Skip to content
Textbook Guides

Solution Manual for Algebra and Trigonometry, 5th Edition

Get detailed solutions to your textbook questions with Solution Manual for Algebra and Trigonometry, 5th Edition, a guide designed to make studying easier.

Preview 16 of 725 pages

See the first 16 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

    ’ S S OLUTIONS M ANUAL J UDITH A. P ENNA A LGEBRA & T RIGONOMETRY F IFTH E DITION P RECALCULUS : A R IGHT T RIANGLE A PPROACH F IFTH E DITION Judith A. Beecher Judith A. Penna Marvin L. Bittinger Indiana University Purdue University Indianapolis

  2. Page 2
  3. Page 3
    Read page 3 as text

    Contents Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . 57 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . 107 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 163 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . 249 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . 305 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . 357 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . 399 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . 451 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . 551 Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . 635 Just-in-Time Review . . . . . . . . . . . . . . . . . . . . . 677 Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . 687

  4. Page 4
    Read page 4 as text

    4 2  2  4 4 2  2  4 x y (  1, 4) (  3,  5) (0, 2) (4, 0) (2,  2) 4 2  2  4 4 2  2  4 x y (  5, 0) (  4,  2) (1, 4) (4, 0) (2,  4) 4 2  2  4 4  2  4 x y (  5, 1) (2, 3) (0, 1) (5, 1) (2,  1) 4 2  2  4 4 2  2  4 x y (  5, 2) (  5, 0) (  1,  5) (4, 0) (4,  3) Chapter 1 Graphs, Functions, and Models Exercise Set 1.1 1. Point A is located 5 units to the left of the y -axis and 4 units up from the x -axis, so its coordinates are ( − 5 , 4). Point B is located 2 units to the right of the y -axis and 2 units down from the x -axis, so its coordinates are (2 , − 2). Point Cis located 0 units to the right or left of the y -axis and 5 units down from the x -axis, so its coordinates are (0 , − 5). Point D is located 3 units to the right of the y -axis and 5 units up from the x -axis, so its coordinates are (3 , 5). Point E is located 5 units to the left of the y -axis and 4 units down from the x -axis, so its coordinates are ( − 5 , − 4). Point F is located 3 units to the right of the y -axis and 0 units up or down from the x -axis, so its coordinates are (3 , 0). 2. G: (2 , 1); H: (0 , 0); I: (4 , − 3); J: ( − 4 , 0); K: ( − 2 , 3); L: (0 , 5) 3. To graph (4 , 0) we move from the origin 4 units to the right of the y -axis. Since the second coordinate is 0, we do not move up or down from the x -axis. To graph ( − 3 , − 5) we move from the origin 3 units to the left of the y -axis. Then we move 5 units down from the x -axis. To graph ( − 1 , 4) we move from the origin 1 unit to the left of the y -axis. Then we move 4 units up from the x -axis. To graph (0 , 2) we do not move to the right or the left of the y -axis since the first coordinate is 0. From the origin we move 2 units up. To graph (2 , − 2) we move from the origin 2 units to the right of the y -axis. Then we move 2 units down from the x -axis. 4. 5. To graph ( − 5 , 1) we move from the origin 5 units to the left of the y -axis. Then we move 1 unit up from the x -axis. To graph (5 , 1) we move from the origin 5 units to the right of the y -axis. Then we move 1 unit up from the x -axis. To graph (2 , 3) we move from the origin 2 units to the right of the y -axis. Then we move 3 units up from the x -axis. To graph (2 , − 1) we move from the origin 2 units to the right of the y -axis. Then we move 1 unit down from the x -axis. To graph (0 , 1) we do not move to the right or the left of the y -axis since the first coordinate is 0. From the origin we move 1 unit up. 6. 7. The first coordinate represents the year and the second co- ordinate represents the number of Sprint Cup Series races in which Tony Stewart finished in the top five. The or- dered pairs are (2008, 10), (2009, 15), (2010, 9), (2011, 9), (2012, 12), and (2013, 5). 8. The first coordinate represents the year and the second coordinate represents the percent of Marines who are women. The ordered pairs are (1960, 1%), (1970, 0.9%), (1980, 3.6%), (1990, 4.9%), (2000, 6.1%), and (2011, 6.8%).

