Presentation is loading. Please wait.

Presentation is loading. Please wait.

FSA/ALGEBRA BELLRINGERS 1st Semester

Similar presentations


Presentation on theme: "FSA/ALGEBRA BELLRINGERS 1st Semester"— Presentation transcript:

1 FSA/ALGEBRA BELLRINGERS 1st Semester
FSA/EOC Review

2 1. Evaluate (-3)² · 2³ · (-4)³

3 1. Solution: -4608 (-3)² = -3 · -3 = 9 2³ = 2 · 2 · 2 = 8
(-4)³ = -4 · -4 · -4 = -64 9 · 8 · -64 = -4608

4 2. Write an algebraic expression for each word phrase:
A. The quotient of x and 8 B. The sum of 20 and x C. The product of r and y D. 8 more than the difference of 15 and x

5 2. Solution: A. x/8 or x ÷ 8 (Quotient represents division)
B x (Sum represents addition) C. r x y (Product represents mult.) D (15 – x) or (15 – x) + 8 (Difference represents subtraction)

6 3. Simplify the expression:
10[ 3(5 – 2 · 2) + 6(3 – 1) ]

7 3. Solution: 150 10 [ 3 (5 - 2 · 2) + 6 (3 – 1)] 10 [ 3 (5 – 4) + 6 (3 – 1)] 10 [ 3 (1) + 6 (2) ] 10 [ ] 10 [ 15 ] 10 · 15 150

8 4. Evaluate the expression given values of the variables:
m = 6 n = 2 y = 1.5 2m + 3n – 3 2y

9 4. Solution: 5 m = 6 n = 2 y = 1.5 2m + 3n – 3 SO 2(6) + 3(2) - 3
3 18 - 3 15 = 5

10 5. Solve: =

11 5. Solution: 6 3 8 = 2 because 2 · 2 · 2 = 8 16 = 4 because 4 · 4 = 16

12 6. Write an algebraic expression:
A state park charges an entrance fee of $20 plus $18 for each night of camping. Write an algebraic expression describing the total cost of camping for n nights? How much would it cost for 2 nights?

13 6. Solution: $18n + $20 or $20 + $18n 2 nights = $56
n = number of nights $20 is a one time fee $18n represents each night $18n + $20 $18(2) + $20 $36 + $20 = $56

14 7. Solve if c = 3 and d = 5 (3c2 - 3d)2 - 21

15 7. Solution: 123 (3c2 – 3d)2 – 21 (3 · 32 – 3 · 5)2 – 21
(3 · 9 – 3 · 5)2 – 21 (27 – 15)2 – 21 (12)2 – 21 144 – 21 123

16 8. Write an algebraic expression:
The table shows how the total cost of a field trip depends on the number of students attending the trip. Write an expression that represents the total cost of the tickets. How much would it cost for 50 students to attend? Number of Students Total Cost 20 (12 x 20) + 150 40 (12 x 40) + 150 60 (12 x 60) + 150

17 8. Solution: N = Number of Students (12 · N) + 150 (12 · 50) + 150
$750 for 50 students to attend

18 9. Define absolute value and complete the following problems:

19 9. Solution: Absolute value is the distance a number is from “0”.
4|3 + 8| = 4|11| = 4 · 11 = 44 C. 2|-7| - 6|9| = 2 · 7 – 6 · 9 = 14 – 54 = -40

20 10. Simplify: A. ( 3 1/2 )² B. ( 6 1/2 )²

21 10. Solution: A. ( 3 1/2 )² = ·2 = 3 B. ( 6 1/2 )² = 6 1/2·2/1 = 6 2/2 = 6

22 11. Solve: A – (-11) = B = C. -1 – 3 = 8 4

23 11. Solution: A.-29 – (-11) = -29 + (+11) = -18 B. -6 + -8 = -14
C. -1 – 3 = -1 – 6 = -7 (You need a common denominator to add or subtract fractions)

24 12. Solve: A. -4 · 3 · -2 = B · -1 = C. -7 – (-1) · =

25 12. Solution: A. -4 · 3 · -2 = -12 · -2 = 24 B · -1 = = -13 C. -7 – (-1) · = -7 + (+1) · = = 2

26 13. Simplify Each Expression:
B / /2 C /3

27 13. Solution: A. 64 1/3 = ∛64 = ∛ 4 3 = 4 B. 32 1/5 - 81 1/2 =
= = 2 – 9 = -7 C /3 = ·2 = ( 125 1/3 )² = ( )² = (∛ 5 3 )² = 5² = 25

