5 . 4 Calculate the final amount if R3200 is invested for 5 years at \( 7,25 \% \) p.a compound interest. Round off the answer to 2 decimal places. \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) (3) [11] QUESTION 6 In this question you are not allowed to use a calculator. SHOW ALL YOUR CALCULATIONS. 6.1 Calculate: 6.1.1 \( \frac{-64}{-4 \times-2} \) \( \qquad \) (2) 6.1 .2 \[ \sqrt{25-16} \] (2) 6.1.3 \[ \sqrt[3]{\sqrt[3]{27}-(-\sqrt{25})}= \] 6.2 Use brackets to evaluate: \[ 30 \div 6+5 \times 2-4 \div 2 \] \( \qquad \) \( \qquad \) \( \qquad \)
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To calculate the final amount after investing R3200 for 5 years at an interest rate of \(7.25\%\) p.a. compounded annually, we can use the formula for compound interest, which is \(A = P(1 + r)^n\). Plugging in \(P = 3200\), \(r = 0.0725\), and \(n = 5\): \[ A = 3200(1 + 0.0725)^5 \approx 3200 \times 1.419 \approx 4539.47 \] So, the final amount after 5 years will be approximately R4539.47. ---- When performing the calculation \( \frac{-64}{-4 \times -2} \), we simplify the denominator first: \(-4 \times -2 = 8\). Therefore, \(\frac{-64}{8} = -8\). For the expression \(\sqrt{25 - 16}\), we first compute \(25 - 16\), which is \(9\). Thus, \(\sqrt{9} = 3\). Now for \(\sqrt[3]{\sqrt[3]{27} - (-\sqrt{25})}\), we know \(\sqrt[3]{27} = 3\) and \(-\sqrt{25} = -5\). Therefore, \(3 - (-5) = 3 + 5 = 8\), leading to \(\sqrt[3]{8} = 2\). Lastly, to evaluate \(30 \div 6 + 5 \times 2 - 4 \div 2\), we perform the operations in this order: \(30 \div 6 = 5\), \(5 \times 2 = 10\), and \(4 \div 2 = 2\). So, it sums up to \(5 + 10 - 2 = 13\).