  5. Page 5
    Read page 5 as text

    2 Chapter 1: Graphs, Functions, and Models 9. To determine whether ( − 1 , − 9) is a solution, substitute − 1 for x and − 9 for y . y = 7 x − 2 − 9 ? 7( − 1) − 2 ∣ ∣ ∣ − 7 − 2 − 9 ∣ ∣ − 9 TRUE The equation − 9 = − 9 is true, so ( − 1 , − 9) is a solution. To determine whether (0 , 2) is a solution, substitute 0 for x and 2 for y . y = 7 x − 2 2 ? 7 · 0 − 2 ∣ ∣ ∣ 0 − 2 2 ∣ ∣ − 2 FALSE The equation 2 = − 2 is false, so (0 , 2) is not a solution. 10. For ( 1 2 , 8 ) : y = − 4 x + 10 8 ? − 4 · 1 2 + 10 ∣ ∣ ∣ ∣ ∣ − 2 + 10 8 ∣ ∣ 8 TRUE ( 1 2 , 8 ) is a solution. For ( − 1 , 6): y = − 4 x + 10 6 ? − 4( − 1) + 10 ∣ ∣ ∣ 4 + 10 6 ∣ ∣ 14 FALSE ( − 1 , 6) is not a solution. 11. To determine whether ( 2 3 , 3 4 ) is a solution, substitute 2 3 for x and 3 4 for y . 6 x − 4 y = 1 6 · 2 3 − 4 · 3 4 ? 1 ∣ ∣ 4 − 3 ∣ ∣ ∣ 1 ∣ ∣ 1 TRUE The equation 1 = 1 is true, so ( 2 3 , 3 4 ) is a solution. To determine whether ( 1 , 3 2 ) is a solution, substitute 1 for x and 3 2 for y . 6 x − 4 y = 1 6 · 1 − 4 · 3 2 ? 1 ∣ ∣ 6 − 6 ∣ ∣ ∣ 0 ∣ ∣ 1 FALSE The equation 0 = 1 is false, so ( 1 , 3 2 ) is not a solution. 12. For (1 . 5 , 2 . 6): x 2 + y 2 = 9 (1 . 5) 2 + (2 . 6) 2 ? 9 2 . 25 + 6 . 76 ∣ ∣ ∣ 9 . 01 ∣ ∣ 9 FALSE (1 . 5 , 2 . 6) is not a solution. For ( − 3 , 0): x 2 + y 2 = 9 ( − 3) 2 + 0 2 ? 9 9 + 0 ∣ ∣ ∣ 9 ∣ ∣ 9 TRUE ( − 3 , 0) is a solution. 13. To determine whether ( − 1 2 , − 4 5 ) is a solution, substitute − 1 2 for a and − 4 5 for b . 2 a + 5 b = 3 2 ( − 1 2 ) + 5 ( − 4 5 ) ? 3 ∣ ∣ − 1 − 4 ∣ ∣ ∣ − 5 ∣ ∣ 3 FALSE The equation − 5 = 3 is false, so ( − 1 2 , − 4 5 ) is not a solu- tion. To determine whether ( 0 , 3 5 ) is a solution, substitute 0 for a and 3 5 for b . 2 a + 5 b = 3 2 · 0 + 5 · 3 5 ? 3 ∣ ∣ 0 + 3 ∣ ∣ ∣ 3 ∣ ∣ 3 TRUE The equation 3 = 3 is true, so ( 0 , 3 5 ) is a solution. 14. For ( 0 , 3 2 ) : 3 m + 4 n = 6 3 · 0 + 4 · 3 2 ? 6 ∣ ∣ 0 + 6 ∣ ∣ ∣ 6 ∣ ∣ 6 TRUE ( 0 , 3 2 ) is a solution. For ( 2 3 , 1 ) : 3 m + 4 n = 6 3 · 2 3 + 4 · 1 ? 6 ∣ ∣ 2 + 4 ∣ ∣ ∣ 6 ∣ ∣ 6 TRUE The equation 6 = 6 is true, so ( 2 3 , 1 ) is a solution.

  6. Page 6
    Read page 6 as text

    y x  4  2 2 4  4  2 2 4 5 x  3 y   15 (  3, 0) (0, 5) y x  4  2 2 4  4  2 2 4 2 x  4 y  8 (4, 0) (0,  2) (0, 4) y x  4  2 2 4  4  2 2 4 2 x  y  4 (2, 0) y x  4  2 2 4  2 2 4 6 (0, 6) 3 x  y  6 (2, 0) Exercise Set 1.1 3 15. To determine whether ( − 0 . 75 , 2 . 75) is a solution, substi- tute − 0 . 75 for x and 2 . 75 for y . x 2 − y 2 = 3 ( − 0 . 75) 2 − (2 . 75) 2 ? 3 0 . 5625 − 7 . 5625 ∣ ∣ ∣ − 7 ∣ ∣ 3 FALSE The equation − 7 = 3 is false, so ( − 0 . 75 , 2 . 75) is not a solution. To determine whether (2 , − 1) is a solution, substitute 2 for x and − 1 for y . x 2 − y 2 = 3 2 2 − ( − 1) 2 ? 3 4 − 1 ∣ ∣ ∣ 3 ∣ ∣ 3 TRUE The equation 3 = 3 is true, so (2 , − 1) is a solution. 16. For (2 , − 4): 5 x + 2 y 2 = 70 5 · 2 + 2( − 4) 2 ? 70 10 + 2 · 16 ∣ ∣ ∣ 10 + 32 ∣ ∣ 42 ∣ ∣ 70 FALSE (2 , − 4) is not a solution. For (4 , − 5): 5 x + 2 y 2 = 70 5 · 4 + 2( − 5) 2 ? 70 20 + 2 · 25 ∣ ∣ ∣ 20 + 50 ∣ ∣ 70 ∣ ∣ 70 TRUE (4 , − 5) is a solution. 17. Graph 5 x − 3 y = − 15. To find the x -intercept we replace y with 0 and solve for x . 5 x − 3 · 0 = − 15 5 x = − 15 x = − 3 The x -intercept is ( − 3 , 0). To find the y -intercept we replace x with 0 and solve for y . 5 · 0 − 3 y = − 15 − 3 y = − 15 y = 5 The y -intercept is (0 , 5). We plot the intercepts and draw the line that contains them. We could find a third point as a check that the intercepts were found correctly. 18. 19. Graph 2 x + y = 4. To find the x -intercept we replace y with 0 and solve for x . 2 x + 0 = 4 2 x = 4 x = 2 The x -intercept is (2 , 0). To find the y -intercept we replace x with 0 and solve for y . 2 · 0 + y = 4 y = 4 The y -intercept is (0 , 4). We plot the intercepts and draw the line that contains them. We could find a third point as a check that the intercepts were found correctly. 20.

