 1 134. y (5x 2 7) 2 25x 4 70x 2 49 d ydx 100x 3 140xf(1) 100(1) 3 140(1) 24035. f(x) x 61F(x) 6 x 61 C1 1 7 x 7 C36. f(x) 3x 41F(x) 3 1 x 11 4x C1 3 2 x 2 4x C37. f(x) 4x 2 6x 71F(x) 4 2 x 21 1 6 1 x 11 7x C1 4 3 x 3 3x 2 7x C38. f(x) 12x 2 6x 11 12 2 x121 1 6 1 x111 x C 4x 3 3x 2 x C39. f(x) 8x 3 5x 2 9x 3F(x) 18 3 x 31 1 5 2 x 21 1 9 1 x 11 3x C11 2x 4 5 3 x 3 9 2 x 2 3x C40. f(x) 1 4 x 4 2 3 x 2 4F(x) 1 4 1 4 x141 2 3 1 2 x121 4x C1 2x05 2 9 x 3 4x C41. f(x) (2x 3)(3x 7) 6x 2 5x 211F(x) 6 2 x 21 1 5 1 x 11 21x C1 2x 3 5 2 x 2 21x C42. f(x) x 4 (x 2) 2 x 6 4x 5 4x 41F(x) 6 x 61 1 4 5 x 51 1 4 4 x 41 C1 1 7 x 7 2 3 x 6 4 5 x 5 C43. f(x) x3 4x 2 xx x 2 4x 11F(x) 2 x 21 1 4 1 x 11 x C1 1 3 x 3 2x 2 x C 5x 344. f(x) 2x2 x 3 2x 11F(x) 2 1 x111 x C x 2 x C45. Any function of the form F(x) 1 6 x 6 1 4 x 4 1 3 x 3 x C, where C is a constant.46a. v(12) 15 4(12) 1 8 (12) 2 81 ft/s111111Chapter 15 480
46b. v(t) 15 4t 1 8 t 249a. h(t) 3 80t 16t 2v(t) 0 4 1t 11 1 h(t) v(t) 0 80 1t 11 16 2t 218 2t 21 80 32t 4 1 4 t49b. v(1) 80 32(1)v(12) 4 1 48 ft/s4 (12)49c. At the ball’s maximum height, the velocity is 0. 7 ft/s 20 80 32t46c. When t 12 the car’s velocity is increasing at a80 32trate of 7 ft/s per second.2.5 t; 2.5 s46d. v(t) 15 4t 1 8 t 249d. h(2.5) 3 80(2.5) 16(2.5) 21s(t) 15t 4 1 t111 1 103 ft8 1 2 t121 C50. f(x) lim f(x h ) f(x) 15t 2t 2 1 2t43 h→0h C lim exh exWhen t 0, s(t) should equal 0, so C 0.h→0hs(t) 15t 2t 2 1 2t43 lim ex e h exh→0h46e. s(12) 15(12) 2(12) 2 1 2(12)43 lim ex (e h 1) 540 fth→0h47. f(x) lim f(x h ) f(x) e x lim eh 1h→0hh→0hh 1 x Acalculator indicates that lim eh 1h→0h 1, so lim h f(x) e x 1 e x . i. e. e x is its own derivative.1 x h→0x x h x(x h) x( x h)h→0 hhx(x h) lim lim h→01 lim h→0 x(x h)1 x(x 0) x1248a. When y 2010, I(2010) 2.75.The total amount spent in 2010 on health carewill be about $2.75 trillion.48b. T(y) is approximately linear near (2010, 2.75).$3.53.02.52.01.51.00.50 hHealth Care Spending2001(Trillions of Dollars)Find the slope of the tangent line at (2010, 2.75).T(2010) 2.75 – 2.562010 – 2009Projections20042007Source: Centers for Medicare & Medicaid Services20102013 0.19In 2010 the amount spent on health care will beincreasing at a rate of about $190 billion peryear.51a. total revenue cost per cup number of cupsr(p) p(100 2p)51b. When r(p) is at a maximum, the derivativeequals zero.r(p) p(100 2p) 100p 2p 2r(p) 100 1p 11 2 2p 21 100 4p0 100 4p100 4p25 p; 25 cents52. lim x2 3x 3x→3x 3 lim (x 3)(x 1)x→3x 3 lim x 1x→3 3 1 or 453a.170 180 190 200 210 220 230 240 250 26053b. See students’ work.54. List all pairs of matching numbers and theirsums.1 1 2; 2 2 4; 3 3 6; 4 4 8;5 5 10; 6 6 12There are 3 sums out of 6 that are greater thanseven.P(sum 7 given that the numbers match) 3 6 55. a n a 1 r n1a 6 9 1 3 619 2 43 or 21756. y 136e 0.06(30) 74 96° 1 2 481 Chapter 15
57. x 2 y 2 Dx Ey F 02 2 (1) 2 2D E F 0→ 2D E F 5 0(3) 2 0 2 3D F 0 → 3D F 9 01 2 4 2 D 4E F 0 → D4E F 17 02D E F 5 0 3D F 9 0(3D E F 9) 0 ( D 4E F 17) 05D E 4 0 4D 4E 8 020D 4E 16 0 4 1 3 4E 8 0(4D 4E 8) 0 4E 2 324D 8 0 E 7 3 D 1 3 3 1 3 F 9 0F 8The solution of the system is D 1 3 , E 7 3 , andF 8x 2 y 2 1 3 x 7 3 y 8 0x 1 6 2 y 7 6 2 1 18 5 i23 1 2 58. 5cos 5 6 i sin 5 6yOy 3 sin( 45˚)90˚69 5 3 5 2 2 i59. x x 1 ta 1 → x 8t 3y y 1 ta 2 → y 3t 260.61.H1253 mAd42˚12180˚ 270˚ 360˚ 450˚H 2319 md 2 253 2 319 2 2(253)(319) cos 42° 12d 165,7 77 161,4 14 cos242° 1d 214.9 m62. p q : 1 2 , 1, 3 2 , 31 2 3 8 32 5 33 2 5 3 06 3 1 2 2 1 012 0The rational roots are 3, 1 2 , and 1.63. 90 x y z 360x y z 270The correct choice is D.8Page 960Mid-Chapter Quiz1. limx→3 (2x2 4x 6) 2(3) 2 4(3) 6 18 12 6 36 x 2 9x 14x→2 2x 2 7x 62. lim( x 7)( x 2) x→2 ( 2x 3)( x 2)x 7 lim x→2 2 x 32 7 2( 2) 3or 5 lim3. xx lim4. f(x) lim f(x h ) f(x)h→0h lim sin x1 0.90930.1 1.98670.01 1.99990.01 1.99990.1 1.98671 0.9093(x h) 2 3 (x 2 3)hh→0 lim h→0 2xh h2h→0h h(2x h)h→0h lim lim 2x 0 or 2x5. f(x) f(x) 0 is a constant. sin 2xx→0x6. f(x) 3x 2 5x 2f(x) 3 2x 21 5 1x 11 0 6x 57. R(M) M 2 C 2 M 3 C 2 M 2 1 3 M 3R(M) C 2 2M 21 1 3 3M 32 2 CM M 28. f(x) x 2 7x 61F(x) 1 2 x 21 1 7 1 x 11 6x C1 1 3 x 3 7 2 x 2 6x C9. f(x) 2x 3 x 2 81F(x) 2 3 x 31 1 2 x 21 8x C1 1 2 x 4 1 3 x 3 8x C10. f(x) 2x 4 6x 3 2x 5F(x) 12 4 x 41 1 6 3 x 31 1 2 1 x 11 5x C1x 2 2xh h 2 3 x 2 3h 2 5 x 5 3 2 x 4 x 2 5x C1111Chapter 15 482
15-3Area Under a CurvePage 966 Check for Understanding1. Sample answer: y x 42. Sample answer: Subdivide the interval from a to binto n equal subintervals, draw a rectangle oneach subinterval that touches the graph at itsupper right corner, add up the areas of therectangles, and then find the limit of the totalarea of the rectangles as n approaches infinity.3. Lorena is correct. If the function is decreasing,then the graph will always be above the tops ofthe rectangles, so the total area of the rectangleswill be less than the area under the graph.4. 22 0x 2 dx lim n→ ni 1 2 in 2 n8 n→ n 3 n(n 1) (2n 1)6 3 2n3 n n 2 n32 lim lim n→ 4 3 lim 4n→3 3 12 n n 8 3 units 25. 3x 2 dx 3x 2 dx 1x 2 dx16. 107. 600 lim n→ ni 1 3 in 2 n lim03 lim n→ ni 1 2 7n→ n3 n(n 1) (2n 1)6 1 lim n→ n 3 n(n 1) (2n 1)6 3 2n3 n n 2 n3 3 2n3 n n 2 n3 lim n→ 9 2 lim n→ 1 6 in 2 1n lim 9n→2 3 12 n n 2 lim 1n→6 3 12 n n 2 9 1 3 or 2 3unit 2x 3 dx lim n→ ni 1 lim6 in 3 n1 1 n→ n 4 n2 (n 1) 24 2 n2 n n 1 2 lim n→ 1 4 lim 1n→4 2 11 n n 2 1 4 unit 2x 2 dx lim n→ ni 1 lim 6 i 2n 6n 2 1n→ n 6 3 n(n 1) (2n 1)6 3n n 2 n3 lim n→36 2n3 3 1 lim 36 2 n→n n 2 72 8. 309a. 6x 3 dx lim n→ ni 109b. 100 3 in 3 3n lim 8 1n→ n4 n2 (n 1) 24 lim 8 1n→4 n2 2n n 1 2 lim 8 1n→4 8 1432tdt lim n→ ni 1 lim 1 n2 n12 32 6 in 6 n 11 5n→ n 2 2 n(n 1)2 n 2 lim 576 n→ n2 n1 lim 5761 n→n 576 ft32tdt lim n→ ni 1 lim32 1 0in 1 0n 32 0n→ n 0 2 n(n 1)2 n 2 lim n→1600 n2 n1 lim 1600 1 n→n 1600 ftYes; integration shows that the ball would fall1600 ft in 10 seconds of free-fall. Since thisexceeds the height of the building, the ball musthit the ground in less than 10 seconds.Pages 966-96810. 2(x 1)dx lim0n→ ni 111. 30 limExercises 2 in 1 2n 1)2n→n 2( n 1 2( n 1 . . . n 1 2 n2 lim n2n→n (1 n 2 ... n)2 lim n2n→n n n(n 1)2 2 lim (2nn→n 1)n 2 lim nn→ 4 n 4 units 2x 2 dx lim n→ ni 1 lim 3 in 2 3n 2 7n→ n3 n(n 1) (2n 1)6 3 2n3 n n 2 n3 lim n→ 9 2 lim 9n→2 3 12 n n 2 9 units 2 2)483 Chapter 15
12. 2 x 2 dx 0 x 2 dx 2x 2 dx17. 4(x 2 6x)dx lim1100n→ n 4 i 2n 6 4 ini 1 1x 2 dx 2x 2 4dx By symmetry lim 00n→ n 4 n 1 2 6 4 n 1 lim n→ n in 2 1n limi 1n→ n 2 in 2 2n 4 n 2 2 6 4 n 2 . . .i 11 lim n→ n 3 n(n 1) (2n 1)6 4 n n 2 6 4 n n4 lim 8 lim n→ n 3 n(n 1) (2n 1)n→ n 1 6(1 2 n2 2 2 . . . n 2 )6 2 4 lim 1n→6 3n 2 n2n3 n3 lim 4n→3 3n 2 n2n3 n n (1 2 . . . n)34 lim n→n 1 6n2 n(n 1) (2n 1)6 lim 1n→6 3 12 n n 2 lim 4n→3 3 12 n n 2 2 4n n(n 1)2 1 3 8 3 or 3 units 2 lim n→ 6 4n3 2n3 3n 2 n6 9 6n2 n2 213. 3xdx 3xdx 1xdx lim100n→ 3 2 3 13(2 n n 2 lim n→ n 3 in 3 n limi 1n→ n in 1 n 6 43 48 or 20 83units 2i 19 lim n→ n 2 n(n 1)2 1 lim n→ n 2 n(n 1)2 18. 3(x 2 x 1)dx0 lim 9n→2 n n2 n2 lim 1n→2 n n2 n lim2n→ n 3 i 2n 3 in 1 3n i 1 lim 9n→2 11 n lim 1 3n→2 11 n lim n→n 3 n 1 2 3 1 n 1 9 2 1 2 or 4 units 23 n 2 2 3 2 n 1 . . .14. 5x 2 dx lim0 n→ n 5 in 2 5n 3 n n 2 3 n n 1i 13 lim 1 25n→ n lim3 n(n 1) (2n 1)6 n→ n 9n(1 2 2 2 2 . . . n 2 )3 lim 12 5n→6 2n3 3n 2 nn n (1 2 . . . n) n33 lim lim 12 5 3 1n→ n 9 n 2 n(n 1) (2n 1)6 3 n n(n 1)2 nn→6 2 n n 2 lim 12 5n→ 2 7n3 2n2 3n 2 n6 9 n 2 n2 n2 33units 2 lim15. 5n→ 9 2 3 12 n n 2 9 2 11 n 32x 2 dx 9 9 2 31 52x 3 12x 3 12 9 2 or 1 52units 20 0 lim n→ n2 5 in 3 5n limi 1n→ n2 in 3 1n 19. lim n→ nsin i n n i 1i 1 lim 2 50n→ n4 n2 (n 1) 24 2 lim n→ n 4 n2 (n 1) 24 20. 28xdx lim0n→ n8 2 in 2 n lim 62 5n→2 n2 2n 1n2 lim 1n→2 2n 1 n2 n i 12 lim 3 2n→ n2 n(n 1)2 lim 62 5 2 1n→2 1 n n 2 lim 1n→2 2 11 n n 2 lim 16 n→ n2 n n 2 62 52 1 2 or 312 units 21 lim 16 1 n→n16. 5x 4 dx lim0 n→ n 5 in 4 5n 16i 1 lim 31 25n→ n 6n 5 15n 4 10n 3 n530 lim 62 5n→6 6 1 5n 1 0 1n2 n 4 625 units 2 n4 n 1) 481 n Chapter 15 484
21. 4(x 2)dx lim1n→ n22. 40i 13 n→n ni 1 lim lim1 3 in 2 3n 3 3 in 3 3n→n 3 1n 3 3 2n . . . n3 3 n lim n→ n3 3n n3 (1 2 . . . n)3 3n3n→n n n(n 1)2 lim9 lim 9 n→ n 2 n2 n2 lim n→9 9 2 1 n1 9 9 2 or 2 2x 2 dx lim n→ ni 17 4 in 2 4n lim 6 4n→ n3 lim 6 4n→ n3 n(n 1) (2n 1)6 lim 3 2n→3 2n3 3n n 2 n3 lim 3 2 3 1n→3 2 n n 2 6 43 (1 2 2 2 3 2 . . . n 2 )23. 58x 3 dx 58x 3 dx 38x 3 dx30 lim n→ ni 108 5 in 3 n5 lim n→ n8 3 in 3 3n i 1 50 00n→ n4 n2 (n 1) 24 lim 6 48n→ n4 n2 (n 1) 24 2n n 1 22n n 1 22 lim lim n→1250 n2 lim n→162 n2 2 1 lim 1250 1 n→n n2 1 lim 162 1 n→n n 2 1250 162 or 108824. 4(x 2 4x 2)dx1 4(x 2 4x 2)dx 1(x 2 4x 2)dx0 lim n→ ni 1limn→ ni 1 4 in lim n→ n4 4 n0i2 4 4 n 2 4n in 2 4 in 2 1n 1 2 4 4 1n 2 4 2n 2 4 4 2n 2 ... 4 nn 2 4 4 n1 2 lim n→n1 n 2 4 1n 2 2n 22 2 . . . n n 2 4 n n 2 4 n n limn→ n4 1 n 1 n lim lim62(1 2 2 2 . . . n 2 )6 (1 2 . . . n) 2n1n→n1n 2(1 2 2 2 . . . n 2 )(1 2 . . . n) 2n4 nn→ 6 4n3 n(n 1) (2n 1)6 lim n→ 1n3 n(n 1) (2n 1)6n→ 3 23 2n3 3n 2 nn3 3n 2 n2n3n3 lim lim n→ 1 6 n→ 3 23 lim 2 n3 n1 6 n n4 lim n→ 1 6 3 12 n n 6 43 32 8 1 3 2 2 24 6 33or 4525. 2(x 5 x 2 )dx lim0n→ n26. 5042 n(n 1)22 n(n 1)2 8 2 32 n2n n 82 2 n2n n 222 1 32 1 n 82 1 21 n 2i 12 n→n ni 1 lim 2 in 5 n 2 in2 2 in 2 2n 5 2 ni 2 2 lim n→n 2 1n 5 2 1n 2 2 2n 5 2 2n 2 2 nn 5 2 nn 2 lim25 . . . (1 5 2 5 . . . n 5 )2n→n 3 n n42 (1 2 2 2 . . . n 2 )n→ 6 4n62n 6 6n 5 5n 4 n212 n83 n(n 1) (2n 1)6 limn→ 1 6 2n3 6 6n 5 5n 4 n2n 6 4 3 3n 2 n2n3n 3n→ 1 6 6 53 2 n n 2 lim limx 3 dx lim n→ ni 12 n3 n12 3 23 8 3 or 4 0 3 5 in 3 n5 lim 6 25n→ n4 n2 (n 1) 24 lim 62 5n→ 4 n2 2n n 1 2 lim 62 5 2 1n→ 4 1 n n 2 62 54 n14 4 3 485 Chapter 15
27. 30 1 2 x 3 dx lim n→ n 1 2 i 1 lim lim lim 3 in 3 3n 81 n→ 2 n4 n2 (n 1) 24 8 1n→ 8 n2 nn 1 2 8 1 1 1n→ 8 1 n n 2 8 18or 10.125 ft 228a. f(20)80 2(20) $4028b. 40(80 2x)dx20 lim n→ n 2 20 i 180 2 n 2 0 n→ n lim 2 0n→ n lim 2 0n→ n ni 1 40 4 n40 4 0in n 0i 2 0n 1 40 4 n 2 ... 40 4 n lim 40n 4 0 n (1 2 . . . n)0 0 n(n 1)2 n2 lim 800 8n→ n2 lim 800 400 n→ n2 n1 lim 800 4001 n→ n 800 400 or $40029. 10(6 0.06x 2 )dx0 lim n→ ni 1 1 0n→ n lim 6 0.06 1 0in 2 1 0n 6 0.06 10 1n 2 26 0.06 10 n6 0.06 10 nn 1 0n→ n lim 2 . . . 2 6n n62 (1 2 2 2 . . . n 2 )03 n(n 1) (2n 1)6 3 n 2 n3 lim 60 6n→ n lim 60 10 n→ 2n3n3 1 lim 60 10 2 n→ n n 60 20 or 40 106 0.06x 2 dx 2(40) or 80 By symmetry10To make a tunnel 100 ft long, multiply 80 by 100.80(100) 8000 ft 330. Setting the two functions equal y2to each other and solving for x,we find that the curves crosswhen x 0 and x 1. x x 21for 0 x 1, so we can findy x 2 y xthe desired area by subtractingthe area between the graph ofxy x 2 and the x-axis from thearea between the graph ofO 1y x and the x-axis. These areas are 1 3 unit 2 and 1 2 unit 2 , respectively, so the answer is 1 2 1 3 1 6 unit 2 .2 31a.31b. 12(50 36t 3t 2 )dt0 lim n→ ni1 1 2n→ n lim31c. $1 4641232a.r(t)1601208040O50 36 1 2in 50 36 12 n 22i 3 1 n 2 1 n2 1 3 12 1n 250 36 12 n 3 12 n50 36 12 nn 3 12 nn 22 lim 50n 4 32 (1n 2 . . . n) 1n→ n 4 32n2 2 2 . . .(12 22 . . . n2 )84 1) 8 4 n(n 1) (2n 1)6 3n n 2 n31600 25921 n→ n3 1 8642 n n 2 lim 600 51n→ n 2 n(n 2 51 n3 n lim 600 n→ 2592 n2 n2 8642n3 lim 600 2592 1728 $1464v(t)15 $12210532b. 10(3.5t 0.25t 2 )dt0 lim n→ ni1 1 0n→ n lim3.5 1 0in 0.25 1 n 10i 2 1 0n 3.5 10 n 0.25 10 n3.5 10 2n 0.25 10 n3.5 10 nn 1 0n→ n 3 n lim . . . n 2 )n→ 3 5n 0 2 n(n 1)2 n2 limr(t) 50 36t 3t 2v(t) 3.5t 0.25t 2 2 1 2 2 . . . n 2 0.25 10 n5 (1 2 . . . n) 2 (1n5 2 2 2 2 n lim n→ 175 n2 n 12 3 limn→ 1751 n1 12 3 175 25 3v(t) 1.2t 0.03t 2O 1 2 3 4 5 6 7 8 9 10t0 or 27 53mt5 0 3 n(n 1) (2n 1)65 2n3 3n n 2 n35 3 1 2 n n 22 Chapter 15 486
100to find the maximum(1.2t 0.03t 2 )dt40. Use a graphing calculator lim n→ nwidth of x 5.4 cm.1.2 1 0in 0.03 1 0in 2 1 0n i1 lim n→ 1 0n 1.2 10 1n 0.03 10 1n 21.2 10 2n 0.03 10 2n 2 . . .Period 2 10 or 5 1.2 10 nn 0.03 10 nn 2 lim 1 0n→ n 1 2 (1n 2 . . . n) n32 (1 2 2 2 . . . n 2 )[5, 10] scl1 by [30, 90] scl1 lim n→ 1 20n2 1)2 3 0n3 n(n 1) (2n 1)6 41. 6 2 8 2 10 2 n lim n→ 60 n2 n2 5 3n n2n3n36 64 10031 lim 601 n→ n3 1 52 n n 2 60 10 or 70 mABC 1The first one results in a greater distance2 (6)(8) or 24 square inches. If thecovered.rectangle has a width of 3 inches, then it has a 2 can be rearranged tolength of 2 4obtain x 2 y 2 r 2 3, which is the circle centered at of the rectangle 2(3) 2(8) or 22 inches. Thethe origin of radius r.Inthe equation y r 2 x , correct choice is C.y must be nonnegative, so the graph is only thetop half of the circle. Therefore, the value of theintegral is 1 2 the area of a circle of radius r,Page 969 History of Mathematicsor 1 2 r 2 .9x 2 2x 7 x 2x→2x 2 2 22 2 0 4 or 015-4The Fundamental Theorem ofCalculusPages 972–973 Check for Understanding1. f(x)dx represents all of the function that havef(x) as their derivative. b f(x)dx is a number; itagives the area under the graph of y f(x) fromx a to x b.2. Sample answer: Let a 0, b 1, f(x) x, and 1 1 825 7 2 5Amplitude 1 2 33. The equation y r 2 x34. f(x) 3x 2 x 2 7xf(x) 3 3x 31 1 2x 21 7 1x 1135. lim36. log1 x 33x 1 3 3x 3 2 or 2737. u u 2, 5, 3 3, 4, 7 2 (3), 5 4, 3 (7) 1, 1, 10i u j u 10k u38. cos 2v 1 2 sin 2 v 1 2 3 5 239. y 1 2 sin 10v100 100ABC is a right triangle with a base of 6 inchesand a height of 8 inches. The area of or 8 inches, since Aw. The perimeter1. See students’ work. The difference in area shoulddecrease as the number of sides of the polygonincreases.2. The roots of the resulting equation are the zeros ofthe derivative of the original function.3. See students’ work.g(x) x. Then b a f(x)g(x)dx 1 0 x2 dx 1 3 , but b a f(x)dx b a g(x)dx 1 0 xdx 1 0 xdx 1 2 1 2 1 4 .3. If the “C ” is included in the antiderivative, itwill appear as a term in both F(b) and F(a) andwill be eliminated when they are subtracted.4. Rose is correct; the order does matter.Interchanging the order multiplies the result by1. In symbols, F(a) F(b) (F(b) F(a)). Sounless the answer is 0, interchanging the orderwill give the opposite of the right answer.5. (2x 2 4x 3)dx 2 1 3 x 3 4 1 2 x 2 3x C 2 3 x 3 2x 2 3x C6. (x 3 3x 1)dx 1 4 x 4 3 1 2 x 2 x C 1 4 x 4 3 2 x 2 x C487 Chapter 15
7. 0 (4 x 2 )dx 4x 1 3 x 3 02240 1 3 0 3 4(2) 1 3 (2) 3 0 8 8 3 8. 2x 4 dx 1 5 x 5 2 00 1 63units 2 1 5 2 5 1 5 0 5 3 25 0 or 3 5units 29. 1 (x 2 4x 4)dx1 1 3 x 3 4 1 2 x 2 4x 1 1 1 3 x 3 2x 2 4x 1 1 1 3 1 3 1 93 1 3 2 1 2 4 12 (1) 3 2 (1) 2 4 (1) 7 3 or 2 3units 210. 32x 3 dx 2 1 4 x 4 31 1 1 2 x 4 3 111. 41 1 2 3 4 1 2 1 4 6 8 12 1 2 or 40(x 2 x 6)dx 1 3 x 3 1 2 x 2 6x 4 1 1 3 4 3 1 2 4 2 6 4 1 3 1 3 1 2 1 2 6 1 1 12 3 3 56 22 46 3 56or 6 3212. 2(2x 2 3x 2)dx 2 1 3 x 3 3 1 2 x 2 2x 2 0013. 42 2 3 x 3 3 2 x 2 2x 2 0 2 3 2 3 3 2 2 2 2 2 2 3 0 3 2 0 2 0 1 43 0 or 1 3(x 3 x 6)dx 1 4 x 4 1 2 x 2 6x 4 2 1 4 4 4 1 2 4 2 6 4 1 4 2 4 1 2 2 2 6 214. 10 cm 1 m1 10 0 cm 0.10 48 (6) or 54 0.1 m500xdx 500 1 2 x 2 0.104 250x 2 0.10 (250 (0.1) 2 ) (250 0 2 ) 2.5 0 or 2.5 JPages 973–976 Exercises15. x 5 dx 1 6 x 6 C16. 6x 7 dx 6 1 8 x 8 C 3 4 x 8 C17. (x 2 2x 4)dx 1 3 x 3 2 1 2 x 2 4x C 1 3 x 3 x 2 4x C18. (3x 2 x 6)dx 3 1 3 x 3 1 1 2 x 2 6x Cx 3 1 2 x 2 6x C19. (x 4 2x 2 3)dx 1 5 x 5 2 1 3 x 3 3x C 1 5 x 5 2 3 x 3 3x C20. (4x 5 6x 3 7x 2 8)dx 4 1 6 x 6 6 1 4 x 4 7 1 3 x 3 8x C 2 3 x 6 3 2 x 4 7 3 x 3 8x C21. (x 2 6x 3)dx 1 3 x 3 6 1 2 x 2 3x C 1 3 x 3 3x 2 3x C22. 32x 2 dx 2 1 3 x 3 3 00 2 3 x 3 3 0 2 3 3 3 2 3 0 3 18 0 or 18 units 223. 3(x 2 2)dx 1 3 x 3 2x 3 22 1 3 3 3 2 3 1 3 2 3 2 23 4 3 or 1 33units 224. 2(4x x 3 )dx 4 1 2 x 2 1 1 4 x 4 2 0025. 40 2x 2 1 4 x 4 2 02 2 2 1 4 2 4 2 0 2 1 4 0 4 4 0 or 4 units 2x 3 dx 1 4 x 4 4 0 1 4 4 4 1 4 0 4 64 0, or 64 units 226. 1 3x 6 dx 3 1 7 x 7 111 3 7 x 7 1 1 3 7 1 7 3 7 (1) 7 3 7 3 7 or 6 7 unit 227. 0 (x 2 2x)dx 1 3 x 3 2 1 2 x 2 022 1 3 x 3 x 2 0 2 1 3 0 3 0 2 1 3 (2) 3 (2) 2 0 2 03 or 2 03units 2Chapter 15 488
