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Solution Manual For Algebra And Trigonometry, 9th Edition

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    1 Chapter R Review Section R.1 1. rational 2. 4 5 6 3 4 30 3 31 + ⋅ − = + − = 3. Distributive 4. ( ) 5 3 6 x + = 5. True 6. False; The Zero-Product Property states that if a product equals 0, then at least one of the factors must equal 0. 7. False; 6 is the Greatest Common Factor of 12 and 18. The Least Common Multiple is the smallest value that both numbers will divide evenly. The LCM for 12 and 18 is 36. 8. True 9. { } { } { } 1, 3, 4,5, 9 2, 4, 6, 7,8 1, 2,3, 4, 5, 6, 7,8, 9 A B ∪ = ∪ = 10. { } { } { } 1, 3, 4,5, 9 1, 3, 4, 6 1, 3, 4, 5, 6, 9 A C ∪ = ∪ = 11. { } { } { } 1, 3, 4,5, 9 2, 4, 6, 7,8 4 A B ∩ = ∩ = 12. { } { } { } 1, 3, 4,5, 9 1, 3, 4, 6 1, 3, 4 A C ∩ = ∩ = 13. { } { } ( ) { } { } { } { } ( ) 1, 3, 4,5, 9 2, 4, 6, 7,8 1,3, 4, 6 1, 2,3, 4,5, 6, 7,8,9 1,3, 4, 6 1, 3, 4,6 A B C ∪ ∩ = ∪ ∩ = ∩ = 14. { } { } ( ) { } { } { } { } ( ) 1, 3, 4,5, 9 2, 4, 6, 7,8 1,3, 4, 6 4 1, 3, 4, 6 1,3, 4, 6 A B C ∩ ∪ = ∩ ∪ = ∪ = 15. { } 0, 2, 6, 7, 8 A = 16. { } 0, 2, 5, 7, 8, 9 C = 17. { } { } { } { } 1, 3, 4, 5, 9 2, 4, 6, 7, 8 4 0, 1, 2, 3, 5, 6, 7, 8, 9 A B ∩ = ∩ = = 18. { } { } { } { } 2, 4, 6, 7, 8 1, 3, 4, 6 1, 2, 3, 4, 6, 7, 8 0, 5, 9 B C ∪ = ∪ = = 19. { } { } { } 0, 2, 6, 7, 8 0, 1, 3, 5, 9 0, 1, 2, 3, 5, 6, 7, 8, 9 A B ∪ = ∪ = 20. { } { } { } 0, 1, 3, 5, 9 0, 2, 5, 7, 8, 9 0, 5, 9 B C ∩ = ∩ = 21. a. { } 2,5 b. { } 6, 2,5 − c. { } 1 6, , 1.333... 1.3, 2,5 2 − − = − d. { } π e. { } 1 6, , 1.333... 1.3, , 2,5 2 π − − = − 22. a. { } 1 b. { } 0,1 c. { } 5 , 2.060606... 2.06,1.25, 0,1 3 − = d. { } 5 e. { } 5 , 2.060606... 2.06,1.25, 0,1, 5 3 − = 23. a. { } 1 b. { } 0,1 c. { } 1 1 1 0,1, , , 2 3 4

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    Chapter R: Review 2 d. None e. { } 1 1 1 0,1, , , 2 3 4 24. a. None b. { } 1 − c. { } 1.3, 1.2, 1.1, 1 − − − − d. None e. { } 1.3, 1.2, 1.1, 1 − − − − 25. a. None b. None c. None d. { } 1 2, , 2 1, 2 π π + + e. { } 1 2, , 2 1, 2 π π + + 26. a. None b. None c. { } 1 10.3 2 + d. { } 2, 2 π − + e. { } 1 2, 2, 10.3 2 π − + + 27. a. 18.953 b. 18.952 28. a. 25.861 b. 25.861 29. a. 28.653 b. 28.653 30. a. 99.052 b. 99.052 31. a. 0.063 b. 0.062 32. a. 0.054 b. 0.053 33. a. 9.999 b. 9.998 34. a. 1.001 b. 1.000 35. a. 0.429 b. 0.428 36. a. 0.556 b. 0.555 37. a. 34.733 b. 34.733 38. a. 16.200 b. 16.200 39. 3 2 5 + = 40. 5 2 10 ⋅ = 41. 2 3 4 x + = ⋅ 42. 3 2 2 y + = + 43. 3 1 2 y = + 44. 2 4 6 x = ⋅ 45. 2 6 x − = 46. 2 6 y − = 47. 6 2 x = 48. 2 6 x = 49. 9 4 2 5 2 7 − + = + = 50. 6 4 3 2 3 5 − + = + = 51. 6 4 3 6 12 6 − + ⋅ = − + = 52. 8 4 2 8 8 0 − ⋅ = − = 53. 4 5 8 9 8 1 + − = − = 54. 8 3 4 5 4 1 − − = − = 55. 1 12 1 13 4 3 3 3 + + = = 56. 1 4 1 3 2 2 2 2 − − = = 57. ( ) ( ) 6 3 5 2 3 2 6 15 2 1 6 17 11 − ⋅ + ⋅ − = − + ⋅         = − = −