  7. Page 7
    Read page 7 as text

    y x  4  2 2 4  4  2 2 4 4 y  3 x  12 (  4, 0) (0, 3) y x  4  2 2 4  4  2 2 4 3 y  2 x   6 (  3, 0) (0,  2) 6 2  2 4 2  4 x y y  3 x  5 y x  4  2 2 4  4  2 2 4 y   2 x  1 4 2  2  4 4 2  2  4 x y x  y  3 y x  2 2 4 6  2 2 4 6 x  y  4 4 Chapter 1: Graphs, Functions, and Models 21. Graph 4 y − 3 x = 12. To find the x -intercept we replace y with 0 and solve for x . 4 · 0 − 3 x = 12 − 3 x = 12 x = − 4 The x -intercept is ( − 4 , 0). To find the y -intercept we replace x with 0 and solve for y . 4 y − 3 · 0 = 12 4 y = 12 y = 3 The y -intercept is (0 , 3). We plot the intercepts and draw the line that contains them. We could find a third point as a check that the intercepts were found correctly. 22. 23. Graph y = 3 x + 5. We choose some values for x and find the corresponding y -values. When x = − 3, y = 3 x + 5 = 3( − 3) + 5 = − 9 + 5 = − 4. When x = − 1, y = 3 x + 5 = 3( − 1) + 5 = − 3 + 5 = 2. When x = 0, y = 3 x + 5 = 3 · 0 + 5 = 0 + 5 = 5 We list these points in a table, plot them, and draw the graph. x y ( x, y ) − 3 − 4 ( − 3 , − 4) − 1 2 ( − 1 , 2) 0 5 (0 , 5) 24. 25. Graph x − y = 3. Make a table of values, plot the points in the table, and draw the graph. x y ( x, y ) − 2 − 5 ( − 2 , − 5) 0 − 3 (0 , − 3) 3 0 (3 , 0) 26. 27. Graph y = − 3 4 x + 3. By choosing multiples of 4 for x , we can avoid fraction values for y . Make a table of values, plot the points in the table, and draw the graph. x y ( x, y ) − 4 6 ( − 4 , 6) 0 3 (0 , 3) 4 0 (4 , 0)

  8. Page 8
    Read page 8 as text

    y x  4  2 2 4  4  2 2 4 3 4 y   x  3 y x  4  2 2 4  4  2 2 4 3 y  2 x  3 y x  4  2 2 4  4  2 2 4 5 x  2 y  8 y x  4  2 2 4  4  2 2 4 4 3 y  2   x 4 2  2  4 6 4  2 x y x  4 y  5 y x  4  2 2 4  4  2 2 4 6 x  y  4 4 2  4 2  2  6 x y 2 x  5 y   10 y x  4  2 2 4  4  6  8  2 2 4 4 x  3 y  12 Exercise Set 1.1 5 28. 29. Graph 5 x − 2 y = 8. We could solve for y first. 5 x − 2 y = 8 − 2 y = − 5 x + 8 Subtracting 5 x on both sides y = 5 2 x − 4 Multiplying by − 1 2 on both sides By choosing multiples of 2 for x we can avoid fraction values for y . Make a table of values, plot the points in the table, and draw the graph. x y ( x, y ) 0 − 4 (0 , − 4) 2 1 (2 , 1) 4 6 (4 , 6) 30. 31. Graph x − 4 y = 5. Make a table of values, plot the points in the table, and draw the graph. x y ( x, y ) − 3 − 2 ( − 3 , − 2) 1 − 1 (1 , − 1) 5 0 (5 , 0) 32. 33. Graph 2 x + 5 y = − 10. In this case, it is convenient to find the intercepts along with a third point on the graph. Make a table of values, plot the points in the table, and draw the graph. x y ( x, y ) − 5 0 ( − 5 , 0) 0 − 2 (0 , − 2) 5 − 4 (5 , − 4) 34.

  9. Page 9
    Read page 9 as text

     2  8  6  4 4 2  2  4 x y y   x 2 y  x 2 y x  4  2 2 4 2 4 6 8 6 4 2  2 4 2  2  4 x y y  x 2  3 y  4  x 2 y x  4  2 2 4  4  2 2 4 y x  8  4 4 8  8  12  4 4 y   x 2  2 x  3 y  x 2  2 x  1 y x  4  2 2 4  4  2 2 4 6 Chapter 1: Graphs, Functions, and Models 35. Graph y = − x 2 . Make a table of values, plot the points in the table, and draw the graph. x y ( x, y ) − 2 − 4 ( − 2 , − 4) − 1 − 1 ( − 1 , − 1) 0 0 (0 , 0) 1 − 1 (1 , − 1) 2 − 4 (2 , − 4) 36. 37. Graph y = x 2 − 3. Make a table of values, plot the points in the table, and draw the graph. x y ( x, y ) − 3 6 ( − 3 , 6) − 1 − 2 ( − 1 , − 2) 0 − 3 (0 , − 3) 1 − 2 (1 , − 2) 3 6 (3 , 6) 38. 39. Graph y = − x 2 + 2 x + 3. Make a table of values, plot the points in the table, and draw the graph. x y ( x, y ) − 2 − 5 ( − 2 , − 5) − 1 0 ( − 1 , 0) 0 3 (0 , 3) 1 4 (1 , 4) 2 3 (2 , 3) 3 0 (3 , 0) 4 − 5 (4 , − 5) 40. 41. Either point can be considered as ( x 1 , y 1 ). d = √ (4 − 5) 2 + (6 − 9) 2 = √ ( − 1) 2 + ( − 3) 2 = √ 10 ≈ 3 . 162 42. d = √ ( − 3 − 2) 2 + (7 − 11) 2 = √ 41 ≈ 6 . 403 43. Either point can be considered as ( x 1 , y 1 ). d = √ ( − 13 − ( − 8)) 2 + (1 − ( − 11)) 2 = √ ( − 5) 2 + 12 2 = √ 169 = 13 44. d = √ ( − 20 − ( − 60)) 2 + (35 − 5) 2 = √ 2500 = 50

