Question
Use the picture below to solve the following problems. Use the π
in your calculator and round your answer to the nearest whole number if necessary. Use the same picture but the measurements will be different for each problem.
Question 1
What is the volume of the shape if the radius is 14?(1 point)
Volume =
3
Question 2
What is the diameter if the volume is 50965 units 3
?(1 point)
diameter =
units
Question 3
Wyatt just used his allowance to buy a soccer ball. At full capacity the diameter of the ball measures 22 cm. How much air can the ball hold if it is pumped to full capacity? You must show all work in order to receive credit. (2 points)
in your calculator and round your answer to the nearest whole number if necessary. Use the same picture but the measurements will be different for each problem.
Question 1
What is the volume of the shape if the radius is 14?(1 point)
Volume =
3
Question 2
What is the diameter if the volume is 50965 units 3
?(1 point)
diameter =
units
Question 3
Wyatt just used his allowance to buy a soccer ball. At full capacity the diameter of the ball measures 22 cm. How much air can the ball hold if it is pumped to full capacity? You must show all work in order to receive credit. (2 points)
Answers
GPT-4o mini
To solve the problems, we need to use the formula for the volume of a sphere, which is given by:
\[
V = \frac{4}{3} \pi r^3
\]
where \( V \) is the volume and \( r \) is the radius of the sphere.
### Question 1
**What is the volume of the shape if the radius is 14?**
Using the formula:
1. Substitute \( r = 14 \) into the volume formula:
\[
V = \frac{4}{3} \pi (14)^3
\]
2. Calculate \( (14)^3 \):
\[
14^3 = 2744
\]
3. Now substitute back into the volume formula:
\[
V = \frac{4}{3} \pi (2744)
\]
4. Calculate \( \frac{4}{3} \times 2744 \):
\[
\frac{4}{3} \times 2744 = 3658.67
\]
5. Finally, multiply by \( \pi \) (approximately 3.14159):
\[
V \approx 3658.67 \times 3.14159 \approx 11448.53
\]
So, rounding to the nearest whole number:
\[
\text{Volume} \approx 11449 \text{ units}^3
\]
### Question 2
**What is the diameter if the volume is 50965 units³?**
First, we will use the volume formula to solve for the radius.
1. Set the volume equal to 50965:
\[
50965 = \frac{4}{3} \pi r^3
\]
2. Rearranging gives:
\[
r^3 = \frac{50965 \times 3}{4 \pi}
\]
3. Calculate \( \frac{50965 \times 3}{4} \):
\[
\frac{50965 \times 3}{4} = \frac{152895}{4} = 38223.75
\]
4. Now divide by \( \pi \):
\[
r^3 = \frac{38223.75}{\pi} \approx 12137.79
\]
5. Take the cube root:
\[
r \approx \sqrt[3]{12137.79} \approx 22.86
\]
6. The diameter \( d \) is twice the radius:
\[
d = 2r \approx 2 \times 22.86 \approx 45.72
\]
Rounding to the nearest whole number:
\[
\text{Diameter} \approx 46 \text{ units}
\]
### Question 3
**Wyatt just used his allowance to buy a soccer ball. At full capacity the diameter of the ball measures 22 cm. How much air can the ball hold if it is pumped to full capacity?**
1. Calculate the radius:
\[
r = \frac{22}{2} = 11 \text{ cm}
\]
2. Use the volume formula:
\[
V = \frac{4}{3} \pi r^3
\]
3. Substitute \( r = 11 \):
\[
V = \frac{4}{3} \pi (11)^3
\]
4. Calculate \( (11)^3 \):
\[
11^3 = 1331
\]
5. Now substitute back into the volume formula:
\[
V = \frac{4}{3} \pi (1331)
\]
6. Calculate \( \frac{4}{3} \times 1331 \):
\[
\frac{4}{3} \times 1331 = 1774.67
\]
7. Now multiply by \( \pi \):
\[
V \approx 1774.67 \times 3.14159 \approx 5567.77
\]
Rounding to the nearest whole number:
\[
\text{Volume} \approx 5568 \text{ cm}^3
\]
So, the answers are:
- **Question 1**: 11449 units³
- **Question 2**: 46 units
- **Question 3**: 5568 cm³
\[
V = \frac{4}{3} \pi r^3
\]
where \( V \) is the volume and \( r \) is the radius of the sphere.
### Question 1
**What is the volume of the shape if the radius is 14?**
Using the formula:
1. Substitute \( r = 14 \) into the volume formula:
\[
V = \frac{4}{3} \pi (14)^3
\]
2. Calculate \( (14)^3 \):
\[
14^3 = 2744
\]
3. Now substitute back into the volume formula:
\[
V = \frac{4}{3} \pi (2744)
\]
4. Calculate \( \frac{4}{3} \times 2744 \):
\[
\frac{4}{3} \times 2744 = 3658.67
\]
5. Finally, multiply by \( \pi \) (approximately 3.14159):
\[
V \approx 3658.67 \times 3.14159 \approx 11448.53
\]
So, rounding to the nearest whole number:
\[
\text{Volume} \approx 11449 \text{ units}^3
\]
### Question 2
**What is the diameter if the volume is 50965 units³?**
First, we will use the volume formula to solve for the radius.
1. Set the volume equal to 50965:
\[
50965 = \frac{4}{3} \pi r^3
\]
2. Rearranging gives:
\[
r^3 = \frac{50965 \times 3}{4 \pi}
\]
3. Calculate \( \frac{50965 \times 3}{4} \):
\[
\frac{50965 \times 3}{4} = \frac{152895}{4} = 38223.75
\]
4. Now divide by \( \pi \):
\[
r^3 = \frac{38223.75}{\pi} \approx 12137.79
\]
5. Take the cube root:
\[
r \approx \sqrt[3]{12137.79} \approx 22.86
\]
6. The diameter \( d \) is twice the radius:
\[
d = 2r \approx 2 \times 22.86 \approx 45.72
\]
Rounding to the nearest whole number:
\[
\text{Diameter} \approx 46 \text{ units}
\]
### Question 3
**Wyatt just used his allowance to buy a soccer ball. At full capacity the diameter of the ball measures 22 cm. How much air can the ball hold if it is pumped to full capacity?**
1. Calculate the radius:
\[
r = \frac{22}{2} = 11 \text{ cm}
\]
2. Use the volume formula:
\[
V = \frac{4}{3} \pi r^3
\]
3. Substitute \( r = 11 \):
\[
V = \frac{4}{3} \pi (11)^3
\]
4. Calculate \( (11)^3 \):
\[
11^3 = 1331
\]
5. Now substitute back into the volume formula:
\[
V = \frac{4}{3} \pi (1331)
\]
6. Calculate \( \frac{4}{3} \times 1331 \):
\[
\frac{4}{3} \times 1331 = 1774.67
\]
7. Now multiply by \( \pi \):
\[
V \approx 1774.67 \times 3.14159 \approx 5567.77
\]
Rounding to the nearest whole number:
\[
\text{Volume} \approx 5568 \text{ cm}^3
\]
So, the answers are:
- **Question 1**: 11449 units³
- **Question 2**: 46 units
- **Question 3**: 5568 cm³