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    Section R.1: Real Numbers 3 58. ( ) ( ) [ ] [ ] 2 8 3 4 2 3 2 8 3 6 3 2 8 18 3 2 10 3 20 3 23 ⋅ − ⋅ + − = ⋅ − ⋅ −         = ⋅ − − = ⋅ − − = − − = − 59. ( ) ( ) 2 3 5 8 2 1 2 2 16 1 4 16 1 12 1 11 ⋅ − + ⋅ − = ⋅ − + − = − + − = − = 60. ( ) ( ) 1 4 3 2 2 1 12 2 2 1 12 11 − ⋅ − + = − − + = − = − 61. ( ) [ ] [ ] [ ] 10 6 2 2 8 3 2 10 6 4 5 2 10 2 5 2 10 7 2 10 14 4 − − ⋅ + − ⋅ = − − + ⋅     = − + ⋅ = − ⋅ = − = − 62. ( ) ( ) [ ] 2 5 4 6 3 4 2 20 6 1 18 6 18 6 12 − ⋅ − ⋅ − = − − ⋅ −         = − − − = − + = − 63. ( ) ( ) 1 1 5 3 2 1 2 2 − = = 64. ( ) ( ) 1 1 5 4 9 3 3 3 + = = 65. 4 8 12 6 5 3 2 + = = − 66. 2 4 2 1 5 3 2 − − = = − − 67. 3 10 3 2 5 3 5 21 5 3 7 ⋅ ⋅ ⋅ = = ⋅ ⋅ 2 5 ⋅ ⋅ 5 3 ⋅ 2 7 7 = ⋅ 68. 5 3 5 3 5 9 10 3 3 5 2 ⋅ ⋅ = = ⋅ ⋅ ⋅ 3 ⋅ 3 3 5 ⋅ ⋅ 1 6 2 = ⋅ 69. 6 10 2 3 5 2 2 3 25 27 5 5 3 9 ⋅ ⋅ ⋅ ⋅ ⋅ = = ⋅ ⋅ ⋅ 5 ⋅ 2 5 ⋅ 5 3 ⋅ ⋅ 4 45 9 = ⋅ 70. 21 100 3 7 4 25 3 25 3 25 3 ⋅ ⋅ ⋅ ⋅ = = ⋅ 7 4 25 ⋅ ⋅ ⋅ 25 3 ⋅ 28 = 71. 3 2 15 8 23 4 5 20 20 + + = = 72. 4 1 8 3 11 3 2 6 6 + + = = 73. 5 9 25 54 79 6 5 30 30 + + = = 74. 8 15 16 135 151 9 2 18 18 + + = = 75. 5 1 10 3 13 18 12 36 36 + + = = 76. 2 8 6 40 46 15 9 45 45 + + = = 77. 1 7 3 35 32 16 30 18 90 90 45 − − = = − = − 78. 3 2 9 4 5 14 21 42 42 − − = = 79. 3 2 9 8 1 20 15 60 60 − − = = 80. 6 3 12 15 3 35 14 70 70 − − = = − 81. 5 18 5 27 5 9 3 5 9 18 11 9 2 11 11 27     ⋅ ⋅ ⋅   = ⋅ = = ⋅ ⋅       3 9 ⋅ 15 22 2 11 = ⋅ ⋅ 82. 5 21 5 35 5 7 5 5 7 21 2 7 3 2 2 35     ⋅ ⋅ ⋅   = ⋅ = = ⋅ ⋅       5 7 ⋅ 25 6 3 2 = ⋅ ⋅ 83. 1 3 7 3 7 3 7 10 1 2 5 10 10 10 10 10 + ⋅ + = + = = =