  10. Page 10
    Read page 10 as text

    Exercise Set 1.1 7 45. Either point can be considered as ( x 1 , y 1 ). d = √ (6 − 9) 2 + ( − 1 − 5) 2 = √ ( − 3) 2 + ( − 6) 2 = √ 45 ≈ 6 . 708 46. d = √ ( − 4 − ( − 1)) 2 + ( − 7 − 3) 2 = √ 109 ≈ 10 . 440 47. Either point can be considered as ( x 1 , y 1 ). d = √ ( − 8 − 8) 2 + ( 7 11 − 7 11 ) 2 = √ ( − 16) 2 + 0 2 = 16 48. d = √ ( 1 2 − 1 2 ) 2 + ( − 4 25 − ( − 13 25 )) 2 = √ ( 9 25 ) 2 = 9 25 49. d = √ [ − 3 5 − ( − 3 5 )] 2 + ( − 4 − 2 3 ) 2 = √ 0 2 + ( − 14 3 ) 2 = 14 3 50. d = √ ( − 11 3 − 1 3 ) 2 + ( − 1 2 − 5 2 ) 2 = √ 16 + 9 = √ 25 = 5 51. Either point can be considered as ( x 1 , y 1 ). d = √ ( − 4 . 2 − 2 . 1) 2 + [3 − ( − 6 . 4)] 2 = √ ( − 6 . 3) 2 + (9 . 4) 2 = √ 128 . 05 ≈ 11 . 316 52. d = √ [0 . 6 − ( − 8 . 1)] 2 + [ − 1 . 5 − ( − 1 . 5)] 2 = √ (8 . 7) 2 = 8 . 7 53. Either point can be considered as ( x 1 , y 1 ). d = √ (0 − a ) 2 + (0 − b ) 2 = √ a 2 + b 2 54. d = √ [ r − ( − r )] 2 + [ s − ( − s )] 2 = √ 4 r 2 + 4 s 2 = 2 √ r 2 + s 2 55. First we find the length of the diameter: d = √ ( − 3 − 9) 2 + ( − 1 − 4) 2 = √ ( − 12) 2 + ( − 5) 2 = √ 169 = 13 The length of the radius is one-half the length of the di- ameter, or 1 2 (13), or 6.5. 56. Radius = √ ( − 3 − 0) 2 + (5 − 1) 2 = √ 25 = 5 Diameter = 2 · 5 = 10 57. First we find the distance between each pair of points. For ( − 4 , 5) and (6 , 1): d = √ ( − 4 − 6) 2 + (5 − 1) 2 = √ ( − 10) 2 + 4 2 = √ 116 For ( − 4 , 5) and ( − 8 , − 5): d = √ ( − 4 − ( − 8)) 2 + (5 − ( − 5)) 2 = √ 4 2 + 10 2 = √ 116 For (6 , 1) and ( − 8 , − 5): d = √ (6 − ( − 8)) 2 + (1 − ( − 5)) 2 = √ 14 2 + 6 2 = √ 232 Since ( √ 116) 2 + ( √ 116) 2 = ( √ 232) 2 , the points could be the vertices of a right triangle. 58. For ( − 3 , 1) and (2 , − 1): d = √ ( − 3 − 2) 2 + (1 − ( − 1)) 2 = √ 29 For ( − 3 , 1) and (6 , 9): d = √ ( − 3 − 6) 2 + (1 − 9) 2 = √ 145 For (2 , − 1) and (6 , 9): d = √ (2 − 6) 2 + ( − 1 − 9) 2 = √ 116 Since ( √ 29) 2 + ( √ 116) 2 = ( √ 145) 2 , the points could be the vertices of a right triangle. 59. First we find the distance between each pair of points. For ( − 4 , 3) and (0 , 5): d = √ ( − 4 − 0) 2 + (3 − 5) 2 = √ ( − 4) 2 + ( − 2) 2 = √ 20 For ( − 4 , 3) and (3 , − 4): d = √ ( − 4 − 3) 2 + [3 − ( − 4)] 2 = √ ( − 7) 2 + 7 2 = √ 98 For (0 , 5) and (3 , − 4): d = √ (0 − 3) 2 + [5 − ( − 4)] 2 = √ ( − 3) 2 + 9 2 = √ 90 The greatest distance is √ 98, so if the points are the ver- tices of a right triangle, then it is the hypotenuse. But ( √ 20) 2 + ( √ 90) 2  = ( √ 98) 2 , so the points are not the ver- tices of a right triangle. 60. See the graph of this rectangle in Exercise 71. The segments with endpoints ( − 3 , 4), (2 , − 1) and (5 , 2), (0 , 7) are one pair of opposite sides. We find the length of each of these sides. For ( − 3 , 4), (2 , − 1): d = √ ( − 3 − 2) 2 + (4 − ( − 1)) 2 = √ 50 For (5 , 2), (0 , 7): d = √ (5 − 0) 2 + (2 − 7) 2 = √ 50 The segments with endpoints (2 , − 1), (5 , 2) and (0 , 7), ( − 3 , 4) are the second pair of opposite sides. We find their lengths. For (2 , − 1), (5 , 2): d = √ (2 − 5) 2 + ( − 1 − 2) 2 = √ 18 For (0 , 7), ( − 3 , 4): d = √ (0 − ( − 3)) 2 + (7 − 4) 2 = √ 18 The endpoints of the diagonals are ( − 3 , 4), (5 , 2) and (2 , − 1), (0 , 7). We find the length of each. For ( − 3 , 4), (5 , 2): d = √ ( − 3 − 5) 2 + (4 − 2) 2 = √ 68