28. 3(x 2 2x 3)dx 1 1 3 x 3 2 1 2 x 2 3x 3 37. 0 (x 4 x 3 )dx11 1 3 x 3 x 2 3x 3 1 115 x 5 1 4 x 4 0 1 13 3 3 3 2 3 3 1 3 1 3 1 2 9 3 1 0 2 9 1 1 3 or 1 0 9or 2 063units 238. 2(x 3 x 1)dx29. 1(x 3 x)dx 1 4 x 4 1 2 x 2 1 000 1 4 1 4 1 2 1 2 1 4 0 4 1 2 0 2 3 4 0 or 3 4 unit 2 8 0 or 830. 3 (x 3 8x 10)dx39. 5 (x 2 3x 8)dx1 1 4 x 4 8 1 2 x 2 10x 3 2 1 3 x 3 3 1 2 x 2 8x 512 1 4 x 4 4x 2 10x 3 1 3 x 3 3 2 x 2 8x 512 1 4 3 4 4 3 2 10 3 1 3 5 3 3 2 5 2 8 5 1 4 (1) 4 4 (1) 2 10 1 1 3 (2) 3 3 2 (2) 2 8 2 34 54 2 34 or 92 units 2 26 56 7 43 31. 76x 2 dx 6 1 3 x 3 7 26 56 14 86or 41 3600 2x 3 7 40. 3(x 3)(x 1)dx01 (2 7 3 ) (2 0 3 ) 3(x 2 2x 3)dx 686 0 or 686132. 43x 4 dx 3 1 5 x 5 4 1 3 x 3 2 1 2 x 2 3x 3 122 3 5 x 5 4 1 3 x 3 x 2 3x 3 12 3 5 4 5 3 5 2 5 30 725 9 65or 29 9 5763 or 3 23533. 3 (x 4)dx 1 2 x 2 4x 3 41. 3(x 1) 3 dx 3(x 3 3x 2 3x 1)dx2211 1 1 2 3 2 4 3 1 2 (1) 2 4 4 1 x 4 3 1 3 x 3 3 1 2 x 2 x 3 2 3 32 7 2 1 4or 20 x 4 x 3 3 2 x 2 x 3 234. 5(3x 2 2x 1)dx 3 1 3 x 3 2 1 2 x 2 x 5 1 4 3 4 3 3 3 2 3 2 311 x 3 x 2 x 5 1 4 2 4 2 3 3 2 2 2 21 1 5 (5 3 5 2 5) (1 3 1 2 4 0 or 1 45 1) 105 1 or 10442. 1 x2 x 2x dx2 1 (x 2)(x 1)x dx20035. 3(x 3 x 2 )dx 1 4 x 4 1 3 x 3 3 1(x 1)dx110 1 4 3 4 1 3 3 3 1 4 1 4 1 3 1 3 1 2 x 2 x 1 4 05 14 1 2 1 1 35112 1 2 or 3 2 1 2 1 1 2 0 2 043 336. 12 (x 4 2x 2 1)dx043. 2x(4x 2 1)dx 2(4x 3 x)dx 1 5 x 5 2 1 3 x 3 x 1 000 4 1 1 5 x 5 2 3 x 3 x 1 4 x 4 1 2 x 2 2 00 1 5 1 5 2 3 1 3 1 1 5 0 5 2 3 0 3 0 2 815 0 or 2 815 18 0 or 18 1 5 0 5 1 4 0 4 1 5 (1) 5 1 4 (1) 4 1 4 x 4 1 2 x 2 x 2 0 1 4 2 4 1 2 2 2 2 1 4 0 4 1 2 0 2 0 1 3 3 3 3 2 3 3 1 3 1 3 1 2 3 1 x 4 1 2 x 2 2 02 4 1 2 2 2 0 4 1 2 0 2 489 Chapter 15
44. 1 (x 1)(3x 2)dx49a.1 1 (3x 2 5x 2)dx1 3 1 3 x 3 5 1 2 x 2 2x 1 1 x 3 5 2 x 2 2x 1 11 3 5 2 1 2 2 1(1) 3 5 2 (1) 2 2 1 1 21 1 2 or 645a. 20.5x 3 dx 1 4 x 4 20.500 1 4 20.5 4 1 4 0 4 44,152.52 0 or 44,152.52 20i 3 202 (20 1)24i1 400 (441)40or 44,10045b. 100.5x 2 dx 1 3 x 3 100.500 1 3 100.5 3 1 3 0 3 338,358.38 0 or 338,358.38100 i 2 100(100 1)(2 100 1) 6i1 100(10 1)(201)6or 338,35046. 2490,000 xdx 490,000 1 2 x 2 2 00 245,000 x 2 2 0 (245,000 2 2 ) (245,000 0 2 ) 980,000 0 or 980,000 J47a. Since the graph is below the x-axis, f(x) isnegative. Each f(x i ) is negative and x isnpositive, so each term in the sum f(xi1 i )x isnegative. Therefore, the sum is negative. since b f(x)dx is a limit of negative sums, it is alsoanegative.47b. 2(x 2 5)dx0 1 3 x 3 5x 2 0 1 3 2 3 5 2 1 3 0 3 5 0 2 2347c. Since the function is negative, the integral inpart b gives the opposite of the area of theregion. The area is 2 23.48. yThe integral represents the108642O2 1 2 3 4 5x4 y 3x 66area of a right triangle. Thevalue is 1 2 (3)(9) 2 72.110100908070605040302010149b. 6 0 60 75 8x 1 2 x 2 dx 1 6 75x 8 1 2 x 2 1 2 1 3 x 3 6 0 1 6 75x 4x 2 1 6 x 3 6 0 1 6 75 6 4 6 2 1 6 6 3 75 0 4 0 2 1 6 1 6 (558 0) $93 0 3 149c. 12 6 1275 8x 1 2 x 2 dx6 1 6 75x 8 1 2 x 2 1 2 1 3 x 3 1250. R ROf(x) 1 6 75x 4x 2 1 6 x 3 12 1 6 75 12 4 12 2 1 6 12 2 75 6 4 6 2 1 6 1 6 (1188 558) $1056 6 3 (R 2 x 2 )dx R 2 x 1 3 x 3 R RR 2 R 1 3 R 3 R 2 R 1 3 (R) 3 2 3 R 3 2 3 R 3 4 3 R 3f(x) 75 8x 1x 2251a. 1000 k(6.4 10 6 ) 211000 k 40.96 10 124.1 10 16 k; 4.1 10 16 Nm 23.8 10851b. 4.1 6.4 10 6 1016 x 2 dx 4.1 10 16 (1)x 1 3.8 4.1 10 16x 10 86 10 86.4 10 6 3.8 6.4 10 6 4 . 1 13.8 016 4 . 1 11086 . 4 016 106 1.1 10 8 64.1 10 8 6.3 10 9 JxChapter 15 490
52. 20 1 2 x 2 dx lim n→ ni1 lim 1 2 2 in 2 2n 4 n→ 2n 2 3 n(n 1) (2n 1)6 3n n 2 n32 lim n→ 4 6 2n3 lim 2n→ 3 3 12 n n 4 3 53. f(x) 2x 6 3x 2 2f(x) 2 6x 61 3 2x 21 0 12x 5 6x54. In a normal distribution, 68.3% of the data liewithin 1 standard deviation of the mean. So,100 68.3 or 31.7% of the data lie outside1 standard deviation of the mean. Thus, 31.7% ofthe test-takers scored more than 100 points aboveor below the mean of 500.55. To begin with 25 out of the 50 numbers are odd.With each consecutive draw, there is 1 fewer oddnumbers and 1 fewer tickets.P(four odd numbers) 2 550 2 449 2 348 2 247 56. A Pe rt 600e 0.06(15) $1475.7657. h 6; k 1h p 36 p 3 or p 3(y k) 2 4p(x h)(y 1) 2 12(x 6)258. r 2 or 22v 2 3 3 or 3 253 4 6062cos 3 i sin 3 ; 2 1 2 i 32 1 3i59. sin 52° 4v45 ft/s52˚V xV yy545 sin 52° v y35.46 v y ; 35.46 ft/s60. mAE+ m∠AXE, so ∠AXE is the smallest angle.If the circle was divided into 5 equal parts, eachangle would measure 72°. Since the circle is notdivided evenly, the smallest angle AXE is lessthan 72°. The correct choice is A.Chapter 15 Study Guide and AssessmentPage 977 Understanding the Vocabulary1. false; sometimes 2. true3. false; indefinite 4. true5. false; secant 6. false; tangent7. false; derivative 8. true9. false; rate of change 10. false; derivativePages 978–980 Skills and Concepts11. There is a point at (2, 1) so f(2) 1. However,the closer x is to 2, the closer y is to 3.So, lim f(x) 3.x→212. limx→2 (x3 x 2 5x 6) (2) 3 (2) 2 5(2) 68 4 10 6 413. lim (2x cos x) 2(0) cos 0x→0 0 11 x2 36x→1x 614. lim 5 x2x→02x15. lim lim lim (x 6)(x 6)x→1x 1 lim x→1x 6 1 6 or 5x→0 5 2 x 5 2 (0) 0x16. lim 2 2xx→4 x 2 lim 3x 10x(x 2)4 x→4 (x 5) (x 2)x lim x→4 x 54 4 5 or 417. lim (x sin x) 0 sin 0x→0 018. lim x2 x cos xx→02x lim x(x x→02x19. lim x→220. lim21. lim x→5 x2 22. f(x) lim lim h→0 lim h→0 lim 2h→0h 2cos x) lim x cos xx→02 0 cos 02h 1 2 x 3 2x 2 4x 8x 2 4 (x 2)(x 2 4)x→2 x2 4 lim lim x→2x 2 2 2 or 4 (x 3) 2 9x→02x lim x2 6x 9 9x→02x lim x(x 6)x→02x lim x 6x→02 0 62or 39x 20x 2 5x lim (x 5) (x 4)x→5x(x 5) lim x 4x→5x 5 45or 1 5 f(x h ) f(x)h→0h2(x h) 1 (2x 1)h2x 2h 1 2x 1h491 Chapter 15
23. f(x) lim f(x h ) f(x)h→0h lim4(x h) 2 3(x h) 5 (4x 2 3x 5)hh→0 lim4x 2 8xh 4h 2 3x 3h 5 4x 2 3x 5hh→0 8xh 4 h 2 3hh→0 h h(8x 4h 3)h→0 h lim lim24. f(x) lim lim 8x 4h 3h→0 8x 3 f(x h ) f(x)h→0h lim h→0 limx 3 3x 2 h 3xh 2 h 3 3x 3h x 3 3xhh→0 lim h→0lim h→0 lim(x h) 3 3(x h) (x 3 3x)h3x 2 h 3xh 2 h 3 3hhh(3x 2 3xh h 2 3)hh→03x 2 3xh h 2 3 3x 2 325. f(x) 2x 6f(x) 2 6x 61 12x 526. f(x) 3x 7f(x) 3 1x 11 0327. f(x) 3x 2 5xf(x) 3 2x 21 5 1x 11 6x 528. f(x) 1 4 x 2 x 4f(x) 1 4 2x 21 1 1x 11 0 1 2 x 129. f(x) 1 2 x 4 2x 3 1 3 x 4f(x) 1 2 4x 41 2 3x 31 1 3 1x 11 0 2x 3 6x 2 1 3 30. f(x) (x 3)(x 4) x 2 7x 12f(x) 2x 21 7 1x 11 0 2x 731. f(x) 5x 3 (x 4 3x 2 ) 5x 7 15x 5f(x) 5 7x 71 15 5x 51 35x 6 75x 432. f(x) (x 2) 3 x 3 6x 2 12x 8f(x) 3x 31 6 2x 21 12 1x 11 0 3x 2 12x 1233. f(x) 8x1F(x) 8 1 x111 C 4x 2 C34. f(x) 3x 2 21F(x) 3 2 x121 2x C x 3 2x C35. f(x) 1 2 x 3 2x 2 3x 2F(x) 1 2 1 3 x 31 1 2 2 x 2111 3 1 x 11 2x C1 1 8 x 4 2 3 x 3 3 2 x 2 2x C36. f(x) x 4 5x 3 2x 61F(x) 4 x 41 1 5 3 x 31 1 2 1 x 111 6x C 1 5 x 5 5 4 x 4 x 2 6x C37. f(x) (x 4)(x 2) x 2 2x 81F(x) 2 x 21 1 2 1 x 11 8x C1 1 3 x 3 x 2 8x C38. f(x) x2 xx x 11F(x) 1 x 11 x C39. 2040. 1041. 431 1 2 x 2 x C2x dx lim n→ ni 18 lim n→n lim 4 n→ n2 n lim112 2 in 2 n 2 n(n 1)2 n 2n→4 1 n1 4 units 2x 3 dx lim n→ n in 3i 1 n1 1) 21 lim n→n 4 n2 (n4 lim 1n→4 2 n2 n n 1 2 lim 1n→4 2 11 n n 1 4 unit 2x 2 dx 4x 2 dx 3x 2 dx0 lim n→ ni 10 4 in 2 n2 14 lim n→ ni 1 lim 6 4n→n3 n(n 1) (2n 1)6 lim 2 7n→n3 n(n 1) (2n 1)6 lim 6 4n→6 2n3 3n n 2 n3 lim 2 7n→6 2n3 3n n 2 n3 lim 6 4 3n→6(2 n n12 ) lim 2 7n→6 6 43 2 73 3 73units 21 3 in 2 3n 2 n3 n12 Chapter 15 492
42. 26x 2 dx 26x 2 dx 16.x 2 dx100 lim n→ n6 2 in 2 2n limi 1n→ n6 in 2 1n i 1 lim 4 8n→n3 n(n 1) (2n 1)6 6 lim n→n 3 n(n 1) (2n 1)6 lim 4 8n→6 2n3 3n 2 nn3 lim 6n→6 3 2n3 n n 2 n33 13 1 lim 8(2 n→n n 2) lim 1 2 n→n n 2 16 2 or 14 units 243. 46xdx 6 1 2 x 2 4 22 3x 2 4 2 (3 4 2 ) (3 2 2 ) 48 12 or 3644. 2 3x 2 dx 3 1 3 x 3 233 x 3 2 3 2 3 (3) 2 8 (27) or 3545. 2 (3x 2 x 3)dx 3 1 3 x 3 1 1 2 x 2 3x 222Page 981 Applications and Problem Solving51. limt→100 1 2 m 501 t 2 2 1 2 m 5021 100 2 0.0000125 m52. c(x) 9x 5 135x 3 10,000c(x) 9 5x 51 135 3x 31 045x 4 405x 2c(2.6) 45(2.6) 4 405(2.6) 2 or $681.4153a. 6 0im1 r h 52 80ft 1 hr1 i m 3 600s 88 ft/sa 88 ft/s5 s 17.6 ft/s 253b. v(t) t17.6dx0 17.6x t 0 17.6t 17.6(0) 17.6t53c. d(t) t17.6xdx01 17.6 1 x111 t 0 8.8x 2 t 0 8.8t 2 8.8(0) 2 8.8t 2 x 3 1 2 x 2 3x 2 2 (2 3 1 2 2 2 3 2) ( 2) 3 1 2 (2) 2 3 (2) 12 (16) or 2846. 4(x 2)(2x 3)dx 4(2x 2 7x 6)dx047. 6x 4 dx 6 1 5 x 5 C 6 5 x 5 C0 2 1 3 x 3 7 1 2 x 2 6x 4 0 2 3 x 3 7 2 x 2 6x 4 0 2 3 4 3 7 2 4 2 6 4 2 3 0 3 7 2 0 2 6 0 36 83 0 or 36 3848. (3x 2 2x)dx 3 1 3 x 3 2 1 2 x 2 Cx 3 x 2 C49. (x 2 5x 2)dx 1 3 x 3 5 1 2 x 2 2x C 1 3 x 3 5 2 x 2 2x C50. (3x 5 4x 4 7x)dx 3 1 6 x 6 4 1 5 x 5 7 1 2 x 2 C 1 2 x 6 4 5 x 5 7 2 x 2 CPage 981 Open-Ended Assessment1. Sample answer: f(x) x 2 2x 2;limx→1 (x2 2x 2) 1 2 2(1) 2 52. Sample answer: g(x) 16x 3 ; 1016x 3 dx 4x 4 1 0 4 1 4 4 0 4 4Chapter 15 SAT & ACT PreparationPage 9831. 1 2 1 3 1 1 2 1 3 SAT and ACT Practice1 3 6 2 6 1 1 6 6The correct choice is A.493 Chapter 15
2. Let x represent the length of the second side of thetriangle. Then the first side has length 2x.6. Since 5 is prime, 5 3(5) or 15.x2xSince 16 is composite, 16 1 2 (16) or 8.Clearly the perimeter must be greater than 3x, soeliminate answer choices A and B. Use theTriangle Inequality Theorem.x 2x ?P x 2x 3x3x ?P 6x? x 2x P x 2x x? x P 4xSince the perimeter cannot equal 6x or 4x,eliminate answer choices C and E as well.The only possible answer choice is D.3. Draw a figure. Begin with the 3 parallel lines.Draw the 3 nonparallel lines in positions that areas general as possible. For example, do not drawperpendicular lines or concurrent lines. Draw thefirst nonparallel line, and mark the intersections.Then draw the second line, making sure itintersects each of the other lines. Then draw thethird line, making sure it intersects each of theothers.There are 12 intersections.The correct choice is D.4. The triangles are right triangles. The verticalangles formed by the two triangles each measure45°, since 180 135 45. The hypotenuse of thetriangles is the radius of the circle, which is 6,since the diameter is 12. Using the relationshipsof 454590 right triangles, the length of each6height and base must be . The area of one6 6triangle is 1 2 22 or 9. The area of bothtriangles is 2 9 or 18. The correct choice is B.5. x 2 x 6x 2 x 6 0(x 3)(x 2) 0At x 3 and x 2,the inequality equals0. Test values that aregreater than and lessthan 3 and 2 todetermine for whichvalues the inequalityis less than 0.?For values of x that are between 2 and 3, theinequality is true.The correct choice is C.2x 2 x 6 0x y3 62 00 63 04 65 16 15 8 or 23.Now, you need to determine which of the choices isequal to 23. Calculate each one.2146 10.5 23 69 31 63 23 69 34.5The correct choice is B.7. The nth term of an arithmetic sequence with firstterm a 1 and common difference d is given bya n a 1 (n – 1)d. a 1 4, and d 3, soa 37 4 (37 – 1)(3) 4 (36)(3) or 112The correct choice is C.8. Draw a figure.Of the 16 cubes on the front, only the 4 in thecenter have just one blue face. It is the same oneach of the faces. There are 6 faces, so there are6 4 or 24 cubes with just one blue face. Thecorrect choice is A.9. To find the value of {{x}}, first find the value of {x}.{x} x 2 1.So {{x}} {x 2 1}.And {x 2 1} (x 2 1) 2 1 (x 4 2x 2 1) 1 x 4 2x 2The correct choice is D.10. There are two distinct prime factors of 20. Theyare 2 and 5 and their product is 10. There is onlyone distinct prime factor of 16. It is 2.2016The answer is 5.Chapter 15 494
Extra PracticeLesson 1-1Page A261. D {2, 1, 2}; R {4, 2, 4}; no2. D {3, 0.5, 0.5, 3}; R {0.5, 3}; yes3. D {1, 0, 2, 5, 7}; R {1, 2, 3, 5, 7}; yes4. D {2, 2.3, 3.2}; R {4, 1, 3, 4}; no5. f(4) 4(4) 216 2 or 146. g(3) 2(3) 2 (3) 5 2(9) 3 5 18 3 5 or 2637. h(1.5) 2(1 .5) 3 3 or 18. k(5m) ⏐3(5m) 2 3⏐ ⏐3(25m 2 ) 3⏐ ⏐75m 2 3⏐Lesson 1-24. [f g](x) f(g(x)) f(2x 3 x 2 x 1)2(2x 3 x 2 x 1)4x 3 2x 2 2x 2[g f](x) g(f(x)) g(2x) 2(2x) 3 (2x) 2 (2x) 116x 3 4x 2 2x 1Lesson 1-3Page A261. x 2 0x 2f(x)Oxf(x) x 2Page A261. f(x) g(x) 2x 1 x 2 3x 1 x 2 5x 2f(x) g(x) 2x 1 x 2 3x 1x 2 xf(x) g(x) (2x 1)(x 2 3x 1) 2x 3 5x 2 5x 1f f( x)(x) g g( x)2x 1 x 2 3x 12. [f g](x) f(g(x)) f(4x 2 ) 3 4x 2[g f](x) f(f(x)) g(3 x) 4(3 x) 2 36 24x 4x 23. [f g](x) f(g(x))f(x 9) 1 3 (x 9) 1 1 3 x 2[g f](x) g (f(x)) g 1 3 x 1 1 3 x 1 9 1 3 x 82. 3x 4 03x 4x 4 3 Of(x) 3x 4f(x)Of(x) 1f(x)3. 1 0, false 4. 4x 0none x 0xxf(x)f(x) 4xOx495 Extra Practice
5. 2x 1 02x 1x 1 2 f(x)f(x) 2x 1Ox6. x 5 0x 5x 5f(x) x 5f(x)Ox15. m 3 or 1 3 y 6 1 3 (x (2))3y 18 x 2x 3y 16 06. x 10 is a horizontal line; perpendicular slope isundefined.y 15 or y 15 07. m 25 or 2 5 y (7) 2 5 (x 3)5y 35 2x 62x 5y 29 0Lesson 1-4Page A261. y mx b → y 2x 12. y 2 1(x 1)y 2 x 1y x 33. y mx b → y 1 4 x 34. y (4) 0(x (2))y 4 0y 43 15. m 2 22 4or 1 2 y 1 1 2 (x 2)y 1 1 2 x 1y 1 2 x 26 06. m 0 (1) 6 1 or 6y mx b → y 6x 67. y 08. y 0 1.5(x 10)y 1.5x 15Lesson 1-5Page A271. None of these; the slopes are neither the same noropposite reciprocals.2. y (2) 1(x 0)y 2 xx y 2 03. y 3 2(x 1)y 3 2x 22x y 1 04. y 1 is a vertical line; parallel slope is undefined.y 12 or y 12 0Lesson 1-6Page A271.Enrollment(millions)y484746454443424140x001990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001Year2. Sample answer: y 0.6091x 1171.647.2 – 40.5m 2 001– 1990 0.6091y 40.5 0.6091(x 1990)y 0.6091x 1171.63. Sample answer: y 0.6125x 1178; r 0.99Enter the School Year data as List 1.Enter the Enrollment data as List 2.Perform a linear regression on the graphingcalculator.4. Sample answer: 53.7 million; yes; the correlationcoefficient shows a strong correlation.f(2011) 0.6125(2011) 1178 53.7 millionExtra Practice 496
Lesson 1-75. y6.1 x y 4yPage A271. f(x)2.g(x)Oxg(x) |x 2|y |x |OxOxOxLesson 2-13. h(x)4.f(x)Page A28Ox1.y2x y 1 0Ox3y 6x 3Ox5. g(x)6.g(x) |5 2x|k(x)y 2x 1; consistent and dependent2.yOxOx(1, 5)y 3x 8Ox y 4xLesson 1-8(1, 5); consistent and independent3. yPage A273x y 11. y2.yOxy 22y 6x 4O x3. y4.O xy x 1Ox y 3yO4x 2y 6xxno solution; inconsistent4. 5x 2y 1 x 2y 5x 2y 5 1 2y 54x 4 2y 6x 1 y 3(1, 3)5. 2x 4y 8 2x 3y 82x 3y 8 2x 3(0) 8y 0 2x 8x 4(4, 0)497 Extra Practice
6. 8x 2y 2 → 16x 4y 43x 4y 233x 4y 2319x 19x 18x 2y 28(1) 2y 22y 10y 5(1, 5)Lesson 2-2Page A281. x y 6 y z 8x z 2 y z 2y z 8 2y 10y 5x y 6x z 2x 5 61 z 2x 1z 3(1, 5, 3)2. 2x 2y z 6 → 4x 4y 2z 12x y 2z 6 x y 2z 63x 5y 18x 2y z 72x 2y z 6x 4y 133x 5y 18 → 3x 5y 18x 4y 13 3x 12y 397y 21y 3x 4y 13x y 2z 6x 4(3) 13x 1(1, 3, 2)3. 2x 3y z 1x y z 43x 2y 31 (3) 2z 62z 4z 2x y z 4 → 2x 2y 2z 83x 2y 2z 3 3x 2y 2z 35x5x 13x 2y 3x y z 43(1) 2y 3 1 0 z 42y 0z 3y 0 z 3(1, 0, 3)Lesson 2-3Page A281. 2x y 1 x 2y 1y 1 2x x 2(1 2x) 13x 2 13x 3x 1y 1 2(1) 1(1, 1)2. x 2y 5 x 2y 52x 2y 2 1 2y 53x 3 2y 4x 1 y 2(1, 2)3. 3x y 7 4y 5x3x 7 y4(3x 7) 5x12x 28 5x7x 28x 4y 3(4) 75(4, 5)4 7 1 (5)4. A C 1 0 5 1 2 8 6 411 6 1 6 10 105. D E4 1 1 (5)2 (3) 3 2556 16. 4B4(2) 4(0) 4(3)4(4)4(3) 4(2)8 0 121612 87. impossible2(7) 2(5) 3(4) 3(1)8. 2C 3A 2(0) 2(1) 3(1) 3(5)2(8) 2(4)14 12 0 316 626 13 3 17 22 269. impossible3(2)10 (3)2 158 183(6)1 5 43 2 21 31(4) (5)(2) 1(1) (5)(3)3(4) 2(2) 3(1) 2(3)10. ED 1416143Extra Practice 498