28 14. Simplify Each Expression:
B. 2³/2⁶ C. (2 · 3)³ D. (3/4)² E. (2³)²

29 14. Solution: A. 4² · 4³ = 4²⁺³ = 4⁶ = 4·4·4·4·4·4=4096
B. 2³/2⁶ = 2¯³ = 1/2³ = 1/2·2·2 = 1/8 C. (2 · 3)³ = 6³ = 6·6·6 = 216 D. (3/4)² = 3²/4² = 9/16 E. (2³)² = 2⁶ = 2·2·2·2·2·2 = 64

30 15. Simplify Each Expression
B. 2¯³ · 5¯² C. 2/4¯² D. 6¯⁴ · 6¯¹

31 15. Solution: A. 3¯² = 1/3² = 1/9 B. 2¯³ · 5¯² = 1/2³ · 1/5² = 1/8 · 1/25 = 1/200 C. 2/4¯² = 2·4² = 2·16 = 32 D. 6¯⁴·6¯¹ = 6¯⁴⁺¯¹ = 6¯⁵ = 1/6⁵ = 1/6·6·6·6·6 = 7776

32 16. Label the Real Numbers: Rational, Irrational, Integers, Whole (Some may have more than 1 Label)
C D. -8/17 E. 8 F. -12 G

33 16. Solution: Label the Real Numbers: Rational, Irrational, Integers, Whole
A. -5 – Rational, Integer B. 0 – Rational, Integer, Whole C Irrational D. -8/17 - Rational E. 8 – Rational, Integer, Whole F – Rational, Integer G ³ = 2 – Rational, Integer, Whole

34 17. Name the property A. -y + 0 = -y B. 13(-11) = (-11)13
C. -5 · (m · 8) = (-5 · m) · 8 D · 1 = 2.6

35 17. Solution: A. -y + 0 = -y Identity Property of Addition
B. 13(-11) = -11(13) Commutative Property of Multiplication C. -5 · (m · 8) = (-5 · m) · 8 Associative Property of Multiplication D · 1 = 2.6 Identity Property of Multiplication

36 18. Use the distributive property:
A. 6(4y – 7) = B. -2(-5y – 8) = C. 4(3 + 4y – 2m +9n) =

37 18. Solution: A. 6(4y – 7) = 24y - 42 B. -2(-5y – 8) = 10y + 16
C. 4(3 + 4y – 2m +9n) = y – 8m + 36n

38 19. Simplify each expression:
A. 6y – – m B. 5mn – 3mn +7mn -1 C. 2(y + 3) – 4(y – 2)

39 19. Solution: Combine “like” terms.
A. 6y – – m -m + 6y - 3 B. 5mn – 3mn +7mn -1 9mn - 1 C. 2(y + 3) – 4(y – 2) Distribute 2y + 6 – 4y + 8 -2y + 14

40 20. Tell whether the given number is a solution of each equation:
A. 5x + 1 = 16; -3 B. 2 = 10 – 4y; 2 C. 14 = 1x + 5; 9

41 20. Solution. Substitute the given number into the equation.
A. 5x + 1 = 16; -3 5(-3) + 1 = 16 = 16 -14 = 16 NO B. 2 = 10 – 4y; 2 2 = 10 – 4(2) 2 = 10 – 8 2 = 2 YES C. 14 = 1x + 5; 9 14 = 1(9) + 5 14 = 9 + 5 14 = 14 YES

42 21. Is the ordered pair a solution of the equation? (Yes or No)
A. y = x – 4; (5, 1) B. y = -x; (-7,7) C. y = -3x + 3; (2, 6)

43 21. Solution: (x,y) A. y = x – 4; (5, 1) 1 = 5 – 4; 1 = 1 YES
B. y = -x; (-7,7) 7 = -(-7) 7 = 7 YES C. y = -3x + 3; (2, 6) 6 = -3(2) + 3 6 = 6 = -3 NO

44 22. Evaluate the expression: h = -1 and k = -3
(2h)3 – (k3 – h2)

45 22. Solution: 20 h = -1 and k = -3 (2h)3 – (k3 – h2)
(2 · -1)3 – ((-3)3 – (-1)2) (-2)3 – ((-27) – 1) -8 – (-28) 20

46 23. Simplify each expression:
A. -8(2x – (4 – 7)) B. -.5( y)

47 23. Solution: A. -8(2x – (4 – 7)) -8(2x – (-3)) -8(2x + 3) -16x – 24
-.5( y) 3 – 6y