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    Chapter R: Review 4 84. 2 4 1 2 2 2 2 2 2 2 2 3 5 6 3 5 3 2 3 5 3 2 3 15 2 5 2 10 2 10 2 12 3 5 15 15 15 15 15 4 3 4 3 4 5 3 5 3 5 ⋅ ⋅ + ⋅ = + = + = + ⋅ ⋅ ⋅ ⋅ + = ⋅ + = + = = ⋅ ⋅ = = = ⋅ ⋅ 85. 3 3 2 3 3 6 3 6 2 3 2 4 8 1 4 8 4 8 4 2 8 12 3 12 3 15 8 8 8 8 ⋅ + = ⋅ + = + = ⋅ + + = + = = 86. 5 1 3 5 1 3 5 1 3 5 1 3 6 2 1 6 2 3 2 2 3 2 2 5 1 5 1 4 2 2 2 2 2 ⋅ ⋅ ⋅ − = ⋅ − = − = − ⋅ ⋅ − = − = = = 87. ( ) 6 4 6 24 x x + = + 88. ( ) 4 2 1 8 4 x x − = − 89. ( ) 2 4 4 x x x x − = − 90. ( ) 2 4 3 4 12 x x x x + = + 91. 3 1 3 1 2 3 2 2 2 2 4 2 4 2 2 2 2 2 3 2 3 1 2 2 2 2 x x x x x ⋅   − = ⋅ − ⋅ = −   ⋅   ⋅ = − = − ⋅ 92. 2 1 2 1 3 2 3 3 3 3 3 6 3 6 3 3 2 3 2 3 1 2 3 3 2 2 x x x x x ⋅   + = ⋅ + ⋅ = +   ⋅   ⋅ = + = + ⋅ 93. ( )( ) 2 2 2 4 4 2 8 6 8 x x x x x x x + + = + + + = + + 94. ( )( ) 2 2 5 1 5 5 6 5 x x x x x x x + + = + + + = + + 95. ( )( ) 2 2 2 1 2 2 2 x x x x x x x − + = + − − = − − 96. ( )( ) 2 2 4 1 4 4 3 4 x x x x x x x − + = + − − = − − 97. ( )( ) 2 2 8 2 2 8 16 10 16 x x x x x x x − − = − − + = − + 98. ( )( ) 2 2 4 2 2 4 8 6 8 x x x x x x x − − = − − + = − + 99. ( ) ( ) 2 3 2 3 2 3 5 5 x x x x x x x + = ⋅ + ⋅ = + ⋅ = ⋅ = 100. 2 3 4 2 12 14 + ⋅ = + = since multiplication comes before addition in the order of operations for real numbers. ( ) 2 3 4 5 4 20 + ⋅ = ⋅ = since operations inside parentheses come before multiplication in the order of operations for real numbers. 101. ( ) ( ) 2 3 4 2 12 24 ⋅ = = ( ) ( ) ( ) ( ) 2 3 2 4 6 8 48 ⋅ ⋅ ⋅ = = 102. 4 3 7 1 2 5 7 + = = + , but 4 3 4 5 3 2 20 6 26 13 2.6 2 5 10 10 10 5 ⋅ + ⋅ + + = = = = = 103. Subtraction is not commutative; for example: 2 3 1 1 3 2 − = − ≠ = − . 104. Subtraction is not associative; for example: ( ) ( ) 5 2 1 2 4 5 2 1 − − = ≠ = − − . 105. Division is not commutative; for example: 2 3 3 2 ≠ . 106. Division is not associative; for example: ( ) 12 2 2 6 2 3 ÷ ÷ = ÷ = , but ( ) 12 2 2 12 1 12 ÷ ÷ = ÷ = . 107. The Symmetric Property implies that if 2 = x, then x = 2.

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    Section R.2: Algebra Essentials 5 108. From the principle of substitution, if 5 x = , then ( )( ) ( ) ( ) 2 2 2 5 5 25 25 5 30 x x x x x x x =  =  + = +  + = 109. There are no real numbers that are both rational and irrational, since an irrational number, by definition, is a number that cannot be expressed as the ratio of two integers; that is, not a rational number Every real number is either a rational number or an irrational number, since the decimal form of a real number either involves an infinitely repeating pattern of digits or an infinite, non- repeating string of digits. 110. The sum of an irrational number and a rational number must be irrational. Otherwise, the irrational number would then be the difference of two rational numbers, and therefore would have to be rational. 111. Answers will vary. 112. Since 1 day = 24 hours, we compute 12997 541.5416 24 = . Now we only need to consider the decimal part of the answer in terms of a 24 hour day. That is, ( )( ) 0.5416 24 13 ≈ hours. So it must be 13 hours later than 12 noon, which makes the time 1 AM CST. 113. Answers will vary. Section R.2 1. variable 2. origin 3. strict 4. base; exponent (or power) 5. 3 1.2345678 10 × 6. True. 7. True 8. False; the absolute value of a real number is nonnegative. 0 0 = which is not a positive number. 9. False; a number in scientific notation is expressed as the product of a number , x , 1 10 x ≤ < or 10 1 x − < ≤ − , and a power of 10. 10. False; to multiply two expressions with the same base, retain the base and add the exponents. 11. −2.5 −1 5 2 1 0 0.25 3 4 12. 1 3 3 2 −2 2 −1.5 0 2 3 13. 1 0 2 > 14. 5 6 < 15. 1 2 − > − 16. 5 3 2 − < − 17. 3.14 π > 18. 2 1.41 > 19. 1 0.5 2 = 20. 1 0.33 3 > 21. 2 0.67 3 < 22. 1 0.25 4 = 23. 0 x >

