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algebraic expression x + 9. ANSWER: x +9 Study Guide and Review -Chapter 1 Write a verbal expression for each algebraic expression. 8. h – 7 SOLUTION: The expression shows h minus seven. So, the verbal expression the difference between h and 7 can be used to describe the algebraic expression h – 7. ANSWER: the difference between h and 7 9. 3x 12. two thirds of a number d to the third power SOLUTION: The words two-thirds of suggest multiplication. So, the verbal expression two-thirds of a number d to the third power can be written as the algebraic expression . ANSWER: 2 13. 5 less than four times a number SOLUTION: The expression shows the product of the factors 3 2 2 and x . The factor x represents a number raised to the second power or squared. So, the verbal expression, the product of 3 and x squared can be used to describe the algebraic expression 3x . SOLUTION: Let x represent a number. The words less than suggest subtraction, and the word times suggests multiplication So, the verbal expression 5 less than four times a number can be written as the algebraic expression 4x – 5. ANSWER: the product of 3 and x squared ANSWER: 4x – 5 2 Evaluate each expression. 3 10. 5 + 6m 5 14. 2 SOLUTION: 3 The expression shows the sum of 5 and 6m . The 3 term 6m represents the product of the factors 6 and 3 SOLUTION: 3 m . The factor m represents a number raised to the third power or cubed. So, the verbal expression five more than the product of six and m cubed can be 3 used to describe the algebraic expression 5 + 6m . ANSWER: five more than the product of six and m cubed ANSWER: 32 3 15. 6 SOLUTION: Write an algebraic expression for each verbal expression. 11. a number increased by 9 SOLUTION: Let x represent a number. The word increased suggests addition. So, the verbal expression a number increased by 9 can be written as the algebraic expression x + 9. ANSWER: 216 4 16. 4 SOLUTION: ANSWER: x +9 12. two thirds of a number d to the third power eSolutions Manual - Powered by Cognero SOLUTION: The words two-thirds of suggest multiplication. So, ANSWER: 256 Page 1 17. BOWLING Fantastic Pins Bowling Alley charges $2.50 for shoe rental plus $3.25 for each game. ANSWER: Study Guide and Review -Chapter 1 216 4 ANSWER: 18 20. 7 + 2(9 – 3) 16. 4 SOLUTION: SOLUTION: ANSWER: 256 17. BOWLING Fantastic Pins Bowling Alley charges $2.50 for shoe rental plus $3.25 for each game. Write an expression representing the cost to rent shoes and bowl g games. SOLUTION: Let g represent the number of games. To find the cost of g games, multiply the cost of one game, $3.25, by g. To find the total cost, add the result to the cost of shoe rental. So, the expression 2.50 + 3.25g represents the cost to rent shoes and bowl g games. ANSWER: 19 21. 8 ∙ 4 – 6 ∙ 5 SOLUTION: ANSWER: 2 5 22. [(2 – 5) ÷ 9]11 SOLUTION: ANSWER: 2.50 + 3.25g Evaluate each expression. 18. 24 – 4 ∙ 5 SOLUTION: ANSWER: 33 ANSWER: 4 2 23. SOLUTION: 19. 15 + 3 – 6 SOLUTION: ANSWER: 18 20. 7 + 2(9 – 3) SOLUTION: ANSWER: 3 eSolutions Manual - Powered by Cognero ANSWER: 19 Evaluate each expression if a = 4, b = 3, and c = 9. Page 2 24. c + 3a SOLUTION: ANSWER: Study Guide and Review -Chapter 1 3 Evaluate each expression if a = 4, b = 3, and c = 9. 