7 52 011. BC 3 0 14 3 282(7) 0(0) (3)(8)4(7) (3)(0) 2(8)10 2244 15BC D10 (4)44 26 2342 18Lesson 2-4Page A2842(5) 0(1) (3)(4)4(5) (3)(1) 2(4)22 115 35 22 1 11. 0.5 122 3 4 6 1 1 2 A2 1 2 , 1 1 2 , B(1, 2), C(1, 3)62354715232. 4 5 6 158 4 2J(4, 5), K(5, 8), L(6, 4), M(1, 2)013804223323233. 1 5 3 02N(5, 2), P(3, 8), Q(0, 4)C(1, 3)yL(4, 7)8L(6, 4) M(1, 5)4M(1, 2)8 4 O 4 8x4 J(6, 2)K(3, 5)J(4, 5)8K(5, 8)52P(3, 8)yB(1, 2)OC(2, 6)8y804B(2, 4)P(3, 8)N(5, 2) 4 N(5, 2)O8 4 4 8x4 Q(0, 4)Q(0, 4)x( )A 2 1 12, 12A(5, 3)001 3 2 1203 6 5 232 1 0W(3, 3), X(6, 2), Y(5, 1), Z(2, 0)4. 1 3 6 5 y8W(3, 3)Z(2, 0)X(6, 2) 4Z(2, 0) Y(5, 1)8 4 O 4 8xY(5, 1)4 X(6, 2)0 0 11 0 1 01100240 1 2 3 7 2F(4, 2), G(2, 3), H(6, 7)5. Rot 90 R 1y x 1 4 6 2 Lesson 2-5Page A293111. 3(2) (11)(7)37 832. 3(2) (7)(5)506 30221 3293. 5(6) 1 2 (0) 1 2 1 04. 1 3572528y8F(4, 2) 4OF(4, 2)8 4 4 8xG(2, 3)4 G(2, 3)H(6, 7) 83671 1 20 3 22 3 111(5) 0(19) 2(2) 11 3 24 2 13 3 45. 3535 11 2 1 3 4 1 2 4 23W(3, 3)H(6, 7)4341(11) 3(19) 2(6) 3433499 Extra Practice
4 0 16. 5 3625 265 2 2 2 4(36) 0(22) 1(19) 163 4 3 0 5 6 (1) 5 33117. 1 1 5 1 2 1 1 5452010313 5 1 5 18. 1 4 10 0 5 44365432 3 2 04519. 1 2 5 6 3 416531110. 3 3 5 6 13 311 0 1 8 6 535 2 1613 32 1 1 1 5 2 5 20 100 1 4 5353 31 1 1 2211. 12 y 61 2 2 1 8 2 2 3 2 1 21312322xxyxy1325226 1 8 2 2 1 8 2 2 1(4, 5)34 112. 263 y171 1 2 1 11 4 2 3 1110133244430x3xyxy45 2 11(2, 1)24133412 41321 0 xy1241313. 1101 2 1 4 324 11013xy 3 1 114 5 2 , 12 0x 5 2y 13434243 61 2127413 Lesson 2-6Page A291.y(0, 1)O(0, 3)O(3, 7)(3, 1)vertices: (3, 1), (0, 1), (3, 7)f(x, y) 4x 3yf(3, 1) 4(3) 3(1) or 15f(0, 1) 4(0) 3(1) or 3 → minimumf(3, 7) 4(3) 3(7) or 33 → maximum2. y(4, 7)x(0, 2) (4, 2)vertices: (0, 3), (4, 7), (4, 2), (0, 2)f(x, y) 2x yf(0, 3) 2(0) 3 or 3 → minimumf(4, 7) 2(4) 7 or 1f(4, 2) 2(4) 2 or 6 → maximumf(0, 2) 2(0) 2 or 23. y(1, 1)(1, 6)Oxx(1, 1)(1, 2)vertices: (1, 1), (1, 6), (1, 2), (1, 1)f(x, y) x yf(1, 1) (1) 1 or 0 → minimumf(1, 6) (1) (6) or 7 → maximumf(1, 2) 1 (2) or 1f(1, 1) 1 (1) or 0 → minimumExtra Practice 500
Lesson 2-7Page A291. Let x the number of cars.Let y the number of buses.6x 30y 600yx y 6060 x y 60x 06x 30y 600y 040(0, 20)20(50, 10)x 0(0, 0)O 20 40 60 xy 0 (60, 0)C(x, y) 3x 8yC(0, 0) 3(0) 8(0) or 0C(0, 20) 3(0) 8(20) or 160C(50, 10) 3(50) 8(10) or 230C(60, 0) 3(60) 8(0) or 180The maximum income is with 50 cars and 5 buses.2. Let x number of gallons of black walnut.Let y number of gallons of chocolate mint.y 2xy 20 thousandx y 45 thousandy4030y 2x(15, 30)(25, 20)20(10, 20)10x y 4510 20 30 40xC(x, y) 2.95x 2.95yC(10, 20) 2.95(10) 2.95(20) or 88.50C(15, 30) 2.95(15) 2.95(30) or 132.75C(25, 20) 2.95(25) 2.95(20) or 132.75alternate optimal solutions3. Let a the number of company A’s cards.Let b the number of company B’s cards.a b 90ba 4080a 40b 2560(40, 50)40a b 90(40, 25) (65, 25)20 b 25C(a, b) 0.30a 0.32bC(40, 25) 0.30(40) 0.32(25) or 20C(40, 50) 0.30(40) 0.32(50) or 28C(65, 25) 0.30(65) 0.32(25) or 27.5The maximum profit is with 40 cards fromcompany A and 50 cards from company B.O204060 80a4. Let x the number of video store hours.Let y the number of landscaping companyhours.y 0y16 x 4x 4x y 1512 (4, 11)C(x, y) 5x 7yC(4, 0) 5(4) 7(0) or 20C(4, 11) 5(4) 7(11) or 97C(15, 0) 5(15) 7(0) or 75The maximum earnings is $97.Lesson 3-1Page A301. f(x) 4xf(x) 4(x)f(x) (4x)f(x) 4xf(x) 4xyes2. f(x) x 2 3f(x) (x) 2 3 f(x) (x 2 3)f(x) x 2 3 f(x) x 2 3no13. f(x) 3x 3f(x) 3(1x) 34(4, 0)O 4 8y 0 f(x) 31x 3x y 151f(x) 3 x 3 1f(x) 3 x 3yes4. xy 2 → ab 2x-axis a(b) 2ab 2ab 2; noy-axis (a)b 2ab 2ab 2; noy x (b)(a) 2ab 2; yesy x (b)(a) 2ab 2; yesy x, y x5. y x 2 3 → b a 2 3x-axis (b) a 2 3b a 2 3; noy-axis b (a) 2 3b a 2 3; yesy x (a) (b) 2 3a b 2 3; noy x (a) (b) 2 3a b 2 3; noy-axis8(15, 0)12 16x501 Extra Practice
6. y 2 2 x 2 17→ b 2 2 a 2 17x-axis (b) 2 2 a 2 17b 2 2 a 2 1; yes7y-axis b 2 2( a) 27 1b 2 2 a 2 1; yes7y x (a) 2 2( b) 2 17a 2 2 b 2 1; no7y x (a) 2 2( b) 2 17a 2 2 b 2 1; no7x-axis, y-axis7. ⏐x⏐ 4y → ⏐a⏐ 4bx-axis⏐a⏐ 4(b)⏐a⏐ 4b; noy-axis⏐(a)⏐ 4b⏐a⏐ 4b; yesy x⏐(b)⏐ 4(a)⏐b⏐ 4a; noy x⏐(b)⏐ 4(a)⏐b⏐ 4a; noy-axis8. y 3x → b 3ax-axis(b) 3ab 3a; noy-axisb 3(a)b 3a; noy x(a) 3(b)a 3b; noy x(a) 3(b)a 3ba 3b; nonone of these9. y x 2 1 → b a2 1x-axis (b) a2 1b a2 1b a; 2 1 yesy-axis b (a) 2 1b a; 2 1 yesy x(a) (b) 2 1a (b) 2 1; noy x (a) (b) 2 1a b2 1a b; 2 1 nox-axis, y-axisLesson 3-2Page A301. g(x) is a translation of f(x) up 2 units.2. g(x) is the graph of f(x) expanded vertically by afactor of 3.3. g(x) is a translation of f(x) left 4 units and down3 units.Extra Practice 5024. g(x) is a translation of f(x) right 1 unit andcompressed vertically by a factor of 2.5. y 1 2 x 5Lesson 3-3Page A301. y2.y x 2 13. y4.OO5. y6.y |x| 2Oy |x 4|7. Case 1 Case 2⏐x 2⏐ 3 ⏐x 2⏐ 3(x 2) 3 x 2 3x 2 3 x 5x 1{x⏐1 x 5}8. Case 1 Case 2⏐4x 2⏐ 18 ⏐4x 2⏐ 18(4x 2) 18 4x 2 184x 2 18 4x 204x 16 x 5x 4{x⏐x 4 or x 5}9. Case 1 Case 2⏐5 2x⏐ 9 ⏐5 2x⏐ 9(5 2x) 9 5 2x 95 2x 9 2x 42x 14x 2x 7{x⏐2 x 7}xxy x 2xy 2|x 1|yOy (x 2) 2 1yOyOxxx
10. Case 1 Case 2⏐x 1⏐ 3 1 ⏐x 1⏐ 3 1(x 1) 3 1 (x 1) 3 1(x 1) 4 x 3x 1 4x 5{x⏐x 5 or x 3}11. Case 1 Case 2⏐2x 3⏐ 27 ⏐2x 3⏐ 27(2x 3) 27 2x 3 272x 3 27 2x 242x 30 x 12x 15{x⏐15 x 12}12. Case 1 Case 2⏐3x 4⏐ 3x 0 ⏐3x 4⏐ 3x 0(3x 4) 3x 0 (3x 4) 3x 06x 4 04 0; true6x 4x 2 3 all real numbersLesson 3-4Page A301. f(x)2.f 1 (x)3.f 1 (x)f (x)Of(x)Of(x)4. f(x) 4x 5y 4x 5x 4y 5x 5 4yy x 545xxf(x)f 1 (x) x 4; Yes, it is a function.5. f(x) 2x 2y 2x 2x 2y 2x 2 2yy x 22f 1 (x) x 2;2Yes, it is a function.Of(x)f 1 (x)x6. f(x) x 2 6y x 2 6x y 2 6x 6 y 2y x 6f 1 (x) x 6; No, it is not a function.7. f(x) (x 2) 2y (x 2) 2x (y 2) 2x y 2y 2 xf 1 (x) 2 x; No, it is not a function.x8. f(x) 2y 2x x 2y 2x yy 2xf 1 (x) 2x; Yes, it is a function.19. f(x) x 41y x 41x y y 4 1 x 4y 1 x 4f 1 (x) 1 x 4; Yes, it is a function.10. f(x) x 2 8x 2y x 2 8x 2x y 2 8y 2x 2 y 2 8yx 2 16 (y 4) 2x 18 y 44 x 18 yf 1 (x) 4 x; 18 No, it is not a function.11. f(x) x 3 4y x 3 4x y 3 4x 4 y 3 3 x 4 yf 1 (x) 3 x; 4 Yes, it is a function.312. f(x) (x 1) 23y (x 1) 23x (y 1) 2(y 1) 2 3 x y 1 3 x y 1 3 x f 1 (x) 1 3 x ; No, it is not a function.503 Extra Practice
Lesson 3-5Page A311. Yes; the function approaches 1 as x approaches 2from both sides.2. No; the function is undefined when x 3.3. No; the function is undefined when x 1.4. Yes; the function approaches 1 as x approaches 3from both sides.5. jump discontinuity6. y → as x → , y → as x → 7. y → as x → , y → as x → 8. y → 0 as x → , y → 0 as x → Lesson 3-6Page A311. abs. max.: (1, 2)2. rel. min.: (3, 0), rel max.:(1, 3), abs. min.: (2, 1)Lesson 3-7Page A3131. x 2 y xy y as x → , y → 3; y 322. x 3 y xy 4. yes y x 5x 3x 2 Lesson 3-8 1 03k6 1 484 kk5. y x 2x 2 3 xx— x x 2 x 31 2 x x 2 3 2 x2x2——x 3 x 2 x 22y 1 x 3 x 2no horizontalasymptotes since asx → , y is undefinedx 5x 53. h(x) x 2 6x 5y x 2 6x 5x 5x 5 x 2 x 2(x 5)(x 1) ——y 2 xx 5, x 1y x x2 6 5x2 x 2 1 x 5 x 2——1 6 x 5 x 2as x → , y → 0; y 0x 5161 2x → y x 5 x 3x 2 3x16As x → , x 3→ 0.5x 1 So, the graph of f(x)5x 15 will approach that of16 y x 5.Page A311. y kx y 4x8 k(2) y 4(9)4 k y 360 w2. g kw g 1 310 k(3) 4 1 0 3 w k 6 5 w3. t k r t r84 rt 84r(7) 84r 124. y kxz y 3xz60 k(5)(4) y 3(5)(10)3 k y 150432y 2 xk27 (3 ) 2 yx 2 243243 k y(5) 2 243y 9.726. a kbc 3 a 1.5bc 336 k(3)(2) 3 a 1.5(5)(3) 31.5 k a 202.5Lesson 4-1Page A321. yes; f(x) x 3 7x 2 2x 40f(2) (2) 3 7(2) 2 2(2) 408 28 4 40 02. no; f(x) x 3 7x 2 2x 40f(1) (1) 3 7(1) 2 2(1) 40 1 7 2 40 363. no; f(x) x 3 7x 2 2x 40f(2) (2) 3 7(2) 2 2(2) 40 8 28 4 40 244. yes; f(x) x 3 7x 2 2x 40f(5) (5) 3 7(5) 2 2(5) 40 125 175 10 40 0Extra Practice 504
5. (x 3)(x 4) 0x 2 7x 12 0; even; 26. (x (2))(x (1))(x 2) 0(x 2)(x 1)(x 2) 0(x 2)(x 2 x 2) 0x 3 x 2 4x 4 0; odd; 37. (x (1.5))(x (1))(x 1) 0(x 1.5)(x 1)(x 1) 0(x 1.5)(x 2 1) 0x 3 1.5x 2 x 1.5 0; odd; 38. (x (2))(x (i))(x i) 0(x 2)(x i)(x i) 0(x 2)(x 2 1) 0x 3 2x 2 x 2 0; odd; 19. (x (3i))(x 3i)(x (i))(x i) 0(x 3i)(x 3i)(x i)(x i) 0(x 2 9)(x 2 1) 0x 4 10x 2 9 0; even; 010. (x (1))(x 1)(x 2)(x 3) 0(x 1)(x 1)(x 2)(x 3) 0(x 2 1)(x 2 5x 6) 0x 4 5x 3 5x 2 5x 6 0; even; 4Lesson 4-2Page A321. x 2 4x 5 0x 2 4x 5x 2 4x 4 5 4(x 2) 2 9x 2 3x 2 3x 2 3x 5x 12. x 2 6x 8 0x 2 6x 8x 2 6x 9 8 9(x 3) 2 1x 3 1x 3 1x 3 1x 2x 43. m 2 3m 2 0m 2 3m 2m 2 3m 9 4 2 9 4 m 3 2 2 1 74m 3 2 17 2m 3 2 17 24. 2a 2 8a 6 0a 2 4a 3 0a 2 4a 3a 2 4a 4 3 4(a 2) 2 7a 2 7a 2 75. h 2 12h 4h 2 12h 36 4 36(h 6) 2 40h 6 210h 6 2106. x 2 9x 1 0x 2 9x 1x 2 9x 8 141 8 1 4x 9 2 2 7 74x 9 2 772x 9 2 77 27. b 2 4ac (3) 2 4(4)(7) or 121; 2 realx (3) 1212(4)x 3 118x 7 4 or x 18. b 2 4ac (2) 2 4(1)(10) or 44; 2 realw 2 442(1)w 1 119. b 2 4ac (5) 2 4(12)(6) or 263; 2 imaginary (5) 263 t 263 t 5 i 2410. b 2 4ac (6) 2 4(1)(13) or 88; 2 realx 6 882(1)x 3 2211. b 2 4ac (4) 2 4(4)(1) or 0; 1 realn ( 4) 02(4)n 1 2 12. b 2 4ac (6) 2 4(4)(15) or 276; 2 realx 6 2762(4)x 6 2 69 869 x 3 4Lesson 4-32(12)Page A321. 2⏐ 1 10 82 161 8⏐ 8x 8, R82. 1⏐ 1 3 4 11 2 21 2 2⏐ 1x 2 2x 2, R13. 1⏐ 1 0 3 51 1 21 1 2⏐ 3x 2 x 2, R3505 Extra Practice
4. 4⏐ 1 2 7 3 44 8 4 41 2 1 1⏐ 0x 3 2x 2 x 15. f(x) x 2 2x 8f(4) (4) 2 2(4) 8 16 8 8 0; yes6. f(x) x 3 12f(1) (1) 3 12 1 12 or 13; no7. f(x) 4x 3 2x 2 6x 1f(1) 4(1) 3 2(1) 2 6(1) 14 2 6 17; no8. f(x) x 4 4x 2 16f(4) (4) 4 4(4) 2 16 256 64 16208; noLesson 4-4Page A321. p: 1, 2, 3, 6q: 1 p q : 1, 2, 3, 6r 1 2 5 61 1 3 2 81 1 1 6 02 1 4 3 02 1 0 5 163 1 5 10 243 1 1 2 0rational roots: 3, 1, 22. p: 1q: 1, 2 p q : 1, 1 2 r 2 1 2 3 11 2 1 3 0 11 2 3 5 8 9 1 2 2 0 2 2 0 1 2 2 2 3 9 2 1 34rational root: 1 2 3. p: 1, 2q: 1 p q : 1, 2r 1 1 0 21 1 2 2 01 1 0 0 22 1 3 6 102 1 1 2 6rational root: 14. p: 1, 2, 4, 8q: 1, 2, 3, 6 p q : 1, 2, 4, 8, 1 2 , 1 3 , 2 3 , 4 3 , 8 3 , 1 6 r 6 1 22 4 81 6 7 29 33 251 6 5 27 23 15 2 3 6 3 24 12 0 1 2 6 4 24 16 0rational roots: 2 3 , 1 2 5. 2 or 0 positivef(x) x 3 4x 2 x 41 negativer 1 4 1 41 1 5 4 0x 2 5x 4 0(x 4)(x 1) 0x 4, x 1rational zeros: 1, 1, 46. 2 or 0 positivef(x) x 4 x 3 3x 2 5x 102 or 0 negativer 1 1 3 5 101 1 2 5 0 10 10 1 9 93 935 9360rational zeros: none7. 2 or 0 positivef(x) 4x 3 7x 31 negativer 4 0 7 3 3 2 4 6 2 04x 2 6x 2 0(4x 2)(x 1) 0x 1 2 , x 1rational zeros: 3 2 , 1 2 , 1Extra Practice 506
8. 3 or 1 positivef(x) x 4 x 3 4x 41 negativerational zero: 1Lesson 4-5Page A331.r 1 1 0 4 41 1 0 0 4 0 4 1 5 20 76 300r 2 4 52 2 8 111 2 6 10 2 4 51 2 2 72 2 0 53 2 2 11 and 0, 2 and 32.r 1 0 0 52 1 2 4 131 1 1 1 60 1 0 0 51 1 1 1 42 1 2 4 31 and 23.r 1 0 1 4 22 1 2 3 2 61 1 1 0 4 20 1 0 1 4 21 1 1 0 4 62 1 2 3 10 222 and 1, 1 and 04-6. Use the TABLE feature of a graphing calculator.4. 0.3, 1.35. 2.2, 0.3, 1.26. 1.3, 1.3Lesson 4-6Page A331. 6 x x 56 x 2 5xx 2 5x 6 0(x 2)(x 3) 0x 2 0 or x 3 0x 2 x 372. y 1 4 y y y 17y 4(y 1) y 23y 4 y 2y 2 3y 4 0(y 4)(y 1) 0y 4 0 or y 1 0y 4y 15 4 13. r 1 r 1 r 2 15(r 1) 4(r 1) 15r 5 4r 4 1r 101 44. 2 2 t 2 t 42t 21 t 22(t 2) 4 12t 4 32t 7t 3.515. 3 w 4 5 w 1155 12 w17 w; w 01 4Test w 1: 3( 1) 5( 1) ? 11 5 1 3 4 5 ? 1 1 5 Test w 1: 3(11) 1 715 ? 1 1 51 5( 1) ? 11 5 1 3 1 5 ? 11 58 1 5 ? 1 1 5 ; false1 1Test w 18: 3( 18) 5( 18) ? 11 5w 0 or w 171 1 5 4 9 0 ? 11 5; true4 1 35 ? 115; true507 Extra Practice
6. x 2x x 4x 6(x 6)(x 2) x(x 4)x 2 8x 12 x 2 4x12 4x3 x; x 0 or 6) 21) 4Test x 1: (1 1 ? ( ( 1) 63 ? 5 7 ; falseTest x 1: 1 1 ? 1 1 61 ? 3 5 ; trueTest x 4: 4 24 ? 4 44 6 1 2 ? 0; false2 4Test x 7: 7 7 ? 7 7 6 5 7 ? 3; true0 x 3 or x 6Lesson 4-72 4Page A331. 2 t 3 4 Check: 2 3143 42 3t 16 16 43t 144 4 t 1 432. 4 x 2 1 Check: 4 11 2 1x 2 3 4 3 1x 2 91 1 x 113. 3 y 7 10 2 Check: 3 505 7 10 2 3 y 7 8 8 10 2y 7 5122 2 y 5054. a 1 5 a 6a 1 10a 1 25 a 610a 1 30a 1 3a 1 9a 10Check: 10 1 5 10 6 3 5 22 2no real solution5. 2x 3 22x 3 42x 1x 1 2 2x 3 02x 3x 3 2 Test x 2: 2(2) 3 ? 21 ? 2; meaninglessTest x 0: 2(0) 3 ? 23 ? 2; trueTest x 1: 2(1) 3 ? 25 ? 2; false<strong>Solution</strong>: 3 2 x 1 2 6. 4 6a 2 46a 2 2566a 258a 436a 2 06a 2a 1 3 Test a 0: 4 6(0) 2 ? 4 4 2 ? 4; meaninglessTest a 1: 4 6(1) 2 ? 4 4 4 ? 4; falseTest a 44: 4 6(44) 2 ? 4 4 262 ? 4; true<strong>Solution</strong>: a 43Lesson 4-8Page A331. f(x) 0.75x 22. f(x) x 3 x 2 x 23. Sample answer: f(x) 0.51x 2 0.02x 0.794a. Sample answer: y 1.632x 99.2754b. Sample answer: about 122.123 thousandf(x) 1.632x 99.275f(15) 1.632(14) 99.275 122.123Lesson 5-1Page A341. 13.75° 13° (0.75 60) 13° 45 13° 45Extra Practice 508
2. 75.72° 75° (0.72 60) 75° 43.275° 43 (0.2 60) 75° 43 12 75° 43 123. 29.44° 29° (0.44 60)29° 26.429° 26 (0.4 60)29° 26 2429° 26 244. 87.81° 87° (0.81 60) 87° 48.6 87° 48 (0.6 60) 87° 48 36 87° 48 365. 144° 12 30 144° 12 1°60 30 136 144.208°12. sin v cs452sin A s ideoppositehypotenuse16sin A 2 1 or 8 11313 113sideoppositetan A s ideadjacenttan A 1 614hypotenusesec A s ideadjacent°00sec A 2 1 13 14or 1 7131°0 10 1°3600BC 2 15938.253°1°0 45 1°3600sin A s ideoppositehypotenuse107.213°sin A 1591°0 32 1°283600sideoppositetan A s ideadjacenttan A 15925hypotenusesec A s ideadjacentsec A 2 82117 (AB) 2117sin A s ideoppositehypotenuse6sin A 3 13 or 2 13 13sideoppositetan A s ideadjacenthypotenusesec A s ideadjacentsec A 3 13 or 139 3Lesson 5-3Page A341. tan v y x Since tan v 0, y 0.cot v x y x 01—or 6 5 5 6 6. 38° 15 10 38° 15 67. 107° 12 45 107° 12 68. 51° 14 32 51° 14 6 51.242°9. 850° 360° 490°490° 360° 130°130°; II10. 65° 360° 295°295°; IV11. 1012° 360° 652°652° 360° 292°292°; IV12. 578° 360° 218°218°; III13. 180° 126° 54°14. 480° 360° 120°120° 360° 240°240° 180° 60°60°15. 642° 360° 282°360° 282° 78°78°16. 1154° 360° 794°794° 360° 434°434° 360° 74°74°Lesson 5-2Page A3411. cot v ta n vcot v c v1sin v 2 .5 or 0.43. (AC ) 2 (BC ) 2 (AB) 214 2 16 2 (AB) 2452 (AB) 2 AB; 452 or 2113 cos A shy 1 cos A 2csc A sh or 8 7 csc A 2 14. (AC ) 2 (BC ) 2 (AB) 225 2 (BC ) 2 28 2BC 159 cot A ss cot A 14 1cos A sh cos A 2 25csc A sh csc A 2cot A ss5 2cot A 5. (AC) 2 (BC) 2 (AB) 29 2 6 2 (AB) 2 AB; 117 or 313cos A s h cos A 39csc A shtan A 6 9 or 2 3 csc A 3 6cot v is undefined.2. Sample answers: 90°, 270°cot A ssideadjacentpotenuse41 or 7 1 13 13 113ypotenuseideopposite1 13 6 or 1 813ideadjacentideopposite6or 7 8 ideadjacentypotenuse8 ypotenuseideopposite81 or 28 15959 159ideadjacentideopposite51 or 25 15959 159ideadjacentypotenuse 13 or 31313 ypotenuseideopposite13 or 13 2ideadjacentideopposite cot A 9 6 or 3 2 cos v x r Since cos v 0, x 0.On the unit circle, x 0 when v 90° or v 270°.509 Extra Practice