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    Chapter R: Review 6 24. 0 z < 25. 2 x < 26. 5 y > − 27. 1 x ≤ 28. 2 x ≥ 29. Graph on the number line: 2 x ≥ − 0 −2 30. Graph on the number line: 4 x < 4 0 31. Graph on the number line: 1 x > − −1 0 32. Graph on the number line: 7 x ≤ 0 7 33. ( , ) (0,1) 1 0 1 1 d C D d = = − = = 34. ( , ) (0, 3) 3 0 3 3 d C A d = − = − − = − = 35. ( , ) (1,3) 3 1 2 2 d D E d = = − = = 36. ( , ) (0,3) 3 0 3 3 d C E d = = − = = 37. ( , ) ( 3,3) 3 ( 3) 6 6 d A E d = − = − − = = 38. ( , ) (1, 1) 1 1 2 2 d D B d = − = − − = − = 39. 2 2 2 3 2 6 4 x y + = − + ⋅ = − + = 40. 3 3( 2) 3 6 3 3 x y + = − + = − + = − 41. 5 2 5( 2)(3) 2 30 2 28 xy + = − + = − + = − 42. 2 2( 2) ( 2)(3) 4 6 2 x xy − + = − − + − = − = − 43. 2( 2) 4 2 4 2 3 5 5 x x y − − = = = − − − − 44. 2 3 1 1 2 3 5 5 x y x y − + + = = = − − − − − 45. 3( 2) 2(3) 6 6 3 2 0 0 2 2 3 5 5 x y y − + − + + = = = = + + 46. 2( 2) 3 4 3 2 3 7 3 3 3 x y − − − − − = = = − 47. 3 ( 2) 1 1 x y + = + − = = 48. 3 ( 2) 5 5 x y − = − − = = 49. 3 2 3 2 5 x y + = + − = + = 50. 3 2 3 2 1 x y − = − − = − = 51. 3 3 1 3 3 x x = = = 52. 2 2 1 2 2 y y − = = = − − − 53. 4 5 4(3) 5( 2) 12 10 22 22 x y − = − − = + = = 54. 3 2 3(3) 2( 2) 9 4 5 5 x y + = + − = − = = 55. 4 5 4(3) 5( 2) 12 10 12 10 2 2 x y − = − − = − − = − = = 56. 3 2 3 3 2 2 3 3 2 2 9 4 13 x y + = + − = ⋅ + ⋅ = + = 57. 2 1 x x − Part (c) must be excluded. The value 0 x = must be excluded from the domain because it causes division by 0.

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    Section R.2: Algebra Essentials 7 58. 2 1 x x + Part (c) must be excluded. The value 0 x = must be excluded from the domain because it causes division by 0. 59. 2 ( 3)( 3) 9 x x x x x = − + − Part (a) must be excluded. The values 3 and 3 x x = − = must be excluded from the domain because they cause division by 0. 60. 2 9 x x + None of the given values are excluded. The domain is all real numbers. 61. 2 2 1 x x + None of the given values are excluded. The domain is all real numbers. 62. 3 3 2 ( 1)( 1) 1 x x x x x = − + − Parts (b) and (d) must be excluded. The values 1, and 1 x x = = − must be excluded from the domain because they cause division by 0. 63. 2 2 3 5 10 5 10 ( 1)( 1) x x x x x x x x x + − + − = − + − Parts (b), (c), and (d) must be excluded. The values 0, 1, and 1 x x x = = = − must be excluded from the domain because they cause division by 0. 64. 2 2 3 2 9 1 9 1 ( 1) x x x x x x x x − − + − − + = + + Part (c) must be excluded. The value 0 x = must be excluded from the domain because it causes division by 0. 65. 4 5 x − 5 x = must be exluded because it makes the denominator equal 0. { } Domain 5 x x = ≠ 66. 6 4 x − + 4 x = − must be excluded sine it makes the denominator equal 0. { } Domain 4 x x = ≠ − 67. 4 x x + 4 x = − must be excluded sine it makes the denominator equal 0. { } Domain 4 x x = ≠ − 68. 2 6 x x − − 6 x = must be excluded sine it makes the denominator equal 0. { } Domain 6 x x = ≠ 69. 5 5 5 ( 32) (32 32) (0) 0 C 9 9 9 C F = − = − = = ° 70. 5 5 5 ( 32) (212 32) (180) 100 C 9 9 9 C F = − = − = = ° 71. 5 5 5 ( 32) (77 32) (45) 25 C 9 9 9 C F = − = − = = ° 72. 5 5 ( 32) ( 4 32) 9 9 5 ( 36) 9 20 C C F = − = − − = − = − ° 73. 2 ( 4) ( 4)( 4) 16 − = − − = 74. 2 2 4 (4) 16 − = − = − 75. 2 2 1 1 4 16 4 − = = 76. 2 2 1 1 4 16 4 − − = − = − 77. 6 4 6 4 2 2 1 1 3 3 3 3 9 3 − − + − ⋅ = = = = 78. 2 3 2 3 1 4 4 4 4 4 − − + ⋅ = = = 79. ( ) ( )( ) 1 2 1 2 2 3 3 3 9 − − − − = = =

