1. What is the radian measure of an angle whose degree measure is 72◦ 180◦ 72◦ = πradians xradians Solving for x 72π x = radians 180 2π x = radians 5 Answer: B
2. What is the length of AC? Using Pythagorean Theorem:
162 + b2 = 202
256 + b2 = 400 b2 = 144 b = 12 Answer: E
3. one solution to z2 + 64 = 0 is
z2 + 64 = 0
z2 = −64 √ √ z2 = −64 Since there is a negative under the radical, we get imaginary roots. Thus
z = −8i
or z = +8i Answer: A
4. Simplifying: p 36x10y12 − 36y12 = p36y12 (x10 − 1) p = 36y12p(x10 − 1) √ = 6y6 x10 − 1
Answer: C
1 5. Simplifying:
1 1 1 1 1 −3 9 6 3 −3 9 6 27a b c = 27 3 a 3 b 3 c 3 = 3a−1b3c2 3b3c2 = a Answer: C
6. Solving for x: √ 8 + x + 14 = 12 √ x + 14 = 12 − 8 √ x + 14 = 4 √ 2 x + 14 = 42 x + 14 = 16 x = 16 − 14 x = 2
Answer: C
7. Simplify x2 − 9 (x + 1)2 2x − 6 × ÷ x2 − 1 (2x + 3)(x + 3) 1 − x Rewrite as a multiplication.
x2 − 9 (x + 1)2 1 − x × x2 − 1 (2x + 3)(x + 3) 2x − 6
Simplify each polynomial
(x − 3)(x + 3) (x + 1)(x + 1) 1 − x × (x − 1)(x + 1) (2x + 3)(x + 3) 2(x − 3)
Canceling like terms (x + 1) − 2(2x + 3) Answer: C
2 8. Simplifying:
1 11 x−5 11 + 2 2 + 2 x−5 (x−5) = (x−5) (x−5) x + 1 x + 1 x−5+11 2 = (x−5) x + 1 x+6 2 = (x−5) x + 1 x + 6 = (x − 5)2(x + 1)
Answer: B
9. What is x? 15◦ + 15◦ + (5x)◦ = 180◦ 30◦ + (5x)◦ = 180◦ (5x)◦ = 180◦ − 30◦ (5x)◦ = 150◦ x = 30◦ Answer: E
10. A number is not a root of x4 − 8x3 − 19x2 + 158x + 168 if it does not make the polynomial equal to 0. So, check each answer until one of them does not make the polynomial 0. In this case the answer is 1. f(1) = 14 − 8(1)3 − 19(1)2 + 158(1) + 168 = 300 6= 0 Answer: D
11. Answer: E
12. Using the distance formula where (x1, y1) = (7, 8) and (x2, y2) = (6, 1):
p 2 2 d = (x2 − x1) + (y2 − y1) = p(6 − 7)2 + (1 − 8)2) = p(−1)2 + (−7)2 √ = 1 + 49 √ = 50
Answer: C
3 13. Solving for x: 16x16x+12 = 163x−4 16x+x+12 = 163x−4 162x+12 = 163x−4 2x+12 3x−4 20 = 20 2x + 12 = 3x − 4 x = 16 Answer: B 14. What is the measure of angle C? A triangle’s angles add to 180◦. Thus 65◦ + 45◦ + d◦ = 180◦ Solving for d. d◦ = 70◦ Since d and c are supplementary angles, they add up to 180◦. 70◦ + c = 180◦ c = 110◦ Answer: C 15. Simplifying: xy2 = (xy2)(3x2y−1)4 (3x2y−1)−4 = (xy2)(81x8y−4) 81x9 = y2 Answer: D
4 16. Using rules of Logarithms: logbc = 4 ⇒ b = c. Answer: D 17. (3, −5) → (3, 5) → (−3, 5) → (5, −3) Answer: A 18. Simplify b8 a3 + a4 4a3 4b13 Simplifying b8 a3(1 + a) 4a3 4b13 1 + a 16b5 Answer: C
4 19. Simplifying: √ √ 3 27x 3 33x √ = √ 81x 92x √ 3 3 x = √ 9 x 1 ! 1 x 3 = 1 3 x 2
1 1 1 = x 3 − x 2 3 1 1 − 1 = x 3 2 3 1 − 1 = x 6 3 1 = 1 3x 6 1 = √ 3 6 x
Answer: D
20. Simplify (cd3)3 × (−2c4d)4 c3d9 × 16c16d4 16c19d13 Answer: E
21.
log11(x + 4) = 2log115 2 log11(x + 4) = log115
log11(x + 4) = log1125 11log11(x+4) = 11log1125 11log11 (x+4) = 11log11 25 x + 4 = 25 x = 25 − 4 x = 21
Answer: E
22. x2 − 2x < 8
5 x2 − 2x − 8 < 0 (x − 4)(x + 2) < 0 Test an x value between -2 and 4, say 0
(0 − 4)(0 + 2) = (−4)(2) < 0 Therefore −2 < x < 4 Answer: D 23. One root of 2x2 − 2x − 1 is Using the quadratic equation √ 2 ± 4 − 4 ∗ 2 ∗ −1 4 √ 2 ± 12 4 √ 2 ± 2 3 4 √ 1 ± 3 2 √ 1 + 3 x = 2 or √ 1 − 3 x = 2 Answer: B 24. Length = θ ∗ radians π L = ∗ 60 = 10π 6 Answer: C 25. From the property of right triangles we have the following equations AB2 = 52 + x2 (1) BC2 = 112 + x2 (2) 162 = AB2 + BC2 (3) By substituting equation (1) and (2) into equation (3) we have that 162 = 52 + x2 + 112 + x2 2x2 = 110 √ x = 55 Answer: B
6 26. By solving for x we have