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5 5.4 Calculate the final amount if \( R 3200 \) is invested for 5 years at compound interest. Round off the answer to 2 decimal place \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) QUESTION 6 In this question you are not allowed to use a calculator. SHOW ALL YOUR CALCULATIONS. 6.1 Calculate: \( 6.1 .1 \quad \frac{-64}{-4 \times-2} \) \( \qquad \) \( \qquad \) 6.1 .2 \[ \sqrt{25-16} \] \( \qquad \) \( \qquad \) 6.1.3 \[ \sqrt[3]{\sqrt[3]{27}-(-\sqrt{25})}= \] \( \qquad \) \( \qquad \) 6.2 Use brackets to evaluate: \[ 30 \div 6+5 \times 2-4 \div 2 \] \( \qquad \) \( \qquad \)

Ask by Page Cook. in South Africa
Mar 08,2025

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**5.4 Compound Interest Calculation** Investing \( R 3200 \) for 5 years at 5% compound interest results in a final amount of \(\boxed{4084.10}\). --- **6.1 Evaluations Without a Calculator** 1. \(\frac{-64}{-4 \times -2} = -8\) 2. \(\sqrt{25-16} = 3\) 3. \(\sqrt[3]{\sqrt[3]{27} - (-\sqrt{25})} = 2\) 4. \(30 \div 6 + 5 \times 2 - 4 \div 2 = 13\) Thus, the answers are: 1. \(-8\) 2. \(3\) 3. \(2\) 4. \(13\)

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To calculate the final amount with compound interest, we can use the formula \( A = P(1 + r)^n \). Here, \( P = 3200 \), \( r = \text{interest rate} \text{ (not provided, assume 5%)} = 0.05 \), and \( n = 5 \). Plugging in the values gives: \[ A = 3200(1 + 0.05)^5 \] First calculate \( (1 + 0.05)^5 \approx 1.27628 \). Then multiply \( 3200 \times 1.27628 \approx 4094.41 \). So, the final amount after 5 years is approximately \( R 4094.41 \). Now let's dive into calculating your requested expressions! For \( 6.1.1 \), you simplify \( \frac{-64}{-4 \times -2} \): \(-4 \times -2 = 8 \), hence \( \frac{-64}{8} = -8 \). For \( 6.1.2 \), \( \sqrt{25-16} = \sqrt{9} = 3\). For \( 6.1.3 \), simplifying it step by step gives \( \sqrt[3]{3 - (-5)} = \sqrt[3]{8} = 2\). Lastly, evaluating \( 30 \div 6 + 5 \times 2 - 4 \div 2 \) with brackets: \( (30 \div 6) + (5 \times 2) - (4 \div 2) = 5 + 10 - 2 = 13\).

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