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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

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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 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 p p 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) p = 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: p 8 + x + 14 = 12 p x + 14 = 12 − 8 p x + 14 = 4 p 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 p = 1 + 49 p = 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: p p 3 27x 3 33x p = p 81x 92x p 3 3 x = p 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 = p 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 p 2 ± 4 − 4 ∗ 2 ∗ −1 4 p 2 ± 12 4 p 2 ± 2 3 4 p 1 ± 3 2 p 1 + 3 x = 2 or p 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 p x = 55 Answer: B 6 26. By solving for x we have 2 log2 x − 33 − log2 (x) = 3 x2 − 33 log = 3 2 x x2 − 33 =) = 23 x 8x = x2 − 33 0 = x2 − 8x − 33 (x + 3) (x − 11) = 0 x = −3; 11 Since we cannot log a negative number we have that the solution is x = 11 Answer: C 27. By solving for x we have that 9x = 31−2x 32x = 31−2x =) 2x = 1 − 2x 4x = 1 1 x = 4 Answer: C 28. By finding common denominators we have that a 6b a b − = − 6 6 6 6b − a = 6 Answer: E 29. By finding common denominators we have that y 4 y (9x − 13) 4 (x + 5y) − = − x + 5y 9x − 13 (x + 5y) (9x − 13) (9x − 13) (x + 5y) 9xy − 13y − 4x − 20y = (9x − 13) (x + 5y) 9xy − 33y − 4x = 9x2 − 13x + 45xy − 65y Answer: B 7 30. By factoring we have that x2 + 8x > −12 x2 + 8x + 12 > 0 (x + 2) (x + 6) > 0 Using x = −2; −6 as critical values we see that the number line can be split into three regions in which we will test numbers in each region. If x = 0 we have that (0 + 2) (0 + 6) > 0 which is a true statement. If x = −3 we have that (−3 + 2) (−3 + 6) > 0 which is a false statement. If x = −10 we have that (−10 + 2) (−10 + 6) > 0 which is a true statement. So we have a true statement in the regions x < −6 and x > −2 Answer: B 31. By solving for x we have that log5 (x) − log5 (x + 2) = log5 (11) x log = log (11) 5 x + 2 5 x =) = 11 x + 2 x = 11x + 22 −10x = 22 11 x = − 5 But since we cannot take the log of a negative number we have no solution, which corresponds to solution E. Answer: E 32. Since we know that the circumference is given by C = πr we can solve for r and then use the formula for the area of a circle that is we have 2πr = C 2πr = 12π r = 6 Now plugging this value into the area of a circle formula we have A = πr2 = π(6)2 = 36π Answer: A 8 33. Since we know that the area of the circle is A, we also know that A = πr2. Because of the increase in the diameter by a factor of 2, we know that the radius will increase by a factor of 2 as well. Thus the knew area of the circle is given by π (2r)2 = π4r2 = 4πr2 = 4A Answer: E 34. Since we know that f(x) = 7x2 − x + 2, then f(c − 4) = 7(c − 4)2 − (c − 4) + 2 = 7(c2 − 8c + 16) − c + 4 + 2 = 7c2 − 56c + 112 − c + 6 = 7c2 − 57c + 118 Answer: A 35. From the problem we know that l1 and l2 are parallel, and l3 is perpendicular to l2. Now we must evaluate each statement to determine which is true and which is false by looking at a simple drawing. From the drawing we can see that l3 is perpendicular to l1 do to the parallel relationship l1 has with l2, thus a) is true. If a line l4 was drawn perpendicular to l3 we can see that l4 would be parallel to l1, thus b) is true. We see that if we draw a line l4 to be parallel to l3 it would in fact intersect l1 meaning they are not parallel, so c) is false. 9 We can easily see that l1 intersects l3, so we know that d) is true. If a lie l4 is no perpendicular to l3, then by the relationship of the lines we can see that it would have to intersect l2, thus we have e) to be true. Evaluating the above statements we see that the false statement is C. Answer: C 36. What is the length of EC The length of AC is 12, the length of DE is 9, and the length of BC is 20. Also the segments of DE and AC are parallel. Using similar triangles: AC DE = BC BE 12 9 = 20 BE BE = 15 EC = BC − BE EC = 20 − 15 EC = 5 Answer: A 10.
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