This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here's the solution to Question 7 and Question 8. QUESTION 7 7.1 Calculate the scrap value of the server at the end of 6 years. The value depreciates using the diminishing-balance method. Step 1: Use the diminishing-balance depreciation formula A = P(1 - i)^n. Given: P = R500\,000, i = 20\% = 0.20, n = 6 years. A = 500\,000(1 - 0.20)^6 A = 500\,000(0.8)^6 A = 500\,000(0.262144) A = 131\,072 The scrap value of the server is R131\,072.00. 7.2 Calculate the cost of the new server if the inflation rate is at 7% per annum. Step 1: Use the compound interest formula for inflation A = P(1 + i)^n. Given: P = R500\,000, i = 7\% = 0.07, n = 6 years. A = 500\,000(1 + 0.07)^6 A = 500\,000(1.07)^6 A = 500\,000(1.50073035) A = 750\,365.175 The cost of the new server is R750\,365.18. 7.3.1 Calculate the value of the monthly installment into the sinking fund. The sinking fund needs to accumulate the cost of the new server (from 7.2), which is R750\,365.18. Interest is 8% per annum compounded monthly. Payments are made immediately, and the last payment is at the end of the 6-year period. This implies an annuity due with 72 payments. Step 1: Identify the variables. FV = R750\,365.18 i = (0.08)/(12) per month n = 6 years × 12 months/year = 72 payments Step 2: Use the future value of an annuity due formula: FV = PMT ((1+i)^n - 1)/(i) (1+i). 750\,365.18 = PMT ((1 + 0.08)/(12))^72 - 1(0.08)/(12) (1 + (0.08)/(12)) Step 3: Calculate the factor. ((1 + 0.08)/(12))^72 - 1(0.08)/(12) (1 + (0.08)/(12)) ≈ (1.006666667)^72 - 10.006666667 (1.006666667) ≈ (1.61303639 - 1)/(0.006666667) (1.006666667) ≈ (0.61303639)/(0.006666667) (1.006666667) ≈ 91.9554585 × 1.006666667 ≈ 92.56914 Step 4: Solve for PMT. PMT = (750\,365.18)/(92.56914) PMT ≈ 8106.00 The monthly installment into the sinking fund is R8106.00. 7.4 After how many months will there be more than R1 000 000 in the fund? The business pays a monthly installment of R15 000. Assuming payments are made at the beginning of each month (annuity due, consistent with 7.3.1). Step 1: Identify the variables. FV = R1\,000\,000 PMT = R15\,000 i = (0.08)/(12) per month Step 2: Use the future value of an annuity due formula: FV = PMT ((1+i)^n - 1)/(i) (1+i). 1\,000\,000 = 15\,000 ((1 + 0.08)/(12))^n - 1(0.08)/(12) (1 + (0.08)/(12)) Step 3: Isolate the term with n. (1\,000\,000 × 0.08)/(12)15\,000 × (1 + (0.08)/(12)) = (1 + (0.08)/(12))^n - 1 (6666.66667)/(15\,000 × 1.006666667) = (1 + (0.08)/(12))^n - 1 (6666.66667)/(15100) = (1 + (0.08)/(12))^n - 1 0.441501 = (1 + (0.08)/(12))^n - 1 1.441501 = (1 + (0.08)/(12))^n Step 4: Solve for n using logarithms. (1.441501) = n (1 + (0.08)/(12)) (1.441501) = n (1.006666667) 0.36569 ≈ n × 0.0066444 n ≈ (0.36569)/(0.0066444) ≈ 55.04 Since the question asks "more than R1 000 000", we round up to the next whole month. After 56 months there will be more than R1 000 000 in the fund. QUESTION 8 8.1 Determine f'(x) from first principles if f(x) = x^3. Step 1: Write down the definition of the derivative from first principles. f'(x) = _h 0 (f(x+h) - f(x))/(h) Step 2: Substitute f(x) = x^3 and f(x+h) = (x+h)^3 into the formula. Recall (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3. f'(x) = _h 0 ((x^3 + 3x^2h + 3xh^2 + h^3) - x^3)/(h) Step 3: Simplify the numerator. f'(x) = _h 0 (3x^2h + 3xh^2 + h^3)/(h) Step 4: Factor out h from the numerator and cancel it with the denominator. f'(x) = _h 0 (3x^2 + 3xh + h^2) Step 5: Apply the limit as h 0. f'(x) = 3x^2 + 3x(0) + (0)^2 f'(x) = 3x^2 8.2 Determine: 8.2.1 f'(x) if f(x) = (x + (1)/(x))^2. Step 1: Expand the function f(x). f(x) = (x + x^-1)^2 f(x) = x^2 + 2(x)(x^-1) + (x^-1)^2 f(x) = x^2 + 2 + x^-2 Step 2: Differentiate f(x) term by term. f'(x) = (d)/(dx)(x^2) + (d)/(dx)(2) + (d)/(dx)(x^-2) f'(x) = 2x + 0 - 2x^-3 f'(x) = 2x - (2)/(x^3) 8.2.2 (dy)/(dx) if y = (3)/(x^3) - x. Step 1: Rewrite the function using negative exponents. y = 3x^-3 - x Step 2: Differentiate y term by term with respect to x. (dy)/(dx) = (d)/(dx)(3x^-3) - (d)/(dx)( x) (dy)/(dx) = 3(-3)x^-3-1 - (1)x^1-1 (dy)/(dx) = -9x^-4 - (dy)/(dx) = -(9)/(x^4) - That's 2 down. 3 left today — send the next one.