48 24. Solve the following equations:
A. 5x = -15 8 B. -1 = x + 5

49 24. Solution A. 5x = -15 8 8 ·5x = -15 ·8 5 8 5 x = -24 B. -1 = x + 5
x = -24 B. -1 = x + 5 -6 = x

50 25. Solve the following equation:
A x = -21 10

51 25. Solution: -360 A. 15 + x = -21 -15 10 -15 x = -36 10
x = -36 10 10 ·x = -36 ·10 1 1 X = -360

52 26. Solve the following equation:
-15 = 5(3x – 10) – 5x

53 26. Solution: 3.5 -15 = 5(3x – 10) – 5x Distribute
-15 = 15x – 50 – 5x Combine -15 = 10x – Solve for x 35 = 10x Divide 3.5 = x

54 27. Define the variable, write an equation and solve:
Breanna’s pizza shop charges $9 for a large cheese pizza. Additional toppings cost $1.25 per topping. Brittany paid $15.25 for her large pizza. How many toppings did she order?

55 27. Solution: 5 toppings t = # of toppings 9 + 1.25t = 15.25 -9 -9 .00
1.25t = 6.25 t = 5 toppings

56 28. Define the variable, write an equation, and solve.
Ms. Seli ate at the same restaurant four times. Each time she ordered a salad and left a $5 tip. She spent a total of $54. Find the cost of each salad.

57 28. Solution: $8.50 C = cost of each salad 4(c + 5) = 54 4c + 20 = 54
4c = 34 c = 8.50

58 29. Solve the equation: 14 + 3x = 8x – 3(x – 4)

59 29. Solution: x = 1 14 +-2x = 12 14 + 3x = 8x – 3(x – 4) Distribute
14 + 3x = 8x – 3x + 12 Combine 14 + 3x = 5x + 12 -5x -5x 14 +-2x = 12 -2x = -2 -2 -2 x = 1

60 30. Solve the literal equation for c:

61 30. Solution: P · S = C + I + G + N ·P 1 P 1 PS = C + I + G + N
- I – G – N = - I - G - N SO C = PS – I – G - N

62 31. What is the value of x in the equation?
100 – 3x = x

63 31. Solution: x = 25 100 – 3x = x -x -x 100 – 4x = 0 -100 -100
-4x = -100 x = 25

64 32. Define the variable and solve:
Mr. Meader is buying lunch for his friends(If he has any). He decides to buy chicken wraps and chips for each person. Chicken wraps cost $3.20 and chips cost $0.75 each. What is the maximum number of people Mr. Meader can feed if he has $20?(He is so cheap)

65 32. Solution: 5 people X = maximum # of people Mr. Meader can feed
3.20x + .75x = 20 3.95x = 20 x ≈ 5.06 so x = 5 people

66 33. Define the variable and solve:
The fare for a taxi is $7.50 plus $0.40 per mile. Coach Woodside paid a total fare of $13.50 to take a taxi from Jackson to his house.(His car broke down & no one would give him a ride). What is the distance, in miles, from Jackson to his house? (Poor Coach)

67 33. Solution: 15 miles X = # of miles 7.50 + .40x = 13.50 -7.50 - 7.50
.40x = 6.00 x = 15 miles

68 34. Solve: 2x – 3(4x+5) = -6(x – 3) - 1

69 34. Solution: -8 2x – 3(4x+5) = -6(x – 3) - 1
-4x = 32 x = -8

70 35. Solve the literal equation for “x”.
x + r + 1 = 0 t

71 35. Solution: x = -1t - r x + r + 1 = 0 t -1 -1 x + r = -1 t
-r -r x = -1t - r

72 36. Solve the proportion: x + 3 = x – 3

73 36. Solution: 39 x + 3 = x – 3 7 6 7(x-3) = 6(x+3) Distribute
7(x-3) = 6(x+3) Distribute 7x – 21 = 6x + 18 -6x -6x 1x – 21 = 18 x = 39

74 37. Which property of equality is illustrated in each equation?
B. (2 + 3) + 4 = 2 + (3 + 4) C = 9 + 3

75 37. Solution A. Distributive Property
B. Associative Property of Addition C. Commutative Property of Addition

76 A. Solve for r in terms of C:
38. A. Solve for r in terms of C: C = 2πr B. Solve for “x”: 1 2 𝑥 = 2 6

77 A. Solve for r in terms of C:
38. Solution: A. Solve for r in terms of C: C = 2πr = C = r 2 π 2 π π B. Solve for “x”: 1 2 𝑥 = − same as ( 4 6 ) 1 2 𝑥 = −2 6 (Multiply ½ by the reciprocal) X = −2 6 · = −4 6 = −2 3

78 39. Choose the correct equation that can be used to find the total earnings (T) for h hours:
Raquel works at the zoo and uses the chart to keep track of her hours: A. T = h + 8 B. T = 10h – 8 C. T = 8h Hours (h) Total Earnings (T) 4 32 6 48 10 80 15 120 h ?