24. c + 3a SOLUTION: Replace c with 9 and a with 4. ANSWER: 10 27. ICE CREAM The cost of a one-scoop sundae is $2.75, and the cost of a two-scoop sundae is $4.25. Write and evaluate an expression to find the total cost of 3 one-scoop sundaes and 2 two-scoop sundaes. SOLUTION: To find the cost of 3 one-scoop sundaes and 2 twoscoop sundaes, multiply the cost of a one-scoop sundae by 3 and add that to the product of 2 and the cost of a two-scoop sundae. So, the expression 2.75 (3) + 4.25(2) can be used to find the total cost of 3 one-scoop sundaes and 2 two-scoop sundaes. ANSWER: 21 2 25. 5b ÷ c SOLUTION: Replace b with 3 and c with 9. The total cost of 3 one-scoop sundaes and 2 twoscoop sundaes is $16.75. ANSWER: 2.75(3) + 4.25(2); $16.75 Evaluate each expression using properties of numbers. Name the property used in each step. 28. 18 ∙ 3(1 ÷ 3) SOLUTION: 18 ∙ 3(1 ÷ 3) ANSWER: 5 Substitution = 18 ∙ (3) 2 26. (a + 2bc) ÷ 7 Multiplicative Inverse Multiplicative Identity = 18 ∙ 1 = 18 SOLUTION: Replace a with 4, b with 3 and c with 9. ANSWER: 18 ∙ 3(1 ÷ 3) Substitution = 18 ∙ (3) Multiplicative Inverse Multiplicative Identity = 18 ∙ 1 = 18 29. ANSWER: 10 SOLUTION: 27. ICE CREAM The cost of a one-scoop sundae is $2.75, and -the cost by of Cognero a two-scoop sundae is $4.25. eSolutions Manual Powered Write and evaluate an expression to find the total cost of 3 one-scoop sundaes and 2 two-scoop sundaes. Substitution Substitution Page 3 = 18 ∙ 1 29. (16 – 4 ) + 9 = 16 – 16 + 9 =0 +9 =9 Inverse = 18 Multiplicative Identity Study Guide and Review -Chapter 1 Substitution Additive Inverse Additive Identity 31. SOLUTION: SOLUTION: Substitution Substitution Substitution Multiplicative Inverse Substitution Multiplicative Inverse Multiplicative Identity Substitution ANSWER: Substitution ANSWER: Substitution Multiplicative Inverse Substitution 2 30. (16 – 4 ) + 9 Substitution Multiplicative Inverse Multiplicative Identity Substitution SOLUTION: 2 (16 – 4 ) + 9 = 16 – 16 + 9 =0 +9 =9 Substitution Additive Inverse Additive Identity 32. 18 + 41 + 32 + 9 SOLUTION: 18 + 41 + 32 + 9 = 18 + 32 + 41 + 9 = (18 + 32) + (41 + 9) = 50 + 50 = 100 ANSWER: 2 (16 – 4 ) + 9 = 16 – 16 + 9 =0 +9 =9 Substitution Additive Inverse Additive Identity Commutative (+) Associative (+) Substitution Substitution 31. ANSWER: 18 + 41 + 32 + 9 = 18 + 32 + 41 + 9 = (18 + 32) + (41 + 9) = 50 + 50 = 100 SOLUTION: Commutative (+) Associative (+) Substitution Substitution Substitution eSolutions Manual - Powered by Cognero 33. Substitution Multiplicative Inverse Multiplicative Identity Substitution SOLUTION: Page 4 Commutative (+) Commutative (+) = 18 + 32 + 41 + 9 Associative (+) = (18 + 32) + (41 + 9) Substitution = 50 + 50 100 and Review -Chapter Substitution Study=Guide 1 = 8 ∙ 5 ∙ 0.5 = (8 ∙ 5) ∙ 0.5 = 40 ∙ 0.5 = 20 Commutative (×) Associative (×) Substitution Substitution 33. 35. 5.3 + 2.8 + 3.7 + 6.2 SOLUTION: SOLUTION: Commutative (+) Associative (+) Substitution Substitution Commutative (+) Associative (+) Substitution Substitution ANSWER: ANSWER: Commutative (+) Associative (+) Substitution Substitution Commutative (+) Associative (+) Substitution Substitution 34. 