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    Chapter R: Review 8 80. ( ) ( )( ) 3 1 3 1 3 2 2 2 8 − − − − = = = 81. 2 25 5 5 = = 82. 2 36 6 6 = = 83. ( ) 2 4 4 4 − = − = 84. ( ) 2 3 3 3 − = − = 85. ( ) ( ) 2 2 3 2 3 6 8 8 64 x x x = = 86. ( ) 1 2 2 2 1 1 4 4 4 x x x − − = = − − 87. ( ) ( ) ( ) 4 2 2 2 2 1 2 1 4 2 2 x x y x y x y y − − − = ⋅ = = 88. ( ) ( ) 3 3 3 1 1 3 3 3 3 y x y x y x y x − − − = ⋅ = = 89. 2 3 2 1 3 4 1 1 4 x y x x y x y y xy − − − = = = 90. 2 2 1 1 2 3 1 2 3 1 x y x y x y x y x y − − − − − − = = = 91. 3 4 2 4 2 2 2 3 3 4 1 2 3 2 1 3 1 1 3 ( 2) ( ) 8 3 9 8 9 8 9 8 9 x y z x y z x y z x y z x y z x y z x z y − − − − − − = − = − = = − 92. 2 1 2 1 1 3 4 4 2 4 1 1 1 6 2 1 6 2 4 ( ) 4 2 8 4 8 1 2 1 2 x y z x y z x y x y x y z x y z x y z − − − − − − − − − − − − − = = = = 93. 2 2 2 1 2 2 2 1 2 2 2 3 3 4 4 16 4 3 4 3 9 y x x x x x y y y y − − − −       = = = =             94. ( ) ( ) 3 3 3 2 2 2 2 2 2 3 3 2 6 3 6 3 2 5 5 6 6 6 5 6 216 125 5 y x x y x y x x y y − − − −       = =             = = 95. ( ) ( ) 1 2 2 2 2 4 1 x xy y − = = = − − 96. ( ) ( ) 1 3 1 3 3 3 2 2 y x y x − − − − − = = = 97. ( ) ( ) 2 2 2 2 2 1 4 1 5 x y + = + − = + = 98. ( ) ( ) 2 2 2 2 2 1 4 1 4 x y = − = ⋅ = 99. ( ) ( ) ( ) ( ) 2 2 2 2 1 2 4 xy = ⋅ − = − = 100. ( ) ( ) ( ) ( ) 2 2 2 2 1 1 1 x y + = + − = = 101. 2 2 2 x x = = = 102. ( ) 2 2 x x = = 103. ( ) ( ) 2 2 2 2 2 1 4 1 5 x y + = + − = + = 104. 2 2 2 1 2 1 3 x y x y + = + = + − = + = 105. 1 1 2 2 y x − = = 106. ( ) 2 1 1 x y = − =

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    Section R.2: Algebra Essentials 9 107. If 2, x = 3 2 3 2 2 3 5 4 2 2 3 2 5 2 4 16 12 10 4 10 x x x − + − = ⋅ − ⋅ + ⋅ − = − + − = If 1, x = 3 2 3 2 2 3 5 4 2 1 3 1 5 1 4 2 3 5 4 0 x x x − + − = ⋅ − ⋅ + ⋅ − = − + − = 108. If 1, x = 3 2 3 2 4 3 2 4 1 3 1 1 2 4 3 1 2 8 x x x + − + = ⋅ + ⋅ − + = + − + = If 2, x = 3 2 3 2 4 3 2 4 2 3 2 2 2 32 12 2 2 44 x x x + − + = ⋅ + ⋅ − + = + − + = 109. 4 4 4 4 (666) 666 3 81 222 (222)   = = =     110. ( ) 3 3 3 3 3 3 3 3 1 (0.1) (20) 2 10 10 1 2 10 10 2 8   = ⋅ ⋅     = ⋅ ⋅ = = 111. 6 (8.2) 304, 006.671 ≈ 112. 5 (3.7) 693.440 ≈ 113. 3 (6.1) 0.004 − ≈ 114. 5 (2.2) 0.019 − ≈ 115. 6 ( 2.8) 481.890 − ≈ 116. 6 (2.8) 481.890 − ≈ − 117. 4 ( 8.11) 0.000 − − ≈ 118. 4 (8.11) 0.000 − − ≈ − 119. 2 454.2 4.542 10 = × 120. 1 32.14 3.214 10 = × 121. 2 0.013 1.3 10 − = × 122. 3 0.00421 4.21 10 − = × 123. 4 32,155 3.2155 10 = × 124. 4 21, 210 2.121 10 = × 125. 4 0.000423 4.23 10 − = × 126. 2 0.0514 5.14 10 − = × 127. 4 6.15 10 61,500 × = 128. 3 9.7 10 9700 × = 129. 3 1.214 10 0.001214 − × = 130. 4 9.88 10 0.000988 − × = 131. 8 1.1 10 110, 000, 000 × = 132. 2 4.112 10 411.2 × = 133. 2 8.1 10 0.081 − × = 134. 1 6.453 10 0.6453 − × = 135. A lw = 136. ( ) 2 P l w = + 137. C d π = 138. 1 2 A bh = 139. 2 3 4 A x = 140. 3 P x = 141. 3 4 3 V r π = 142. 2 4 S r π =