79 39. Solution: C 32/4 = 8 48/6 = 8 80/10 = 8 120/15 = 8 Therefore Raquel makes $8 an hour so her Total earnings would be represented by T = 8h

80 40. Write an equation and solve:
A company that refinishes our gym floor charges $200 for materials plus $35 per hour for labor. What will be the total cost if it takes two 8- hour days to re-do the gym floor?

81 40. Solution: $760 T = Total Cost Materials: $200
Labor: 16 hours at $35 per hour 35(16) = T = $760

82 41. Solve: 5x + 1 = 3x – 5

83 41. Solution: 11 5x + 1 = 3x – 5 8 4 4(5x + 1) = 8(3x – 5) Distribute
8 4 4(5x + 1) = 8(3x – 5) Distribute 20x + 4 = 24x – 40 -24x x -4x + 4 = -40 -4x = -44 x = 11

84 42. Are the numbers given a solution of the inequality?
3x + 1 > -3 A. -3 B. -1

85 42. Solution: 3x + 1 > -3 A. 3(-3) + 1 > -3 -9 + 1 > -3
-8 > -3 NO B. 3(-1) + 1 > -3 > -3 -2 > -3 YES

86 43. Write an inequality for each graph:
B.

87 43. Solution: A. x ≤ 3 (A closed circle includes the number)
B. X > -1 (An open circle does not include the number)

88 44. Solve the following inequalities:
A. X – ≥ 10 B. 5x + 5 – 4x < 8 C. -4x ≤ 20

89 44. Solution: A. X – 4 + 2 ≥ 10 x – 2 ≥ 10 +2 +2 x ≥ 12
x ≥ 12 B. 5x + 5 – 4x < 8 1x + 5 < 8 x < 3 C. -4x ≤ 20 (Reverse the sign) x ≥ -5

90 45. Solve and graph the inequality:
3x + 6 ≤ -5(x + 2)

91 45. Solution: 3x + 6 ≤ -5(x + 2) Distribute 3x + 6 ≤ -5x + -10 +5x +5x
8x ≤ -16 x ≤ -2

92 46. Solve:

93 46. Solution: A. 0.40 x + (2x + 0.15) + (x + 0.05) = 1.8
4x = 1.6 x = .4

94 47. Remember: Distance = Rate · Time Distance =
47. Remember: Distance = Rate · Time Distance = ?, Rate = 30, Time = Solve for “t”

95 47. Solution: 75 miles 50(t – 1) = 30t Distribute
50t – 50 = 30t Solve for “t” -30t t 20t – 50 = 0 20t = 50 Time = 2.5 hours Remember: distance = rate · time So distance = 30 · 2.5 Distance = 75 miles

96 48. Solve for “a”:

97 48. Solution: a ≤ 19.56 1.2(a + 0.065a) ≤ 25 Distribute
1.2a a ≤ Combine “Like” Terms 1.278a ≤ 25 a ≤ 19.56

98 49. Solve and Graph: 15 ≤ 7n – 2(n – 10) < 35

99 49. Solution: -1 ≤ n < 3 15 ≤ 7n – 2(n – 10) < 35
-5 ≤ 5n < 15 -1 ≤ n < 3

100 50. Solve and Graph: 84 ≤ x ≤ 86 4

101 50. Solution: 85 ≤ x ≤93 4 ·84 ≤ 86 + 85 + 80 + x ≤ 86 · 4 4
85 ≤ x ≤ 93

102 51. Solve and Graph 3x + 2 < -7 or -4x + 5 < 1

103 51. Solution: x < -3 or x > 1
3x < or -4x < -4 The sign reverses x < or x > 1

104 50. Solve

105 50. Solution: 12 T = 4 breeds of cattle Z = 3 breeds of cattle
T x Z = 3 x 4 = 12 elements

106 51. Solve

107 51. Solution: 11 K U P = 2,10,11,14,18 (K U P) ∩ D = { 11 } Remember that union “unites” everything together and intersection is what is in common between the sets. The GREATEST age would be 11.