3. r x 2 y 2 (1) 2 ) (22 5sin v y r 2sin v or 2 5 5 5tan v y x 2cos v x r 51cos v or 5 5rcsc v ytan v 1or 2 csc v 2rsec v xsec v 5 or 514. r x 2 y 2 (2) 2 (2) 2 22sin v y r sin v 22tan v y x or 22 2cot v x y cot v cos v x r 512or 1 2 2cos v or 2 22rcsc v y2tan v 2or 1 csc v 2 2 or 22sec v xr 2cot v x y sec v 2 2or 2 cot v 2or 15. r x 2 y 2 (5) 2 (2) 2 29sin v y r cos v x r 2sin v 29 or 2 29 5 cos v 29 29tan v y x rcsc v ytan v 2 5 csc v 29 2rsec v xsec v 2956. r x 2 y 2 (4) 2 (3) 2 5sin v y r sin v 3 5 tan v y x tan v 3 4 sec v xr sec v 5 4 Lesson 5-4Page A341. tan A a b tan 38° 1a5 15 tan 38° a11.7 acot v x y cot v 5 2 cos v x r 2cos v 4 5 csc v yr csc v 5 3 cot v x y cot v 4 3 2 or 529292. cos B a c acos 87° 1 9 19 cos 87° a1.0 a3. cos B a c cos 65.4° 16 .5c16.c cos565.4°c 39.6b4. tan B atan 42.5° 1 2 a1a tana 13.15. sin B b c bsin 75° 5 .8 5.8 sin 75° b5.6 b6. tan 48° 20 bx20 bx tan48°242.5°tan 42° b x tan 42° 20 tan 42° b tan 42° b tan 48°20 tan 42° b(tan 48° tan 42°)Lesson 5-5Page A35 b85.7 b; 85.7 ft1. Let A arcsin 3 4 . Then sin A 3 4 .sin arcsin 3 4 3 4 2. Let A cos 1 1 2 . Then cos A 1 2 .1sec A co s Asec A or 2sec cos3. Let A tan 1 1. Then tan A 1.tan(tan 1 1) 14. tan A a b 5. sin B b c tan A 3 220 tan 42°(tan 48° tan 42°)1 1 2 85A tan 1 3 2A 56.7°85 48˚42˚b––20 bt an48°xsin B 1 179B sin 1 1 1B 63.5°7920 feetbExtra Practice 510
6. cos B a c cos B 2 430 B cos 1 2 3B 36.9°8. tan A a b tan A 2 38.46.540 A tan 1 2 3A 37.9°8.46.57. cos B a c 9.cos B 122.6B cos 1 9. 1B 43.1°22.67. A 180° (60° 75°) or 45°K 1 2 a 2 sin sK 1 2 (8) 2 sin 6 siK 37.9 units 2B sin Cin A0°sin75°n 45°8. K 1 2 bc sin AK 1 2 (16)(12) sin 43°K 65.5 units 2Lesson 5-6Page A351. C 180° (75° 50°) or 55°a b sin A sin B7 sin 75°b 7 sin50°sin75° b sin 50° sin7a c sin A sin Cc75° sin 55°c 7 sin55°sin75°b 5.551472956 c 5.936340197C 55°, b 5.6, c 5.92. B 180° (97° 42°) or 41°c b sin C sin B12 b sin 42° sin 41°b 12 sin41°sin42° sin12c a sin C sin Aa42° sin 97°a 12 sin97°sin42°b 11.76557801 a 17.80004338B 41°, a 17.8, b 11.83. A 180° (49° 32°) or 99°a b sin A sin B10 b sin 99° sin 49°b 10 sin49°sin99° sin10a c sin A sin Cc99° sin 32°c 10 sin32°sin99°b 7.641171301 c 5.365247745A 99°, b 7.6, c 5.44. B 180° (22° 41°) or 117°b ab c sin B sin Asin B sin C25a25c sin 117° sin 22°sin 117° sin 41°a 2 5 sin22°sin117°c 2 5 sin41°sin117°a 10.51077021 c 18.40780654B 117°, a 10.5, c 18.45. K 1 2 bc sin AK 1 2 (12)(6) sin 34°K 20.1 units 26. C 180° (87° 56.8°) or 36.2°K 1 2 c 2 sin A sin BsinCK 1 2 (6.8) 2 sin 8 siK 32.7 units 27°sin56.8°n 36.2°Lesson 5-7Page A351. Since 145° 90°, consider Case II.5 10; no solution2. Since 25° 90°, consider Case I.b sin A 10 sin 25° 4.2261826179 4.226182617; 2 solutionsa b sin A sin B6 10 sin 25° si n B6 sin B 10 sin 25°B sin 1 10 si n25° 6B 44.77816685B 180° 44.8° or 135.2°<strong>Solution</strong> 1C 180° (25° 44.8°) or 110.2°a c sin A sin C6c sin 25° sin 1 10.2°c 6s in110.2°sin25°c 13.32398206B 44.8°, C 110.2°, c 13.3<strong>Solution</strong> 2C 180° (25° 135.2°) or 19.8°a c sin A sin C6c sin 25° sin 1 9.8°c 6s in19.8°sin25°c 4.809133219B 135.2°, C 19.8°, c 4.83. Since 56° 90°, consider Case I.C sin B 50 sin 56° 41.4518786334 41.5; no solution511 Extra Practice
4. C 180° (45° 85°) or 50°a c sin A sin Ca 15 sin 45° sin 50°a 15 sin45°sin50°a 13.8459352b c sin B sin Cb 15 sin 85° sin 50°b 15 sin85°sin50°b 19.50659731C 50°, a 13.8, b 19.5Lesson 5-8Page A351. a 2 b 2 c 2 2bc cos Aa 2 6 2 8 2 2(6)(8) cos 62°a 2 54.93072997a 7.411526831a b sin A sin Ba 6 sin 62° sin B62°sin B 6sin aB sin 1 6sin 62°a B 45.62599479C 180° (62° 45.6°) or 72.4°a 7.4, B 45.6°, C 72.4°2. a 2 b 2 c 2 2bc cos A9 2 7 2 12 2 2(7)(12) cos A112 168 cos AA cos 1 112168 A 48.1896851a b sin A sin B9 7 sin A sin Bn Asin B 7si 9B sin 1 7si n A9 B 35.43094469C 180° (48.2° 35.4°) or 96.4°A 48.2°, B 35.4°, C 96.4°3. b 2 a 2 c 2 2ac cos Bb 2 14 2 18 2 2(14)(18) cos 48°b 2 182.7581744b 13.51880817a b sin A sin B14b si n A sin 48°n48°sin A 14 si bA sin 1 14 si n48°b A 50.31729382C 180° (50.3° 48°) or 81.7°b 13.5, A 50.3°, C 81.7°4. c 2 a 2 b 2 2ab cos Cc 2 (14.2) 2 (24.5) 2 2(14.2)(24.5) cos 85.3°c 2 744.8771857c 27.29243825a c sin A sin C14.2c s inA sin 8 5.3°n 85.3°sin A 14.2 si cA sin 1 14.2 si n 85.3°c A 31.23444201B 180° (31.2° 85.3°) or 63.5°c 27.3, A 31.2°, B 63.5°5. s 1 2 (a b c)s 1 2 (4 7 10)s 10.5K s(s a)(s b)(s c)K 10.5(1 0.5 .5 4)(10)(10.5 7) 10K 119.43 75K 10.9 units 26. s 1 2 (a b c)s 1 2 (4 6 5)s 7.5K s(s a)(s b)(s c)K 7.5(7. 5 )(7.5 46)(7.5 5)K 98.43 75K 9.9 unit 27. s 1 2 (a b c)s 1 2 (12.4 8.6 14.2)s 17.6K s(s a)(s b)(s c)K 17.6(1 7.6 (17.6 12.4) )(17.68.6.2) 14K 2800. 512 K 52.9 units 28. s 1 2 (a b c)s 1 2 (150 124 190)s 232K s(s a)(s b)(s c)K 232(2 32 150)(2 32 124)(2 32 190) K 86,29 2,864 K 9289.4 units 2Extra Practice 512
9a.110˚d 2 28 2 35 2 2(28)(35) cos 110°d 2 2679.359481d 51.8 in.9b. Area 2 1 2 (28)(35) sin 110° 920.9 in 2Lesson 6-1Page A361. 120° 120° 18 2 30°2. 280° 280° 18 14 90°3. 440° 440° 18 22 94. 150° 150° 18 5 65. 8 3 8 3 18 0° 480°6. 5 12 5 12 18 0°0°0° 75°7. 2 2 18 0°114.6°8. 10.5 10.5 18 0° 601.6°9. reference angle: 5 6 6 ; Quadrant IIsin 5 6 1 2 10. reference angle: 4 3 3 ; Quadrant IIIsin 4 3 2311. 9 4 is coterminal with 4 ; Quadrant Icos 9 4 2212. 3 2 is coterminal with 2 cos 3 2 013. If the diameter 10 in., the radius 5 in.80° 80° 1 80 4 9s rvs 5 4 9 s 7.0 in.35 in.28 in.Lesson 6-2Page A361. 5 2 10 or about 31.4 radians2. 3.8 2 7.6 or about 23.9 radians3. 14.2 2 28.4 or about 89.2 radians4. 2.1 2 4.2q v t q 4. 2 5q 2.6 radians/s5. 1.5 2 3q v t q 3 2q 4.7 radians/min6. 15.8 2 31.6q v t . 6q 31 18q 5.5 radians/s7. 140 2 280q v t 0q 28 20q 44.0 radians/min8. q v t q 2 30 q about 0.2 radian/sLesson 6-3Page A361. 12. 03. 14.5.4 3 24y sin xy cos x3y1O1y1O1xx513 Extra Practice
Lesson 6-4Page A361. ⏐2⏐ 2; 2 1 2y3. 1 2 1 22 ; 8 1 4 Lesson 6-5Page A37y2y 2 cos 11O 2 3 O12122. ⏐3⏐ 3; 0 . 5 4c2. k 2 1yy32y 3 sin 0.5 21O1 2 3 O232y11. kc 2 or 2 y sin(2 ) 2 31 y 2 cos 1 y sin4( 2 2 ) or 2y 2 cos( 2) 2 3c3. k 2 or 1 2 yO2 4 6 8O 2 3114. A ⏐0.5⏐ 2 k 6A 0.5 k 2 6 or 1 3 vy 0.5 sin 35. A ⏐2⏐ 2 k 3 A 2y 2 sin 6v2k or 66. A 3 5 2 k 4A 3 5 k 2 4 or 1 2 y 3 5 vcos 27. A ⏐0.25⏐ 2 k 8A 0.25 k 2 8 or 4 y 0.25 cos 4 v3 4. A ⏐2⏐ 2 k 2 c 1 h 1A 2 2 or 1 c k 2 y 2 sin (v ) 15. A ⏐0.5⏐ 2 k 4 c 8 0 h 3A 0.5 k 2 4or 8 c 0y 0.5 sin 8v 36. A ⏐20⏐ 2 k 2 c 4 2 h 4A 20 2 or 4 c 8k 2 y 20 cos (4v 8) 47. A ⏐ 3 4 ⏐ 2 c 10 0 h k 1 2 5 A 3 4 k 2 10 or 5 c 0y 3 4 cos 5 v 1 2 Extra Practice 514
Lesson 6-6 62Page A373. cos 13 cos1. Let v Cos 1 6 13 620. 2. Let v Arcsin 0. cos Cos v 06 Sin v 0v sin( 315°)2 4. tan (315°) v 0c os( 315°)4. If y tan 3 ,4 then y 1.Cos 1 tan 3 4 Cos 1 yPage A371a. 12.1 2.7 9.4 h Cos 1 (1) 1b. 12.1 2.7 14.8 h5. Let a Cos 1 1 2 and b Sin 1 0.1c. m 10 represents the middle of October.Cos a 1d 2.7 sin (0.5(10) 1.4) 12.12 Sin b 0d 10.9 ha 2. A 6 2 or 3; 2 14 3 b 07 sin Cos 1 1 2 Sin 1 0 sin (a b)y 3 cos 7 t sin 3 0 sin 3 Lesson 6-7 326. Let v sin 1 3. cos2 2 Sin 1 32 cos (2v)sin v 3 cos22 3Page A37v 31. cos 2 y38 1 y cot( 4 )2 42.8y4O 2 3 4Lesson 7-1Page A3811. csc v si n v1 1 s cov21y sec 2 1 1 4 21 2 3O 1 51 63.1y 4 15 424 15 5 115y csc(2 2)O 2 3 4 1515212. tan v 4co t v 1 633 Lesson 6-86 6 63. Let v Tan 1 1. cos(Tan 1 1) cos 4 sin (45° (360°)) cos (45° (360°))Tan v 1v 4 2 2 sin45°c os45° tan 45°515 Extra Practice
5. csc (930°) sin (16. t sin vanv sinv c osvsinv930°)1sin (360°(2) 210°) 1sin (210°) 1sin (30°) 1 sin 30° csc 30° 1co s v sec v7. cot v tan v sin v sec v c osv sinv1sinv c sin v osvco 1 sinvc osv 1 tan v8. (1 sin x)(1 sin x) 1 sin 2 x cos 2 x c osx sin x9. cot xsinxsinxc scx cosx 1 cos xsin xcos x c osxsinxLesson 7-2 sin xPage A381. csc 2 v cot 2 v sin v csc vcsc 2 v cot 2 1v sin v si n vcsc 2 v cot 2 v 1csc 2 v csc 2 v2. se c v cscvcscv secvsecv cscv csc v sec v csc v sec v1 1 cs c v se c v sin v cos v sin v cos v sin v cos vsin v cos v sin v cos v3. sin 2 x cos 2 x sec 2 x tan 2 x1 tan 2 x 1 tan 2 x1 14. sec A cos A tan A sin A1 co s A cos A tan A sin A1 co s A c os2AcosA 1 cos 2 AcosA s in2AcosAsinA c osA tan A sin A tan A sin A tan A sin A sin A tan A sin Atan A sin A tan A sin As v5. Sample answer: cos x 1cot x c 1scx c osxsinx1sin x 1cos x 16. Sample answer: cot x 22 tan x sin x 2 cos x csc x2 sinxcosx sin x cos x 1sin xx cos2 2 xsin2 1sinLesson 7-3cos x2 1 cos x 1sin x2 c osxsinx2 cot xPage A381. cos 75° cos (45° 30°) cos 45° cos 30° sin 45° sin 30° 2 3 2 1 2 2 2 2 6 242. sin 105° sin (60° 45°) sin 60° sin 45° cos 60° cos 45° 32 22 6 24 1 2 23. tan 12 tan 4 6 4. tan 7 1tan 4 tan p 6 1 tan 4 tan 6 1 331 1 33 2 32 tan 3 4 1 331 331 331 33tan 3 tan 4 1 tan 3 tan 4 3 1 11 3 1 13 1 1 3 2 32xExtra Practice 516
95. sec 2 12 sec 5 1 26. cot 375° cot 15°7. sin x 1cos 4 6 1cos 4 cos 6 sin 4 sin 6 1 2 3 2 1 2 2 2 2 1 6 244 6 26 2 6 2 6 21tan (45 ° 30°)1tan 45° tan 30°1 tan 45° tan 30° 1 2 2 2 3 2 5 2cos y 3 4 21 115 7 1 61 331 1 331 33——1 33 2 15sin (x y) sin x cos y cos x sin y 21 5 47 2 5 3 4 6 7320 7 4 8. 12 2 5 2 119 12 2 11 2 23sin x 1 19 12sin y 3122cos (x y) cos x cos y sin x sin y5 11 1 2 1 2 1 19 12 3122 55 2737 1441 33——1 339. If cot x 4 3 , then tan x 3 4 .sec 2 y 1 tan 2 y 5 4 2 1 tan 2 y9 1 6 tan 2 y 3 4 tan ytatan (x y) 1n x tany tanx tany 3 4 3 4 ——1 3 4 3 4 2 478510. 7 2 6 2 85; cos x 6 ; sin x 7 8585csc y 8 5 8 2 5 2 39 sin1y 8 5 cos y 8sin y 5 8 1sec (x y) cos (x y)Lesson 7-4 6 4 —71 6Page A391. sin 15° sin 3 0° 239 851cos x cos y sin x sin y1 6 85 39 7 85 5 85 8 85 8 680 63315 3585 63315 3585 63315 3585 os 30° 1 c 21 32 2 2 3 2Since 15° is in Quadrant I, sin 15° 2 3 .22. cos 75° cos 15 0°2 1 c483315 28085 179 o s150 ° 21 32 2 2 3 2Since 75° is in Quadrant I, cos 75° 2 3 .2517 Extra Practice
3. tan 12 tan 6 7. 7 2 2 2 35; sin v 3— 7; tan v 3 22sin 2v 2 sin v cos v1 cos —6 —1 cos — 6 —1 323 2 21 2 3 3 2 2 33 3)2 (2 1Since 1 2 is in Quadrant I, tan 14. cos 22.5° cos 4 5° 2 1 c os 45°2 1 222 2 2 2or 2 32 2 3.Since 22.5° is in Quadrant I, cos 22.5° 2 2 .25. sin 5 12 sin 5 6—2 1 cos 5 6 2 2 2 3 2Since 5 12is in Quadrant I, sin 5 p12 2 3 2.6. tan 112.5° tan 225° 1 c os1 co 225°s 225°1 222 221 321 2 (2 2 2 2 2 22 2 2)2Since 112.5° is in Quadrant II,tan 112.5° 1 2.or 1 2 2 3 57 2 7 12 549cos 2v cos 2 v sin 2 v5 2 tan 2v 12 2 7 2 3 7 4 149tanv tan2 v 23 52 5 2 1 3 2 12 5418. 3 2 2 2 5; cos v , tan v 2 35sin 2v 2 sin v cos v 2 2 3 3 4 595 cos 2v cos 2 v sin 2 v 53 1 9 tan 2v 12 2 2 3 2tanv tan2 v 2 2 551 2 5 455 2 5510559. (3) 2 (1) 2 10; sin v 3 10; cos v 1sin 2v 2 sin v cos v 2 3 10 10 1 3 5 cos 2v cos 2 v sin 2 vtan 2v 12 1010 4 5 tanv tan2 v2 ( 3) 1 ( 3 ) 2 3 4 100 2 3 10 10 2 10 0Extra Practice 518
10. csc v 3 2 3 2 ( 2) 2 5 si1n v 3 2 cos v 35sin v 2 3 tan v 2 5sin 2v 2 sin v cos v 2 2 3 3 4 595 cos 2v cos 2 v sin 2 v 53 1 9 tan 2v 12Lesson 7-5Page A39 2 2 3 2tanv tan2 v 2 2 551 2 5 25 451. 4 cos 2 x 2 0 cos 2 x 1 2 cos x 22x 45°, 135°, 225°, 315°2. sin 2 x csc x 1 0sin 2 1x s in x 1 0sin x 1x 90°3. 3 cos x 2 cos x3 c osxsinx 2 cos x3 c osxsinx 2 cos x 0cos x 3 sinx2 03cos x 0 or s 2 0x 90°, 270°in x3 s in x 32 25 sin xx 60°, 120°4. 3 cos 2 x 6 cos x 33 cos 2 x 6 cos x 3 03(cos 2 x 2 cos x 1) 03(cos x 1)(cos x 1) 0cos x 1 0cos x 1x 0°Lesson 7-6Page A391. x cos 30° y sin 30° 12 0 32x 1 2 y 12 03x y 24 02. x cos 3 y sin 3 2 0 1 2 x 3 y 2 02x 3y 4 03. x cos 150° y sin 150° 1 2 0 3 x 1 2 2 y 1 2 0 3x y 1 04. AB 2 2 40 2 1 2 2294x 2 29 1 0y 2 29 1 0 2 29 2 29 x 5 29 y 5 29 29 29 29p 5 2929 29 0 029 sin f 5 29, cos f 2 29tan f 5 2 f Arctan 5 2 180°f 248°5. A B 2 2 11) 2 ( 2 2 22x y 2 2 2 22 0 x y 2 02p 2sin f 2, cos f 222tan f 1f Arctan (1) 360°f 315°6. A B 2 2 2 2 3 2 13 2 13 13 3x 13 2y 11 03 21313 13 x 3 13y 12 13 0p 12 13 13sin f 3 13 , cos f 2131313tan f 3 2 f Arctan 3 2 f 56°519 Extra Practice
Lesson 7-7Page A391. 3 (2) 2( 1) 2 2 3(2 ) 2 2 13 2 13132. 2(3 ) (0) 2 4 244 2 2 20 4 2 5 2 553(1) (4) 1 2 3. (3) 2 1 2 10 105 ; ⏐d⏐ 54. y 2 3 x 2 → 2x 3y 6 0 2( 4) 3(2) 6 8 22 ( ) 3 2 135. (0, 1)d 0 65 6 556. (0, 3)d 3 (3)2 13 2 13137. (2, 1)d 2(2) 2(1) 4122 2 8 1313; ⏐d⏐ 6 5 2( 0) 73(2 ) 2 25(1) 45 22 25 2 95 0.9; ⏐d⏐ 0.9 unitLesson 8-1Page A401. 2.1.7 cm3. 330˚70˚3.6 cm2.1 cm10 104˚4. 3.6 cm; 89° 5. 2.6 cm; 23°6. 3.7 cm; 357° 7. 1.2 cm; 342°8. 7.2 cm; 330° 9. 8.8 cm; 340°10.hcos 70° 1 .7 xsin 70° 1 .7 1.7 cos 70° h 1.7 sin 70° v0.58 h 1.60 v11.hcos 76° 2 .1 vsin 76° 2 .1 2.1 cos 76° h 2.1 sin 76° v0.51 h 2.04 v12.hcos 30° 3 .6 vsin 30° 3 .6 3.6 cos 30° h 3.6 sin 30° v3.12 h 1.8 vLesson 8-2Page A40u1. AB 4 3, 1 6 1, 5u⏐AB ⏐ 1 5) 2 ( 2 26u2. AB 2 (1), 2 3 1, 1u⏐AB ⏐ (1) 2) (12 2u3. AB 1 0, 8 (4) 1, 4u⏐AB ⏐ (1) 2) (42 17u4. AB 3 1, 9 10 2, 19u⏐AB ⏐ 2 19) 2 ( 2 365 u5. AB 3 (6), 6 0 3, 6u⏐AB ⏐ 3 6) 2 ( 2 45 35u6. AB 0 4, 7 (5) 4, 12u⏐AB ⏐ (4) 212 2 160 4107. ⏐5, 6⏐ 5 2 6 2 615i u 6j u8. ⏐2, 4⏐ (2) 2 4 2 20 252i u 4j u9. ⏐10, 5⏐ (10) 2 5) (2 125 5510i u 5j u10. ⏐2.5, 6⏐ (2.5) 2 6 2 42.25 6.52.5i u 6j uExtra Practice 520
11. ⏐2, 6⏐ 26) 2 ( 2 40 2102i u 6j u12. ⏐15, 12⏐ (15) 12) 2 (2 369 34115i u 12j uLesson 8-3Page A401. u p 21, 2, 1 3(4, 3, 0 2, 4, 2 12, 9, 0 10, 5, 22. u p 1, 2, 1 1 2 2, 2, 4 4, 3, 0 1, 2, 1 1, 1, 2 4, 3, 0 2, 2, 33. u p 22, 2, 4 4, 3, 0 4, 4, 8 4, 3, 0 0, 7, 84. u p 3 4 4, 3, 0 21, 2, 1 3, 9 4 , 0 2, 4, 2 1, 1 3 4 , 25. 2, 4, 1 5, 4, 3 u v 0, 0, 07, 0, 4 u v 0, 0, 0uv 0 7, 0 0, 0 4uv 7, 0, 48. 3, 0, 4 1, 5, 2 ⏐ iu u j u k3 0 41 5 2⏐20i u 10j u 15k u 20, 10, 1520, 10, 15 3, 0, 4 20 3 (10) 0 15 4 0; yes20, 10, 15 1, 5, 2 20 (1) (10) 5 15 2 0; yes9. 1, 1, 0 2, 1, 3 ⏐ iu u j u k1 1 02 1 3⏐ 3i u 3j u 3k u 3, 3, 33, 3, 3 1, 1, 0 3 (1) 3 1 1(3) 0 0; yes3, 3, 3 (2, 1, 3 3 2 3 1 (3) 3 0; yes10. 1, 3, 2 6, 1, 2 ⏐ iu u j u k1 3 26 1 ⏐ 2 8i u 10j u 19k u 8, 10, 198, 10, 19 1, 3, 2 8 (1) 10 (3) 19 2 0; yes8, 10, 19 6, 1, 2 8 6 10 (1) 19 (2) 0; yesLesson 8-5Lesson 8-4Page A411. 3, 4 2, 5 3 2 4 5 26; no2. 3, 2 4, 6 3 4 2 6 0; yes3. 5, 3 2, 3 5 2 3 (3)19; no4. 8, 6 2, 3 8 (2) 6 (3)34; no5. 3, 4, 0 4, 3, 6 3 4 4 (3) 0 6 0; yes6. 4, 5, 1 1, 2, 3 4 (1) 5 (2) 1 311; no7. 1, 0, 3 1, 1, 2 1 0 3⏐ iu u j u k1 1 2⏐3i u j u k u 3, 1, 13, 1, 1 (1, 0, 3 3 1 1 0 1 3 0; yes3, 1, 1 1, 1, 2 3 1 1 1 1 2 0; yesPage A411. magnitude 200 2 220 2 297.32 Ndirection: tan v 2 00220v 42.3°2. ⏐r u ⏐ 2 50 2 50 2 2(50)(50) cos 120°⏐r u ⏐ 2 7500⏐r u ⏐ 86.60 mph; 30°3. ⏐r u ⏐ 2 350 2 280 2 2(350)(280) cos 135°⏐r u ⏐ 339,492.9291⏐r u ⏐ 582.66 N582.66280 s in 135° s invsin v 280 sin135°582.66sin v 0.34v 19.9°4. ⏐r u ⏐ 90 2 110 2⏐r u ⏐ 142.13 N90 N110 N521 Extra Practice
Lesson 8-6Page A411. x 2, y 3 t1, 0x 2 t y 3 0x 2 t y 32. x (1), y (4) t5, 2x 1, y 4 t5, 2x 1 5ty 4 2tx 1 5ty 4 2t3. x (3), y 6 t2, 4x 3, y 6 t2, 4x 3 2ty 6 4tx 3 2ty 6 4t4. x 3, y 0 t0, 1x 3 0y 0 tx 3y t5. x 3t →xt 3y 2 t → t y 2xy 2 3y 1 3 x 26. x 1 2t → t x 2y 4t →yt 4y x 4 2 1 2 1y 2x 27. x 3t 10 → t x 103y t 1 → t y 1y 1 1 3 x 1 03y 1 3 x 7 3 2c. y 5 75t sin 25° 16t 2 5 75 2. 132 sin 25° 16 2. 21. 90˚ 2.120˚ 60˚K150˚30˚180˚210˚240˚270˚1 2 3 4300˚0˚330˚3. 90˚ 4.120˚ 60˚150˚ 20.6 ftLesson 8-8N30˚567656234313 2Page A411. The figure is 4 times the original size andreflected over the xy-plane.2. The figure is half the original size.3. The figure is 1.5 times the original size andreflected over the yz-plane.Lesson 9-1Page A422323223M1 2 3 4533601166Lesson 8-7Page A411. v y 70 sin 34° 39.14 yd/s70 yd/sv x 70 cos 34° 58.03 yd/s34˚2a. x t ⏐v u ⏐ cos v 75t cos 25°y t ⏐v u ⏐ sin v 1 2 gt 2 h 75t sin 25° 1 2 (32)t 2 5 5 75t sin 25° 16t 22b. y 0 when t ?t 75 sin 25° (75 n si25°)2 16)( 4( 5)2(16)0.1468595989 or t 2.127882701t 2.13 sx 75t cos 25° 75(2.13) cos 25° 145 ftt180˚210˚240˚270˚1 2 3 4300˚0˚330˚5.2 2 336.5667643321 2 3 45301167.2 2 338.5667643321 2 3 453011676150˚180˚43120˚210˚240˚567623433290˚270˚23201 2 3 4P1165360˚1 2 3 4300˚31 2 3 45330˚0˚330˚60116Extra Practice 522