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    Chapter R: Review 10 143. 3 V x = 144. 2 6 S x = 145. a. If 1000, x = 4000 2 4000 2(1000) 4000 2000 $6000 C x = + = + = + = The cost of producing 1000 watches is $6000. b. If 2000, x = 4000 2 4000 2(2000) 4000 4000 $8000 C x = + = + = + = The cost of producing 2000 watches is $8000. 146. 210 80 120 25 60 32 5 $98 + − + − − − = His balance at the end of the month was $98. 147. We want the difference between x and 4 to be at least 6 units. Since we don’t care whether the value for x is larger or smaller than 4, we take the absolute value of the difference. We want the inequality to be non-strict since we are dealing with an ‘at least’ situation. Thus, we have 4 6 x − ≥ 148. We want the difference between x and 2 to be more than 5 units. Since we don’t care whether the value for x is larger or smaller than 2, we take the absolute value of the difference. We want the inequality to be strict since we are dealing with a ‘more than’ situation. Thus, we have 2 5 x − > 149. a. 110 108 110 2 2 5 x − = − = − = ≤ 108 volts is acceptable. b. 110 104 110 6 6 5 x − = − = − = > 104 volts is not acceptable. 150. a. 220 214 220 6 6 8 x − = − = − = ≤ 214 volts is acceptable. b. 220 209 220 11 11 8 x − = − = − = > 209 volts is not acceptable. 151. a. 3 2.999 3 0.001 0.001 0.01 x − = − = − = ≤ A radius of 2.999 centimeters is acceptable. b. 3 2.89 3 0.11 0.11 0.01 x − = − = − = ≤ / A radius of 2.89 centimeters is not acceptable. 152. a. 98.6 97 98.6 1.6 1.6 1.5 x − = − = − = ≥ 97˚F is unhealthy. b. 98.6 100 98.6 1.4 1.4 1.5 x − = − = = < 100˚F is not unhealthy. 153. The distance from Earth to the Moon is about 8 4 10 400, 000, 000 × = meters. 154. The height of Mt. Everest is about 3 8848 8.848 10 = × meters. 155. The wavelength of visible light is about 7 5 10 0.0000005 − × = meters. 156. The diameter of an atom is about 10 1 10 0.0000000001 − × = meters. 157. The smallest commercial copper wire has a diameter of about 4 0.0005 5 10 − = × inches. 158. The smallest motor ever made is less than 2 0.05 5 10 − = × centimeters wide.

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    Section R.3: Geometry Essentials 11 159. ( )( ) ( )( ) 5 1 1 2 2 1.86 10 6 10 2.4 10 3.65 10 186, 000 60 60 24 365 × × × × ⋅ ⋅ ⋅ ⋅ = 10 12 586.5696 10 5.865696 10 = × = × There are about 12 5.9 10 × miles in one light- year. 160. 7 2 5 93, 000, 000 9.3 10 5 10 186, 000 1.86 10 500 seconds 8 min. 20 sec. × = = × × = ≈ It takes about 8 minutes 20 seconds for a beam of light to reach Earth from the Sun. 161. 1 0.333333 ... 0.333 3 = > 1 3 is larger by approximately 0.0003333 ... 162. 2 3 0.666666 ... 0.666 = > 2 3 is larger by approximately 0.0006666 ... 163. No. For any positive number a , the value 2 a is smaller and therefore closer to 0. 164. We are given that 2 1 10 x < < . This implies that 1 10 x < < . Since 10 3.162 x < ≈ and 3.142 x π > ≈ , the number could be 3.15 or 3.16 (which are between 1 and 10 as required). The number could also be 3.14 since numbers such as 3.146 which lie between π and 10 would equal 3.14 when truncated to two decimal places. 165. Answers will vary. 166. Answers will vary. 5 < 8 is a true statement because 5 is further to the left than 8 on a real number line. Section R.3 1. right; hypotenuse 2. 1 2 A bh = 3. 2 C r π = 4. similar 5. True. 6. True. 2 2 2 6 8 36 64 100 10 + = + = = 7. False; the volume of a sphere of radius r is given by 3 4 3 V r π = . 8. True. The lengths of the corresponding sides are equal. 9. True. Two corresponding angles are equal. 10. False. The sides are not proportional. 11. 2 2 2 2 2 5, 12, 5 12 25 144 169 13 a b c a b c = = = + = + = + =  = 12. 2 2 2 2 2 6, 8, 6 8 36 64 100 10 a b c a b c = = = + = + = + =  = 13. 2 2 2 2 2 10, 24, 10 24 100 576 676 26 a b c a b c = = = + = + = + =  = 14. 2 2 2 2 2 4, 3, 4 3 16 9 25 5 a b c a b c = = = + = + = + =  = 15. 2 2 2 2 2 7, 24, 7 24 49 576 625 25 a b c a b c = = = + = + = + =  =

