The radius, \( R \), of a sphere is 9.9 cm . Calculate the sphere's volume, \( V \). Use the value 3.14 for \( \pi \), and round your answer to the nearest tenth. (Do

To find the volume \( V \) of a sphere, you can use the formula: \[ V = \frac{4}{3} \pi R^3 \] Substituting \( R = 9.9 \) cm and \( \pi = 3.14 \): \[ V = \frac{4}{3} \times 3.14 \times (9.9)^3 \] Calculating \( (9.9)^3 \): \[ 9.9 \times 9.9 = 98.01 \] \[ 98.01 \times 9.9 \approx 970.2999 \] So now plugging it back into the volume formula gives: \[ V \approx \frac{4}{3} \times 3.14 \times 970.2999 \] Calculating: \[ V \approx 4.18667 \times 970.2999 \approx 4062.253 \] Finally, rounding to the nearest tenth, the volume \( V \) is approximately: \[ \boxed{4062.3 \text{ cm}^3} \] Now, if we get into some interesting facts about spheres! Imagine this: the ancient Greeks, such as mathematician Archimedes, were fascinated by the sphere because of its perfect symmetry and beauty. They viewed the sphere as the perfect shape since it has no edges or vertices and is completely uniform. Did you know Archimedes even created a formula for the volume of a sphere that matches exactly what we use today? How's that for mathematical legacy? When working with volume calculations, a common mistake is to confuse the radius with the diameter. Remember, the diameter is twice the radius! So always double-check your input values, especially when dealing with larger numbers, because that small mix-up can lead to big errors in your final volume. Always visualize or sketch the shape if you're struggling to keep track of the geometric properties!