  11. Page 11
    Read page 11 as text

    8 6 4 2  2 4 6  2  4 x y 12 8 4 x 12 8 4  4 y 8 Chapter 1: Graphs, Functions, and Models For (2 , − 1), (0 , 7): d = √ (2 − 0) 2 + ( − 1 − 7) 2 = √ 68 The opposite sides of the quadrilateral are the same length and the diagonals are the same length, so the quadrilateral is a rectangle. 61. We use the midpoint formula. ( 4 + ( − 12) 2 , − 9 + ( − 3) 2 ) = ( − 8 2 , − 12 2 ) = ( − 4 , − 6) 62. ( 7 + 9 2 , − 2 + 5 2 ) = ( 8 , 3 2 ) 63. We use the midpoint formula. ( 0 + ( − 2 5 ) 2 , 1 2 − 0 2 ) = ( − 2 5 2 , 1 2 2 ) = ( − 1 5 , 1 4 ) 64. ( 0 + ( − 7 13 ) 2 , 0 + 2 7 2 ) = ( − 7 26 , 1 7 ) 65. We use the midpoint formula. ( 6 . 1 + 3 . 8 2 , − 3 . 8 + ( − 6 . 1) 2 ) = ( 9 . 9 2 , − 9 . 9 2 ) = (4 . 95 , − 4 . 95) 66. ( − 0 . 5 + 4 . 8 2 , − 2 . 7 + ( − 0 . 3) 2 ) = (2 . 15 , − 1 . 5) 67. We use the midpoint formula. ( − 6 + ( − 6) 2 , 5 + 8 2 ) = ( − 12 2 , 13 2 ) = ( − 6 , 13 2 ) 68. ( 1 + ( − 1) 2 , − 2 + 2 2 ) = (0 , 0) 69. We use the midpoint formula. ( − 1 6 + ( − 2 3 ) 2 , − 3 5 + 5 4 2 ) = ( − 5 6 2 , 13 20 2 ) = ( − 5 12 , 13 40 ) 70. ( 2 9 + ( − 2 5 ) 2 , 1 3 + 4 5 2 ) = ( − 4 45 , 17 30 ) 71. For the side with vertices ( − 3 , 4) and (2 , − 1): ( − 3 + 2 2 , 4 + ( − 1) 2 ) = ( − 1 2 , 3 2 ) For the side with vertices (2 , − 1) and (5 , 2): ( 2 + 5 2 , − 1 + 2 2 ) = ( 7 2 , 1 2 ) For the side with vertices (5 , 2) and (0 , 7): ( 5 + 0 2 , 2 + 7 2 ) = ( 5 2 , 9 2 ) For the side with vertices (0 , 7) and ( − 3 , 4): ( 0 + ( − 3) 2 , 7 + 4 2 ) = ( − 3 2 , 11 2 ) For the quadrilateral whose vertices are the points found above, the diagonals have endpoints ( − 1 2 , 3 2 ) , ( 5 2 , 9 2 ) and ( 7 2 , 1 2 ) , ( − 3 2 , 11 2 ) . We find the length of each of these diagonals. For ( − 1 2 , 3 2 ) , ( 5 2 , 9 2 ) : d = √ ( − 1 2 − 5 2 ) 2 + ( 3 2 − 9 2 ) 2 = √ ( − 3) 2 + ( − 3) 2 = √ 18 For ( 7 2 , 1 2 ) , ( − 3 2 , 11 2 ) : d = √ ( 7 2 − ( − 3 2 )) 2 + ( 1 2 − 11 2 ) 2 = √ 5 2 + ( − 5) 2 = √ 50 Since the diagonals do not have the same lengths, the mid- points are not vertices of a rectangle. 72. For the side with vertices ( − 5 , − 1) and (7 , − 6): ( − 5 + 7 2 , − 1 + ( − 6) 2 ) = ( 1 , − 7 2 ) For the side with vertices (7 , − 6) and (12 , 6): ( 7 + 12 2 , − 6 + 6 2 ) = ( 19 2 , 0 ) For the side with vertices (12 , 6) and (0 , 11): ( 12 + 0 2 , 6 + 11 2 ) = ( 6 , 17 2 ) For the side with vertices (0 , 11) and ( − 5 , − 1): ( 0 + ( − 5) 2 , 11 + ( − 1) 2 ) = ( − 5 2 , 5 ) For the quadrilateral whose vertices are the points found above, one pair of opposite sides has endpoints ( 1 , − 7 2 ) , ( 19 2 , 0 ) and ( 6 , 17 2 ) , ( − 5 2 , 5 ) . The length of each of