108 52. Solve

109 52. Solution: 35 15 Soccer Players + 11 Basketball Players + 9 Soccer and Basketball Players = 35 students

110 53. Find the range of the function for the given domain.
f(x) = 3x² {2, 4, -3}

111 53. Solution: Range {11, 47, 26} f(x) = 3x² - 1
f(2) = 3 · 2² - 1 f(-3) = 3 · (-3)² - 1 f(2) = 3 · 4 – 1 f(-3) = 3 · 9 - 1 f(2) = 12 – 1 f(-3) = f(2) = 11 f(-3) = 26 f(4) = 3 · 4² - 1 f(4) = 3 · 16 – 1 f(4) = 48 – 1 f(4) = 47

112 54. Solve and Graph: 1 ≤ -2x + 7 < 9

113 54. Solution: 1 ≤ -2x + 7 < 9 -7 -7 -7
-6 ≤ -2x < 2 Switch the inequality symbols 3 ≥ x > -1 Rewrite: -1 < x ≤ 3

114 55. Solve and Graph: 3x – 1 < -7 or 4x + 9 ≥ 13

115 55. Solution: 3x – 1 < -7 or 4x + 9 ≥ 13 +1 +1 -9 -9
3x < -6 or 4x ≥ 4 X < -2 or x ≥ 1

116 56. Solve

117 56. Solution: 9 Domain (x-values): { 0, 1, 2, 3, 4, 5, 6 }
Range (y-values): { 0, 5, 8, 9, 8, 5, 0 } So the greatest element in the range is 9.

118 57. Solve if C = 3 and d = 5 (3c2 - 3d)2 - 21

119 57. Solution: 123 (3c2 – 3d)2 – 21 (3 x 32 – 3 x 5)2 – 21
(27 – 15)2 – 21 (12)2 – 21 144 – 21 123

120 58. Choose the correct answer:

121 58. Solution: C Domain would represent the days on the x-axis.
Range would represent the Dollars on the y-axis. The range in dollars is from $10 - $60 inclusive. Therefore: 10 ≤ y ≤ 60

122 59. Simplify the expression:
10[ 3(5 – 2 · 2) + 6(3 – 1) ]

123 59. Solution: 150 10 [ 3 (5 - 2 · 2) + 6 (3 – 1)] 10 [ 3 (5 – 4) + 6 (3 – 1)] 10 [ 3 (1) + 6 (2) ] 10 [ ] 10 [ 15 ] 10 · 15 150

124 60. Define absolute value and complete the following problems:

125 60. Solution: Absolute value is the distance a number is from “0”.
B. 4|3 + 2 x 4| = 4|3 + 8| = 4|11| = 4 x 11 = 44 C. 2|-7| - 6|9| = 2 x 7 – 6 x 9 = 14 – 54 = -40

126 61. Choose the correct answer:

127 61. Solution: B 4.3/10 = 0.43 8.6/20 = 0.43 12.9/30 = 0.43 17.2/40 = 0.43 21.5/50 = 0.43 So the equation would be p = 0.43d OR you could substitute each number into the equation to see which one works. P = 0.43d so 4.3 = 0.43 x 10 (Yes), 8.6 = 0.43 x 20 (yes) etc.

128 62. Solve: 14 + 3x = 8x – 3(x – 4)

129 62. Solution: x = 1 14 +-2x = 12 14 + 3x = 8x – 3(x – 4) Distribute
14 + 3x = 8x – 3x + 12 Combine 14 + 3x = 5x + 12 -5x -5x 14 +-2x = 12 -2x = -2 -2 -2 x = 1

130 63. Solve the proportion: x + 3 = x – 3

131 63. Solution: 39 x + 3 = x – 3 7 6 7(x-3) = 6(x+3) Distribute
7(x-3) = 6(x+3) Distribute 7x – 21 = 6x + 18 -6x -6x 1x – 21 = 18 x = 39

132 64. Solve:

133 64. Solution: 96.5 $ = h = 3.5h 96.5 = h Total number of hours = 96.5

134 65. Solve for “a”:

135 65. Solution: a ≤ 19.56 1.2(a + 0.065a) ≤ 25 Distribute
1.2a a ≤ Combine “Like” Terms 1.278a ≤ 25 a ≤ 19.56

136 66. Solve for “x” x = 5 9x

137 66. Solution: x = 3 (x + 12) = 5 (9x) 9 9(x + 12) = 45x 9x + 108 = 45x
-36x = -108 X = 3