8 ∙ 0.5 ∙ 5 SOLUTION: 8 ∙ 0.5 ∙ 5 = 8 ∙ 5 ∙ 0.5 = (8 ∙ 5) ∙ 0.5 = 40 ∙ 0.5 = 20 Commutative (×) Associative (×) Substitution Substitution ANSWER: 8 ∙ 0.5 ∙ 5 = 8 ∙ 5 ∙ 0.5 = (8 ∙ 5) ∙ 0.5 = 40 ∙ 0.5 = 20 36. SCHOOL SUPPLIES Monica needs to purchase a binder, a textbook, a calculator, and a workbook for her algebra class. The binder costs $9.25, the textbook $32.50, the calculator $18.75, and the workbook $15.00. Find the total cost for Monica’s algebra supplies. SOLUTION: To find the total cost for Monica’s Algebra supplies, find the sum of the costs of the binder, the textbook, the calculator and the workbook. $9.25 + $32.50 + $18.75 + $15.00 = $75.50 So, the total cost for Monica’s Algebra supplies is $75.50. ANSWER: $75.50 Commutative (×) Associative (×) Substitution Substitution Use the Distributive Property to rewrite each expression. Then evaluate. 37. (2 + 3)6 SOLUTION: 35. 5.3 + 2.8 + 3.7 + 6.2 SOLUTION: eSolutions Manual - Powered by Cognero ANSWER: Commutative (+) Associative (+) Substitution Substitution ANSWER: 2(6) + 3(6); 30 38. 5(18 + 12) SOLUTION: Page 5 ANSWER: Study Guide and 2(6) + 3(6); 30Review -Chapter 1 38. 5(18 + 12) ANSWER: –2(5) – (–2)(3); –4 42. (8 – 3)4 SOLUTION: SOLUTION: ANSWER: 5(18) + 5(12); 150 ANSWER: 8(4) – 3(4); 20 39. 8(6 – 2) SOLUTION: Rewrite each expression using the Distributive Property. Then simplify. 43. 3(x + 2) SOLUTION: ANSWER: 8(6) – 8(2); 32 ANSWER: 3(x) + 3(2); 3x + 6 40. (11 – 4)3 SOLUTION: 44. (m + 8)4 SOLUTION: ANSWER: 11(3) – 4(3); 21 41. –2(5 – 3) SOLUTION: ANSWER: m(4) + 8(4); 4m + 32 45. 6(d − 3) SOLUTION: ANSWER: 6(d) – 6(3); 6d – 18 ANSWER: –2(5) – (–2)(3); –4 46. –4(5 – 2t) SOLUTION: 42. (8 – 3)4 SOLUTION: ANSWER: –4(5) – (–4)(2t); –20 + 8t ANSWER: 8(4) – 3(4); 20 eSolutions Manual - Powered by Cognero Rewrite each expression using the Distributive Property. Then simplify. 47. (9y – 6)(–3) SOLUTION: Page 6 So, Mrs. Green gives 48 tutoring lessons in 4 weeks. ANSWER: Study Guide and Review -Chapter 1 –4(5) – (–4)(2t); –20 + 8t ANSWER: 4(3 + 5 + 4); 48 Find the solution set of each equation if the replacement sets are x: {1, 3, 5, 7, 9} and y: {6, 8, 10, 12, 14} 50. y – 9 = 3 47. (9y – 6)(–3) SOLUTION: ANSWER: (9y)(–3) – (6)(–3); –27y + 18 SOLUTION: y y–9=3 6 8 10 12 14 48. –6(4z + 3) SOLUTION: True or False? False False 6– 9=3 8– 9=3 10 – 9 = 3 12 – 9 = 3 14 – 9 = 3 False True False The solution set is {12}. ANSWER: –6(4z) + (–6)(3); –24z – 18 49. TUTORING Write and evaluate an expression for the number of tutoring lessons Mrs. Green gives in 4 weeks. ANSWER: {12} 51. 14 + x = 21 SOLUTION: x 14 + x = 21 1 14 + 1 = 21 3 14 + 3 = 21 5 14 + 5 = 21 7 14 + 7 = 21 9 14 + 9 = 21 SOLUTION: To find the number of tutoring lessons Mrs. Green gives in 4 weeks, multiply 4 by the sum of the number of students Mrs. Green tutors on Monday, Tuesday, and Wednesday. So, the expression 4(3 + 5 + 4) can be used to find the number of tutoring lessons Mrs. Green gives in 4 weeks. The solution set is {7}. ANSWER: {7} 52. 4y = 32 SOLUTION: y 4y = 32 6 8 10 12 14 So, Mrs. Green gives 48 tutoring lessons in 4 weeks. ANSWER: 4(3 + 5 + 4); 48 SOLUTION: eSolutions Manual - Powered by Cognero y–9=3 4(6) = 32 4(8) = 32 4(10) = 32 4(12) = 32 4(14) = 32 True or False? False True False False False Find the solution set of each equation if the replacement sets are x: {1, 3, 5, 7, 9} and y: {6, 8, 10, 12, 14} 50. y – 9 = 3 y True or False? False False False True False True or False? The solution set is {8}. ANSWER: {8} 53. 3x – 11 = 16 SOLUTION: x 3x – 11 = 16 Page 7 True or The solution set is {8}. The solution set is {6}. ANSWER: Study Guide and Review -Chapter 1 {8} 53. 3x – 11 = 16 55. 