9. r 5 or r 5Lesson 9-2Page A421. 90˚ 2.120˚ 60˚150˚180˚210˚240˚circlespiral of Archimedes3. 90˚ 4. Sample answer:120˚ 60˚ r sin 5v150˚30˚180˚210˚240˚cardioidLesson 9-3Page A421. r 11) 2 ( 2 v Arctan 11 2 2 7 42, 7 4 2. r 3 2 0 2 9 or 3 0(3, 0)3. r 22) 2 ( 2 v Arctan 2 6 or about 2.45 0.62(2.45, 0.62)4. x 2 cos 4 y 2 sin 4 2 22 22, 2v Arctan 0 3 2 2 22 5. x 1 4 cos 2 y 1 4 sin 2 1 4 (0) 1 4 (1) 00, 1 4 270˚1 2 3 4300˚1 2 3 4270˚300˚30˚0˚330˚0˚330˚5676 1 4 23432322 3510152053601166. x 5 cos 240° y 5 sin 240° 5 1 2 5 32 5 2 or 2.5 5 32or about 4.33(2.5, 4.33)7. x 2r cos v 22r co s vr 2 sec v9. x 2 y 2 36(r cos v) 2 (r sin v) 2 36r 2 (cos 2 v sin 2 v) 36r 2 36r 6 or r 610. x 2 y 2 3y(r cos v) 2 (r sin v) 2 3r sin vr 2 (cos 2 v sin 2 v) 3r sin vr 2 3r sin vr 3 sin v11. r 4r 2 16x 2 y 2 16Lesson 9-48. y 6r sin v 66r si n vr 6 csc v12. r 4 cos vr 2 4r cos vx 2 y 2 4xPage A421. AB 2 2 65) 2 ( 2616 61 5x 61 6y 61 06cos f 61 5, sin f 61 , p 6 61f Arctan 5 6 180° 140° 6 61 61 r cos (v 140°)2. A B 2 2 3 2 9 2 3103 x 9 y 90 03 3 310 10 10010 1061 cos f ,1 sin f 3 , p 31010f Arctan 3 72°310 r cos (v 72°)3. 8 r cos (v 30°)0 r(cos v cos 30° sin v sin 30°) 830 r cos v 1 22 r sin v 80 3 x 1 2 2 y 80 3x 2y 164. 1 r cos (v )0 r (cos v cos sin v sin ) 10 r cos v 0 10 x 1x 1523 Extra Practice
5.120˚90˚60˚Lesson 9-6150˚180˚210˚240˚1 2 3 430˚0˚330˚270˚300˚Page A431. 4x 6yi 14 12i4x 14 6y 12x 1 44y 1 26x 3.5y 22. i3.iLesson 9-5(4, 1)OPage A431. i 10 (i 4 ) 3 i 2 1 3 (1)13. i 1000 (i 4 ) 250 1 250 12. i 17 (i 4 ) 4 i 1 4 i i4. i 12 i 4 (i 4 ) 3 (i 4 ) 1 1 3 1 1 25. (4 i) (3 5i) (4 (3)) (i 5i) 1 4i6. (6 6i) (2 4i) (6 (2)) (6i (4i)) 4 2i7. (3 i)(5 3i) 15 4i 3i 2 18 4i8. (2 5i) 2 (2 5i)(2 5i) 4 20i 25i 221 20i9. (1 2i)(3 8i) 3 8i 32i 16i 23 22i 32i 4i 27 2i10. 4 i1 i 4 i1 i 1 i1 i 4 5i i21 i2 3 5i2 3 2 5 2 i6 2i 6 2i11. 2 i 2 i 2 i2 i 12 10i 2i24 i2 10 10i52 2i12. ( i 2)24 2i i2 4i 44 2i 3 4i4 2i 4 2i4 2i 12 22i 8i216 4i 2 4 22i20 1 5 11 1 0iO(0, 5)⏐z⏐ 4 2 1 ⏐z⏐ 05) 2 ( 2 17 25 or 54. i⏐z⏐ 23)2 ( 2 7O5. r 4 2 4 2 32 or 42 4 4 4i 42cos 4 i sin 4 6. r (2) 2 1 2 v Arctan 12 5 2.682 i 5(cos 2.68 i sin 2.68)7. r 42)2 ( 2 v Arctan 24 2 18 or 32 5.944 2i 32 (cos 5.94 i sin 5.94)8a. 5(cos 0.9 i sin 0.9) 3.11 3.92j8(cos 0.4 j sin 0.4) 7.37 3.12j8b. (3.11 3.92j) (7.37 3.12j) (3.11 7.37) (3.92j 3.12j) 10.48 7.04j ohms8c. r (10.48 ) .04) 2 (7 2 v Arctan 7.0410.48 12.63 0.5910.48 7.04j 12.63 (cos 0.59 j sin 0.59) ohmsLesson 9-7(2, 3)v Arctan 4 4 Page A431. r 6 4 or 24 v 2 4 3 424cos 3 4 i sin 3 4 24 22 i 22 122 122iExtra Practice 524
32. r or 6 v 3 4 1 2 6(cos 7 4 i sin 7 )4 6 22 7 4 i 22 32 32i3. r 5 2 or 10 v 135° 45° 180°10(cos 180° i sin 180°) 10(1 i(0))10Lesson 9-8Page A431. 4 4 cos (4) 2 i sin (4) 2 256(cos 2 i sin 2) 256(1 i(0)) 2562. r 12 2 (5) 2 v Arctan 5 13 1.96558744613 3 (cos (3)(v) i sin (3)(v)) 2035 828i3. r 1 2 1 2 v Arctan 1 1 2 4 2 1 3 cos 1 3 4 i sin 1 3 12 2 1 3 cos 1 2 i sin 1 2 1.08 0.29i4. r (1) 2 0 2 v 11 1 5 cos 1 5 () i sin 1 5 () 1cos 5 i sin 5 0.81 0.59iLesson 10-1Page A441. d (x x 2 1 ) 2 (y 2 y 1 ) 2d (4 2)) ( 2 (5 2) 2d 6 2 3 2 d 45 or 35 x 1 x 22, y 1 y 22 2 2 (1, 3.5)2. d (x x 2 1 ) 2 (y 2 y 1 ) 2 4, 2 52 d (8 3)) ( 2 (1 6) 2d 11 2 (7) 2d 170 x 1 x 22, y 1 2y 2 3 2 (2.5, 2.5) 8, 6 (1)2 123. d (x x 2 1 ) 2 (y 2 y 1 ) 2d (r ) r2 2 () 62d 0 8) 2 ( 2d 64 or 8 x 1 x 224. 6 2 6 x 22, y 1 2y 2 r 2 (r, 2)r, 6 (2)2 x 2, 2 y 22 (5, 8) 5 2 y 22 86 x 2 10 2 y 2 16x 2 16 y 2 14Then A has coordinates (16, 14).5. A quadrilateral is aparallelogram if bothpairs of opposite sidesare parallel. Since onlyone pair of opposite sidesare parallel, thequadrilateral is not aparallelogram.6a. 0 502, 0 402 (25, 20)6b. d (x x 2 1 ) 2 (y 2 y 1 ) 2d (25 0) 2 (20 0) 2d 25 2 20 2d 1025 d 541 or about 32 ftLesson 10-2Page A441. (x h) 2 (y k) 2 r 2[x (2)] 2 (y 2) 2 2 2(x 2) 2 (y 2) 2 2y(x 2) 2 (y 2) 2 2O2. (x h) 2 (y k) 2 r 2(x 0) 2 (y (4)) 2 4 2x 2 (y 4) 2 16yO xx 2 (y 4) 2 16xy (5, 10)O(2, 8)(2, 5)(5, 2)x525 Extra Practice
3. x 2 49 y 2 → x 2 y 2 49y8 x 2 y 2 49844 O484 8x4. x 2 y 2 6x 8y 18 0(x 2 6x 9) (y 2 8y 16) 18 9 16(x 3) 2 (y 4) 2 7y(x 3) 2 (y 4) 2 7O5. x 2 y 2 Dx Ey F 02 2 (2) 2 2D 2E F 0→ 2D 2E F 80 2 (4) 2 0(D) 4E F 0→ 4E F 16(2) 2 (2) 2 2D 2E F 0→ 2D 2E F 82D 2E F 8 2(2 0 2E F ) 2(8)2D 2E F 8 4E F ) 164D 0 F 0D 04E (0) 164E 16E 4x 2 y 2 4y 0 center: (0, 2)x 2 (y 2 4y 4) 0 4 radius: 2x 2 (y 2) 2 46. x 2 y 2 Dx Ey F 0(1) 2 3 2 D 3E F 0→ D 3E F 10(4) 2 6 2 4D 6E F 0→ 4D 6E F 52(7) 2 3 2 7D 3E F 0→ 7D 3E F 582(D 3E F ) 2(10) 4D 6E F ) 2(524D 6E F ) 2(52 2(7D 3E F ) 2(58)2D F ) 2(32 10D F ) 2(642D F 32 (8) 3E F 1010D F 64 4(8) 6E F 5212D 96 24) 3E 42D 8 3E 18E 6(8) 3(6) F 1026 F 10F 16x 2 y 2 8x 6y 16 0(x 2 8x 16) (y 2 6y 9) 16 16 9(x 4) 2 (y 3) 2 9center: (4, 3)radius: 3x7. r (4 ) 2 2 (05)) (2r 2 2 5 2 r 29; r 2 29(x 4) 2 (y 0) 2 29(x 4) 2 y 2 29Lesson 10-3Page A441. center: (h, k) (0, 0)a 1 02or 5b 6 2 or 3 (x h) 2a 2 (y k) 2b 2 1 (x 5 2 0) 2 (y 3 2 0) 2 1x2 2 5 y 29 1c 2 a 2 b 2c 2 25 9c 2 16c 4foci: (4, 0)2. center: (h, k) (2, 1)a 8 2 or 4b 4 2 or 2 (y a 2 k) 2 (x b 2 h) 2 1 (y 1) 24 2 [x (2)] 222 (y 1) 216 (x 2) 24 1 1c 2 a 2 b 2c 2 16 4c 2 12c 12 or 23foci: (2, 1 23)3. The major axis contains the foci and it is locatedon the x-axis.ycenter: (h, k) (4, 1)c 2 a 2 b 28 (x 4) 2 (y 1)2 164 16c 2 64 164c 2 80c 80 or 454 O 4 8 12x4foci: (h c, k) (4 45, 1)major axis vertices: (h a, k) (4 8, 1) (12, 1) and (4, 1)minor axis vertices: (h, k b) (4, 1 4) (4, 5) and (4, 3)Extra Practice 526
Lesson 10-4Page A451.y8 (1, 2 58)y 2 7 (x 1)3 y 2 7 (x 1)43(1, 2)4 O 4 8 x4(1, 2 58)2. y84Lesson 10-5Page A451. vertex (h, k) (0, 0)4p 4p 1focus: (h p, k) (0 1, 0) or (1, 0)directrix: x h px 0 1x 1axis of symmetry: y ky 0yy 2 4x8 4 O 4 8x4xy 168Ox3. center: (h, k) (4, 3) (x h) 2a 2 (y k) 2 1b 2 [x (4)] 232 (y 3) 22 2 (x 4) 29 (y 3) 24 1 14. transverse axis: x h 2foci: (h, k c) (2, 7) k c 7(h, k c) (2, 3) k c 32k 4k 2; c 5vertices: (h, k a) (2, 5) 2 a 5(h, k a) (2, 1) a 3a 2 b 2 c 23 2 b 2 5 2b 2 16b 4 (y a 2 k) 2 (x b 2 h) 2 1 (y 3 2 2) 2 (x 4 2 2) 2 1 (y 2) 29 (x 2) 216 12. x 2 4x 4 12y 12(x 2) 2 12(y 1)vertex (h, k) (2, 1)4p 12p 3focus: (h, k p) (2, 1 3) or (2, 4)directrix: y k py 1 3y 2axis of symmetry: x hx 2yx 2 4x 4 12y 12O3. vertex: (h, k) (2, 3)focus: (h p, k) (0, 3)h p 0, k 32 p 0p 2(y k) 2 4p(x h)(y 3) 2 4(2)[x (2)](y 3) 2 8(x 2)4. directrix: y k p 3focus: (h, k p) (0, 2)k p 32.5 p 2k p 2 p 0.52k 5k 2.5(x h) 2 4p(y k)(x 0) 2 4(0.5)[y (2.5)]x 2 2(y 2.5)x527 Extra Practice
Lesson 10-6Page A451. A 1, C 1; since A C, the conic is a circle.x 2 y 2 8x 2y 13 0(x 2 8x 16) (y 2 2y 1) 13 16 1(x 4) 2 (y 1) 2 42. A 1, C 4; since A and C have differentsigns, the conic is a hyperbola.x 2 4y 2 10x 16y 5(x 2 10x 25) 4(y 2 4y 4) 5 25 16(x 5) 2 4(y 2) 2 4 (x 5) 24 (y 2) 21 13. A 0, C 1; since A 0, the conic is a parabola.y 2 5x 6y 9 0(y 2 6y 9) 5x 9 9(y 3) 2 5x4. A 1, C 2; since A and C have the same signx 2 2y 2 2x 8y 15(x 2 2x 1) 2(y 2 4y 4) 15 1 8(x 1) 2 2(y 2) 2 24Lesson 10-7 (x 1) 224 (y 2) 212 1Page A451. B 2 4AC 0 2 4(1)(1) 0 4 or 4Since 4 0 and A C, the graph is a circle.x 2 y 2 9(x 1) 2 [y (1)] 2 9(x 1) 2 (y 1) 2 9x 2 2x 1 y 2 2y 1 9x 2 y 2 2x 2y 7 02. B 2 4AC 0 2 4(4)(1) 0 16 or 16Since 16 0 and A C, the graph is an ellipse.4x 2 y 2 164[x (3)] 2 [y (2)] 2 164(x 3) 2 (y 2) 2 164(x 2 6x 9) (y 2 4y 4) 164x 2 y 2 24x 4y 24 03. B 2 4AC 0 2 4(49)(16) 0 3136 or 3136Since 3136 0, the graph is a hyperbola.49x 2 16y 2 78449 x cos 4 y sin 4 2 16 x sin 4 49 222 y cos 4 2 784x y2 2 16 x2 y2 2 78449 1 2 (x ) 2 x y 1 2 (y ) 2 16 1 2 (x ) 2 x y 1 2 (y ) 2 784 4 92(x ) 2 49x y 4 92(y ) 2 8(x ) 2 16x y 8(y ) 2 78449(x ) 2 98x y 49(y ) 2 16(x ) 2 32x y 16(y ) 2 156833(x ) 2 130x y 33(y ) 2 1568 0224. B 2 4AC 0 2 4(4)(25) 0 400 or 400Since 400 0, the graph is a hyperbola.4x 2 25y 2 644(x cos 90° y sin 90°) 2 25(x sin 90° y cos 90°) 2 644(0 y ) 2 25(x 0) 2 644(y ) 2 25(x ) 2 64 05. B 2 4AC 22 2 4(1)(2) 8 8 or 0parabolaBtan 2v A C22tan 2v 1 2tan 2v 222v 70.52877937°v 35, 35°6. B 2 4AC 5 2 4(15)(5) 25 300 or 275Since 275 0 and A C, the graph is an ellipse.Btan 2v A C5tan 2v 15 5tan 2v 1 2 Lesson 10-82v 26.56505118°v 13°Page A451. xy 3 x 2 y 2 8y 3 x x 2 3 x 2 8x 2 9 x 2 8x 4 9 8x 2x 4 8x 2 9 0(x 2 9)(x 2 1) 0x 2 9 0 or x 2 1 0x 2 9 x 2 1x 3 x 1 or i3If x 3, then y ( 3) or 1.3If x 3, then y ( 3)or 1.Since x 1 is an imaginary number, disregardthis solution.(3, 1), (3, 1)y(3, 1)(3, 1)OxExtra Practice 528
2. x y 4 x 2 10y 2 10x y 4 (y 4) 2 10y 2 10y 2 8y 16 10y 2 109y 2 8y 6 0y (8) (8)2 )(6)4(9y 8 1y 4 92 70 870 70 70 y 4 9or y 4 9y 1.4 y 0.5If y 1.4, then x (1.4) 4 or 5.4.If y 0.5, then x (0.5) 4 or 3.5.(5.4, 1.4), (3.5, 0.5)Lesson 11-1Page A46y848 4 O 4 8x48(5.4, 1.4)(3.5, 0.5)1. (12) 2 (112) 2 112 21 1 44 44 1 681 44 2. 12 2 113. (4 6) 3 4 3 6 364 216 13,824165. 16(4 2 ) 1 216 1 2 1 64or 46. 27 1 2 20 1 2 (3 2 3) 1 2 (2 2 5) 1 2 (3 2 ) 1 2 3 1 2 (2 2 ) 1 2 5 1 2 3 2 15 1 2 6157. 4 625 2 625 2 4 625 1 2 625 or 2518. 1 3 (15) 615 6 3 115 2 2125 9. (2a 4 ) 2 2 2 (a 4 ) 2 4a 810. (x 4 ) 3 x 5 x 12 x 5 x 174. 2 3 4 2 311. ((3f) 2 ) 3 (3f) 63a1 (3 f) 61 3 6 f 61 72 9f 66a8a12. c c4 2 a c c c 14a1 c 14a13. (2n 1 3 3n 1 2 ) 6 2 6 n 6 3 3 6 n 6 2 14. h61216h3 64n 2 799n 3 46,656n 59 1 3 3 2h16h9 3 216 1 3 216 3 h3 h6315. 3 z 4 (z 4)1 2 3 z 4 z 2 3 z6 z 6 3 z 216. (4r 2 t 5 )(16r 4 t 8 ) 1 4 4r 2 t 5 (16 1 4 r 4 4 t 8 4 ) (4r 2 t 5 )(2rt 2 ) 8r 3 t 717. a 3 b 5 a 3 2 b 5 2 18. 3 64m 9 n 6 64 1 3 m 9 3 n 6 3 (4 3 ) 1 3 m 3 n 2 4m 3 n 219. 15 3 r 12 t 2 15r 1 32t 2 3 15r 4 t 2 3 20. 8 256x y 2 16 256 1 8 x 2 3 y 1 68Lesson 11-2Page A46216 1 3 h 3 (2 8 ) 1 8 x 1 4 y 2 2x 1 4 y 21. y2.y 3 xy 3 xyOxOx529 Extra Practice
3.yOLesson 11-3xy 3 x 1Page A461. p (100 a)e bt ap (100 18)e 0.6(2) 18 42.7%2. y ae kt cy 140e 0.01(10) 70 197° F3a. y 6.7e 4 t8.18.1y 6.7e 4 15y 0.271292 millions of cubic feety 271,292 ft 33b. y 6.7e 4 8.1t8.1y 6.7e 4 50y 2.560257 millions of cubic feety 2,256,275 ft 34. Continuously SemiannuallyA Pe rtrA P1 n ntA 5000e 0.058(20) A 50001 0.0 582 2(20)A $15,949.67 A $15,688.63The account that compounds continuously wouldearn $261.04 more than the account compoundedsemiannually.Lesson 11-4Page A471. 16 1 4 22. 1 2 3 83. 4 1 1 4 4. log 8 x 25. log x 32 56. log116 247. log 1 5 5 x5 x 1 5 5 x 5 1x 18. log 3 27 x3 x 273 x 3 3x 39. log 36 6 x36 x 6(6 2 ) x 66 2x 6 12x 1x 1 2 10. log 3 y 43 4 y81 y11. log 5 r log 5 8r 812. log 5 35 log 5 d log 5 5log 3 5 5 d log5 5 3 5 d 57 d13. log 4 4 xlog 4 4 1 2 x 1 2 log 4 4 x 1 2 x14. log 4 (2x 3) log 4 152x 3 152x 12x 615. 4 log 8 2 1 3 log 8 27 log 8 alog 8 2 4 log 8 27 1 3 log 8 alog 8 16 log 8 3 log 8 alog 8 48 log 8 a48 aLesson 11-5Page A471. log 5000 log (5 1000) log 5 log 10 3 log 5 3 log 10 0.6990 3 3.69902. log 0.0008 log (0.0001 8) log 10 4 log 84 log 10 log 84 0.90313.09693. log 0.14 log (0.01 14) log 10 2 log 142 log 10 log 142 1.14610.85394. log 3 81 l og81log3 45. log 6 12 l og12log6 1.38696. log 5 29 l og29log5 2.0922Extra Practice 530
7. 3 x 45x log 3 log 45x l og45log3x 3.46508. 6 x 2 x 1x log 6 (x 1) log 2x log 6 x log 2 log 2x log 6 x log 2 log 2x (log 6 log 2) log 2log2x log 6 log 2x 0.63099. 5 log y log 32log y 5 log 32y 5 32y 5 2 5y 2Lesson 11-6Page A471. 3.55532. 0.57633. 3.43984. log 15 10 l n 10ln15 0.85035. log 3 14 l n 14ln3 2.40226. log 8 350 ln 350ln 8 2.81717. 5 x 90x ln 5 ln 90x ln 90ln5x 2.79598. 7 x 2 5.25(x 2) ln 7 ln 5.25x ln 7 2 ln 7 ln 5.25x ln 7 ln 5.25 2 ln 7x ln 5.25 2ln7ln 7x 1.14789. 4 x 43x ln 4 ln 4343x ln ln 4x 1.396210. 6e x 48e x 8x ln e ln 8x 2.079411. 50.2 e 0.2xln 50.2 0.2x ln e ln 50.20.2 xx 19.580112. 16 10(1 e x )1.6 1 e x0.6 e xln 0.6 x ln e0.5108 xLesson 11-7Page A47ln1. t 02. 045 15.40 yrln2. t 20 .06 11.55 yrln 23. t 0.0 8125 8.53 yr4a. y 5.2449(1.5524) x4b. y 5.2449(e ln 1.5524 ) xy 5.2449e 0.4398x4c. Use t ln k; k 0.4398n 2t 0ḷ 4398 1.58 hrLesson 12-12Page A481. d 3 7 or 41 (4) 5, 5 (4) 9,9 (4) 13, 13 (4) 175, 9, 13, 172. d 1 0.5 or 1.52.5 (1.5) 4, 4 (1.5) 5.5,5.5 (1.5) 7, 7 (1.5) 8.54, 5.5, 7, 8.53. d 8 (14) or 62 6 4, 4 6 10,10 6 16, 16 6 224, 10, 16, 224. d 2.8 3 or 0.22.6 (0.2) 2.4, 2.4 (0.2) 2.2,2.2 (0.2) 2, 2 (0.2) 1.82.4, 2.2, 2, 1.85. d x 4x or 5x6x (5x) 11x, 11x (5x) 16x,16x (5x) 21x, 21x (5x) 26x11x, 16x, 21x, 26x6. d (2y 2) (2y 4) 2y 2y (2) (4) 22y 2, 2y 2 2 2y 4,2y 4 2 2y 6, 2y 6 2 2y 82y 2, 2y 4, 2y 6, 2y 8531 Extra Practice
7. a n a 1 (n 1)da 16 2 (16 1)5 778. 20 6 (n 1)(2)26 2n 228 2n14 n9. 42 a 1 (12 1)442 a 1 442 a 110. 30 7 (13 1)d23 12d1 1121 d11. d 10 10.5 or 0.5a 24 10.5 (24 1)(0.5)a 24 112. S n n 2 [2a 1 (n 1)d]; d 2.8 2 or 0.8S 12 1 22[2 2 (12 1)(0.8)] 6(4 8.8) 76.813. S n n 2 [2a 1 (n 1)d]80 n 2 [2 (4) (n 1)4]160 n(8 4n 4)160 12n 4n 20 4(n 2 3n 40)0 4(n 8)(n 5)n 8 or n 5Since n cannot be negative, n 8.Lesson 12-2Page A4871. r 1 4or 0.53.5(0.5) 1.75, 1.75(0.5) 0.875,0.875(0.5) 0.43751.75, 0.875, 0.437542. r 2or 28(2) 16, 16(2) 32, 32(2) 6416, 32, 64 3 8 1 2 3. r or 3 4 3 8 3 4 9 9 3, 2 3 29 2781 3, 2 1 2, 8 5 125 3 4 127 272, 8 1 28 3 4 581124. r 10or 0.52.5(0.5) 1.25, 1.25(0.5) 0.625,0.625(0.5) 0.31251.25, 0.625, 0.31255. r 8 2 or 28162 162, 1622 32, 322 322162, 32, 322a6. r a 10 or a1218a 6 a 2a 4 , a 2 , 17. a n a 1 r n1a 6 9(2) 61 2888. 100 a 1 (4) 81100 16,384 a 125 4 0112 a 4 , a 4 a 2 a 2 , a 2 a96 a 19. 10 a 1 1 2 51110 1a6 1160 a 1a 2 160 1 2 or 80a 3 80 1 2 or 40160, 80, 4010. 256 4r 4164 r 34 r4(4) 16, 16(4) 644, 16, 64, 25611. S n a 1 a1 rn; r 9 3 or 31r 3( 3)6 3 3 21872S 6 3 1 1 109212a. There are four 15-minute periods in an hour, sor 2 4 or 16.b t b 0 16 t12b. b 4 12 16 4 786,432Lesson 12-3Page A49 4 2nn→ 3n1. lim limn→ 34 0 2 3 2 3 2 nn→ 3n n lim2. Limit does not exist. lim n4 3n3n→ n3 lim n n→ n 2 lim n. But as n approaches infinity, n becomesn→increasingly large, so there is no limit. 8n2 6n 2n→ 4n23. lim 4n2 2n 1n→ n 2 24. lim lim 8 n3n→ 4n26nn→ 4 n2 lim 2 0 0 2 lim ––––2n→ 4 0 01 0 4 4 nn 2 2 2 n 1n2 n 2 n 2n2 n 22n→ 4 n2 lim Extra Practice 532