138 67. Solve for “C” S = C + I + G +N P

139 67. Solution: C = PS – I – G - N P · S = C + I + G +N ·P 1 P 1
I – G – N I - G – N PS – I – G – N = C So C = PS – I – G - N

140 68. Solve for “x” x + r + 1 = 0 t

141 68. Solution: x = -1t - r x + r + 1 = 0 t -1 -1 t · x+r = -1 · t 1 t 1
- r r x = -1t - r

142 69. Solve: =

143 69. Solution: 6 3 8 = 2 because 2 · 2 · 2 = 8

144 70. Simplify Each Expression:
B / /2 C /3

145 70. Solution: A. 64 1/3 = ∛64 = ∛ 4 3 = 4 B. 32 1/5 - 81 1/2 =
= = 2 – 9 = -7 C /3 = ·2 = ( 125 1/3 )² = ( )² = (∛ 5 3 )² = 5² = 25

146 71. Simplify Each Expression:
B. 2³/2⁶ C. (2 · 3)³ D. (3/4)² E. (2³)²

147 71. Solution: A. 4² · 4³ = 4²⁺³ = 4⁶ = 4·4·4·4·4·4=4096
B. 2³/2⁶ = 2¯³ = 1/2³ = 1/2·2·2 = 1/8 C. (2 · 3)³ = 6³ = 6·6·6 = 216 D. (3/4)² = 3²/4² = 9/16 E. (2³)² = 2⁶ = 2·2·2·2·2·2 = 64

148 72. Simplify Each Expression:
B. 2¯³ · 5¯² C. 2/4¯² D. 6¯⁴ · 6¯¹

149 72. Solution: A. 3¯² = 1/3² = 1/9 B. 2¯³ · 5¯² = 1/2³ · 1/5² = 1/8 · 1/25 = 1/200 C. 2/4¯² = 2·4² = 2·16 = 32 D. 6¯⁴·6¯¹ = 6¯⁴⁺¯¹ = 6¯⁵ = 1/6⁵ = 1/6·6·6·6·6 = 7776

150 73. Label the Real Numbers: Rational, Irrational, Integers, Whole (Some may have more than 1 Label)
C D. -8/17 E. 8 F. -12 G

151 73. Label the Real Numbers: Rational, Irrational, Integers, Whole (Some may have more than 1 Label)
A. -5 – Rational, Integer B. 0 – Rational, Integer, Whole C Irrational D. -8/17 - Rational E. 8 – Rational, Integer, Whole F – Rational, Integer G ³ = 2 – Rational, Integer, Whole

152 2nd Semester Bellwork FSA/EOC Review Algebra Dr. Sorensen

153 1. Solve for “x” x + (2x ) + (x ) = 1.8

154 1. Solution: x = 0.4 x + (2x + 0.15) + (x + 0.05) = 1.8 Distribute
x + 2x x = 1.8 Combine 4x = 1.8 Solve 4x = 1.6 Divide x = 0.4

155 2. Solve Breanna has a total of $25 to spend on dinner, which includes a 6.5% sales tax and a 20% tip. Taylor used the inequality shown below to calculate the amount in dollars, a , she can spend before tax and tip. How much would that be? 1.2(a a) ≤ 25

156 2. Solution: a ≤ 19.56 1.2(a + 0.065a) ≤ 25 Distribute
1.2a a ≤ 25 Combine “Like” Terms 1.278a ≤ 25 Divide a ≤ $ approximately

157 3. Solve for “x” 2x – 3(4x+5) = -6(x – 3) - 1

158 3. Solution: x = -8 2x – 3(4x+5) = -6(x – 3) - 1
-4x = 32 x = -8

159 4. Solve for “x” 3x + 6 ≤ -5(x + 2)

160 4. Solution: x ≤ -2 3x + 6 ≤ -5(x + 2) Distribute 3x + 6 ≤ -5x + -10
8x ≤ -16 x ≤ -2

161 5. Solve: 3x + 2 < -7 or -4x + 5 < 1

162 5. Solution: x < -3 or x > 1
3x < or -4x < -4 The sign reverses x < or x > 1

163 6. Solve 84 ≤ x ≤ 86 4

164 6. Solution: 85 ≤ x ≤ 93 4 ·84 ≤ 86 + 85 + 80 + x ≤ 86 · 4 1 4 1
336 ≤ x ≤ 344 85 ≤ x ≤ 93

165 7. List the domain and range of the following table. Is this a function?
X y 2 3 5 4 6 7

166 7. Domain are the “x” values – {2,3,4,5} Range are the “y” values –
{3,5,6,7} To be a function, every “x” value must be different. (The “y” values can repeat) THIS IS A FUNCTION