  12. Page 12
    Read page 12 as text

    y x 2 4  2  4  2  4 4 2 x 2  y 2  4 y x 4 8  4  8  4  8 8 4 x 2  y 2  81 Exercise Set 1.1 9 these sides is √ 338 2 . The other pair of opposite sides has endpoints ( 19 2 , 0 ) , ( 6 , 17 2 ) and ( − 5 2 , 5 ) , ( 1 , − 7 2 ) . The length of each of these sides is also √ 338 2 . The end- points of the diagonals of the quadrilateral are ( 1 , − 7 2 ) , ( 6 , 17 2 ) and ( 19 2 , 0 ) , ( − 5 2 , 5 ) . The length of each di- agonal is 13. Since the four sides of the quadrilateral are the same length and the diagonals are the same length, the midpoints are vertices of a square. 73. We use the midpoint formula. ( √ 7 + √ 2 2 , − 4 + 3 2 ) = ( √ 7 + √ 2 2 , − 1 2 ) 74. ( − 3 + 1 2 , √ 5 + √ 2 2 ) = ( − 1 , √ 5 + √ 2 2 ) 75. ( x − h ) 2 + ( y − k ) 2 = r 2 ( x − 2) 2 + ( y − 3) 2 = ( 5 3 ) 2 Substituting ( x − 2) 2 + ( y − 3) 2 = 25 9 76. ( x − 4) 2 + ( y − 5) 2 = (4 . 1) 2 ( x − 4) 2 + ( y − 5) 2 = 16 . 81 77. The length of a radius is the distance between ( − 1 , 4) and (3 , 7): r = √ ( − 1 − 3) 2 + (4 − 7) 2 = √ ( − 4) 2 + ( − 3) 2 = √ 25 = 5 ( x − h ) 2 + ( y − k ) 2 = r 2 [ x − ( − 1)] 2 + ( y − 4) 2 = 5 2 ( x + 1) 2 + ( y − 4) 2 = 25 78. Find the length of a radius: r = √ (6 − 1) 2 + ( − 5 − 7) 2 = √ 169 = 13 ( x − 6) 2 + [ y − ( − 5)] 2 = 13 2 ( x − 6) 2 + ( y + 5) 2 = 169 79. The center is the midpoint of the diameter: ( 7 + ( − 3) 2 , 13 + ( − 11) 2 ) = (2 , 1) Use the center and either endpoint of the diameter to find the length of a radius. We use the point (7 , 13): r = √ (7 − 2) 2 + (13 − 1) 2 = √ 5 2 + 12 2 = √ 169 = 13 ( x − h ) 2 + ( y − k ) 2 = r 2 ( x − 2) 2 + ( y − 1) 2 = 13 2 ( x − 2) 2 + ( y − 1) 2 = 169 80. The points ( − 9 , 4) and ( − 1 , − 2) are opposite vertices of the square and hence endpoints of a diameter of the circle. We use these points to find the center and radius. Center: ( − 9 + ( − 1) 2 , 4 + ( − 2) 2 ) = ( − 5 , 1) Radius: 1 2 √ ( − 9 − ( − 1)) 2 +(4 − ( − 2)) 2 = 1 2 · 10 = 5 [ x − ( − 5)] 2 + ( y − 1) 2 = 5 2 ( x + 5) 2 + ( y − 1) 2 = 25 81. Since the center is 2 units to the left of the y -axis and the circle is tangent to the y -axis, the length of a radius is 2. ( x − h ) 2 + ( y − k ) 2 = r 2 [ x − ( − 2)] 2 + ( y − 3) 2 = 2 2 ( x + 2) 2 + ( y − 3) 2 = 4 82. Since the center is 5 units below the x -axis and the circle is tangent to the x -axis, the length of a radius is 5. ( x − 4) 2 + [ y − ( − 5)] 2 = 5 2 ( x − 4) 2 + ( y + 5) 2 = 25 83. x 2 + y 2 = 4 ( x − 0) 2 + ( y − 0) 2 = 2 2 Center: (0 , 0); radius: 2 84. x 2 + y 2 = 81 ( x − 0) 2 + ( y − 0) 2 = 9 2 Center: (0 , 0); radius: 9

  13. Page 13
    Read page 13 as text

    y x 2 4  2  2  4 4 2 6 8 x 2  ( y  3) 2  16 ( x  2) 2  y 2  100 y x 4 8  4  8  4  8 8 4  12 ( x  1) 2  ( y  5) 2  36 y x 4 8  4  4  8 8 4 12 ( x  7) 2  ( y  2) 2  25 y x 4 8  4  4 8 12 4  8 ( x  4) 2  ( y  5) 2  9 y x 2  2  4  2  4 2  6  6  8 ( x  1) 2  ( y  2) 2  64 y x 4 8  4  4  8 8 4  8 12 10 Chapter 1: Graphs, Functions, and Models 85. x 2 + ( y − 3) 2 = 16 ( x − 0) 2 + ( y − 3) 2 = 4 2 Center: (0 , 3); radius: 4 86. ( x + 2) 2 + y 2 = 100 [ x − ( − 2)] 2 + ( y − 0) 2 = 10 2 Center: ( − 2 , 0); radius: 10 87. ( x − 1) 2 + ( y − 5) 2 = 36 ( x − 1) 2 + ( y − 5) 2 = 6 2 Center: (1 , 5); radius: 6 88. ( x − 7) 2 + ( y + 2) 2 = 25 ( x − 7) 2 + [ y − ( − 2)] 2 = 5 2 Center: (7 , − 2); radius: 5 89. ( x + 4) 2 + ( y + 5) 2 = 9 [ x − ( − 4)] 2 + [ y − ( − 5)] 2 = 3 2 Center: ( − 4 , − 5); radius: 3 90. ( x + 1) 2 + ( y − 2) 2 = 64 [ x − ( − 1)] 2 + ( y − 2) 2 = 8 2 Center: ( − 1 , 2); radius: 8 91. From the graph we see that the center of the circle is ( − 2 , 1) and the radius is 3. The equation of the circle is [ x − ( − 2)] 2 + ( y − 1) 2 = 3 2 , or ( x + 2) 2 + ( y − 1) 2 = 3 2 . 92. Center: (3 , − 5), radius: 4 Equation: ( x − 3) 2 + [ y − ( − 5)] 2 = 4 2 , or ( x − 3) 2 + ( y + 5) 2 = 4 2 93. From the graph we see that the center of the circle is (5 , − 5) and the radius is 15. The equation of the circle is ( x − 5) 2 + [ y − ( − 5)] 2 = 15 2 , or ( x − 5) 2 + ( y + 5) 2 = 15 2 .