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    Chapter R: Review 12 16. 2 2 2 2 2 14, 48, 14 48 196 2304 2500 50 a b c a b c = = = + = + = + =  = 17. 2 2 2 5 3 4 25 9 16 25 25 = + = + = The given triangle is a right triangle. The hypotenuse is 5. 18. 2 2 2 10 6 8 100 36 64 100 100 = + = + = The given triangle is a right triangle. The hypotenuse is 10. 19. 2 2 2 6 4 5 36 16 25 36 41 false = + = + = The given triangle is not a right triangle. 20. 2 2 2 3 2 2 9 4 4 9 8 false = + = + = The given triangle is not a right triangle. 21. 2 2 2 25 7 24 625 49 576 625 625 = + = + = The given triangle is a right triangle. The hypotenuse is 25. 22. 2 2 2 26 10 24 676 100 576 676 676 = + = + = The given triangle is a right triangle. The hypotenuse is 26. 23. 2 2 2 6 3 4 36 9 16 36 25 false = + = + = The given triangle is not a right triangle. 24. 2 2 2 7 5 4 49 25 16 49 41 false = + = + = The given triangle is not a right triangle. 25. 2 4 2 8 in A l w = ⋅ = ⋅ = 26. 2 9 4 36 cm A l w = ⋅ = ⋅ = 27. 2 1 1 (2)(4) 4 in 2 2 A b h = ⋅ = = 28. 2 1 1 (4)(9) 18 cm 2 2 A b h = ⋅ = = 29. 2 2 2 (5) 25 m 2 2 (5) 10 m A r C r = π = π = π = π = π = π 30. 2 2 2 (2) 4 ft 2 2 (2) 4 ft A r C r = π = π = π = π = π = π 31. 3 8 4 7 224 ft V l w h = = ⋅ ⋅ = ( ) ( ) ( )( ) ( )( ) 2 2 2 2 2 8 4 2 8 7 2 4 7 64 112 56 232 ft S lw lh wh = + + = + + = + + = 32. 3 9 4 8 288 in V l w h = = ⋅ ⋅ = ( )( ) ( )( ) ( )( ) 2 2 2 2 2 9 4 2 9 8 2 4 8 72 144 64 280 in S lw lh wh = + + = + + = + + = 33. 3 3 3 2 2 2 4 4 256 4 cm 3 3 3 4 4 4 64 cm V r S r = π = π⋅ = π = π = π ⋅ = π 34. 3 3 3 2 2 2 4 4 3 36 f 3 3 4 4 3 36 ft V r t S r = π = π⋅ = π = π = π ⋅ = π

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    Section R.3: Geometry Essentials 13 35. 2 2 3 (9) (8) 648 in V r h = π = π = π ( ) ( )( ) 2 2 2 2 2 2 9 2 9 8 162 144 306 in S r rh π π π π π π π = + = + = + = 36. 2 2 3 (8) (9) 576 in V r h = π = π = π ( ) ( ) ( ) 2 2 2 2 2 2 8 2 8 9 128 144 272 in S r rh π π π π π π π = + = + = + = 37. The diameter of the circle is 2, so its radius is 1. 2 2 (1) square units A r = π = π = π 38. The diameter of the circle is 2, so its radius is 1. 2 2 2 (1) 4 square units A = − π = − π 39. The diameter of the circle is the length of the diagonal of the square. 2 2 2 2 2 4 4 8 8 2 2 2 2 2 2 2 d d d r = + = + = = = = = = The area of the circle is: ( ) 2 2 2 2 square units A r = π = π = π 40. The diameter of the circle is the length of the diagonal of the square. 2 2 2 2 2 4 4 8 8 2 2 2 2 2 2 2 d d d r = + = + = = = = = = The area is: ( ) 2 2 2 2 2 4 square units A = π − = π − 41. Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get 8 4 2 8 2 4 4 x x x = ⋅ = = In addition, corresponding angles must have the same angle measure. Therefore, we have 90 A = ° , 60 B = ° , and 30 C = ° . 42. Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get 6 12 16 6 16 12 8 x x x = ⋅ = = In addition, corresponding angles must have the same angle measure. Therefore, we have 30 A = ° , 75 B = ° , and 75 C = ° . 43. Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get 30 20 45 30 45 20 135 or 67.5 2 x x x x = ⋅ = = = In addition, corresponding angles must have the same angle measure. Therefore, we have 60 A = ° , 95 B = ° , and 25 C = ° . 44. Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get 8 10 50 8 50 10 40 x x x = ⋅ = = In addition, corresponding angles must have the same angle measure. Therefore, we have 50 A = ° , 125 B = ° , and 5 C = ° .

