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Calculate the number of learners in June and solve other related problems.

Mathematics

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5 parts
3.2

594 learners\text{594 learners}

3.3

120\text{120}

4.2

2.466 units long\text{2.466 units long}

4.3

R85.50\text{R}85.50

5.1

343^4

Step-by-step solution

12 steps

Here are the solutions to the visible questions:

Part 3.2· 3 steps

  1. Identify the number of learners in March.

    The number of learners in March is given as 540.

  2. Calculate the 10% increase from the March number.

    10% of 540=10100×540=5410\% \text{ of } 540 = \frac{10}{100} \times 540 = 54

  3. Add the increase to the March number to find the number of learners in June.

    540+54=594540 + 54 = 594

    The number of learners in June is 594 learners\boxed{\text{594 learners}}.

Part 3.3

120\text{120}

  1. Find the full number (100%) if 30% of the number is 45.

    Let the number be xx.

    0.30x=450.30x = 45 x=450.30x = \frac{45}{0.30} x=150x = 150

  2. Calculate 80% of that number.

    80% of 150=80100×15080\% \text{ of } 150 = \frac{80}{100} \times 150 0.80×150=1200.80 \times 150 = 120

    80% of that number is 120\boxed{\text{120}}.

Part 4.1

The expression to calculate is not provided in the image.

Part 4.2

2.466 units long\text{2.466 units long}

  1. Divide the length of the ribbon by 4.

    9.8644=2.466\frac{9.864}{4} = 2.466

    Each piece of ribbon will be 2.466 units long\boxed{\text{2.466 units long}}.

Part 4.3

R85.50\text{R}85.50

  1. Convert the mass of apples from grams to kilograms.

    3750 g=37501000 kg=3.75 kg3750 \text{ g} = \frac{3750}{1000} \text{ kg} = 3.75 \text{ kg}

  2. Calculate the cost of apples.

    Cost of apples=3.75 kg×R12.40/kg=R46.50\text{Cost of apples} = 3.75 \text{ kg} \times \text{R}12.40/\text{kg} = \text{R}46.50

  3. Calculate the cost of bananas.

    Cost of bananas=2.5 kg×R15.60/kg=R39.00\text{Cost of bananas} = 2.5 \text{ kg} \times \text{R}15.60/\text{kg} = \text{R}39.00

  4. Calculate the total amount the hawker pays.

    Total cost=R46.50+R39.00=R85.50\text{Total cost} = \text{R}46.50 + \text{R}39.00 = \text{R}85.50

    The hawker pays R85.50\boxed{\text{R}85.50} all together.

Part 5.1

343^4

  1. Find the prime factors of 81.

    81=3×2781 = 3 \times 27 81=3×3×981 = 3 \times 3 \times 9 81=3×3×3×381 = 3 \times 3 \times 3 \times 3

  2. Write 81 in exponential form using the smallest base.

    The smallest base is 3.

    81=3481 = 3^4

    In exponential form, 81 is 34\boxed{3^4}.

Part 5.2

The expression to calculate is not provided in the image.

Part 6.1

The expression to calculate is not provided in the image.

Part 6.2

The statement to complete is not provided in the image.

Part 7.1

The figures and the question related to them are not provided in the image.

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