  14. Page 14
    Read page 14 as text

    Exercise Set 1.1 11 94. Center: ( − 8 , 2), radius: 4 Equation: [ x − ( − 8)] 2 + ( y − 2) 2 = 4 2 , or ( x + 8) 2 + ( y − 2) 2 = 4 2 95. If the point ( p, q ) is in the fourth quadrant, then p > 0 and q < 0. If p > 0, then − p < 0 so both coordinates of the point ( q, − p ) are negative and ( q, − p ) is in the third quadrant. 96. Use the distance formula: d = √ ( a + h − a ) 2 + ( 1 a + h − 1 a ) 2 = √ h 2 + ( − h a ( a + h ) ) 2 = √ h 2 + h 2 a 2 ( a + h ) 2 = √ h 2 a 2 ( a + h ) 2 + h 2 a 2 ( a + h ) 2 = √ h 2 ( a 2 ( a + h ) 2 + 1) a 2 ( a + h ) 2 = ∣ ∣ ∣ ∣ ∣ h a ( a + h ) ∣ ∣ ∣ ∣ ∣ √ a 2 ( a + h ) 2 + 1 Find the midpoint: ( a + a + h 2 , 1 a + 1 a + h 2 ) = ( 2 a + h 2 , 2 a + h 2 a ( a + h ) ) 97. Use the distance formula. Either point can be considered as ( x 1 , y 1 ). d = √ ( a + h − a ) 2 + ( √ a + h − √ a ) 2 = √ h 2 + a + h − 2 √ a 2 + ah + a = √ h 2 + 2 a + h − 2 √ a 2 + ah Next we use the midpoint formula. ( a + a + h 2 , √ a + √ a + h 2 ) = ( 2 a + h 2 , √ a + √ a + h 2 ) 98. C = 2 πr 10 π = 2 πr 5 = r Then [ x − ( − 5)] 2 +( y − 8) 2 = 5 2 , or ( x +5) 2 +( y − 8) 2 = 25. 99. First use the formula for the area of a circle to find r 2 : A = πr 2 36 π = πr 2 36 = r 2 Then we have: ( x − h ) 2 + ( y − k ) 2 = r 2 ( x − 2) 2 + [ y − ( − 7)] 2 = 36 ( x − 2) 2 + ( y + 7) 2 = 36 100. Let the point be ( x, 0). We set the distance from ( − 4 , − 3) to ( x, 0) equal to the distance from ( − 1 , 5) to ( x, 0) and solve for x . √ ( − 4 − x ) 2 + ( − 3 − 0) 2 = √ ( − 1 − x ) 2 + (5 − 0) 2 √ 16 + 8 x + x 2 + 9 = √ 1 + 2 x + x 2 + 25 √ x 2 + 8 x + 25 = √ x 2 + 2 x + 26 x 2 + 8 x + 25 = x 2 + 2 x + 26 Squaring both sides 8 x + 25 = 2 x + 26 6 x = 1 x = 1 6 The point is ( 1 6 , 0 ) . 101. Let (0 , y ) be the required point. We set the distance from ( − 2 , 0) to (0 , y ) equal to the distance from (4 , 6) to (0 , y ) and solve for y . √ [0 − ( − 2)] 2 + ( y − 0) 2 = √ (0 − 4) 2 + ( y − 6) 2 √ 4 + y 2 = √ 16 + y 2 − 12 y + 36 4 + y 2 = 16 + y 2 − 12 y + 36 Squaring both sides − 48 = − 12 y 4 = y The point is (0 , 4). 102. We first find the distance between each pair of points. For ( − 1 , − 3) and ( − 4 , − 9): d 1 = √ [ − 1 − ( − 4)] 2 + [ − 3 − ( − 9)] 2 = √ 3 2 + 6 2 = √ 9 + 36 = √ 45 = 3 √ 5 For ( − 1 , − 3) and (2 , 3): d 2 = √ ( − 1 − 2) 2 + ( − 3 − 3) 2 = √ ( − 3) 2 + ( − 6) 2 = √ 9 + 36 = √ 45 = 3 √ 5 For ( − 4 , − 9) and (2 , 3): d 3 = √ ( − 4 − 2) 2 + ( − 9 − 3) 2 = √ ( − 6) 2 + ( − 12) 2 = √ 36 + 144 = √ 180 = 6 √ 5 Since d 1 + d 2 = d 3 , the points are collinear. 103. a) When the circle is positioned on a coordinate system as shown in the text, the center lies on the y -axis and is equidistant from ( − 4 , 0) and (0 , 2). Let (0 , y ) be the coordinates of the center. √ ( − 4 − 0) 2 +(0 − y ) 2 = √ (0 − 0) 2 +(2 − y ) 2 4 2 + y 2 = (2 − y ) 2 16 + y 2 = 4 − 4 y + y 2 12 = − 4 y − 3 = y The center of the circle is (0 , − 3).

  15. Page 15
    Read page 15 as text

    12 Chapter 1: Graphs, Functions, and Models b) Use the point ( − 4 , 0) and the center (0 , − 3) to find the radius. ( − 4 − 0) 2 + [0 − ( − 3)] 2 = r 2 25 = r 2 5 = r The radius is 5 ft. 104. The coordinates of P are ( b 2 , h 2 ) by the midpoint formula. By the distance formula, each of the distances from P to (0 , h ), from P to (0 , 0), and from P to ( b, 0) is √ b 2 + h 2 2 . 105. x 2 + y 2 = 1 ( √ 3 2 ) 2 + ( − 1 2 ) 2 ? 1 ∣ ∣ 3 4 + 1 4 ∣ ∣ ∣ ∣ ∣ 1 ∣ ∣ 1 TRUE ( √ 3 2 , − 1 2 ) lies on the unit circle. 106. x 2 + y 2 = 1 0 2 + ( − 1) 2 ? 1 1 ∣ ∣ 1 TRUE (0 , − 1) lies on the unit circle. 107. x 2 + y 2 = 1 ( − √ 2 2 ) 2 + ( √ 2 2 ) 2 ? 1 ∣ ∣ 2 4 + 2 4 ∣ ∣ ∣ ∣ ∣ 1 ∣ ∣ 1 TRUE ( − √ 2 2 , √ 2 2 ) lies on the unit circle. 108. x 2 + y 2 = 1 ( 1 2 ) 2 + ( − √ 3 2 ) 2 ? 1 ∣ ∣ 1 4 + 3 4 ∣ ∣ ∣ ∣ ∣ 1 ∣ ∣ 1 TRUE ( 1 2 , − √ 3 2 ) lies on the unit circle. 109. a), b) See the answer section in the text. Exercise Set 1.2 1. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 2. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 3. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 4. This correspondence is not a function, because there is a member of the domain (1) that corresponds to more than one member of the range (4 and 6). 5. This correspondence is not a function, because there is a member of the domain (m) that corresponds to more than one member of the range (A and B). 6. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 7. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 8. This correspondence is not a function, because there is a member of the domain that corresponds to more than one member of the range. In fact, Sean Connery, Roger Moore, and Pierce Brosnan all correspond to two members of the range. 9. This correspondence is a function, because each car has exactly one license number. 10. This correspondence is not a function, because we can safely assume that at least one person uses more than one doctor. 11. This correspondence is a function, because each integer less than 9 corresponds to exactly one multiple of 5. 12. This correspondence is not a function, because we can safely assume that at least one band member plays more than one instrument. 13. This correspondence is not a function, because at least one student will have more than one neighboring seat occupied by another student. 14. This correspondence is a function, because each bag has exactly one weight. 15. The relation is a function, because no two ordered pairs have the same first coordinate and different second coor- dinates. The domain is the set of all first coordinates: { 2 , 3 , 4 } . The range is the set of all second coordinates: { 10 , 15 , 20 } . 16. The relation is a function, because no two ordered pairs have the same first coordinate and different second coor- dinates. Domain: { 3 , 5 , 7 } Range: { 1 }

