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Math test english 8
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Nội dung Text: Math test english 8
- – ACT MATH TEST PRACTICE –
61. Simplify ( 21 2 )−3.
x
6
a. 6x
b. 8x6
1
c. 6x6
3
d. 8x5
1
e. 8x5
62. If 4x = 3y + 15 and 2y − x = 0, find x.
f. 6
g. 3
h. 2
i. −1
j. 5
−3
63. Simplify 36 2 .
a. −6
b. −216
c. −12
1
d. 216
− 21
e. 16
64. If x3 = −50, the value of x is between which two integers?
f. 3 and 4
g. 7 and 8
h. −3 and −4
i. −2 and −3
j. −7 and −8
65. Find the value of x.
112°
x° x°
a. 25°
b. 136°
c. 112°
d. 68°
e. 34°
179
- – ACT MATH TEST PRACTICE –
66. Line l is parallel to line m. Find the measure of angle x.
l
21°
x°
m
120°
f. 99°
g. 39°
h. 21°
i. 121°
j. 106°
67. Find the radius of the circle with center (4, −2) that is tangent to the y-axis.
a. 2
b. 6
c. 1
d. 4
e. 10
68. Find the area, in square units, of the circle represented by the equation (x − 5)2 + (y − 2)2 = 36.
f. 6π
g. 36π
h. 25π
i. −2π
j. 4π
69. m∠ABC = 120° and m∠CDE = 110°. Find the measure of ∠BCD.
A
B
120°
D
E
110°
C
a. 70°
b. 50°
c. 60°
d. 150°
e. 40°
180
- – ACT MATH TEST PRACTICE –
70. The ratio of the side lengths of a right triangle is 1:1: 2. Find the sine of the smallest angle.
f. 1
2
2
g. 2
h. 2
i. 1
j. 2
71. What is the minimum value of 9cos x?
a. 9
b. 0
c. −90
d. −2
e. −9
72. A triangle with angles measuring 30°, 60°, and 90° has a smallest side length of 7. Find the length of
the hypotenuse.
f. 14
g. 7 3
h. 2
i. 12
j. 18
73. The Abrams’ put a cement walkway around their rectangular pool. The pool’s dimensions are 12 feet
by 24 feet and the width of the walkway is 5 feet in all places. Find the area of the walkway.
a. 748 square feet
b. 288 square feet
c. 460 square feet
d. 205 square feet
e. 493 square feet
74. Triangle XYZ is an equilateral triangle. YW is an altitude of the triangle. If YX is 14 inches, what is the
length of the altitude?
f. 7 3 inches
g. 7 inches
h. 7 2 inches
i. 6 3 inches
j. 12 inches
181
- – ACT MATH TEST PRACTICE –
75. What is the sum of the solutions to the equation 2x2 = 2x + 12?
a. 4
b. 7
c. 1
d. 9
e. −1
9
76. Find the value of sin A if angle A is acute and cos A = 10 .
11
f. 10
5
g. 4
10
h. 9
19
i. 100
19
j. 10
77. Find the value of x.
¯¯¯
√5
45°
x
a. 2
b. 1
c. 7
d. 10
e. 2 5
182
- – ACT MATH TEST PRACTICE –
78. Which equation corresponds to the graph below?
(0,3)
(−5,0) (5,0)
(0,−3)
x2 y2
f. + =1
25 9
g. 25x2 + 9y 2 = 1
x2 y2
h. =1
−
25 9
y2 x2
i. + =1
25 9
j. 5x2 + 3y2 = 3
79. What is the inequality that corresponds to the graph below?
a. y > 3x + 2
b. y ≤ − 3x + 2
c. y ≥ − 3x + 2
d. y < 3x + 2
e. y < − 3x + 2
4x − 5
80. What is the domain of the function f (x) = x2 + 3x − 4 ?
f. {x | x ≠ 0}
g. Ø
h. All real numbers
i. {x | x ≠ 3}
j. {x | x ≠ −4 and x ≠ 1}
183
- – ACT MATH TEST PRACTICE –
P ractice Questions Answers and Explanations
1. Choice a is correct. The word and indicates a decimal point. Therefore, the decimal point should go
after five hundred twelve and before sixteen thousandths. The number 16 must end in the thousandths
place, which is three digits to the right of the decimal. The correct answer is 512.016.
Choice b is “five hundred twelve and sixteen hundredths.”
Choice c is “five hundred twelve thousand, one hundred sixty.”
Choice d is “fifty one and two hundred sixteen thousandths.”
Choice e is “five hundred twelve and sixteen ten thousandths.”
2. Choice f is correct. First, change the fractional parts of the problem to have the common denominator
of 12.
4 142 − 1 192
Subtract the numerators. Since 4 is less than 9, you must borrow one whole from the whole number 4.
This means that you are adding 12 to the first fraction.
12
3 16 − 1 192 .
12
Subtract the fractional parts then the whole numbers. The final answer is 2 172 .
3. Choice b is correct. The correct order of operations must be used to simplify the expression. You may
remember this as PEMDAS or “Please Excuse My Dear Aunt Sally.” The P stands for parentheses or any
grouping symbol. Absolute value is a grouping symbol, so it will be done first.
|−8| + 4 × 23 =
8 + 4 × 23
Next, perform the exponent part.
8+4×8
Then, the multiplication.
8 + 32
Last, the addition.
The final answer is 40.
4. Choice g is correct. This problem can be approached a couple of different ways. The simplest way
might be to look at multiples of 4 and 5 until the multiples add to 18. If both 4 and 5 are multiplied by
2, they become 8 and 10. 8 plus 10 is 18. Therefore, there are 8 boys and 10 girls in the class.
The problem can also be done with an equation.
4x + 5x = 18
When solved, x = 2. Multiply 4 by 2 to find that there are 8 boys.
184
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