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    Chapter R: Review 14 45. The total distance traveled is 4 times the circumference of the wheel. Total Distance 4 4( ) 4 16 64 201.1 inches 16.8 feet C d = = π = π ⋅ = π ≈ ≈ 46. The distance traveled in one revolution is the circumference of the disk 4 π . The number of revolutions = dist. traveled 20 5 1.6 revolutions circumference 4 = = ≈ π π 47. Area of the border = area of EFGH – area of ABCD 2 2 2 10 6 100 36 64 ft = − = − = 48. FG = 4 feet; BG = 4 feet and BC = 10 feet, so CG= 6 feet. The area of the triangle CGF is: 2 1 (4)(6) 12 ft 2 A = ⋅ = 49. Area of the window = area of the rectangle + area of the semicircle. 2 2 1 (6)(4) 2 24 2 30.28 ft 2 A = + ⋅ π⋅ = + π ≈ Perimeter of the window = 2 heights + width + one-half the circumference. 1 2(6) 4 (4) 12 4 2 2 16 2 22.28 feet P = + + ⋅ π = + + π = + π ≈ 50. Area of the deck = area of the pool and deck – area of the pool. 2 2 2 2 (13) (10) 169 100 69 ft 216.77 ft A = π − π = π − π = π ≈ The amount of fence is the circumference of the circle with radius 13 feet. 2 (13) 26 ft 81.68 ft C = π = π ≈ 51. We can form similar triangles using the Great Pyramid’s height/shadow and Thales’ height/shadow: h 126 114 240 { { 2 3 This allows us to write 2 240 3 2 240 160 3 h h = ⋅ = = The height of the Great Pyramid is 160 paces. 52. Let x = the approximate distance from San Juan to Hamilton and y = the approximate distance from Hamilton to Fort Lauderdale. Using similar triangles, we get 1046 58 53.5 1046 53.5 58 964.8 x x x = ⋅ = ≈ 1046 58 57 1046 57 58 1028.0 y y y = ⋅ = ≈ The approximate distance between San Juan and Hamilton is 965 miles and the approximate distance between Hamilton and Fort Lauderdale is 1028 miles. 53. Convert 20 feet to miles, and solve the Pythagorean Theorem to find the distance: 2 2 2 sq. miles 1 mile 20 feet 20 feet 0.003788 miles 5280 feet (3960 0.003788) 3960 30 5.477 miles d d = ⋅ = = + − = ≈ 3960 3960 20 ft d 54. Convert 6 feet to miles, and solve the Pythagorean Theorem to find the distance: 2 2 2 sq. miles 1 mile 6 feet 6 feet 0.001136 miles 5280 feet (3960 0.001136) 3960 9 3 miles d d = ⋅ = = + − = ≈ 3960 3960 6 ft d

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    Section R.3: Geometry Essentials 15 55. Convert 100 feet to miles, and solve the Pythagorean Theorem to find the distance: 1 mile 100 feet 100 feet 0.018939 miles 5280 feet = ⋅ = 2 2 2 sq. miles (3960 0.018939) 3960 150 12.2 miles d d = + − ≈ ≈ Convert 150 feet to miles, and solve the Pythagorean Theorem to find the distance: 1 mile 150 feet 150 feet 0.028409 miles 5280 feet = ⋅ = 2 2 2 sq. miles (3960 0.028409) 3960 225 15.0 miles d d = + − ≈ ≈ 56. Given 0, 0 and m n m n > > > , if 2 2 2 2 , 2 and a m n b mn c m n = − = = + , then ( ) ( ) 2 2 2 2 2 2 4 2 2 4 2 2 4 2 2 4 2 2 4 2 a b m n mn m m n n m n m m n n + = − + = − + + = + + and ( ) 2 2 2 2 4 2 2 4 2 c m n m m n n = + = + + 2 2 2 , and a b c a b c ∴ + = → represent the sides of a right triangle. 57. Let l = length of the rectangle and w = width of the rectangle. Notice that 2 2 ( ) ( ) [( ) ( )][( ) ( )] (2 )(2 ) 4 4 l w l w l w l w l w l w l w lw A + − − = + + − + − − = = = 2 2 1 4 So [( ) ( ) ] A l w l w = + − − Since 2 ( ) 0 l w − ≥ , the largest area will occur when l – w = 0 or l = w ; that is, when the rectangle is a square. But 1000 2 2 2( ) 500 2 250 l w l w l w l l w = + = + = + = = = The largest possible area is 2 250 62500 = sq ft. A circular pool with circumference = 1000 feet yields the equation: 500 2 1000 r r π π =  = The area enclosed by the circular pool is: 2 2 2 2 500 500 79577.47 ft A r π π π π   = = = ≈     Thus, a circular pool will enclose the most area. 58. Consider the diagram showing the lighthouse at point L, relative to the center of Earth, using the radius of Earth as 3960 miles. Let P refer to the furthest point on the horizon from which the light is visible. Note also that 362 362 feet miles. 5280 = Apply the Pythagorean Theorem to CPL Δ : ( ) ( ) 2 2 2 1 362 3960 3960 5280 d   + = +     ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 2 1 362 5280 362 5280 3960 3960 3960 3960 23.30 mi. d d = + − = + − ≈ Therefore, the light from the lighthouse can be seen at point P on the horizon, where point P is approximately 23.30 miles away from the lighthouse. Brochure information is slightly overstated. Verify the ship information: Let S refer to the ship’s location, and let x equal the height, in feet, of the ship. We need 1 2 40 d d + ≥ . Since 1 23.30 miles d ≈ we need 2 40 23.30=16.70 miles. d ≥ −

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