To fill in the blanks, we can calculate the volume of the basketball using the formula for the volume of a sphere.
Given the radius \( r = 5 \) inches:
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First, we calculate \( (5)^3 \): \[ (5)^3 = 125 \]
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Then we substitute this value into the volume formula: \[ V = \frac{4}{3} \pi (5)^{3} = \frac{4}{3} \pi (125) \] \[ V = \frac{500}{3} \pi \]
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Next, we can approximate \( \pi \approx 3.14 \): \[ V \approx \frac{500}{3} \times 3.14 \] \[ V \approx \frac{1570}{3} \approx 523.33 \text{ in}^3 \]
Now, filling in the blanks:
- \( V = \frac{500}{3} \pi \)
- \( V \approx 523.33 \) in\(^3\)
Final answer:
- V = \( \frac{500}{3} \pi \)
- V ~ 523.33 in.\(^3\)