Math Review of Exponents D5 Solution

HPC Notes & HW Review of Exponents [Day 5] Name: Factor Completely 1. x³y³ - x⁴y² x³y²(y - x²) 2. x⁵y² - x³y⁷ x³y²(x² - y⁵) 3. x^(3/2)y^(1/2) - x^(1/2)y^(3/2) x^(1/2)y^(1/2)(x - y) 4. x^(1/2)y^(-1/2) - x^(3/2)y^(1/2) x^(1/2)y^(-1/2)(1 - xy) 5. (x - 1)^(3/2) - 4(x - 1)^(1/2) (x - 1)^(1/2)(x - 1 - 4) (x - 1)^(1/2)(x - 5) 6. (x + 1)^(3/2) + 4(x + 1)^(1/2) (x + 1)^(1/2)(x + 1 + 4) (x + 1)^(1/2)(x + 5) 7. (x² + 1)^(3/2) - x²(x² + 1)^(1/2) (x² + 1)^(1/2)(x² + 1 - x²) (x² + 1)^(1/2) 8. (x² + 2)^(1/2) - x²(x² + 2)^(-3/2) (x² + 2)^(-1/2)(x² + 2 - x²) 2(x² + 2)^(-1/2) 9. (2x + 1)^(2/3) - 4(2x + 1)^(-1/3) (2x + 1)^(-1/3)(2x + 1 - 4) (2x + 1)^(-1/3)(2x - 3) 10. (1 + x²)^(-3/2) - (1 + x²)^(-1/2) (1 + x²)^(-3/2)(1 - (1 + x²)) (1 + x²)^(-3/2)(-x²) -x²(1 + x²)^(-3/2) Exponential Function: y = ab^x a = initial amount b = growth/decay factor x = time y = final amount Exponential Growth: b > 1 Domain: (-∞, ∞) Range: (0, ∞) Exponential Decay: 0 < b < 1 Domain: (-∞, ∞) Range: (0, ∞) 1. Complete the table below. The cost of each item grows exponentially. | Item | Annual Rate of Increase | Cost Now | Cost in 10 years | Cost in 20 years | |---|---|---|---|---| | a. Airplane Ticket | 15% | $300 | $1213.67 | $4909.96 | | b. Jar of Mustard | 7% | $1 | $1.97 | $3.87 | | c. College Tuition | 10% | $12000 | $31124.91 | $80730.00 | 2. Complete the table below. The value of each item decays exponentially. | Item | Annual Rate of Decrease | Cost Now | Value in 3 years | Value in 6 years | |---|---|---|---|---| | a. Farm Tractor | 25% | $65000 | $27421.88 | $11568.60 | | b. Value of the Dollar | 6% | $1 | $0.83 | $0.69 | | c. Value of the Dollar | 8% | $1 | $0.78 | $0.61 | 3. The value of a car in dollars t years from now is V(t) = 4000(0.85)^t. a. What is the annual rate of depreciation (the rate at which the car loses value)?   r = 0.15 or 15% b. In how many years will the value of the car be approximately half what it is now? 2000 = 4000(0.85)^t t ≈ 4.3 years 4. The cost of a certain brand of camera has been increasing at 8% per year. If a camera now costs $150, find the cost: a. 2 years and 6 months from now   150(1 + 0.08)^(2.5) ≈ $181.82 b. 4 years and 3 months ago 150(1 + 0.08)^(-4.25) ≈ $108.15 5. The value of a computer depreciates at the rate of 25% per year. If a computer is now worth $2400, find its approximate value: a. 3 years and 6 months   from now 2400(0.75)^(3.5) ≈ $876.85 b. 20 months ago 2400(0.75)^(-20/12) ≈ $3756.52 6. Suppose the population of a nation grows at 3% per year. If the population was 30,000,000 people in 1990, what will be the population, to the nearest million, in the year 2000?   30,000,000(1 + 0.03)^10 ≈ 40,317,491.38 ≈ 40,000,000 7. If $1000 is invested so that it grows at the rate of 10% per year, what will the investment be worth in 20 years? 1000(1 + 0.10)^20 = 1000(1.1)^20 ≈ $6727.50 Half-Life [8-11] A(t) = A₀(½)^(t/k) A₀ = initial amount t = time k = half-life interval 8. A colony of bacteria decays so that the population t days from now is given by A(t) = 1000(½)^(t/4) a. What is the amount present when t = 0? A₀ = 1000 b. How much will be present in 4 days? 1000(½)^(4/4) = 500 c. What is the half-life? 4 days 9. The half-life of a radioactive isotope is 4 days. If 3.2 kg are present now, how much will be present after: a. 4 days? 3.2(½)^(4/4) = 1.6 kg b. 8 days? 3.2(½)^(8/4) = 0.8 kg c. 20 days? 3.2(½)^(20/4) = 0.1 kg d. t days? 3.2(½)^(t/4) 10. The half-life of radium is about 1600 years. If 1 kg is present now, how much will be present after: a. 3200 years? 1(½)^(3200/1600) = 0.25 kg b. 16000 years? 1(½)^(16000/1600) = 0.00098 kg c. 800 years? 1(½)^(800/1600) = 0.707 kg d. t years? 1(½)^(t/1600) 11. The table shows the amount A(t) (in grams) of a radioactive element present after t days. Suppose that A(t) decays exponentially. | t (days) | A(t) | |---|---| | 0 | 320 | | 4 | 160 | | 8 | 80 | | 10 | 57 | a. What is the half-life of the element? 4 days b. About how much will be present after 16 days? 320(½)^(16/4) = 20 g c. Find an equation for A(t). A(t) = 320(½)^(t/4)