167 8. Solve f(x)=x² + 4; Domain or x = {0,1,2,3}

168 8. f(x) = {4,5,8,13} f(x) = x² + 4 or y = x² + 4 X (Domain) x² + 4
F(x) or y (Range) 0² + 4 4 1 1² + 4 5 2 2² + 4 8 3 3² + 4 13

169 9. Which of the following relations is NOT a function
9. Which of the following relations is NOT a function? Why is it not a function? A. {(0,1),(1,2)(2,3),(3,4)} B. {(1,1),(2,4)(3,3),(4,4)} C. {(1,6),(2,2)(1,3),(3,4)} D. {(1,1),(2,5)(3,6),(4,7)}

170 9. Solution: C C. {(1,6),(2,2)(1,3),(3,4)} A. {(0,1),(1,2)(2,3),(3,4)}
B. {(1,1),(2,4)(3,3),(4,4)} C. {(1,6),(2,2)(1,3),(3,4)} D. {(1,1),(2,5)(3,6),(4,7)} The Domain (x) must be different to be a function and the letter “C” has repeating Domains. The Range (y) can repeat and still be a function.

171 10. Solve Which equation represents the line that passes through the points (5,-4) and (-2,6)? A. 10x + y = 22 B. 10x + 7y = 22 C. 10x – 7y = -22

172 10. Solution: B Which equation represents the line that passes through the points (5,-4) and (-2,6)? A. 10x + y = 22 NO B. 10x + 7y = 22 YES C. 10x – 7y = -22 NO ***Try each ordered pair until you find an equation that works for both ordered pairs. 10(5) + -4 = 22 NO 10(5) + 7(-4) = 22 YES 10(-2) + 7(6) = 22 YES

173 11. Solve The slope of a line is ¾. Write an equation of a line that is PARALLEL to the given line and passes through the point (-2,4).

174 11. Solution: y = 3/4x + 5.5 Remember that y = mx + b
The original line is y = 3/4x + 0 The slope of a parallel line would be the same slope so “m” = ¾ Substitute the ordered pair of the new line in for x and y. (-2,4) y = mx + b 4 = ¾ (-2) + b Solve for “b” 4 = -6/4 + b (Add 6/4 to cancel) +6/ /4 5.5 = b So the new line is y = 3/4x + 5.5

175 12. Solve The formula for finding the circumference of a circle is C = 2Πr. Which of the following is the same equation solved for r in terms of C? A. r = C Π 2 B. r = C 2 Π

176 12. Solution: B C = 2Πr To solve for “r”, we need to divide each side by 2Π . C = 2Πr 2Π 2Π r = C

177 13. Write an equation that is perpendicular to
y = x - 2 and goes through the point (-4,-1).

178 y = -1/2x - 3 13. Solution: y = -1/2x - 3 y = 2x – 2 : m = 2 b = -2
A perpendicular slope would by the reciprocal slope with the opposite sign, which would be -1/2. Now we need to find the new “b” using (-4,-1) as the point the line goes through. y = mx + b so -1 = (-1/2)-4 + b -1 = 2 + b Solve for “b” -3 = b so the new line would be y = -1/2x - 3

179 14. Choose the correct equation.
Breanna tutors students in math. She uses the chart below to keep track of her total earnings. Which equation can Breanna use to find her total earnings, T, for h hours? A. T = h + 8 B. T = 8h C. T = 6h + 8 Hours (h) Total Earnings (T) 4 32 6 48 10 80 15 120 h ?

180 14. Solution: B Total Earnings would be $8 times the total hours. T = 8h 32 = 8(4) 48 = 8(6) 80 = 8(10) 120 = 8(15)

181 15. Name the property. Which property of equality is illustrated by the following equation? (84 x 25) + (84 x 75) = 84( ) A. Commutative B. Associative C. Distributive D. Property of 0

182 15. Solution: C - Distributive
Review: A. Commutative: (3 + 4) = (4 + 3) B. Associative: (3 + 4) + 5 = 3 + (4 + 5) C. Distributive: 84( ) = (84 x 25) + (84 x 75) D. Property of 0: 5 x 0 = 0

183 16. Solve Which of the following is an equation of a horizontal line?
A. 3x + 6y = 0 B. -3y = 21 C. -3x = 21

184 16. Solution: B: -3y = 21 y = -7 C.-3x = 21 -3 -3 x = -7
A. 3x + 6y = 0 -3x x 6y = -3x + 0 so y = -1/2x + 0 B y = 21 y = -7 C.-3x = 21 x = -7

185 17. Which of the following is an equation of a vertical line? A. 3x + 6y = 0 B. -3y = 21 C. -3x = 21

186 17. Solution: C -3x = 21 A. 3x + 6y = 0 -3x x 6y = -3x so y = -1/2x + 0 (Direct Variation) B. -3y = 21 y = -7 (Horizontal line across the y- axis at -7) C. .-3x = 21 x = -7 (Vertical line across the x axis at -7)

187 18.

