write a verbal expression for each...

14
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 2 SOLUTION: The expression shows the product of the factors 3 and x 2 . The factor x 2 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 2 . ANSWER: the product of 3 and x squared 10. 5 + 6m 3 SOLUTION: The expression shows the sum of 5 and 6m 3 . The term 6m 3 represents the product of the factors 6 and m 3 . The factor m 3 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 used to describe the algebraic expression 5 + 6m 3 . ANSWER: five more than the product of six and m cubed 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: x + 9 12. two thirds of a number d to the third power 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: 13. 5 less than four times a number 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: 4x 5 Evaluate each expression. 14. 2 5 SOLUTION: ANSWER: 32 15. 6 3 SOLUTION: ANSWER: 216 4 eSolutions Manual - Powered by Cognero Page 1 Study Guide and Review - Chapter 1

Upload: others

Post on 22-Jul-2020

7 views

Category:

Documents


1 download

TRANSCRIPT

Page 1: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 1

Study Guide and Review -Chapter 1

Page 2: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 2

Study Guide and Review -Chapter 1

Page 3: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 3

Study Guide and Review -Chapter 1

Page 4: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 4

Study Guide and Review -Chapter 1

Page 5: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 5

Study Guide and Review -Chapter 1

Page 6: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 6

Study Guide and Review -Chapter 1

Page 7: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 7

Study Guide and Review -Chapter 1

Page 8: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 8

Study Guide and Review -Chapter 1

Page 9: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 9

Study Guide and Review -Chapter 1

Page 10: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 10

Study Guide and Review -Chapter 1

Page 11: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 11

Study Guide and Review -Chapter 1

Page 12: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 12

Study Guide and Review -Chapter 1

Page 13: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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.

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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 13

Study Guide and Review -Chapter 1

Page 14: Write a verbal expression for each algebraickennettmath.weebly.com/uploads/4/1/9/1/4191431/ch_1_review_answers.pdfcost of a two -scoop sundae. So, the expression 2.75 (3) + 4.25(2)

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. 3x2

SOLUTION:  The expression shows the product of the factors 3

and x2. The factor x

2 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 3x2.

ANSWER:  the product of 3 and x squared

10. 5 + 6m3

SOLUTION:  

The expression shows the sum of 5 and 6m3. The

term 6m3 represents the product of the factors 6 and

m3. The factor m

3 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

used to describe the algebraic expression 5 + 6m3.

ANSWER:  five more than the product of six and m cubed

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:  

x + 9

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:  

13. 5 less than four times a number

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:  4x – 5

Evaluate each expression.

14. 25

SOLUTION:  

ANSWER:  32

15. 63

SOLUTION:  

ANSWER:  216

16. 44

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:  2.50 + 3.25g

Evaluate each expression.

18. 24 – 4 ∙ 5

SOLUTION:  

ANSWER:  4

19. 15 + 32 – 6

SOLUTION:  

ANSWER:  18

20. 7 + 2(9 – 3)

SOLUTION:  

ANSWER:  19

21. 8 ∙ 4 – 6 ∙ 5

SOLUTION:  

ANSWER:  2

22. [(25 – 5) ÷ 9]11

SOLUTION:  

ANSWER:  33

23. 

SOLUTION:  

ANSWER:  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:  21

25. 5b2 ÷ c

SOLUTION:  Replace b with 3 and c with 9.  

ANSWER:  5

26. (a2 + 2bc) ÷ 7

SOLUTION:  Replace a with 4, b with 3 and c with 9. 

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 two-scoop 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.  

  The total cost of 3 one-scoop sundaes and 2 two-scoop 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)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse = 18 Multiplicative Identity

ANSWER:  

 

18 ∙ 3(1 ÷ 3)  

= 18 ∙ (3) Substitution

= 18 ∙ 1 Multiplicative Inverse

= 18 Multiplicative Identity

29. 

SOLUTION:  

 

 

  Substitution

Substitution

Multiplicative Inverse

ANSWER:  

 

 

 

  Substitution

Substitution

Multiplicative Inverse

30. (16 – 42) + 9

SOLUTION:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

ANSWER:  

 

(16 – 42) + 9  

= 16 – 16 + 9 Substitution

= 0 + 9 Additive Inverse = 9 Additive Identity

31.   

SOLUTION:  

 

 

 

Substitution

  Substitution

Multiplicative Inverse

Multiplicative Identity

Substitution

ANSWER:  

 

 

Substitution

  Substitution

Multiplicative Inverse Multiplicative Identity

Substitution

32. 18 + 41 + 32 + 9

SOLUTION:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

ANSWER:  

 

18 + 41 + 32 + 9   = 18 + 32 + 41 + 9 Commutative (+)

= (18 + 32) + (41 + 9) Associative (+)

= 50 + 50 Substitution = 100 Substitution

33. 

SOLUTION:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

34. 8 ∙ 0.5 ∙ 5

SOLUTION:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

ANSWER:  

 

8 ∙ 0.5 ∙ 5   = 8 ∙ 5 ∙ 0.5 Commutative (×) = (8 ∙ 5) ∙ 0.5 Associative (×) = 40 ∙ 0.5 Substitution = 20 Substitution

35. 5.3 + 2.8 + 3.7 + 6.2

SOLUTION:  

 

  Commutative (+)

Associative (+)

Substitution Substitution

ANSWER:  

 

 

 

  Commutative (+)

Associative (+)

Substitution Substitution

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

Use the Distributive Property to rewrite each expression. Then evaluate.