2(x – 1) = 8 SOLUTION: x 3x – 11 = 16 1 3 5 7 9 3(1) – 3(2) – 3(5) – 3(7) – 3(9) – ANSWER: {6} 11 = 16 11 = 16 11 = 16 11 = 16 11 = 16 True or False? False False False False True SOLUTION: x 2(x – 1) = 8 1 3 5 7 9 2(1 – 2(3 – 2(5 – 2(7 – 2(9 – 1) = 8 1) = 8 1) = 8 1) = 8 1) = 8 True or False? False False True False False The solution set is {9}. The solution set is {5}. ANSWER: {9} ANSWER: {5} 54. Solve each equation. 56. a = 24 – 7(3) SOLUTION: y 6 True or False? True 8 False 10 False 12 False 14 False SOLUTION: ANSWER: 3 2 57. z = 63 ÷ (3 – 2) SOLUTION: The solution set is {6}. ANSWER: {6} 55. 2(x – 1) = 8 SOLUTION: x 2(x – 1) = 8 1 3 5 7 9 2(1 – 1) = 8 2(3 – 1) = 8 2(5 – 1) = 8 2(7 – 1) = 8 2(9 – 1) = 8 True or False? False False True False False The solution set is {5}. ANSWER: {5} eSolutions Manual - Powered by Cognero Solve each equation. 56. a = 24 – 7(3) ANSWER: 9 58. AGE Shandra’s age is four more than three times Sherita’s age. Write an equation for Shandra’s age. Then solve the equation if Sherita’s is 3 years old. SOLUTION: Let K = Sherita’s age. Let E = Shandra’s age. The words more than suggest addition and the word times suggests multiplication. So, 3K + 4 = E. To find Shandra’s age when Sherita is 3, replace the K in the equation with 3 and solve for E. Page 8 ANSWER: Study Guide and Review -Chapter 1 9 58. AGE Shandra’s age is four more than three times Sherita’s age. Write an equation for Shandra’s age. Then solve the equation if Sherita’s is 3 years old. SOLUTION: Let K = Sherita’s age. Let E = Shandra’s age. The words more than suggest addition and the word times suggests multiplication. So, 3K + 4 = E. To find Shandra’s age when Sherita is 3, replace the K in the equation with 3 and solve for E. Mapping: List the x-values in the domain and the yvalues in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range. The domain is {1, 2, 3, 4}, and the range is {3, 4, 5, 6}. ANSWER: So, Shandra is 13 years old. ANSWER: 3K + 4 = E; 13 Express each relation as a table, a graph, and a mapping. Then determine the domain and range. 59. {(1, 3), (2, 4), (3, 5), (4, 6)} SOLUTION: Table: Place the x-coordinates into the first column of the table. Place the corresponding y-coordinates in the second column of the table. Graph: Graph each ordered pair on a coordinate plane. Mapping: List the x-values in the domain and the yvalues in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the eSolutions Manual - Powered by Cognero D = {1, 2, 3, 4} R = {3, 4, 5, 6} 60. {(–1, 1), (0, –2), (3, 1), (4, –1)} SOLUTION: Table: Place the x-coordinates into the first column of the table. Place the corresponding y-coordinates in the second column of the table. Graph: Graph each ordered pair on a coordinate plane. Page 9 Mapping: List the x-values in the domain and the yvalues in the range. Draw arrows from the x-values Graph: Graph each ordered pair on a coordinate plane. Study Guide and Review -Chapter 1 Mapping: List the x-values in the domain and the yvalues in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range. Mapping: List the x-values in the domain and the yvalues in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range. The domain is {–1, 0, 3, 4}, and the range is {–2, –1, 1}. ANSWER: The domain is {–2, –1, 0}, and the range is {2, 3, 4}. ANSWER: D = {–1, 0, 3, 4} R = {–2, –1, 1} 61. {(–2, 4), (–1, 3), (0, 2), (–1, 2)} SOLUTION: Table: Place the x-coordinates into the first column of the table. Place the corresponding y-coordinates in the second column of the table. Express the relation shown in each table, mapping, or graph as a set of ordered pairs. Graph: Graph each ordered pair on a coordinate plane. 