n3 n2 4n→ 5 2n35. lim n 3n3 n 2 4n3 n 3n→ 5 n 3 2 n3n3 1 0 00 2 lim 1 2 6. Limit does not exist.2lim — n n—n→ 2 n lim 2 n n—— n→ 2 n lim 2 n 12 n→ — — n 1limn→ (2n 1) or lim 2 n As n approaches infinity, 2 nn→becomes increasingly large, so there is no limit.97. 0.09 — 1 00 — 9 —— 10, 000 . . .9a 1 — 1 00 — 1, r — 1 00 —9 1 00 S n —11 — 1 00 — — 111 —1 38. 0.13 1 0 1 00 10300 . . .3a 1 1 00 1, r 1 01S n 1 02 1 5 1300 —11 1 007407000 1,0 00,000004071 00049. 7.407 7 14071a 1 1, r 10000S n 7 ——11 10 0010. a 1 210S n 11 7 2 714 0, r —or 1 2 12 0—1 1 2 2 1 2 0or 1 012 0 . . .11. S n does not exist. This series is geometric with acommon ratio of 2. Since this ratio is greater than1, the sum of the series does not exist.Lesson 12-4Page A491. a n (2 n ) 2 , a n1 (2 n1 ) 2 2 2n 2 2n2n→22n 22 n 2n→222nr lim lim 4divergent 2 2n22. a n n 3 , a n1 n 3r lim n→ n 1n→n lim n 13— n 3 1 lim nn→n 1 lim n→n 1 0 1test provides no information13. a n 3 n (n ,1) 3 a n1 1 3 n1 (n 2) 3––––n→1 3 n (n 1) 3r lim lim3 n (n 3 3n 2 3n 1)3 n 3(n 3 6n 2 12n 8)n→ lim n→ n 3n3 3 nn 23 3 nn 3 n13 ———n3n22n 1 0 0 03 0 0 0 1 3 3 n 3 6 n 3 1 n 3 n83 convergent44. a n 2 n, a n1 2 n4 2 n1—4 n→ 2 nr lim2 nn→2 n 2 lim 1 2 41 13 n1 (n 1 1) 3convergent75. The general term is 6n .17 1 6n 1 n for all n, so divergent1 16. The general term is (2 n) 2 or 4 n 2 . 41n 2 n1 for all n, so convergent.1 11 2 n1 1——n→ 1 2 n 12 n 1 2 1 2 n 12n 2 n——n→ 2n 2 12 n 2 n7. a n 2 n, a n1 2 n1r lim lim n→ 2 n lim 1 02 0 1 2 convergent1 1533 Extra Practice
n228. a n n ,2a n1 n (n 12 n 1 n 3—n→ 2nn 22(n 2)( n 3)r lim lim 2n n→2 n 2 n 2n→n 3 2 n 2 n n—n→ n n 3 n lim limLesson 12-5Page A491. 52. 63. 74. 81) 2Lesson 12-6Page A501. (2 x) 4 16 32x 24x 2 8x 3 x 42. (n m) 5 n 5 5n 4 m 10n 3 m 2 10n 2 m 4 5nm 4 m 53. (4a b) 3 64a 3 48a 2 b 12ab 2 b 3m4(3) 24. (m 3) 6 m 6 (3) 0 6 m 5 (3) 1 6 5 2 1 6 5 4 m 3 (3) 3 6 5 4 3 m2 (3)43 2 14 3 2 1 2 0 6 5 4 3 2 m1 (3)55 4 3 2 11 06 5 4 3 2 1 m 20 (3)6 6 5 4 3 2 1divergent m 6 18m 5 135m 4 540m 3 1215m 2 1458m 7295. (2r s) 4 (2r) 4 (s) 0 4(2r) 3 (s) 1 4 3 (2r) 2 (s)22 1 4 3 2 (2r) 1 (s) 3 4 3 2 1(2r)3 2 10 (s)44 3 2 1 16r 4 32r 3 s 24r 2 s 2 8rs 3 s 46. (5x 4y) 3 (5x) 3 (4y) 0 3(5x) 2 (4y) 1(3n 1) (3 1 1) (3 2 1) 3 2(5 x) 1(4y) 2 3 2 1(5x)n 12 10 (4y)33 2 1(3 3 1) (3 4 1) 125x(3 5 1)3 300x 2 y 240xy 2 64y 3 2 5 8 11 148!7. 5!(8 5)!x 85 y 5 8 7 6 5 4 3 2 1 5 4 3 2 1 3 2 1 x 3 y 5 40 56x 3 y 54a 4 3 4 4 4 5 4 67!k 3 12 16 20 248. 4!(7 4)!b 74 3 4 7 6 5 4 3 2 1 4 3 2 1 3 2 1 b 3 9 72 315b 3(k 2 2) (3 2 2) (4 2 2) (5 2 10! 2)9. 2!(1k 3 (6 2 2) (7 2 0 2)!(4z) 102 (w) 2 2)10 9 8 7 6 5 4 3 2 12 1 8 7 6 5 4 3 2 1 7 14 23 34 47 45 65,536z 1258 w 2j 4 5 6 7 2,949,120z 8 w 2 j 3 4 3 5 3 6 3 7 312!j 410. 87!(1 2 7)!(2h) 127 (k) 7 12 11 10 9 8 7 6 5 4 3 2 18 3 4 7 5 8 6 7 6 5 3 2 1 5 4 3 2 19 7 8 1 0 1 1 792 32h 5 (k) 7 3 2 68325,344h 5 k 792403 p 3 0 3 1 3 2 3 3 3 4p 0 1 3 9 27 81 121Lesson 12-72 3 4 n 2 3 4 1 2 3 4 2 2 3 4 3 . . .n 1 2 3 4 Page A50 1 1 2 1 1 8 7 2 3 2 . . . 2 3 4 1. ln (3) ln (1) ln (3) i 1.09861 1 2 2. ln (4.6) ln (1) ln (4.6)S —1 3 4 5. 46. 7. 4n 19. 9m 0 6(3n 2) 8. 10r 24 m1 10. w 14rw! (4z) 8 (w) 2 (2h) 5 (k) 7 i 1.52613. ln (0,75) ln (1) ln (0.75) i 0.2877.2) 2.2) 3.2) 44. e 1.2 1 1.2 (1 2 ! (1 3 ! (1 4 ! 1 1.2 0.72 0.288 0.0864 3.295. e 0.7 1 (0.7) (0 .2 7) 2! (0 . 7) 33 ! 1 0.7 0.245 0.057 0.010 0.50 (0 . 7) 4 4!Extra Practice 534
65) 265) 365) 46. e 3.65 1 3.65 (3. 2 ! (3. 3 ! (3. 4 ! 1 3.65 6.661 8.105 7.395 26.817. cos x 1 x 22 ! x 44 ! x 66 ! x 88 !cos 4 cos 0.7854 1 (0.7 2!8 54) 2 (0.7 84 54) 4! (0.7 86 54) 6! (0.7 854) 8 8!169 0.3 80524 0. 237 2 470 0 . 14484 0,320 1 0.6 2 0.7071actual value: cos 4 2 0.707128. sin x x x 3 ! x 5 !sin 6 sin 0.52363 0.5236 (0.5 3! (0.5 236) 99! 5 x 77 ! x 99 ! 0.5236 0.1 6 0.5000actual value: sin 6 0.59. cos 3 1 x 2 ! x 4 !cos 3 cos 1.04722 1 (1.0 2!2 36) 3 (0.5 25 36) 5! (0.5 236) 7 7!435 0. 0312 940 0 .01080.00305 040 3 62,8804 x 66 ! x 88 ! 1 1.0 2 0.5000actual value: cos 3 0.5Lesson 12-84 72) 2 (1.0 44 72) 4! (1.0 46 72) 6! (1.0 472) 8 8!966 1.2 02624 1. 317 2 880 1 . 44624 0,320Page A501. f(2) 2 (2) or 4f(4) 2 (4) or 8f(8) 2 (8) or 16f(16) 2 (16) or 324, 8, 16, 322. f(4) 4 2 or 16f(16) 16 2 or 256f(256) 256 2 or 65,536f(65,536) 65,536 2 or 4,294,967,29616, 256, 65,536, 4,294,967,2963. z 0 2iz 1 0.5(2i) i 2iz 2 0.5(2i) i 2iz 3 0.5(2i) i 2i2i, 2i, 2i4. z 0 4 4iz 1 0.5(4 4i) i 2 2i i2 3iz 2 0.5(2 3i) i 1 1.5i i 1 2.5iz 3 0.5(1 2.5i) i 0.5 1.25i i 0.5 2.25i2 3i, 1 2.5i, 0.5 2.25i5. p 1 p 0 rp 0p 1 4000 (0.054)4000 4216p 2 4216 (0.054)4216 4443.66p 3 4443.66 (0.054)4443.66 4683.62p 4 4683.62 (0.054)4683.62 4936.54p 5 4936.54 (0.054)4936.54 5203.11$4216, $4443.66, $4683.62, $4936.54, $5203.11Lesson 12-9Page A501. Step 1: Verify that the formula is valid for n 1.Since S 1 2 and 1(1 1) 2, the formula is validfor n 1.Step 2: Assume the formula is valid for n k andshow that it is also valid for n k 1.2 4 6 . . . 2k k(k 1)Derive the formula for n k 1 by adding2(k 1) to each side.2 4 6 . . . 2k 2(k 1) k(k 1) 2(k 1)2 4 6 . . . 2k 2(k 1) (k 1)(k 2)Apply the original formula for n k 1.S k1 (k 1)((k 1) 1) or (k 1)(k 2)The formula gives the same result as adding the(k 1) term directly. Thus, if the formula is validfor n k, it is also valid for n k 1. Since S n isvalid for n 1, it is also valid for n 2, n 3,and so on indefinitely. Hence the formula is validfor all positive integral values of n.535 Extra Practice
2. Step 1: Verify that the formula is valid for n 1.Since S 1 1 and 1(1 1 )(1 2)6 1, the formula isvalid for n 1.Step 2: Assume the formula is valid for n k andshow that it is also valid for n k 1.1 3 6 . . . k(k 1)2 k(k 1 )(k 2)6Derive the formula for n k 1 by adding (k 1) (k 2)2 to each side.1 3 6 . . . k(k 1)2 (k 1) (k 2)2 k(k 1 )(k 2)6 (k 1) (k 2)21 3 6 . . . k(k 1)2 (k 1) (k 2)2k(k 1)(k 2) 3(k 1)(k 2)61 3 6 . . . k(k 1)2(k 1)(k 2)(k 3) (k 1) (k 2)24. dependent5. dependent5!6. P(5, 5) (5 5)! 5 4 3 2 11 1208!7. P(8, 3) (8 3)! 3364!8. P(4, 1) (4 1)! 4 3 2 13 2 1 419. P(10, 9) (108 7 6 5 4 3 2 15 4 3 2 10! 9)!10 9 8 7 6 5 4 3 2 11 3,628,80069!10. P(9, 6) (9 6)!9 8 7 6 5 4 3 2 1 3 2 1 60,4807!611. P(7, 3) (7 3)!7 6 5 4 3 2 1 4 3 2 1 2105!12. P ( 5,2)(5 2)!P(2,1)—2! (2 1)! 5 ! 1!3!2!5 4 3 2 1 1 3 2 1 2 1 108!13. P ( 8,6)(8 6)!P(7,4)—7! (7 4)! 8 ! 3!7!2!8 7 6 5 4 3 2 1 3 2 1 7 6 5 4 3 2 1 2 1 2414. P(5, 2) P(8,4)P(10,1)5! 8! (5 2)! (8 4)!———10! (10 1)! 9!8!5! 1 0!4!3!Apply the original formula for n k 1.S k1 orThe formula gives the same result as adding thek 1 term directly. Thus, if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2,n 3, and so on indefinitely. Hence, the formulais valid for all positive integral values of n.3. S n : 5 n 1 2r for some integer r.Step 1: Verify that S n is valid for n 1.S 1 5 1 1 or 4. Since 4 2 2, S n is valid forn 1.Step 2: Assume that S n is valid for n k and showthat it is also valid for n k 1.S k → 5 k 1 2r for some integer tS k1 → 5 k1 1 2t for some integer t5 k 1 2r5(5 k 1) 5(2r)5 k1 5 10r5 k1 1 10r 45 k1 1 2(5r 2)Thus, 5 k1 1 2t, where t (5r 2) is aninteger. Thus if S k is valid, then S k1 is alsovalid. Since S n is valid for n 1, it is also valid forn 2, n 3, and so on indefinitely. Hence, 5 n 1is even for all positive integral values of n.Lesson 13-1Page A511. Using the Basic Counting Principle,6 6 6 216.8!2. P(8, 8) (8 8)! 40,3203. independent(k 1)((k 1) 1)((k 1) 2)6(k 1)(k 2)(k 3)8 7 6 5 4 3 2 11Extra Practice 5369 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 5 4 3 2 110 9 8 7 6 5 4 3 2 1 4 3 2 1 3 2 1 33604!15. C(4, 2) (4 2)! 2! 4 3 2 12 1 2 1 610!16. C(10, 7) (10 7)! 7!10 9 8 7 6 5 4 3 2 13 2 1 7 6 5 4 3 2 1 120
6!17. C(6, 5) (6 5)! 5! 64!18. C(4, 3) C(7, 3) (4 19. C(3, 1) C(8, 7) (3 3 249!20. C(9, 5) C(4, 3) (9 4!5)! 5! (4 3)! 3!9 8 7 6 5 4 3 2 1 4 3 2 14 3 2 1 5 4 3 2 1 1 2 1 504Lesson 13-2Page A517!1. 0! 1 50407 6 5 4 3 2 18! 8 7 6 5 4 3 2 1 2! 2! 2 1 2 17! 7 6 5 4 3 2 1 3! 3 2 110! 10 9 8 7 6 5 4 3 2 1 2! 2 12. 10,0803. 8404. 1,814,4005. 4 !2! 4 3 2 12 1 126. circular; (4 1)! or 67. circular; since the bracelet can be turned overthere are (9 1)!2or 20,160 permutations8. linear; 5! or 120Lesson 13-36 5 4 3 2 11 5 4 3 2 1Page A514 11. P(ace) 5 2or 1 32. P(a card of 5 or less) 3. P(a red face card) 562 2 052 5or 1 33 or 24. P(not a queen) 1 P(a queen)4 1 55. P(blue) 8 24 27!3)! 3! (7 3)! 3! 4 3 2 1 7 6 5 4 3 2 11 3 2 1 4 3 2 1 3 2 1 140!8!1)! 1! (8 7)! 7!3 2 1 8 7 6 5 4 3 2 1 2 1 1 1 7 6 5 4 3 2 162 4 852 or 1 2132 1 4or 1 7 4 4 4 4 4 526. P(green) 8 47. P(not red) 8 4 8. P(red or green) 889. P(s) 1461 2 01 23 1 544 2 1 4or 2 7 2 64 2 1 4or 3 7 4 4 2 1 214or 6 7 P(f ) 1 P(s)1 1 20 1odds — or 110. P(s) 166odds 1 9205 1 511. P(s) C(2, C9P(f ) 1 P(s)0or 1 920 1 8 1 1 8 or 7 8 1 8 — or 1 7 7 8 1) C(4, 1)( 16,2) 2 41201 —— 1 5 1 41511 5— 4 1 415odds or 11P(f ) 1 P(s) 1 1112. P(s) P(1 white, 1 yellow) P(1 yellow, 1 red) P(1 white, 1 red) C(2, 1)C(16 C(4, 1), 2) C(4, 1) CC ( 16 ( 10, 1), 2) C(2, 1) C(10, 1)C ( 16,2) 2 12 40 4 11 2 00 2 101 201 1 5 1 3 1 6 1 730 P(f) 1 P(s) 1 1 73 1 730 1 3300 or 1 330 odds — or 1 1Lesson 13-473Page A521. dependent; 6 8 2 7 3 12. independent; 1 13. independent; 1 2 6 1 0408 1 018 2 581 3 1 011 3 64 8 2 5 2 5 2or 1 34. inclusive; 1 6 1 6 3165. exclusive; 5425537 Extra Practice
Lesson 13-5Page A521. P(second clip is blue⏐first clip was red)8 1 2 141 —81 2 141 2. P(second clip is blue⏐first clip was blue)4 1 2 131 —41 2 131 3. P(numbers on dice match⏐sum is greater than 7)33 6— 1 536 3 1 5or 1 5 4. P(sum is greater than 7⏐numbers match)33 6 —63 6 3 6 or 1 2 2004. P(exactly 1 hit) C(5, 1) 1000 5 1 5 2 5662556 or 0.4096 2 6255. P(exactly 3 hits) C(5, 3) 11 10 132 6 256. P(at least 4 hits) P(4 hits) P(5 hits)200 1 120000025 1 625 3 1 or 0.051280080040008002000 C(5, 4) 10004 11000 C(5, 5) 11 5 6 25 4 5 1 31125 11 or 0.006722 3 125Lesson 14-1200000Page A531. range 70 22 or 482. Sample answer: 103. Sample answer: 20, 30, 40, 50, 60, 70, 804. Sample answer: 25, 35, 45, 55, 65, 755.20-30 630-40 740-50 850-60 360-70 570-80 1Grams of Fat Frequency5. P(ball is from second box⏐ball is white)P(2nd box and white) P(white) 1 2 3 8 —— 1 2 4 7 1 2 3 8 2 153 6.Grams of Fat Consumedby Adults86Frequency 4 1 1 8 12 1 8 2. P(exactly 2 tails) C(3, 2) 1 2 2 1 2 100 10 20 30 40 50 60 70 80 3 1 4 1 2 Grams of Fat 3 8 7. Sample answer: 40–503. P(at least 2 heads) P(2 heads) P(3 heads) C(3, 2) 1 2 2 1 2 1 C(3, 3) 1 2 3 1 2 0 3 1 4 1 2 1 1 8 1 3 8 1 8 Lesson 13-6Page A521. P(all heads) C(3, 3) 1 2 3 1 2 0 5 18000000Extra Practice 538
Lesson 14-2Page A531. X 1 4 (130 150 180 190) 162.5M d 150 1802or 165Mode: none2. X 1 6 (15 16 17 18 18 19) 17.2M d 17 182or 17.5Mode 183. X 1 6 (25 28 30 36 38 42) 33.2M d 30 362or 33Mode: none14. X 1(10 2 3 4 5 5 6 9 9 10) 5.4M d 5Mode 5 and 915. X 1(2.32 2.5 4 2(5.6) 6 6.4 6.5 2(7) 8 10) 5.9M d 6 6.42or 6.2Mode 5.6 and 716. X 1(142 2(15) 16 20 21 24 27 28 36 2(39)) 24.5M d 21 242or 22.5Mode 15 and 3917. X 1(3.08 3.4 3.6 5.2 2(5.4) 2(5.6)5.7 6.2 6.3 6.8 7.0 7.1 7.6 7.7 8.2) 5.9M d 5.7Mode 5.4 and 5.618. X 1(8004 820 830 890 960 970 980 1040 2(1050) 1080 1110 1170 1180) 995M d 980 10402or 1010Mode 10509. stem leaf1 ⏐ 2792 ⏐ 34468993 ⏐ 4456794 ⏐ 0245 ⏐ 51⏐2 1210. 8 1 6 (4 5 6 9 10 x)48 34 x14 x11. Order the values from least to greatest. Themedian lies between the fourth and fifth terms.3, 4, 4, 7, x, 12, 16, 197.5 7 x215 7 x8 xLesson 14-3Page A541. interquartile range Q 3 Q 1 58 39 19semi-interquartile range 1 92or 9.540 50 60 70 802. interquartile range Q 3 Q 1 7.65 2.75 4.9semi-interquartile range 4 .9 2 or 2.451 2 3 4 5 6 7 8 9 103. X 1 9 (150 175 180 180 195 200 212 220 250) 195.7MD 1 9 (⏐45.8⏐ ⏐20.8⏐ ⏐15.8⏐ . . . ⏐54.2⏐) 21.98(45.8)j 2 (20.8) 2 . . . 54.2 2 27.5614. X 1 1 (1.4 2 2.4 2.9 3 3.5 3.7 4.2 4.6 5.3 5.5) 3.51MD 1 1 (⏐2.1⏐ ⏐1.5⏐ ⏐1.1⏐ . . . ⏐2⏐) 1.05(2.1)j 2 (1.5) 2 (1.1) 2 . . . 2 2911 1.265a. M d 185b. Q 1 16 182or 17Q 3 205c. interquartile range Q 3 Q 1 20 17 35d. semi-interquartile range 3 2 or 1.55e. Any points less than 17 1.5(3) or 12.5 and anypoints greater than 20 1.5(3) or 24.5 areconsidered outliers. There are no such points.539 Extra Practice
5f.Lesson 14-4Page A541a. 25% corresponds to t 0.3.10 0.3(2) 9.4 10.61b. 10 8 2, 14 10 4tj 2 tj 4t(2) 2 t(2) 4t 1 t 2 68 .3% 2 34.15% 95 .5% 2 47.75%34.15% 47.75% 81.9%1c. 10 7 3, 10 10 0tj 3 tj 0t(2) 3 t(2) 0t 1.5 t 0.6% 43.3% 86 21d. 80% corresponds to t 1.3.10 1.3(2) 7.4 12.62a. 0.683(400) 273.22b. 0.955(400) 3822c. 0.6 83215 16 17 18 19 20 21 22 23 24(400) 136.6Lesson 14-5Page A541. 21. j X 9 or about 0.1303.42. j X 1 or 0.340012.43. j X 2 or about 0.80404. A1% confidence level is given when P 99% andt 2.58.4. 2j X 4 0 0.6640783086interval: X tj X 150 2.58j X 148.29 151.715. A1% confidence level is given when P 99% andt 2.58.10j X 7 8 1.132277034interval: X tj X 320 2.58j X 317.08 322.92Lesson 15-1Page A551. lim (x 2 2x 2) 4 2 2(4) 2x→4 222. limx→1 (x4 x 3 2x 1) (1) 4 (1) 3 2(1) 1 13. lim (x sin x) 0 sin 0x→0 04. lim x2 16x 4 lim (x 4)(xx 4x→4x→4 lim (x 4)x→44 482 5x 6x 3)(5. lim lim xx→2 x2 x 23x 9x→2 x 2 5x 246. limLesson 15-2Page A551. f(x) lim 4)x→2 ( x 2)( x 1)( x 2) lim x 3x 1x→2 2 32 1 1 3 3(x 3) x→2 (x 8) (x 3)3 lim x→2 x 83 2 8 3 6 or 1 2 lim f(x h ) f(x)h→0h 5(x h ) 5xh→0h 5x 5 h 5xh→0hh lim lim lim 5h→0h 52. f(x) lim f(x h ) f(x)h→0h lim h→0 lim h→0 lim 9 h h→0h 93. f(x) 1 2 x 2 3 f(x) 1 2 1x 11 0 1 2 9(x h) 2 (9x 2)h9x 9h 2 9x 2h4. f(x) x 2 4x 8f(x) 2x 21 4 1x 11 0 2x 4Extra Practice 540
5. f(x) x 51F(x) 5 x 51 C1 1 6 x 6 C6. f(x) 2x 2 8x 21F(x) 2 2 x 21 1 8 1 x 11 2x C1 2 3 x 3 4x 2 2x C7. f(x) 1 5 x 3 3 4 x 1F(x) 1 5 1 3 x 31 3 4 1 1 x 11 x C 210 x 4 3 8 x 2 x C8. f(x) x3 2x 2 xx1 x 2 2x 11F(x) 2 x 21 1 2 1 x 11 x C1Lesson 15-3Page A551. 30 1 3 x 3 x 2 x C5x dx lim n→ ni115 3 in 3 n lim 4 5n→ n2 n(n 1)2 45 lim n→ 2 n2(n 2 n) lim 4 5n2n→ 2n2 lim 4 5nn→ 2n2 4 52 0 4 522. 5(x 1)dx 5(x 1)dx 1(x 1)dx1 lim n→ ni1 lim n→ n5 5 n0 5 in 1 n5 lim110n→ n ni1i 1 1n 1 1 5 2 n 1 . . . 5 n1lim n→n1 n 1 2 n 1 . . . n n 1 lim n→ n5 n n5 (1 2 . . . n) lim n→ n1 n n1 (1 2 . . . n) 1)5 lim n5n→n n n(n 2 1 lim n1n→n n n(n 25 lim n→n 7 2 n 5 2 1 lim n→n 3 2 n 1 2 lim n→ 3 5 252 2 n 3 52 0 3 2 0 3 22or 16 lim n→ 3 2 1 2n n 1 1) 3. 2(x 2 4x 4)dx0 lim n→ ni1 lim 2 in2 4 2 in 4 2n 2n→n 2 1n 2 4 2 1n 4 2 2n 2 2 4 . . . 2 3n 2 4 2 3n 44 2 n lim (1 2 2 2 . . . n 2 )8 (1n 2 . . . n) 4n2n→n4n 2 lim2n→ n4n 2 lim n(n 1) (2n 1)6n→ 8n3 2n3 3n 2 n6 lim n→ 4 3 3 12 n n 2 8 3 8 8 5 634. 2000(6 0.002x)dx1500 lim n→ ni1 lim n→ 5 n n8 n(n 2 1 n62 n2 n2 1) 4n 8 81 n1 86 0.0021500 50 0in 5 00n 00 ni13 ni 00 3 n1 3 n2 . . . 3 n n lim 5n→n lim 5 00 3n1n→n (1 n 2 . . . n) lim 1500 5n→ n0 0 2 n(n 1)2 lim n→1500 2501 n1 1500 250 or $1250Lesson 15-4Page A551. x 6 dx 1 7 x 7 C2. 5x 4 dx 5 1 5 x 5 C x 5 C3. (x 2 x 5)dx 1 3 x 3 1 2 x 2 5x C4. (4x 4 x 2 6)dx 4 1 5 x 5 1 3 x 3 6x C5. 2 14x 6 dx 14 1 7 x 7 222 4 5 x 5 1 3 x 3 6x C 2x 7 2 2 (2 2 7 ) (2 (2) 7 ) 256 (256) 512541 Extra Practice
6. 60(x 2)dx 1 2 x 2 2x 6 0 1 2 6 2 2 6 1 2 0 2 2 0 30 0 307. 4(x 2 4)dx 1 3 x 3 4x 4 22 1 3 4 3 4 4 1 3 2 3 4 2 1 63 1 63 3 238. 5(x 4)(x 2)dx4 5(x 2 2x 8)dx4 1 3 x 3 2 1 2 x 2 8x 5 4 1 3 x 3 x 2 8x 5 4 1 3 5 3 5 2 8 5 1 3 4 3 4 2 8 4 7 03 8 03 1 03Extra Practice 542
Chapter TestsChapter 1 TestPage A561. D [1, 0, 2, 3}; R {2, 4, 5}; yes2. D {5, 4, 6, 7}; R {3, 2, 0, 2, 7}; no3. f(4) 4 3 4 2 4 48 or 444. f(7) 7 3(7) 27 147 or 1545. f(a 2) a 2 3(a 2) 2 a 2 3(a 2 4a 4) a 2 3a 2 12a 123a 2 11a 106a. E 4Pd 21200 4 ( 0.9) 2 117.89 Im/m 26b. d cannot be negative or zero.7. f(x) g(x) x 2 7 x 3 x 2 x 4f(x) g(x) x 2 7 (x 3) x 2 x 10f(x) g(x) (x 2 7)(x 3) x 3 3x 2 7x 21f( x) g ( x) x 2 7x 38. [f g](x) f(g(x)) f(4x 5) 4x 5 1 4x 4[g f ](x) g(f(x)) g(x 1) 4(x 1) 5 4x 4 5 4x 19. [f g](x) f(g(x)) f(2x 2 6) 5(2x 2 6) 10x 2 30[g f ](x) g(f(x)) g(5x) 2(5x) 2 6 50x 2 610. y11.y12. 13.y16141210y 4x 12 8642O54321 x2 123 45414. y 3 5 3 (x (1))y 3 5 3 (x 1)y 3 5 3 x 5 3 y 5 3 x 1 4315. m 816. parallel: y 2 4(x 0)y 2 4x4x y 2 0perpendicular: y 2 1 4 (x 0)y 2 1 4 x4y 8 xx 4y 8 017. x 5y 3 ⇒ m 1 5 parallel: y 2 1 5 (x (1))yy |x| 2O2 4 0 68or 3 4 y 4 3 4 (x 0)y 4 3 4 xy 2 1 5 (x 1)5y 10 x 1x 5y 11 0perpendicular: y 2 5 1 (x (1))y 2 5(x 1)y 2 5x 55x y 3 018. g(x)19. f(x)g(x) x 1Oxy 3 4 x 4OxxOxy 3x 6Ox2x y 1543 Chapter Tests
