To solve each of the volume problems step by step:
-
Volume of a sphere with a radius of 9 cm: \[ V = \frac{4}{3} \pi r^3 \] Here, \( r = 9 , \text{cm} \). \[ V = \frac{4}{3} \pi (9)^3 = \frac{4}{3} \pi (729) = \frac{2916}{3} \pi = 972 \pi \] Therefore, the volume is 972π cm³.
-
Volume of a sphere with a radius of 3 m: \[ V = \frac{4}{3} \pi r^3 \] Here, \( r = 3 , \text{m} \). \[ V = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = \frac{108}{3} \pi = 36 \pi \] Using \( \pi \approx 3.14 \): \[ V \approx 36 \times 3.14 = 113.04 \] Thus, the volume is approximately 113.04 m³.
-
Volume of a sphere with a diameter of 12 m: First, we find the radius: \[ r = \frac{d}{2} = \frac{12}{2} = 6 , \text{m} \] Now, calculate the volume: \[ V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = \frac{864}{3} \pi = 288 \pi \] Thus, the volume is 288π m³.
-
Volume of a spherical chocolate truffle with a radius of 12 mm: \[ V = \frac{4}{3} \pi r^3 \] Here, \( r = 12 , \text{mm} \). \[ V = \frac{4}{3} \pi (12)^3 = \frac{4}{3} \pi (1728) = \frac{6912}{3} \pi = 2304 \pi \] Using \( \pi \approx 3.14 \): \[ V \approx 2304 \times 3.14 \approx 7238.56 \] Rounding to the nearest hundredth, the volume is approximately 7238.56 mm³.
-
Volume of the container and the space left: The volume of one tennis ball: \[ V = \frac{4}{3} \pi r^3 \] Where \( r = 2.5 , \text{in} \): \[ V = \frac{4}{3} \pi (2.5)^3 = \frac{4}{3} \pi (15.625) = \frac{62.5}{3} \pi \approx 20.83 \pi \] Using \( \pi \approx 3.14 \): \[ V \approx 20.83 \times 3.14 \approx 65.48 , \text{in}^3 \] Total volume for a dozen tennis balls (12 balls): \[ 12 \times 65.48 \approx 785.76 , \text{in}^3 \]
The container volume is \( 3200 , \text{in}^{3} \): \[ \text{Empty space} = 3200 - 785.76 \approx 2414.24 \] Rounding to the nearest whole number, the empty space is approximately 2414 in³.
Summary of the answers:
- 972π cm³
- 113.04 m³
- 288π m³
- 7238.56 mm³
- 2414 in³
5 answers