62. eSolutions Manual - Powered by Cognero SOLUTION: To express the relation as a set of ordered pairs, write the x-coordinates followed by the Page 10 corresponding y-coordinates. So, the ordered pairs are {(5, 3), (3, –1), (1, 2), (–1, 0)}. and then multiply by 7. Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140 Study Guide and Review -Chapter 1 Express the relation shown in each table, mapping, or graph as a set of ordered pairs. The domain is the number of seeds planted, {50, 100, 150, 200}. The range is the number of plants growing, {35, 70, 105, 140}. Graph the number of seeds planted on the x-axis and the number of plants growing on the y-axis. Then, graph the ordered pairs from the table. 62. SOLUTION: To express the relation as a set of ordered pairs, write the x-coordinates followed by the corresponding y-coordinates. So, the ordered pairs are {(5, 3), (3, –1), (1, 2), (–1, 0)}. ANSWER: {(5, 3), (3, –1), (1, 2), (–1, 0)} ANSWER: 63. SOLUTION: To express the relation as a set of ordered pairs, write the values in the domain as the x-coordinates and the corresponding range values as the ycoordinates. So, the ordered pairs are {(–2, –2), (0, – 3), (2, –2), (2, 0), (4, –1)}. ANSWER: {(–2, –2), (0, –3), (2, –2), (2, 0), (4, –1)} 64. GARDENING On average, 7 plants grow for every 10 seeds of a certain type planted. Make a table to show the relation between seeds planted and plants growing for 50, 100, 150, and 200 seeds. Then state the domain and range and graph the relation. SOLUTION: To find the number of plants that grow for a certain number of seeds, divid the number of seeds by 10 and then multiply by 7. Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140 eSolutions Manual - Powered by Cognero The domain is the number of seeds planted, {50, 100, 150, 200}. The range is the number of plants Determine whether each relation is a function. 65. SOLUTION: A function is a relationship between input and output. Page 11 In a function, there is exactly one output for each input. So, this relation is a function. input. For this function the x value of 6 has two different y outputs: 3 and 0, so it is not a function. Study Guide and Review -Chapter 1 Determine whether each relation is a function. ANSWER: not a function If f (x) = 2x + 4 and g(x) = x 2 – 3, find each value. 68. f (–3) SOLUTION: 65. SOLUTION: A function is a relationship between input and output. In a function, there is exactly one output for each input. So, this relation is a function. ANSWER: function ANSWER: –2 69. g(2) SOLUTION: 66. SOLUTION: A function is a relationship between input and output. In a function, there is exactly one output for each input. So, this relation is a function. ANSWER: function ANSWER: 1 70. f (0) SOLUTION: 67. {(8, 4), (6, 3), (4, 2), (2, 1), (6, 0)} SOLUTION: A function is a relationship between input and output. In a function, there is exactly one output for each input. For this function the x value of 6 has two different y outputs: 3 and 0, so it is not a function. ANSWER: 4 71. g(–4) ANSWER: not a function SOLUTION: If f (x) = 2x + 4 and g(x) = x 2 – 3, find each value. 