Math test english 8

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– 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