188

189 19.

190 19. Solution: D (2,-2) According to the graph, the ordered pairs include the following points: (‐4, ‐3), (‐2, 1), (0, 0), (2, 3), (2, ‐2), (3, 1), (4, 3) The x‐coordinate 2 is repeated in two different points, so this does not represent a function. If (2,2) is removed, then this would be considered a function.

191 20. An architect designed an outdoor staircase for a house. The relationship between the height of the steps and the length of the tread is modeled by the equation 57x – 95y = 0. Which of the following represents the slope of the equation? A. 5/3 B. 3/2 C. 2/3 D. 3/5

192 20. Solution: m=57/95 = 3/5 y = mx + b 57x – 95y = 0 Solve for Y
-57x x -95y =-57x y = 57x 95 So the slope = 57 ÷ 19 = 3

193 21. Brianna plotted the two points (20, 75) and (45, 150) on a graph. What is the x- coordinate of the x-intercept of the line that contains these two points? Hint: Use y=mx + b and find the slope and y-intercept. Then find the x-intercept ( ,0). ***This is a bonus buck problem!

194 21. Solution: The x-coordinate of the x-intercept is -5.
y=mx + b m = = 75 = 3 SO y=3x + b Find b by substituting one of the ordered pairs in for x and y. I will use (20,75). 75 = 3(20) + b 75 = 60 + b 15 = b SO y = 3x To find the x- intercept we need to place a “0” in for y and solve for x. It’s easier if the equation is in standard form: -3x + y = 15

195 21. Continued: ( ,0) -3x + y = 15 -3x + 0 = 15 -3x = 15 x = -5 So the x-intercept is (-5,0)

196 22.

197 22. Solution: 25 degrees

198 23.

199 23. Solution: 96.5

200 24.

201 24. Solution: B p=0.43d Substitute the numbers from the table into the equations to see which one works. A = (4.3)(10) NO B = (.43)(20) YES

202 25.

203 25. Solution: C

204 26.

205 26. Solution: 504

206 27.

207 27. Solution: A

208 28.

209 28. Solution: C 5x + 2y ≤ 50 -5x -5x 2y ≤ -5x + 50 2 2 2
y ≤ m = -5 b = 25

210 29.

211 29. Solution: A

212 30. Solution

213 30. Solution: 12.6 y = mx + b 70x + 50y = 630 -70x -70x
y = -7x m = -7 b = 12.6

214 31.

215 31. Solution: C Since line PQ goes down, we know the slope must be negative, so it has to be A or C. (6,12)(11,10) 10 – 12 = -2 11 – y = mx + b 12 = -2 ·6 + b 5 12 = b 14.4 = b so y = -2x

216 32.

217 32. Solution: 2

218 33.

219 33. Solution: 6 Slope of the original line: (-6,1)(4,-4)
-4 – 1 = -5 = -1 Reciprocal and opp. m = 2 4-(-6) y = mx + b (4,-4) is the intersection point -4=2(4) + b -4 = 8 + b -12 = b so y = 2x ( ,0) Standard Form -2x + y = -12 -2x + 0 = -12 x = 6 (6,0)

220 34.

221 34. Solution: -8.4 or -42/5

222 35. **Bonus Buck Problem – you must show your work!

223 35. Solution: C X = Medium Sandwich: $5.39 Y = Large Sandwich: $6.89
Choice “C”: Russ: 3x + 2y = 29.95 3(5.39) + 2(6.89) = 29.95 = YES Stacy: 4x + 1y = 28.45 4(5.39) + 1(6.89) = 28.45 = YES

224 36. Bonus Buck Problem – You must show your work.

225 36. Solution: 9 Songs

226 37.

227 37. Solution: A

228 38.

229 38. Solution: B 16x 2x

230 39.

231 39.

232 40.

233 40. Solution: 35

234 41.

235 41. Solution:


Download ppt "FSA/ALGEBRA BELLRINGERS 1st Semester"

Similar presentations


Ads by Google