68. f (–3) SOLUTION: ANSWER: 13 72. f (m + 2) eSolutions Manual - Powered by Cognero ANSWER: –2 SOLUTION: Page 12 ANSWER: Study Guide and Review -Chapter 1 13 72. f (m + 2) SOLUTION: ANSWER: 2 9p – 3 74. GRADES A teacher claims that the relationship between number of hours studied for a test and test score can be described by g(x) = 45 + 9x, where x represents the number of hours studied. Graph this function. SOLUTION: To graph the function, first make a table of values. ANSWER: 2m + 8 73. g(3p ) SOLUTION: ANSWER: 2 x 1 2 3 4 5 g(x) = 45 + 9x g(1) = 45 + 9(1) = 54 g(2) = 45 + 9(2) = 63 g(3) = 45 + 9(3) = 72 g(4) = 45 + 9(4) = 81 g(5) = 45 + 9(5) = 90 Graph the hours studied, x, on the x-axis and the test scores, g(x), on the y-axis. Then, graph the ordered pairs in the table. Draw a line through the points. 9p – 3 74. GRADES A teacher claims that the relationship between number of hours studied for a test and test score can be described by g(x) = 45 + 9x, where x represents the number of hours studied. Graph this function. SOLUTION: To graph the function, first make a table of values. x 1 2 3 4 5 g(x) = 45 + 9x g(1) = 45 + 9(1) = 54 g(2) = 45 + 9(2) = 63 g(3) = 45 + 9(3) = 72 g(4) = 45 + 9(4) = 81 g(5) = 45 + 9(5) = 90 ANSWER: Graph the hours studied, x, on the x-axis and the test scores, g(x), on the y-axis. Then, graph the ordered pairs in the table. Draw a line through the points. eSolutions Manual - Powered by Cognero 75. Identify the function graphed as linear or nonlinear. Then estimate and interpret the intercepts of the graph, any symmetry, where the function is positive, negative, increasing, and decreasing, the x-coordinate Page 13 of any relative extrema, and the end behavior of the graph. Study Guide and Review -Chapter 1 75. Identify the function graphed as linear or nonlinear. Then estimate and interpret the intercepts of the graph, any symmetry, where the function is positive, negative, increasing, and decreasing, the x-coordinate of any relative extrema, and the end behavior of the graph. SOLUTION: Linear or Nonlinear: The graph is not a line, so the function is nonlinear. y-Intercept: The graph intersects the y-axis at about (0, 56), so the y-intercept is about 5.6. This means that about 56,000 U.S. patents were granted in 1980. x-Intercept: The graph does not intersect the xaxis, so there is no x-intercept. This means that in no year were 0 patents granted. Symmetry: The graph has no line symmetry. Positive/Negative: The function is positive for all values of x, so the number of patents will always have a positive value. Increasing/Decreasing: The function is increasing for all values of x. Extrema: The y-intercept is a relative minimum, so the number of patents granted was at its lowest in 1980. End Behavior: As x increases, y increases. As x decreases, y decreases. ANSWER: Nonlinear; the graph intersects the y-axis at about (0, 56), so the y-intercept is about 56. This means that about 56,000 U.S. patents were granted in 1980. The graph has no symmetry. The graph does not intersect the x-axis, so there is no x-intercept. This means that in no year were 0 patents granted. The function is positive for all values of x, so the number of patents will always have a positive value. The function is increasing for all values of x. The yintercept is a relative minimum, so the number of patents granted was at its lowest in 1980. As x increases, y increases. As x decreases, y decreases. eSolutions Manual - Powered by Cognero Page 14