20a.20b. Sample answer: Using (47, 67) and (95, 88),y 0.44x 46.44m 8 8 6795 47 0.4375y 67 0.4375(x 47)y 67 0.4375x 20.5625y 0.4375x 46.4375y 0.44x 46.44Chapter 2 TestPage A571.EconomicsGrade10080604020y2x y 10O2. 3x y 7 → 6x 2y 145x 2y 12 5x 2y 12x 23x y 73(2) y 7y 1(2, 1)3. 4x 5y 2 → 12x 15y 63x 2y 13 12x 8y 524x 5y 24x 5(2) 24x 12x 3(3, 2)20 40 60 80 100Statistics Grade(1, 3)x y 4x23y 46y 24. 4x 6y 3z 20 → 4x 6y 3z 20x 5y z 15 3x 15y 3z 457x 9y 25x 5y z 15 → 2x 10y 2z 307x y 2z 1 7x y 2z 19x 11y 317x 9y 25 → 63x 81y 2259x 11y 31 63x 77y 2174y 8y 27x 9y 25x 5y z 157x 9(2) 251 5(2) z 157x 711 z 15x 1z 4( 1, 2, 4)5. y 12 2x 3x 4 2y3x 4 2(12 2x)3x 4 24 4x7x 28x 4y 12 2(4) 4(4, 4)6. A B 5 (2) 1 3 4 (3) 2 07 4 1 2 7. 3C 3(1) 3(0) 3 (3)3 (3) 3(2) 3(4) 3 0 99 6 12 8. 2B 2(2) 2(3)2(3) 2(0)4 6602B A 4 (5)6 4 1 510 2 6 1 0 22 39. BC 1 0 33 0 32(1) 3(3)3(1) 0(3)11 6 18 3 0 9242(0) 3(2)3(0) 0(2)10. 1.5 1 4 6 1.5 6 93 3 2 4.5 4.5 3A(1.5, 4.5), B(6, 4.5), and C(9, 3)y8B 64CB 2O C642 A 2 4 6 8 10x46 A2(3) 3(4)3(3) 0(4) Chapter Tests 544
11. 1 0 3 1 6 3 3 1 6 3 0F(3, 2), G(1, 5), H(6, 1), J(3, 5)HH5312. 5(6) 3(7)210 913. 2(9) (10)(4)12 5814. 1 0 1 2 3 1 (1) 3 03 0 1 41761 1(1) 2(10) 1(3) 161 3 12 1 2115.3 1 5 2 1 1 31 3 4 4 2J49J121242yO4351GG16. 111FF2x54 3 5 1 5 1 5 2 5 0244 3 1 5 2 53 2 5 1 0225117. The inverse does not exist since 0.15 122 118. x 71 2 1 3 1313 21y133121 1 5 5 1 5 1 1 2 1 x 1 5 1 1 73 2 3 1 y 323x 2 y3445119.1y 2x 3y(2, 4)(6, 0) y 0 Oy 3x 4x 2(2, 2)( 4 3, 0)xf(x, y) 2x yf(6, 0) 2(6) 0 or 12 c minimumf(2, 4) 2(2) 4 or 0f(2, 2) 2(2) 2 or 2f 4 3 , 0 2 4 3 0 or 8 3 c maximum20. Let x number of ads.Let y number of commercial minutes.x 3yy 4100x 200y 1300100x 200y 1300(3, 5)y 4(5, 4)(3, 4)f(x, y) 12,000x 16,000yf(3, 5) 12,000(3) 16,000(5) 116,000f(3, 4) 12,000(3) 16,000(4) 100,000f(5, 4) 12,000(5) 16,000(4) 124,000The company reaches the most people with 5 adsand 4 commercial minutes.Chapter 3 TestPage A581. y 2x 1 → b 2a 1x-axis b 2a 1b 2a 1y-axis b 2(a) 1b 2a 1y x a 2b 1b a 12noy x a 2(b) 1a 2b 1b a 12nonone of theseOx 3nonox545 Chapter Tests
222. y x 2 → b a 2x-axis b a22b a22 no2y-axis b (a ) 2b a22 yes2y x a b 29. f(x) 5x 4y 5x 4x 5y 4x 4 5yy x 45f 1 (x) 1 5 x 4 5 Yes, it is a function.f(x)f 1 (x) 1 4x 5 5Oxf(x) 5x 42b ano2y x a ( b) 22a b 2b 2 ano; y-axis3. x y 2 3 → a b 2 3x-axis a (b) 2 3a b 2 3 yesy-axis a b 2 3a b 2 3 noy x b a 2 3a b 3 noy x b (a) 2 3b a 2 3a b 3 no; x-axis4. xy 5 → ab 5x-axis a(b) 5ab 5 noy-axis (a)b 5ab 5 noy x ba 5ab 5 yesy x (b)(a) 5ab 5 yesy x, y x5. The graph of g(x) is the graph of f(x) translatedleft 3 units and reflected over the x-axis.6. The graph of g(x) is the graph of f(x) stretchedvertically by a factor of 4 and then translateddown 2 units.7. yOy |x 4|x10. f(x) x 3y x 3x y 3y 2 3 x 222y 3 x 2f 1 (x) 3 x 2; Yes, it is a function.2211. f(2) 2 2 4 0 No; the function is undefined when x 2.12. Yes; the function is defined when x 0, thefunction approaches 1 as x approaches 0 from bothsides, and f(0) 1.13. y → as x → , y → as x → 14. At x 4, y is at a minimumx yvalue, since f(4) f(3) and3 11f(4) f(5).4 125 1115. At x 1, y is at a point ofx yinflection, since0 1f(0) f(1) f(2).1 22 316. x 1 y x4as x → , y → 4; y 417. x 2 4 0(x 2)(x 2) 0x 2y y y x 2y x 1 4 xx x x 1 x 41 1 x x 4xx 2 x 2 4x2 x 28.50y45403530252015105y 2x 2 35432 O 1 2 3 4 5xas x → , y → 0; y 0y 1 x 41 x 2Chapter Tests 546
18a. y 0 0.y y . 25 0.004x25 0.001x 0. 25 x 0.0 04xx 0. 25 x 0.0 01xx. 004as x → , y → 0 0.001or 4; y 418b. As the amount of 4 molar solution addedincreases, the molarity of the mixtureapproaches 4.19. y kx y 0.25x0.5 k(2) y 0.25(10)0.25 k y 2.5k20. y x 2y 72x2k8 (318 ) 2 72x272 k x 2 4x 2Chapter 4 TestPage A591. (x 4)(x i)(x (i) (x 4)(x i)(x i) (x 4)(x 2 i 2 ) (x 4)(x 2 1) x 3 4x 2 x 42. b 2 4ac (5) 2 4(1)(4) 9Since b 2 4ac 0, there are 2 real roots.b bacn 2 42an (5 ) 92(1)3n 5 2 0. 25 x 0.004 0. 25 x 0.0013n 5 or n 5 22n 4 n 13. b 2 4ac (7) 2 4(1)(3) 61Since b 2 4ac 0, there are 2 real roots.b bacz 2 42az (7 ) 61 2(1)61 3z 7 24. b 2 4ac (5) 2 4(2)(4)7Since b 2 4ac 0, there are 2 imaginary roots.b bacx 2 4 2ax (5 ) 7 2(2)7ix 5 45. 2⏐ 2 3 3 44 2 102 1 5 ⏐ 62x 2 x 5, R66. 1⏐ 1 5 13 53 601 6 7 601 6 7 60 ⏐ 0x 3 6x 2 7x 607. f(x) x 3 8x 2 2x 11f(2) (2) 3 8(2) 2 2(2) 118 32 4 11 9; no8. f(x) 4x 4 2x 2 x 3f(1) 4(1) 4 2(1) 2 1 3 4 2 2 0; yes9. 1 positivef(x) 6x 3 11x 2 3x 22 or 0 negativer 6 11 3 2 1 2 6 14 4 0 1 3 6 9 6 02 6 1 1 0rational zeros: 2, 1 3 , 1 2 10. 1 positivef(x) x 4 x 3 9x 2 17x 83 or 1 negativer 1 1 9 17 88 1 9 63 487 38888 1 7 47 393 3136 1 1 2 7 24 321 1 0 9 8 0rational zero: 111. Use the TABLE function of a graphing calculator;0.8, 3.8.12. Use the TABLE function of a graphing calculator:1.3.13. Sample answer: 2; 5r 1 3 5 91 1 4 1 102 1 5 5 1upper bound: 2f(x) x 3 3x 2 5x 9r 1 3 5 91 1 2 7 22 1 1 7 53 1 0 5 64 1 1 1 55 1 2 5 34lower bound: 5547 Chapter Tests
14. Sample answer: 1; 219. 2x 2 3x 52x 2 3x 5r 2 3 1 1 17 x1 2 5 4 5 6Check: 2(7) 2 3(7) 516 16 upper bound: 1f(x) 2x 4 3x 3 x 2 x 120. 11 0m 1 911 10m 81r 2 3 1 1 110m 701 2 1 2 3 2m 711 10m 02 2 1 1 1 310m 1111 lower bound: 2m 1 01 1 115. 8 0 a 1 0Test m 8: 11 0(8) 1 ? 91 7 91 ? 9a 8 09.54 ? 9 truea 8 07Test m 0: 11 0(0) 1 ? 94 316. x 2 x 2 1 4 4 11 ? 943 x 2 (x 2) (x 2) 1 3.32 ? 9 false4 <strong>Solution</strong>: m 74(x 2) 3 (x 2) (x 2)5z 115z 1121. 42z 2 z 6 (z 2) (2z 3)16(x 2) 12 (x 2)(x 2)5z 11A B16x 32 12 x 2 42z 2 z 6 z 2 2z 3x 2 16x 16 05z 11 A(2z 3) B(z 2)16 16 x 2 6)4(1)(1Let z 3 2 .2(1)x 16 835 3 2 11 A2 3 2 3 B 3 2 22x 8 43 7 2 7 2 B517. x 2 5 x 2 1 B3 xLet z 2.5(2) 11 A(2 (2) 3) B(2 2)21 7A3 A15x 15(x 2) 2(x 2)15x 15x 30 2x 42x 34x 17; x 2 or 05Test x 18: 18 2 ? 5 18 3(218)5 1 6 ? 5 1 1 8 2 70.3125 ? 0.3148 true5Test x 3: 3 2 ? 5 2 3 3( 3)5 ? 5 3 2 9 5 ? 1 8 9 false5Test x 1: 1 2 ? 5 2 1 3( 1)5 ? 5 2 3 5 ? 5 2 3 true5Test x 1: 1 ? 5 1 2 3(2 5 3 ? 5 2 3 1 2 3 ? 5 2 3 falsex 17, 2 x 018. y 2 3 0 Check: 11 2 3 0y 2 3 9 3 0y 2 9 3 3 0y 110 0 1)5z 113 1 2z 2 z 6 z 2 2z 37x2 18x 122. 7x( x22 18x 1 1)(x 2)(x 1)(x 1)(x 2)7x2 18x 1 A B C ( x2 1)(x 2)x 1 x 2 x 17x 2 18x 1 A(x 1)(x 2) B(x 1)(x 1) C(x 1)(x 2)Let x 1.7(1) 2 18(1) 1 A(1 1)(1 2) B(1 1)(1 1) C(1 1)(1 2)24 6C4 CLet x 2.7(2) 2 18(2) 1 A(2 1)(2 2) B(2 1)(2 1) C(2 1)(2 2)9 3B3 BLet x 1.7(1) 2 18(1) 1 A(1 1)(1 2) B(1 1)(1 1) C(1 1)(1 2)12 2A6 A7x2 18x 1 6 3 4 ( x2 1)(x 2)x 1 x 2 x 1Chapter Tests 548
23. quadratic24. Let x the height.Then 6x the length and 6x 7 the width.V x(6x)(6x 7)120 36x 3 42x 20 6x 3 7x 2 20Use a graphing calculator to graph the relatedfunction.V(x) 6x 3 7x 2 20It appears as if the zero is at x 2.Use the Factor Theorem to checkV(2) 6(2) 3 7(2) 2 20 0.The height is 2 cm, the length is 6(2) or 12 cm andthe width is 12 7 or 5 cm.12 cm by 5 cm by 2 cm25. Let s the speed of the freight train.Then 30 s the speed of the car.500Car: 500 (30 s)t ⇒ t 30 sTrain: 350 st ⇒ t 35 0s500 30 s 35 0s500 s 10,500 350s150s 10,500s 70 km/hChapter 5 TestPage A601. 9 95°360° 2.76a 360°(2) 995°a 720° 995°x 275°; IV2. 234°360°3. 4 3 0.65a 360°(1) 234°a 360° 234°a 126°; II10°60° 1.14a 360°(1) 410°a 360° 410°a 50°; I4. 1245°360° 3.46a 360° (4) 1245°a 1440° 1245°a 195°; III5. (RS) 2 (ST ) 2 (RT ) 2(RS) 2 12 2 15 2RS 2 225 144RS 81 or 9 in.sin R s ideoppositehypotenusecos R s idhysin R 1 215or 4 5 9cos R 1 5sideoppositehytan R scsc R sideadjacent2tan R 1 9or 4 3 csc R 1 1hypotenusesec R s ideadjacentcot R s ssec R 1 59or 5 3 9cos R 16. tan 60° y x tan 60° 3 or 317. sec 270° xr sec 270° 1 0 ; undefined8. sin (405°) sin (45°)sin (45°) ysin (45°) 229. r x 2 y 2 r 3 2 5 2 r 34e adjacentpotenuse or 3 5 potenuseideopposite52or 5 4 ideadjacentideopposite2or 3 4 sin v y r cos v x r tan v y x 5sin v 34 3cos v 34 tan v 5 3 sin v 5 34 cos v 3343434csc v yr sec v xr cot v x y csc v 34 5 sec v 310. r x 2 y2r (4) 2 2 2r 20 or 2534 cot v 3 5 sin v y r cos v x r tan v y x 242sin v 2cos v 52tan v 5 4sin v 5 cos v 2 5 tan v 1 552 r rcsc v y sec v x cot v x y csc v 2 525 sec v cot v 4 242csc v 5 sec v 52cot v 211. r x 2 y r 03) 2 ( 2r 9 or 3sin v y r cos v x r tan v y x sin v 33cos v 0 3 tan v 30sin v 1 cos v 0 undefinedrcsc v yrsec v x cot v x y 3csc v 3sec v 3 0 0cot v csc v 1 undefined cot v 03549 Chapter Tests
12. cos A b c cos ° 4 c2 42c cos 77°c 186.7b13. tan B abtan 27° 1 3b 13 tan 27°b 6.614. sin A a c asin 32° 17 1 4 a 14 sin 32° 17°a 7.515. cos B a c cos B 2 337 B cos 1 2 337 B 51.6°16. tan A a b tan A 131 A tan 1 131 A 15.3°17. Let h the height.hsin 70° 6 5 h 65 sin 70°h 61.1 m18. B 180° 36° 87° or 57°K 1 2 a 2 sin B sin CsinAK 1 2 (24) 2 sin 5 sinK 410.4 units 219. K 1 2 bc sin A7°sin87°36°K 1 2 (56.4)(92.5) sin 58.4°K 2221.7 units 220. 180° 36°2 72° sin 36°22 sin 72°xx sin 36° 22 sin 72°s 22 sin72°sin36°x 35.6 cmPerimeter 22 35.6 35.6 93.2 cm65 mh70˚36˚x x72˚22 cm21. Since 98° 90°, consider Case II.c a; one solution sin 98°90 si n A6464 sin 98° 90 sin AA sin 1 64 s in 98°90A 44.8°B 180° 98° 44.8° or 37.2° sin 37.2b sin 98°9090 sin 37.2° b sin 98°b 90 sin37.2°sin98°b 54.9A 44.8°, B 37.2°, b 54.922. Since 31° 90°, consider Case I.a b sin A9 20 sin 31°9 10.3; none23. a 2 b 2 c 2 2bc cos A13 2 7 2 15 2 2(7)(15) cos A169 274 210 cos A105 210 cos AA cos 1 1 05210A 60° sin 60°13 sin B77 sin 60° 13 sin BB sin 1 7si n 60°13B 27.8°C 180° 60° 27.8° or 92.2°A 60°, B 27.8°, C 92.2°24. b 2 a 2 c 2 2ac cos Bb 20 2 2(20)(24 2 24) s co47°b 17.92432912 sin 47°b si n C2424 sin 47° b sin CC sin 1 24 si n47°bC 78.3°A 180° 47° 78.3° or 54.7°b 17.9, C 78.3°, A 54.7°25. Let x the distance between the transmitters.70 miles 130 miles130˚xx 2 70° 130 2 2(70)(130) cos 130°x 21,80 0 8,200 130°cos x 183.0 milesChapter Tests 550
Chapter 6 TestPage A611. 225° 225° 18v 85.2 cm/s6. y7.1y sin x 5 42. 480° 480° 18 8 321O20°0°3. reference angle: 5 6 6 ; Quadrant 2sin 5 6 1 2 4. reference angle: 5 4 4 ; Quadrant 3tan 5 4 15. v r v t v 12 7 .11 8. ⏐3⏐ 3; 2 4 or 2 2 3 29. ⏐2⏐ 2; or 4 3 x10. ⏐A⏐ 4 2 k 4A 4 k 2 4 or 0.5y 4 sin 0.5v 211. ⏐A⏐ 0.5 2 k 2 c 4 4 A 0.52k or 4 c y 0.5 cos (4v ) 112. 13.y2 3O3y 3 cos 22 2 21O21yy cos x2xy5y tan (2 )44321 O 14 4 2234514. y Arccsc xx Arccsc yCsc x y or y Csc x15. y tan xx tan yarctan x y or y arctan x16. sin Arccos 1 2 sin 60° 3217. tan sin 1 1 2 tan 6 18. v r v t 2300 240,000 2 t ,000(2) tan 7 61 or 33 3t 240 2300t 656 hours or 27.3 days19. ⏐A⏐ 73 212h 73 21 2 2k 12⏐A⏐ 26 h 47 k 212A 26y 26 sin 6 t c 4721 26 sin 6 1 c 47 c c26 26 sin 6 1 sin 6 sin 1 (1) 6 c 2 6 c 4 6 c 2 3y2y Arccsc xOy2 O22 cy Csc x222xy arctan xxy tan xk 6 Sample answer: y 26 sin 6 t 2 3 47551 Chapter Tests
20. 6 5 mi1 ho lesur 52 80feet1 hour1 mile 360 0 seconds 95.3 ft/s8. s ecx sinxsinx c osx cot xv2tan v sec x cos x sin 2 xr gv tan 1 ( 95.3) 2 sin x cos x cot x 1 2001 sin2x(32) s in x cosx cot xv 0.23 radians c osxsinx cot xcot x cot xcos x cos x9. Chapter 7 Test1 sin x 1 sin xcos x cos x sin x cos x sin x cos x1 sin 2 x2 cosxPage A621 sin2 x1. sin 2 v cos 2 v 1 2 cosx 1 3 2 cos2x cos 2 v 1cos 2 v 8 9 cos v 2 2sec B10. csc (A B) 3sin A cos A tan B2. tan 2 v 1 sec 2 1vtan 2 v 1 (2) 2csc (A B) co s Btan 2 sinv 3sin A cos A Bc osB1tan v 3csc (A B) 3. sin 2 v cos 2 sin A cos B cos A sin Bv 1 4 5 2 1 cos 2 csc (A B) v 1sin (A B)cos 2 9csc (A B) csc (A B)v 2 511. cot 2v cos v 3 15 2 cot v 1 2 tan vcosv sinvcos 2v 12 sinv 2 cosvsec v co s v1sec v or 5 cot 2v co s2v sin2v3 2 sinv cosv 3 5 cot 2v c os2vsin2v14. sin v cs c vsin v 1 5 3 or 3 5 sin 2 v cos 2 v 1 3 5 2 cos 2 v 1cos 2 v 1 625 cos v 4 5 sin( 45. tan (420°) c20°)os( 420°)sin (360° 60°)cos (360° 60°) sin60°cos60°tan 60°6. tan v(cot v tan v) sec 2 vtan v cot v tan 2 v sec 2 v1 tan 2 v sec 2 vsec 2 v sec 2 v7. sin 2 A cos 2 A (1 sin A)(1 sin A)sin 2 A c os2Asin2 A (1 sin A)(1 sin A)cos 2 A (1 sin A)(1 sin A)1 sin 2 A (1 sin A)(1 sin A)(1 sin A)(1 sin A) (1 sin A)(1 sin A) 2 sec x 2 sec x 2 sec x 2 sec x2 co s x 2 sec x2 sec x 2 sec xcot 2v cot 2v12. sin 255° sin (225° 30°) sin 225° cos 30° cos 225° sin 30° 2 62 3 22 2 1 2 24 2 6413. tan 5 12 tan 6 4 tan 6 tan 4 1 tan 6 tan 4 13 11 13111 1 3 3111 1 3 321 13 3 1 1 3 1 3 2 4 3 2 33 2 3Chapter Tests 552
s14. sin 2 v cos 2 v 1 tan v csin 2 v 3 4 2 1sin 2 7v 1 6sin v 4sin 2v 2 sin v cos v 2 74 3 78 3 4 cos 2v cos 2 v sin 2 v 3 4 2 7 24 9 7 1 6 1 62 1 6or 1 8 2 tanvtan 2v 1 tan2 v2 73 7 2 1 3 2 73 2 9 3715. cos (22.5°) cos 4 275° 1 c os 45°2 1 22 2 2 2 2 4 2 216. tan 2 x 3 tan xtan 2 x 3 tan x 0tan x(tan x 3) 0tan x 0ortan x 3 0x 0° tan x 3x 60°17. cos 2x cos x 02 cos 2 x cos x 1 0(2 cos x 1)(cos x 1) 02 cos x 1 0orcos x 1 0cos x 1 2 cos x 1x 120° x 0°18. sin x cos x 0sin x cos xsinx c osx 1tan x 1x 45° or x 225°19. 2 cos 2 x 3 sin x 32(1 sin 2 x) 3 sin x 3 02 sin 2 x 3 sin x 1 0(2 sin x 1)(sin x 1) 02 sin x 1 0 or sin x 1 0sin x 1 2 sin x 1x 30°, 150° x 90°invosv 74 3 4 7320. y x 3x y 3 0AB 2 2 (1) 2 1 2 or 21 1 3 x 2 y 2 02 2x 2y 3 2 02 2 2sin f 2, cos f 2, p 3 2; Quadrant II22 2 22tan f or 1 22f tan 1 (1) 135°21. 5 5y 10x0 10x 5y 5A B 2 2 10 2 (5) 2 or 55105 5 5x 55y 5 5 05 2 5x 5y 5 05 5 5sin f 5, cos f 2 5, p 5; Quadrant IV555 55tan f or 1 2 2 55f tan 1 1 2 2 333°22. 2x y 6 → 2x y 6 0d Ax 1 By 1 CAB2 2d 2( 5) 1(8) 6212 28d or 8 55 5 8 5523. d A x 1 By1 CA2 B 2d 3( 6) (8) 2344 2 2d 1 65 1 6524. 5x 2y 7 → 5x 2y 7 0y 3 4 x 1 → 3x 4y 4 0d 1 5x 1 2y 1 7 d52 222 3 x 1 4y 1 432 4 2 5x 1 2y1 7 3x 1 4y 1 492525x 1 10y 1 35 329x 1 429y 1 429(25 329)x (10 429)y 35 429 025. R v 0 2 gsin 2vR v 0 2 g 2 sin v cos v82R 8 32R 232.32 ft 2 3 5 4 5 553 Chapter Tests
Chapter 8 TestPage A631. 2.5 cm, 60° 2. 1.6 cm, 25°3.a3.9 cm46˚b3.9 cm, 46°4.a334˚1.8 cma b2b a2b1.8 cm, 334°u5. AB 1 3, 9 6 4, 3⏐ABu ⏐ (4) 2(3) 2 25 or 56.uAB 3 (2), 10 7 5, 3⏐ABu ⏐ 5 2 3 2 347.uAB 9 2, 3 (4), 7 5 7, 1, 2u⏐AB ⏐ 7 2 1 2 2 2 54 or 36u8. AB 8 (4), 10 (8), 2 (2) 4, 2, 4u⏐AB ⏐ (4) 2) (2 2 4 2 36 or 69. u r u s 1 4, 3 3, 4 (6) 5, 0, 1010. 3 u s 3 4, 3 3, 3 6 12, 9, 182 u r 2 (1), 2 3, 2 4 2, 6, 83 u s 2 u r 12 (2), 9 6, 18 8 14, 3, 2611. u r 3 u s 1 12, 3 9, 4 (18) 11, 12, 1412. ⏐ u r ⏐ (1) 2 3 2 4 2 2613. ⏐ u s ⏐ 4 2 3 2 6) (2 6114. u r i u 3 u j 4 u k15. u s 4 u i 3 u j 6 u k16. u r u s 1(4) 3(3) 4(6)193 4 1 43 6 4 630 u i 10 u j 15 u k 30, 10, 1518. No; since u r u s 19 and not 0.19.125˚17. u r u s u i u j 1 3 u k60 lb165˚x40˚55˚125 lbx 2 125 2 60 2 2(125)(60) cos 55°x 19,22 5 5,000 15°cos 5x 103.06 lb s in5103 5°.06 si n v6 0103.06 sin v 60 sin 55°v sin 1 60 sin55°103.06v 28.48°v 40° 28.48° 40° or 68.48°20. x x 1 ta 1 y y 1 ta 2x 3 t(2)y 11 t(5)x 2t 3 y 5t 1121. x 2t 3 y t 1 x 32 ty 1 x 23y 1 ty 1 2 x 1 2 22. v x 100 cos 2° v y 100 sin 2°v x 99.94 mph v y 3.49 mph23. The figure is four times the original size andreflected over the yz-plane.u24. AB 1.5 cos 60°, 0, 1.5 sin 60° 0.75, 0, 0.753uF 0, 0, 110110 lbu u uT AB Fuiu juk 0.75 0 0.7530 0 110 0 0.753 u i 0.75 0.753 u j 0.75 0 u k0 110 0i u 82.5j u 0k u⏐T u ⏐ 02.5 2 8 2 0 2 82.5 lb-ft01101.5 ft460˚30 0Chapter Tests 554
