Question
Drag and drop the words into the correct locations
A writer can use ____ (general words, **precise language**) to achieve a more exact meaning and help the audience experience the story.
A writer can use ____ (general words, **precise language**) to achieve a more exact meaning and help the audience experience the story.
Answers
Answer
Use the formula for the volume of a sphere to find the volume of a basketball that has a diameter of 9.2 inches. Use 3.14 as an approximation for pi and round your answer to the nearest hundredth, if necessary.
1. 115.55 in.^3
2. 1,222.54 in^3
3. 407.51 in.^3
4. 3,260.11 in.^3
1. 115.55 in.^3
2. 1,222.54 in^3
3. 407.51 in.^3
4. 3,260.11 in.^3
Answer
Find the volume of a sphere with a radius of 1.8 yards. Leave your answer in terms of pi. Round your answer to the nearest hundredth, if necessary.
7.2pi yds^3
7.78pi yds^3
24.43pi yds^3
23.34pi yds^3
7.2pi yds^3
7.78pi yds^3
24.43pi yds^3
23.34pi yds^3
Answer
Use the formula for the volume of a sphere to solve the following problem. How much air is needed to fill a basketball with a diameter of 9 inches? Use 3.14 as an approximation for pi. Round your answer to the nearest tenth.
3,052.1 in.^3
381.5 in.^3
113 in.^3
189.8 in.^3
3,052.1 in.^3
381.5 in.^3
113 in.^3
189.8 in.^3
Answered by
GPT-4o mini
A writer can use **precise language** to achieve a more exact meaning and help the audience experience the story.
Answered by
GPT-4o mini
To find the volume of a sphere, you can use the formula:
\[
V = \frac{4}{3} \pi r^3
\]
where \( V \) is the volume and \( r \) is the radius of the sphere.
1. First, find the radius from the diameter. The diameter of the basketball is 9.2 inches, so the radius \( r \) is:
\[
r = \frac{9.2}{2} = 4.6 \text{ inches}
\]
2. Next, substitute the radius into the volume formula. Using \( \pi \approx 3.14 \):
\[
V = \frac{4}{3} \times 3.14 \times (4.6)^3
\]
3. Calculate \( (4.6)^3 \):
\[
4.6^3 = 4.6 \times 4.6 \times 4.6 = 97.336
\]
4. Now substitute that back into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 97.336
\]
5. Calculate \( 3.14 \times 97.336 \):
\[
3.14 \times 97.336 \approx 305.80064
\]
6. Now calculate \(\frac{4}{3} \times 305.80064\):
\[
V = \frac{4 \times 305.80064}{3} \approx \frac{1223.20256}{3} \approx 407.7342
\]
7. Finally, round to the nearest hundredth:
\[
V \approx 407.73 \text{ in}^3
\]
Comparing this result to your options, the closest is **3. 407.51 in.^3**.
\[
V = \frac{4}{3} \pi r^3
\]
where \( V \) is the volume and \( r \) is the radius of the sphere.
1. First, find the radius from the diameter. The diameter of the basketball is 9.2 inches, so the radius \( r \) is:
\[
r = \frac{9.2}{2} = 4.6 \text{ inches}
\]
2. Next, substitute the radius into the volume formula. Using \( \pi \approx 3.14 \):
\[
V = \frac{4}{3} \times 3.14 \times (4.6)^3
\]
3. Calculate \( (4.6)^3 \):
\[
4.6^3 = 4.6 \times 4.6 \times 4.6 = 97.336
\]
4. Now substitute that back into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 97.336
\]
5. Calculate \( 3.14 \times 97.336 \):
\[
3.14 \times 97.336 \approx 305.80064
\]
6. Now calculate \(\frac{4}{3} \times 305.80064\):
\[
V = \frac{4 \times 305.80064}{3} \approx \frac{1223.20256}{3} \approx 407.7342
\]
7. Finally, round to the nearest hundredth:
\[
V \approx 407.73 \text{ in}^3
\]
Comparing this result to your options, the closest is **3. 407.51 in.^3**.
Answered by
GPT-4o mini
To find the volume of a sphere using the formula in terms of \( \pi \):
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere. Given that the radius \( r = 1.8 \) yards, we can substitute this value into the formula.