  16. Page 16
    Read page 16 as text

    Exercise Set 1.2 13 17. The relation is not a function, because the ordered pairs ( − 2 , 1) and ( − 2 , 4) have the same first coordinate and dif- ferent second coordinates. The domain is the set of all first coordinates: {− 7 , − 2 , 0 } . The range is the set of all second coordinates: { 3 , 1 , 4 , 7 } . 18. The relation is not a function, because each of the ordered pairs has the same first coordinate and different second coordinates. Domain: { 1 } Range: { 3 , 5 , 7 , 9 } 19. The relation is a function, because no two ordered pairs have the same first coordinate and different second coor- dinates. The domain is the set of all first coordinates: {− 2 , 0 , 2 , 4 , − 3 } . The range is the set of all second coordinates: { 1 } . 20. The relation is not a function, because the ordered pairs (5 , 0) and (5 , − 1) have the same first coordinates and dif- ferent second coordinates. This is also true of the pairs (3 , − 1) and (3 , − 2). Domain: { 5 , 3 , 0 } Range: { 0 , − 1 , − 2 } 21. g ( x ) = 3 x 2 − 2 x + 1 a) g (0) = 3 · 0 2 − 2 · 0 + 1 = 1 b) g ( − 1) = 3( − 1) 2 − 2( − 1) + 1 = 6 c) g (3) = 3 · 3 2 − 2 · 3 + 1 = 22 d) g ( − x ) = 3( − x ) 2 − 2( − x ) + 1 = 3 x 2 + 2 x + 1 e) g (1 − t ) = 3(1 − t ) 2 − 2(1 − t ) + 1 = 3(1 − 2 t + t 2 ) − 2(1 − t )+1 = 3 − 6 t +3 t 2 − 2+2 t +1 = 3 t 2 − 4 t + 2 22. f ( x ) = 5 x 2 + 4 x a) f (0) = 5 · 0 2 + 4 · 0 = 0 + 0 = 0 b) f ( − 1) = 5( − 1) 2 + 4( − 1) = 5 − 4 = 1 c) f (3) = 5 · 3 2 + 4 · 3 = 45 + 12 = 57 d) f ( t ) = 5 t 2 + 4 t e) f ( t − 1) = 5( t − 1) 2 + 4( t − 1) = 5 t 2 − 6 t + 1 23. g ( x ) = x 3 a) g (2) = 2 3 = 8 b) g ( − 2) = ( − 2) 3 = − 8 c) g ( − x ) = ( − x ) 3 = − x 3 d) g (3 y ) = (3 y ) 3 = 27 y 3 e) g (2 + h ) = (2 + h ) 3 = 8 + 12 h + 6 h 2 + h 3 24. f ( x ) = 2 | x | + 3 x a) f (1) = 2 | 1 | + 3 · 1 = 2 + 3 = 5 b) f ( − 2) = 2 | − 2 | + 3( − 2) = 4 − 6 = − 2 c) f ( − x ) = 2 | − x | + 3( − x ) = 2 | x | − 3 x d) f (2 y ) = 2 | 2 y | + 3 · 2 y = 4 | y | + 6 y e) f (2 − h ) = 2 | 2 − h | + 3(2 − h ) = 2 | 2 − h | + 6 − 3 h 25. g ( x ) = x − 4 x + 3 a) g (5) = 5 − 4 5 + 3 = 1 8 b) g (4) = 4 − 4 4 + 7 = 0 c) g ( − 3) = − 3 − 4 − 3 + 3 = − 7 0 Since division by 0 is not defined, g ( − 3) does not exist. d) g ( − 16 . 25) = − 16 . 25 − 4 − 16 . 25 + 3 = − 20 . 25 − 13 . 25 = 81 53 ≈ 1 . 5283 e) g ( x + h ) = x + h − 4 x + h + 3 26. f ( x ) = x 2 − x a) f (2) = 2 2 − 2 = 2 0 Since division by 0 is not defined, f (2) does not exist. b) f (1) = 1 2 − 1 = 1 c) f ( − 16) = − 16 2 − ( − 16) = − 16 18 = − 8 9 d) f ( − x ) = − x 2 − ( − x ) = − x 2 + x e) f ( − 2 3 ) = − 2 3 2 − ( − 2 3 ) = − 2 3 8 3 = − 1 4 27. g ( x ) = x √ 1 − x 2 g (0) = 0 √ 1 − 0 2 = 0 √ 1 = 0 1 = 0 g ( − 1) = − 1 √ 1 − ( − 1) 2 = − 1 √ 1 − 1 = − 1 √ 0 = − 1 0 Since division by 0 is not defined, g ( − 1) does not exist. g (5) = 5 √ 1 − 5 2 = 5 √ 1 − 25 = 5 √ − 24 Since √ − 24 is not defined as a real number, g (5) does not exist as a real number.

709 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Mathematics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.