25. y t⏐v u ⏐ sin v 1 2 gt 2 x t⏐v u ⏐ cos vy 28t sin 35° 1 2 (32)t 2 x 28t cos 35°0 4t(7 sin 35° 4t) x 23.02 feet4t 0 or 7 sin 35° 4t 0t 0t 7s in 35°4t 1.003758764Chapter 9 TestPage A641. 90˚ 2.120˚ 60˚150˚180˚A210˚240˚270˚1 2 3 4300˚30˚0˚330˚3.2 2 334.5667643321 2 3 4C5301165.2 2 336.5667.76150˚180˚43120˚210˚240˚321 2 3 490˚270˚5360˚01161 2 3 4300˚30˚0˚330˚56765676150˚180˚23432343120˚210˚240˚232B31 2 3 42328. r 2 2 2 2 v Arctan 2 2 8 or 22 4 22, 4 9. r (6) 2 0 2 36 or 6Since x 0 and y 0, v .(6, )5331 2 3 490˚270˚5360˚60116601162 4 6 8300˚30˚0˚330˚10. r (2) 2 ) (3 2 v Arctan 2 13 4.12 3.61(3.61, 4.12)11. x r cos v y r sin v 3 cos 5 4 3 sin 5 4 3 2212. x r cos v y r sin v 2 cos 7 6 2 sin 7 6 2 32 1 22 31(3, 1)13. x r cos v y r sin v4 cos 1.4 4 sin 1.4 0.68 3.94(0.68, 3.94)14. y 3r sin v 33r s in v15. x 2 y 2 3xr 2 3r cos vr 3 cos vr 3 csc v16. r 7r 2 49x 2 y 2 4917. 5 r cos (v 45°)0 r cos (v 45°) 50 r (cos v cos 45° sin v sin 45°) 50 2r cos v 2r sin v 5220 2x 2y 52 20 2x 2y 1018. AB 2 2 5 2 3345 34 3x 34 3y 34 0cos f 5 34 , sin f 3 34 , p 334343434f Arctan 35 180° 211°p r cos (v f) 3 34 34 r cos (v 211°)19. AB 2 2 24) 2 ( 2 20 or 254 1x 2y 2 0 2255cos f 5, sin f 2 5, p 555 10f Arctan 21 360° 297°p r cos (v f) 5 r cos (v 297°)1020. i 93 (i 4 ) 23 i 1 23 i i 3 2 3 2 3 222 3 2, 3 22 2 52 3555 Chapter Tests
21. (2 5i) (2 4i) (2 (2)) (5i 4i) 0 (i)i22. 6i (3 2i) 6i 3 2i 8i 323. (3 5i)(3 2i) 9 9i 10i 2 19 9i24. (1 3i)(2 i)(1 2i) (2 7i 3i 2 )(1 2i) (1 7i)(1 2i)1 9i 14i 2 13 9i25. 6 22 ii 6 22 ii 2 2 i i 12 10i 2i24 i2 10 10i5 2 2i26. r (4) 2 4 2 v Arctan 4 32 or 42 3 442cos 3 4 i sin 3 4 27. r (5) 2 0 2 v Arctan 5 25 or 5 5(cos i sin )28. r 4 3 v 3 2 4 12 6 4or 7 412cos 7 4 i sin 7 4 12 22340 i 22 62 62i 6 29. r 2 v 233 2 3 6 or 2 2cos 2 i sin 2 2(0 i) 2i30. r 11) 2 ( 2 v Arctan 11 2 2 7 4(1 i) 8 (2) 8 cos (8) 7 4 i sin (8) 7 4 16 (cos 14 i sin 14) 16 (1 0) 1631. r 027) 2 ( 2 v 2 729 or 27 3 27i (0 27i) 1 3 27 1 3 cos 1 3 2 i sin 1 3 2 3 cos 6 i sin 6 3 3 i2 1 2 3 3 3 2 2 i32. x 3 i 0 → x 3 iFind the cube roots of i.r 0 2 1 2 1 or 1v 2 (0 i) 1 3 1 cos 2 2n i sin 2 2n 1 3 1 1 5 cos 6 2n 3 cos 4n6 i sin 6 i sin 6 2n 3 4nx 1 cos 6 i sin 6 32 1 2 ix 2 cos 5 6 i sin 5 6 23 1 2 ix 3 cos 9 6 i sin 9 6 i31 i2 2133. E I Z 8 (cos 307° j sin 307°) 20 (cos 115° j sin 115°) 8 20 [cos (307° 115°) j sin (307° 115°)] 160 (cos 422° j sin 422°) 160 (cos 62° j sin 62°)Chapter 10 TestPage A651. d (x y 2 1 ) 2 (y 2 y 1 ) 2d 3 1)) ( 2 2) (1 2d 4 1) 2 ( 2d 17 x 1 x 22y 2 3, 2 12 , y 1 2 1 21, 3 2 2. d (x y 2 1 ) 2 (y 2 y 1 ) 2d (2k 3k) 2 1 (k (k 1)) 2 d (k) 2 ) (22d k 2 4 x 1 x 22, y 1 21O1y 2ii31 i2 21 3k 2 5 2 k, k2k, k 1 k 12 3. r (8 (6) 2 (3)) (42r (2) 2 (7) 2r 53(x h) 2 (y k) 2 r 2(x (8) 2 (y 3) 2 53 2(x 8) 2 (y 3) 2 53Chapter Tests 556
4. center: (h, k) (0, 0)a 2 10 b 2 6 c a 2 ba 10 b 6 c 10 6 or 2foci: (h, k c) (0, 2)major axis vertices: (h, k a) 0, 10minor axis vertices: (h b, k) (6, 0)c5. e a 1 2 (x h) 2 a 2 (y k) 2 1b 2If c 1 2 , then a 1. (x 11c a 2 b 2 34 1 2 1 2 b 2 x 2 4 y 2 3 1 1 4 1 b 20) 2(y 0) 2b 2 3 4 6. center: (h, k) (4, 2)a 2 16 b 2 7 b 2 c 2 a 2a 4 b 7 7 c 2 16 → c 23foci: (h, k c) 4, 2 23vertices: (h, k a) (4, 2 4) (4, 6) and (4, 2)asymptotes: y k a b (x h)4y 2 (x (4))7y 2 4 77(x 4)7. foci: (h, k c) ⇒ h 5 e 3 2 c ak c 4 b 2 c 2 a 2k c 2 b 2 3 2 2 22k 2 b 2 5k 1; c 3 (y k) 2 a 2 (x h) 2 1b 2 (y 1) 2 2 2 (x ( 5)) 25 (y 1) 2 (x 5) 245 1 18. vertex: (h, k) (0, 3)4p 8 ⇒ p 2focus: (h p, k) (0 2, 3) (2, 3)directrix: x h px 0 2x 2axis of symmetry: y ky 39. 2k p 5 foci: (h, k p) (3, 5)2k p 2directrix: y k p 22k p 7k 7 2 , p 3 2 (x h) 2 4p(y k)(x 3) 2 4 3 2 y 7 2 (x 3) 2 6y 7 2 10. A 1, C 0; since C 0, the conic is a parabola.x 2 6x 8y 7 0x 2 6x 9 8y 7 9(x 3) 2 8y 16(x 3) 2 8(y 2)11. A 4, C 1; since A and C have opposite signs,the conic is a hyperbola.4x 2 y 2 1x 2 1 4 y 21 1yOyO12. A 1, C 1; since A C, the conic is a circle.x 2 y 2 4x 12y 36 0x 2 4x 4 y 2 12y 36 4(x 2) 2 (y 6) 2 4yOxxx557 Chapter Tests
13. A 3, C 16; since A and C have oppositesigns, the conic is a hyperbola.3x 2 16y 2 18x 128y 37 03(x 2 6x 9) 16(y 2 8y 16) 37 27 2563(x 3) 2 16(y 4) 2 192 (y 4) 212 (x 3) 264y1412108642x64 2O22 4 6 8 1012 1446 114. A 9, C 1; since A and C have opposite signs,the conic is a hyperbola.9x 2 y 2 90x 8y 200 09(x 2 10x 25) (y 2 8y 16) 200 225 169(x 5) 2 (y 4) 2 9 (x 5) 21 (y 4) 29 116. A 0, C 1; since A 0, the conic is a parabola.y 2 2x 10y 27 0y 2 10y 25 2x 27 25(y 5) 2 2x 2(y 5) 2 2(x 1)O y17. A 1, C 2; since A and C have the same signand A C, the conic is an ellipse.x 2 2y 2 2x 12y 11 0(x 2 2x 1) 2(y 2 6y 9) 11 1 18(x 1) 2 2(y 3) 2 8yx (x 1) 28 (y 3) 24 1yOxO15. A 2, C 13; since A and C have oppositesigns, the conic is a hyperbola.2x 2 13y 2 5 02x 2 13y 2 5y 2 x 2 1 13 5 2 5y 251 3x 2 1 5 2 yOxx18. y 2x 2 x19. x 2 cos t y 2 sin tx y 2 cos t 2 sin tcos 2 t sin 2 t 1 2x 2 y2 2 1 x 24 y 24 1x 2 y 2 420. B 2 4AC 0 2 4(4)(1)16A C; ellipse4(x 1) 2 (y 3) 2 364(x 1 3) 2 (y 3 (5)) 2 364(x 2) 2 (y 2) 2 364x 2 16x 16 y 2 4y 4 364x 2 y 2 16x 4y 16 0Chapter Tests 558
21. B 2 4AC 0 2 4(2)(1) 8hyperbolaReplace x with x cos 60° y sin 60° or 1 2 x 3y.2Replace y with x sin 60° y cos 60° or 3x 1 2 2 y.2 1 4 (x) 2 322 1 2 x 32xy 3 4 (y) 2 y 2 3x 1 2 2 y 2 8 3 4 (x) 2 32xy 1 4 (y) 2 8 1 2 (x) 2 3xy 3 2 (y) 2 3 4 (x) 2 32xy 1 4 (y) 2 8(x) 2 63xy 5(y) 2 32 0(x) 2 63xy 5(y) 2 32 022. (x 1) 2 y 2 1 x 2 4y 2 4x 2 2x 1 y 2 1 x 2 4(x 2 2x) 4x 2 y 2 2x 0 x 2 4x 2 8x 4y 2 x 2 2x 3x 2 8x 4 0(3x 2)(x 2) 0x 2 3 or x 2If x 2 3 , y 2 3 2 2 3 0.9.If x 2, y (2) 2 ) 2(2 0.(2, 0), (0.7, 0.9)23. y25. x 2 y 2 90x 2 (2x 3) 2 9x 2 4x 2 12x 9 905x 2 12x 81 0(5x 27)(x 3) 0x 2 75If a person is walking northeast, the person willfirst hit the motion detector at (3, 9).Chapter 11 TestPage A66 or x 3 21. 343 2 3 3 343 7 2 493. ((2a) 3 ) 2 (2a) 61 (2 a) 61 2 6a 6641a 65. 3 27a 6 b 12 (3 3 ) 1 3 a 6 3 b 1 23 3a 2 b 46. m 1 2 n 2 3 m 3 6 n 4 6 6 m3 n 47.y164 1 4 2. 64 1 3 3y 2x 34. x 3 2 y 2 z 5 4 4 x 1 22y 8 z 2 047y 2 2 5 x 6 y 8 z 59 3 or 3 5y 2(3) 3 or 9Ox24. x 2 y 2 Dx Ey F 00 2 1 2 D(0) E(1) F 0⇒ E F 1(2) 2 3 2 D(2) E(3) F 0⇒ 2D 3E F 134 2 5 2 D(4) E(5) F 0⇒ 4D 5E F 414D 6E 2E 26 3E 3F 634D 5E F 41 11E 3F 6711E 3F 67 8E 64E 8E F 12D 3E F 138 F 1 2D 3(8) 7 13F 7 2D 4D 2x 2 y 2 2x 8 y 7 0x 2 2x 1 y 2 8y 16 7 1 16(x 1) 2 (y 4) 2 10center: (h, k) (1, 4)radius 108.9. 4 1 2 2 10. 1 6 3 21611. log 5 625 4 12. log 8 m 513. log x 32 5x 5 321 x 5 321 3 2 x 5 1 2 5 x 5 1 2 xOyOxx559 Chapter Tests
14. log 5 (2x) log 5 (3x 4)2x 3x 44 x15. 3.6 x 72.4x log 3.6 log 72.4x l og72.4log3.6x 3.343016. 6 x1 8 2x(x 1) log 6 (2 x) log 8x log 6 log 6 2 log 8 x log 8x log 6 x log 8 2 log 8 log 6x(log 6 log 8) 2 log 8 log 6x 2 log8 log6log6 log8og15og417. log 4 15 l l 1.953418. log 3 0.9375 log 0.9375log3 0.0587log319. log 81 3 log81 1 4 20. log 542 2.7340 21. ln 0.248 1.394322. antiln (1.9101) 0.148123. t ln 2kln2t 0 . 054t 12.84 yr24. Let x the original number of bacteria.3x xe k(6)3 e 6kln 3 6kk ln 36k 0.18310204818x xe 0.1831t8 e 0.1831tln 8 0.1831tn 8t 0ḷ 1831t 11.3568626 hours0.3568626(60) 21 minutes11 hours, 21 minutes25. Q(t) Qe tBC 1 10 6 (5 10 6 )e t3(2 10 6 ) 1 5 e 6 t10 6tln 0.2 6 10 6t ln 0.2(6 10 6 )t 9.66 10 6 sChapter 12 TestPage A671. d 4.5 2 or 2.57 2.5 9.5, 9.5 2.5 12, 12 2.5 14.5,14.5 2.5 179.5, 12, 14.5, 172. d 1 (6) or 5a 24 6 (24 1)5a 24 1093. 8 4 (5 1)d12 4d3 d4 3 1, 1 3 2, 2 3 54, 1, 2, 5, 84. 345 n 2 (2(12) (n 1)5)690 24n 5n 2 5n0 5n 2 19n 6900 (5n 69)(n 10)n 6 95or n 10Since there cannot be a fractional number ofterms, n 10.11 05. r or 2 5 1 4 1 2 5 2 5 2 1 25 2, 1 25 2 5 4 6 25 4, 6 25 2 5 8 31 25 2 1 25 4, 6 25 , 318256. 1 16r 511 1 6 r 4 r 1 2 16 1 2 8, 8 1 2 4, 4 1 2 216, 8, 4, 2, 157. r or 2 5 2 S 10 5 2 5 2 (2) 10 1 2 5 2 51 202 1 51 215 8. does not exist;limn→n3 33 n2 1 lim n→, lim n→3 3, and lim1n→ n 2 0,so the denominator approaches 3. As n approachesinfinity, the n term in the numerator makes thewhole numerator approach infinity, so the entirefraction has no limit.n9. lim n→ 2n 33 4 3n lim n→ 1 02 0 1 2 n n323 n12 n 3 4n3 n 3 2 nn 33 3nn3Chapter Tests 560
10. The general term is 3131n 2 n1 n 2 .for all n, so convergent11. The series is arithmetic, so it is divergent.12. Sample answer: 19 1 5kk13. Sample answer: k16 3 2 k114. (2a 3b) 5 (2a) 5 (3b) 0 5(2a) 4 (3b) 1 5 4(2 25 4 3(2a) 2 (3b) 33 2 15 4 3 2 (2a) 1 (3b) 44 3 2 1a) 3(3h) 21 32a 5 240a 4 b 720a 3 b 2 1080a 2 b 3 810ab 4 243b 510!15. 5!(1 0 5)!816. 4!(8 4)!5 4 3 2 1(2a) 0 (3b) 55 4 3 2 1 a 105 2 5 5 4 3 2 1 5 4 3 2 1 a 5 32 8064a 5!(3x) 84 (y) 48 7 6 5 4 3 2 14 3 2 1 4 3 2 1 (3x) 4 (y) 4 70(81x 4 )(y 4 ) 5670x 4 y 417. r (2) 2 2 2 or 22210 9 8 7 6 5 4 3 2 1v Arctan 2 or 3 42 2i 22cos 3 4 i sin 3 4 22 e i 3 418. z 0 2iz 1 3(2i) (2 i) 6i 2 i 2 5iz 2 3(2 5i) (2 i) 6 15i 2 i 8 14iz 3 3(8 14i) (2 i) 24 42i 2 i 26 41i19. Step 1: Verify that the formula is valid for n 1.1(1 1)(4 1 5)Since 2 1(2 1 1) 6 and 3 6,the formula is valid for n 1.Step 2: Assume that the formula is valid for n kand prove that it is valid for n k 1.2 3 4 5 6 7 . . . 2k(2k 1) k(k 1) (4k 5)32 3 4 5 6 7 . . . 2k(2k 1) 2(k 1)(2k 3) k(k 1) (4k 5)3 2(k 1)(2k 3) k(k 1) (4k 5)3 6(k 1) (2k 3)3k(k 1)(4k 5) 6(k 1)(2k 3) 3(k 1)(k(4k 5) 6(2k 3) 3(k 1)(4k2 17k 18)3(k 1)(k 2)(4k 9) 3 Evaluate the original formula for n k 1.(k 1)[(k 1) 1][4(k 1) 5]3The formula gives the same result as adding the(k 1) term directly. Thus, if the formula is validfor n k, it is also valid for n k 1. Since theformula is valid for n 1, it is also valid for n 2.Since it is valid for n 2, it is also valid for n 3,and so on, indefinitely. Thus, the formula is validfor all positive integral values of n.20. There are 4 groups of three months in a year.There are 4 10 or 40 groups of three months inten years. So, n 40. Since the interest iscompounded quarterly, the rate per period is 0. 4or 0.02. The common ratio r is 1.02.a 1 200(1.02) 204S n a 1 a1 rn1 r204 204(1.02)401 1.02S 40 S 40 $12,322.00Chapter 13 TestPage A686!1. P(6, 2) (6 2)! 307!2. P(7, 5) (7 25208!3. C(8, 3) (8 565!4. C(5, 4) (5 4)! 4! 5 4 3 2 11 4 3 2 1 55. Using the Basic Counting Principle,5 4 3 2 1 120.5!6. P(5, 3) (5 3)! 5 4 3 2 12 1 607. (7 1)! 6! 7208. C(3, 1) C(12, 8)6 5 4 3 2 14 3 2 15)!7 6 5 4 3 2 12 13!12! (3 1)! 1! (12 8)! 8! 3 2 1 12 11 10 9 8 7 6 5 4 3 2 12 1 1 4 3 2 1 8 7 6 5 4 3 2 1 14854!9. C(4, 2) C(6, 3) (4 3)! 3!8 7 6 5 4 3 2 15 4 3 2 1 3 2 16!2)! 2! (6 3)! 3! 4 3 2 12 1 2 1 120(k 1)(k 2)(4k 9)36 5 4 3 2 13 2 1 3 2 108 561 Chapter Tests
610. P(2 white) 1 0 5 9 4. Sample answer: 1.5, 4.5, 7.5, 10.5, 13.5, 16.5, 19.5 1 5. Sample answer:3 Number 1 1 4 or 3 of Tallies Frequency4 Absences 1 4 or 1 0–3 ||| 33 3 4 11. P(s) 1 4 P(f ) 1 P(s)odds 12. P(5 clubs or 5 hearts or 5 spades or 5 diamonds) P(5 clubs) P(5 hearts) P(5 spades) P(5 diamonds) 4 1 352 1 251 1 510 1 409 9 4 8 33 16 ,66036 3 6155 1 296or 4 32 13. P(sum of 8) P(sum of 4) 3514. P(3 odd digits) 1 2 1 2 1 2 1 8 15. P(all red or all blue) P(all red) P(all blue)3 2 1 2 1 1 1 5 4 1 0 1 2 11 2 20 2121 2 01 13016. P(ace or black card) P(ace) P(black card) P(black ace)4 5 2 2 652 522 2 852 7or 1 317. P(3’stogether⏐oddnumber)P(3’s together and odd number)P(odd number)5 1 818. P(both even⏐even product)P(both even and even product) P(even product)6 3 2 6or 1 319. P(no more than two heads) P(0 heads) P(1 head) P(2 heads) C(5, 0) 2 3 0 1 3 5 C(5, 1) 2 3 1 1 3 4 C(5, 2) 2 3 2 1 3 31 2 43 1040 2 43 2 4351 2 43or 1 781 7925,62520. C(7, 4) 4 5 4 1 5 3 11Chapter 14 Test or 0.114688Page A691. range 20 1 or 192. Sample answer: 33. Sample answer: 0, 3, 6, 9, 12, 15, 18, 21Chapter Tests 56217. X 8 0(6 16 12 7 . . . 8 9 7 9) 10.448. Order the data from least to greatest. Since thereare 80 terms, the median is the average of the40th term and the 41st term.M d 10 102or 109. Order the values from least to greatest. Sincethere are 15 terms, the median is the 15 12or8th term.M d 2.34410. Q 1 2.341; Q 3 2.34711. interquartile range Q 3 Q 1 2.347 2.341 0.006semi-interquartile range 0.0 06212.3–6 ||||| 66–9 |||||||||||| | 169–12 |||| |||| |||| |||| |||| 2512–15 |||| |||| |||| || 1715–18 |||| ||| 818–21 |||| 56.27242118Frequency 151296300 3 6 9 12 15 18 21Number of Absences 0.0032.338 2.340 2.342 2.344 2.346 2.348 2.350113. X 1(2.3345 2.338 . . . 2.350) 2.343933331MD 1(2.343933335 2.334 . . . 2.34393333 2.350) 0.002514. j (2.34393333 2.334) 2 . . . (2.34393333 2.350)215 0.003115. 68.3% of the data lie within 1 standard deviationof the mean.24 2.8 21.2 26.816. 90% corresponds to t 1.65.24 1.65(2.8) 19.3828.62
17. 24 18.4 5.6 32.4 24 8.4tj 5.6 tj 8.4t(2.8) 5.6 t(2.8) 8.4t 2 t 3 95 .5% 2 47.75% 99 .7% 2 49.85%47.75% 49.85% 97.6%18. 29.6 24 5.6tj 5.6t(2.8) 5.6t 2 95 2j19. j X N3.6j X 4 00j X 0.18.5% 47.75%20. A5% level of confidence is given when P 95%.95% corresponds to t 1.96.X tj X 57 1.96(0.18) 56.6557.35Chapter 15 TestPage A701. The closer x is to 1, the closer y is to 1.So, lim f(x) 1. Also, f(1) 1.x→12. The closer x is to 2, the closer y is to 4.So, lim f(x) 4. However, there is a point at (2, 1).x→2So f(2) 1.3. lim x2 3xx→0x lim (x 3)x→0 34. lim x2 3x 2x→1x 1 lim (x 2)(x 1)x→1x 1 lim (x 2)x→115. lim x→36. lim7. f(x) lim(x 3)(x 3)(x 3)(x 2 3x 9)x 2 9 limx 3 27 x→3x 3 lim x→3 x 2 3x 93 33 2 3(3) 96 2 7or 2 9 x2 x 3x→1 3x 2 2 5 12 2 ) 33 (1) (1 2 52 2or 1 f(x h ) f(x)h→0h lim(x h) 2 2(x h) (x 2 2x)hh→0x lim2 2xh h 2 2x 2h x 2 2xhh→0 lim 2xh h 2 2hh→0h 12x 2 lim h→0(2x h 2) 2x 28. f(x) 1 2 x 2 7x 1f(x) 1 2 2x 21 7 1x 11 0 x 79. f(x) 4x 3 4f(x) 4 3x 31 010. f(x) 6x 4 2x 2 30f(x) 6 4x 41 2 2x 21 0 24x 3 4x11. f(x) 2x 5 4x 3 2 5 x 2 6f(x) 2 5x 51 4 3x 31 2 5 2x 21 0 10x 4 12x 2 4 5 x12. f(x) 2x 4 (x 3 3x 2 ) 2x 7 6x 6f(x) 2 7x 71 6 6x 61 14x 6 36x 513. f(x) (x 3) 2 x 2 6x 9f(x) 2x 21 6 1x 11 0 2x 614. f(x) 2x 62F(x) 1 x111 6x Cx 2 6x C15. f(x) x 3 4x 2 x 41F(x) 3x 31 4 2 x 21 1 1 x 11 4x C 1 1 4 x 4 4 3 x 3 1 2 x 2 4x C16. f(x) 1 2 x 3 2 7 x 5F(x) 1 2 1 3 x131 2 7 1 1 x111 5x C 1 8 x 4 2 1x42 5x C 1 8 x 4 1 7 x 2 5x C17. f(x) x3 4x 2 xx x 2 4x 11F(x) 2 x 21 4 1 x 11 x C1 1 3 x 3 2x 2 x C18. 2x 3 dx lim0 n→ n 2 in 3 n lim112 i 1 1 6n→ n4 n2 (n 1) 24 2n→4nn2 n 1 2 lim2 1 lim 41 n→ n n 2 4 units 219. 33x 2 dx 3 1 3 x 3 311 x 3 3 1 3 3 1 3 26 units 220. 1(2x 3)dx 2 1 2 x 2 3x 100 x 2 3x 1 0 [1 2 3(1)] [0 2 3(0)] 4 1563 Chapter Tests
21. 3(x 2 x 6) dx 1 3 x 3 1 2 x 2 6x 311 1 3 (3) 3 1 2 (3) 2 6(3) 1 3 (1) 3 1 2 (1) 2 6(1) 2 72 3 76 2 2322. (1 2x)dx x 2 2 x 2 C x x 2 C23. (3x 2 4x 7) 3 3 x 3 4 2 x 2 7x C x 3 2x 2 7x C24. h(t) 3 95t 16t 2h(t) v(t) 0 95 1t 11 16 2t 21 95 32th(2) v(2) 95 32(2) 31 ft/s25. V 2 r b r x 2 hxdx0 2 h r 1 3 x 3 h 1 2 x 2 r 0h 2 3 r r 3 h 2 2 h r 2 3 h 2V(h, r) V(2, 3) 2 2 3 23 6 units 3r 2 r 2 h 3 r 0 3 h 2 0 2 2 3 22 Chapter Tests 564
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