37. (2 + 3)6

SOLUTION:  

ANSWER:  

2(6) + 3(6); 30

38. 5(18 + 12)

SOLUTION:  

ANSWER:  

5(18) + 5(12); 150

39. 8(6 – 2)

SOLUTION:  

ANSWER:  

8(6) – 8(2); 32

40. (11 – 4)3

SOLUTION:  

ANSWER:  

11(3) – 4(3); 21

41. –2(5 – 3)

SOLUTION:  

ANSWER:  –2(5) – (–2)(3); –4

42. (8 – 3)4

SOLUTION:  

ANSWER:  

8(4) – 3(4); 20

Rewrite each expression using the Distributive Property. Then simplify.

43. 3(x + 2)

SOLUTION:  

ANSWER:  

3(x) + 3(2); 3x + 6

44. (m + 8)4

SOLUTION:  

ANSWER:  

m(4) + 8(4); 4m + 32

45. 6(d − 3)

SOLUTION:  

ANSWER:  

6(d) – 6(3); 6d – 18

46. –4(5 – 2t)

SOLUTION:  

ANSWER:  –4(5) – (–4)(2t); –20 + 8t

47. (9y – 6)(–3)

SOLUTION:  

ANSWER:  

(9y)(–3) – (6)(–3); –27y + 18

48. –6(4z + 3)

SOLUTION:  

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.  

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.  

  So, Mrs. Green gives 48 tutoring lessons in 4 weeks.

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

SOLUTION:  

  The solution set is {12}.

y y – 9 = 3 True or False?

6 6 – 9 = 3 False 8 8 – 9 = 3 False

10 10 – 9 = 3 False 12 12 – 9 = 3 True 14 14 – 9 = 3 False

ANSWER:  {12}

51. 14 + x = 21

SOLUTION:  

  The solution set is {7}.

x 14 + x = 21

True or False?

1 14 + 1 = 21 False 3 14 + 3 = 21 False 5 14 + 5 = 21 False 7 14 + 7 = 21 True 9 14 + 9 = 21 False

ANSWER:  {7}

52. 4y = 32

SOLUTION:  

  The solution set is {8}.

y 4y = 32 True or False?

6 4(6) = 32 False 8 4(8) = 32 True 10 4(10) = 32 False 12 4(12) = 32 False 14 4(14) = 32 False

ANSWER:  {8}

53. 3x – 11 = 16

SOLUTION:  

  The solution set is {9}.

x 3x – 11 = 16 True or False?

1 3(1) – 11 = 16 False 3 3(2) – 11 = 16 False 5 3(5) – 11 = 16 False 7 3(7) – 11 = 16 False 9 3(9) – 11 = 16 True

ANSWER:  {9}

54. 

SOLUTION:  

  The solution set is {6}.

y True or False?

6 True

8 False

10 False

12 False

14 False

ANSWER:  {6}

55. 2(x – 1) = 8

SOLUTION:  

  The solution set is {5}.

x 2(x – 1) = 8 True or False?

1 2(1 – 1) = 8 False 3 2(3 – 1) = 8 False 5 2(5 – 1) = 8 True 7 2(7 – 1) = 8 False 9 2(9 – 1) = 8 False

ANSWER:  {5}

Solve each equation.

56. a = 24 – 7(3)

SOLUTION:  

ANSWER:  3

57. z = 63 ÷ (32 – 2)

SOLUTION:  

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 findShandra’s age when Sherita is 3, replace the K in theequation with 3 and solve for E.  

  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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values 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:  

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 inthe second column of the table.

  Graph: Graph each ordered pair on a coordinate plane.

  Mapping: List the x-values in the domain and the y-values in the range. Draw arrows from the x-values in the domain to the corresponding y-values in the range.

The domain is {–2, –1, 0}, and the range is {2, 3, 4}.

ANSWER:  

Express the relation shown in each table, mapping, or graph as a set of ordered pairs.

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)}  

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 y-coordinates. 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 every10 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.

  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.

Planted Growing 50 50 ÷ 10 × 7 = 35 100 100 ÷ 10 × 7 = 70 150 150 ÷ 10 × 7 = 105 200 200 ÷ 10 × 7 = 140

ANSWER:  

Determine whether each relation is a function.

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

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

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:  not a function

If f (x) = 2x + 4 and g (x) = x2 – 3, find each value.

68. f (–3)

SOLUTION:  

ANSWER:  –2

69. g(2)

SOLUTION:  

ANSWER:  1

70. f (0)

SOLUTION:  

ANSWER:  4

71. g(–4)

SOLUTION:  

ANSWER:  13

72. f (m + 2)

SOLUTION:  

ANSWER:  2m + 8

73. g(3p )

SOLUTION:  

ANSWER:  

9p2 – 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.  

  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.  

x g (x) = 45 + 9x 1 g(1) = 45 + 9(1) = 54 2 g(2) = 45 + 9(2) = 63 3 g(3) = 45 + 9(3) = 72 4 g(4) = 45 + 9(4) = 81 5 g(5) = 45 + 9(5) = 90

ANSWER:  

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-coordinateof any relative extrema, and the end behavior of the graph.

SOLUTION:  Linear or Nonlinear: The graph is not a line, so thefunction 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 x-axis, 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 increasingfor 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 nox-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 y-intercept 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

Study Guide and Review -Chapter 1