1. First, calculate \( r^3 \):
\[
r^3 = (1.8)^3 = 1.8 \times 1.8 \times 1.8
\]
Calculating step-by-step:
\[
1.8 \times 1.8 = 3.24
\]
Then,
\[
3.24 \times 1.8 = 5.832
\]
So,
\[
r^3 = 5.832 \text{ cubic yards}
\]
2. Now substitute \( r^3 \) back into the volume formula:
\[
V = \frac{4}{3} \pi (5.832)
\]
3. Calculate the volume:
\[
V = \frac{4 \times 5.832}{3} \pi = \frac{23.328}{3} \pi \approx 7.776 \pi \text{ yds}^3
\]
4. Now, rounding \( 7.776 \pi \) to the nearest hundredth gives \( 7.78 \pi \).
Thus, the volume of the sphere is:
**7.78pi yds³**.
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere. Given that the radius \( r = 1.8 \) yards, we can substitute this value into the formula.
1. First, calculate \( r^3 \):
\[
r^3 = (1.8)^3 = 1.8 \times 1.8 \times 1.8
\]
Calculating step-by-step:
\[
1.8 \times 1.8 = 3.24
\]
Then,
\[
3.24 \times 1.8 = 5.832
\]
So,
\[
r^3 = 5.832 \text{ cubic yards}
\]
2. Now substitute \( r^3 \) back into the volume formula:
\[
V = \frac{4}{3} \pi (5.832)
\]
3. Calculate the volume:
\[
V = \frac{4 \times 5.832}{3} \pi = \frac{23.328}{3} \pi \approx 7.776 \pi \text{ yds}^3
\]
4. Now, rounding \( 7.776 \pi \) to the nearest hundredth gives \( 7.78 \pi \).
Thus, the volume of the sphere is:
**7.78pi yds³**.
Answered by
GPT-4o mini
To find the volume of a sphere, we can use the formula:
\[
V = \frac{4}{3} \pi r^3
\]
where \( V \) is the volume and \( r \) is the radius of the sphere.
1. First, we need to calculate the radius from the diameter. The diameter of the basketball is 9 inches, so the radius \( r \) is:
\[
r = \frac{9}{2} = 4.5 \text{ inches}
\]
2. Next, substitute the radius into the volume formula with \( \pi \approx 3.14 \):
\[
V = \frac{4}{3} \times 3.14 \times (4.5)^3
\]
3. Calculate \( (4.5)^3 \):
\[
(4.5)^3 = 4.5 \times 4.5 \times 4.5 = 20.25 \times 4.5 = 91.125
\]
4. Now substitute that back into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 91.125
\]
5. Calculate \( 3.14 \times 91.125 \):
\[
3.14 \times 91.125 \approx 285.12875
\]
6. Now calculate \(\frac{4}{3} \times 285.12875\):
\[
V = \frac{4 \times 285.12875}{3} \approx \frac{1140.515}{3} \approx 380.17167
\]
7. Finally, round this value to the nearest tenth:
\[
V \approx 380.2 \text{ in}^3
\]
After cross-verifying with the options provided, the closest option in this context is **381.5 in.^3**.
\[
V = \frac{4}{3} \pi r^3
\]
where \( V \) is the volume and \( r \) is the radius of the sphere.
1. First, we need to calculate the radius from the diameter. The diameter of the basketball is 9 inches, so the radius \( r \) is:
\[
r = \frac{9}{2} = 4.5 \text{ inches}
\]
2. Next, substitute the radius into the volume formula with \( \pi \approx 3.14 \):
\[
V = \frac{4}{3} \times 3.14 \times (4.5)^3
\]
3. Calculate \( (4.5)^3 \):
\[
(4.5)^3 = 4.5 \times 4.5 \times 4.5 = 20.25 \times 4.5 = 91.125
\]
4. Now substitute that back into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 91.125
\]
5. Calculate \( 3.14 \times 91.125 \):
\[
3.14 \times 91.125 \approx 285.12875
\]
6. Now calculate \(\frac{4}{3} \times 285.12875\):
\[
V = \frac{4 \times 285.12875}{3} \approx \frac{1140.515}{3} \approx 380.17167
\]
7. Finally, round this value to the nearest tenth:
\[
V \approx 380.2 \text{ in}^3
\]
After cross-verifying with the options provided, the closest option in this context is **381.5 in.^3**.