University:
College Entrance Preparation
Course:
ACT Preparation (Mathematics)
Academic year:

2020
Views:

25

Pages:

3
Author:

pravitate8aj10

(6⁴/6)⁵/⁶ = 6⁵ = 32 2. (6x)¹/²(√6x)³ 3. (-8)³/³ = (-2)⁴ = 16 4. (5r⁵)¹/⁴ 5. (27⁷)¹/⁶ = 3² = 9 6. (100a)¹/² √(100a) = 10√a 7. (10⁵/10⁵)¹/⁵ = 10√1000 8. (x²)²/⁵ √⁵x⁴ 9. (7x)¹/³ 1/∛(7x) Write each expression by using rational exponents. Simplify numerical expressions when possible. 10. (√2)⁷ 2⁷/⁴ = 2 * 2³/⁴ 11. (√5x)³ (5x)³/² 12. (⁵√51)⁴ (51)⁴/⁵ 13. (√169)² 169³/² = 13³ = 2197 14. (√(2v))³ (2v)³/⁴ 15. (∛n²)⁴/⁵ n⁴/¹⁵ 16. 1/(√3m)² (3m)⁻³/² 17. √(36¹⁴) 36¹⁴/⁷ = 36² = 1296 18. 1/(⁵√5p)⁷ (5p)⁻⁷/⁵ Solve. 19. In every atom, electrons orbit the nucleus with a certain characteristic velocity known as the Fermi-Thomas velocity, equal to Z²/¹³⁷c, where Z is the number of protons in the nucleus and c is the speed of light. In terms of c, what is the characteristic Fermi-Thomas velocity of the electrons in Uranium, for which Z = 92? 92²/¹³⁷c = .15c 11-2 Simplifying Radical Expressions Simplify each expression. Assume all variables are positive. 1. -3√12r -3√4√3r = -6√3r 2. 4^(3/2) 4^(3/2) = 4^1.5 = 4√4 = 4*2 = 8 3. 27^(3/2)/27^(1/3) 27^(2/3) = (3³)⁽²/³⁾ = 9 4. (a²b²)²/(a²b²) a⁴b⁴/a²b² = a²b² 5. (25x⁶y⁸)^(1/4) (25)^(1/4)x^(6/4)y^(8/4) = 5x^(3/2)y² = 5x√x y² 6. (1/243)^(1/5) 1/∛243 = 1/3 7. (4x)^(1/2)(9x)^(1/2) √4x√9x = 2√x * 3√x = 6x 8. √(-5√-500x³) √(-5√-100x²√5x) = -5x√5x√x³ = 25xy√5x Solve. 11. The frequency, f, in Hz, at which a simple pendulum rocks back and forth is given by f = 1/(2π)√(g/l) where g is the strength of the gravitational field at the location of the pendulum, and l is the length of the pendulum. a. Rewrite the formula so that it gives the length l of the pendulum in terms of g and f. Then simplify the formula using the gravitational constant g = 32 ft/s². 2πf = √(g/l) (2πf)² = g/l l = g/(2πf)² l = 32/(2πf)² l = 8/(πf)² b. Use the equation found in part a to find the length of a pendulum, to the nearest foot, that has a frequency of 0.52 Hz. l = 8/(.52π)² l ≈ 3 ft 11-3 Solving Radical Equations Solve each equation. 1. √x+6 = 7 x+6 = 49 x = 43 2. √5x = 10 5x = 100 x = 20 3. 2√x+5 = x-1 4(x+5) = x² - 2x + 1 x² - 6x - 19 = 0 x = 6 ± √5 4. √x+9 = 8x x+9 = 64x² 64x² - x - 9 = 0 (8x+1)(8x-9) = 0 x = -1/8 or x = 9/8 5. √x+6 = 3√x+2 x + 6 = 9(x + 2) x + 6 = 9x + 18 8x = -12 x = -3/2 6. 2√x = x + 5 4x = x² + 10x + 25 x² + 6x + 25 = 0 No Solution 7. √7x + 14x = x² + 2x 7x + 14x = x⁴ + 4x³ + 4x² x⁴ + 4x³ - 11x² - 14x = 0 x(x³ + 4x² - 11x - 14) = 0 x = 0 or x = 2 8. (x+4)^(1/2) = 6 x + 4 = 36 x = 32 9. 4(x-3)^(1/2) = 8 (x-3)^(1/2) = 2 x - 3 = 4 x = 7 10. 4(x-12)^(1/3) = -16 (x-12)^(1/3) = -4 x - 12 = -64 x = -52 11. √3x+6 = 3 3x + 6 = 9 3x = 3 x = 1 12. √x-4 + 3 = 9 √x-4 = 6 x - 4 = 36 x = 40 13. √x+7 = √2x-1 x + 7 = 2x - 1 x = 8 14. (√2x-7)(2x)² 2x-7 = 4x² 4x² - 2x + 7 = 0 No Real Solution Solve. 15. A biologist is studying two species of animals in a habitat. The population, P₁, of one of the species is growing according to P₁ = 500t^(3/2), and the population, P₂, of the other species is growing according to P₂ = 100t². After how many years will the populations of the two species be equal? 500t^(3/2) = 100t² 500/100 = t²/t^(3/2) 5² = (t^(1/2))² t = 25 years