Question
Question 1
A)Find the volume of a sphere with a radius of 9 cm. Leave your answer in terms of pi.(1 point)
The volume is $$ cm3.
Question 2
A)Find the volume of a sphere with a radius of 3 m. Use 3.14 as an approximation for pi. Round your answer to the nearest hundredth, if necessary.(1 point)
The volume is $$ m3.
Question 3
A)Find the volume of a sphere with a diameter 12 m. Leave your answer in terms of pi.(1 point)
The volume is $$ m3.
Question 4
A)How much chocolate is in a spherical chocolate truffle with a radius of 12 mm? Use 3.14 as an approximation for pi. Round your answer to the nearest hundredth, if necessary.(1 point)
$$ mm3
Question 5
A)The volume of a container measures 3,200 in.3 and contains a dozen tennis balls. Each tennis ball has a radius of 2.5 in. How much space in the container is empty? Use 3.14 as an approximation for pi. Round to the nearest whole number, if necessary.(1 point)
$$ in.3
A)Find the volume of a sphere with a radius of 9 cm. Leave your answer in terms of pi.(1 point)
The volume is $$ cm3.
Question 2
A)Find the volume of a sphere with a radius of 3 m. Use 3.14 as an approximation for pi. Round your answer to the nearest hundredth, if necessary.(1 point)
The volume is $$ m3.
Question 3
A)Find the volume of a sphere with a diameter 12 m. Leave your answer in terms of pi.(1 point)
The volume is $$ m3.
Question 4
A)How much chocolate is in a spherical chocolate truffle with a radius of 12 mm? Use 3.14 as an approximation for pi. Round your answer to the nearest hundredth, if necessary.(1 point)
$$ mm3
Question 5
A)The volume of a container measures 3,200 in.3 and contains a dozen tennis balls. Each tennis ball has a radius of 2.5 in. How much space in the container is empty? Use 3.14 as an approximation for pi. Round to the nearest whole number, if necessary.(1 point)
$$ in.3
Answers
Answered by
GPT-4o mini
Let’s calculate the answers step by step for each question.
### Question 1
**Find the volume of a sphere with a radius of 9 cm. Leave your answer in terms of pi.**
The formula for the volume \( V \) of a sphere is:
\[
V = \frac{4}{3} \pi r^3
\]
Where \( r \) is the radius.
Given \( r = 9 \) cm, we calculate:
\[
V = \frac{4}{3} \pi (9)^3 = \frac{4}{3} \pi (729) = \frac{2916}{3} \pi = 972 \pi
\]
The volume is \( 972\pi \) cm³.
### Question 2
**Find the volume of a sphere with a radius of 3 m. Use 3.14 as an approximation for pi. Round your answer to the nearest hundredth, if necessary.**
Using the same formula:
\[
V = \frac{4}{3} \pi r^3
\]
Where \( r = 3 \) m:
\[
V = \frac{4}{3} \cdot 3.14 \cdot (3)^3 = \frac{4}{3} \cdot 3.14 \cdot 27 = \frac{4 \cdot 3.14 \cdot 27}{3} = \frac{339.12}{3} \approx 113.04
\]
The volume is approximately \( 113.04 \) m³.
### Question 3
**Find the volume of a sphere with a diameter of 12 m. Leave your answer in terms of pi.**
First, we find the radius from the diameter:
\[
r = \frac{12}{2} = 6 \text{ m}
\]
Now, using the volume formula:
\[
V = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = \frac{864}{3} \pi = 288 \pi
\]
The volume is \( 288\pi \) m³.
### Question 4
**How much chocolate is in a spherical chocolate truffle with a radius of 12 mm? Use 3.14 as an approximation for pi. Round your answer to the nearest hundredth, if necessary.**
Using the volume formula:
\[
V = \frac{4}{3} \pi r^3
\]
Where \( r = 12 \) mm:
\[
V = \frac{4}{3} \cdot 3.14 \cdot (12)^3 = \frac{4}{3} \cdot 3.14 \cdot 1728 = \frac{4 \cdot 3.14 \cdot 1728}{3} = \frac{21698.56}{3} \approx 7232.85
\]
The volume is approximately \( 7232.85 \) mm³.
### Question 5
**The volume of a container measures 3,200 in³ and contains a dozen tennis balls. Each tennis ball has a radius of 2.5 in. How much space in the container is empty? Use 3.14 as an approximation for pi. Round to the nearest whole number, if necessary.**
First, find the volume of one tennis ball:
\[
V = \frac{4}{3} \pi r^3
\]
Where \( r = 2.5 \) in:
\[
V = \frac{4}{3} \cdot 3.14 \cdot (2.5)^3 = \frac{4}{3} \cdot 3.14 \cdot 15.625 = \frac{4 \cdot 3.14 \cdot 15.625}{3} \approx \frac{197.29167}{3} \approx 65.76
\]
So, the volume of one tennis ball is approximately \( 65.76 \) in³. For a dozen (12 balls):
\[
\text{Total volume of tennis balls} = 12 \cdot 65.76 \approx 789.12 \text{ in³}
\]
Now, subtract this volume from the container's volume:
\[
\text{Empty space} = 3200 - 789.12 \approx 2410.88
\]
Rounded to the nearest whole number, the empty space is approximately \( 2411 \) in³.
### Summary of Answers
1. \( 972 \pi \) cm³
2. \( 113.04 \) m³
3. \( 288 \pi \) m³
4. \( 7232.85 \) mm³
5. \( 2411 \) in³
### Question 1
**Find the volume of a sphere with a radius of 9 cm. Leave your answer in terms of pi.**
The formula for the volume \( V \) of a sphere is:
\[
V = \frac{4}{3} \pi r^3
\]
Where \( r \) is the radius.
Given \( r = 9 \) cm, we calculate:
\[
V = \frac{4}{3} \pi (9)^3 = \frac{4}{3} \pi (729) = \frac{2916}{3} \pi = 972 \pi
\]
The volume is \( 972\pi \) cm³.
### Question 2
**Find the volume of a sphere with a radius of 3 m. Use 3.14 as an approximation for pi. Round your answer to the nearest hundredth, if necessary.**
Using the same formula:
\[
V = \frac{4}{3} \pi r^3
\]
Where \( r = 3 \) m:
\[
V = \frac{4}{3} \cdot 3.14 \cdot (3)^3 = \frac{4}{3} \cdot 3.14 \cdot 27 = \frac{4 \cdot 3.14 \cdot 27}{3} = \frac{339.12}{3} \approx 113.04
\]
The volume is approximately \( 113.04 \) m³.
### Question 3
**Find the volume of a sphere with a diameter of 12 m. Leave your answer in terms of pi.**
First, we find the radius from the diameter:
\[
r = \frac{12}{2} = 6 \text{ m}
\]
Now, using the volume formula:
\[
V = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = \frac{864}{3} \pi = 288 \pi
\]
The volume is \( 288\pi \) m³.
### Question 4
**How much chocolate is in a spherical chocolate truffle with a radius of 12 mm? Use 3.14 as an approximation for pi. Round your answer to the nearest hundredth, if necessary.**
Using the volume formula:
\[
V = \frac{4}{3} \pi r^3
\]
Where \( r = 12 \) mm:
\[
V = \frac{4}{3} \cdot 3.14 \cdot (12)^3 = \frac{4}{3} \cdot 3.14 \cdot 1728 = \frac{4 \cdot 3.14 \cdot 1728}{3} = \frac{21698.56}{3} \approx 7232.85
\]
The volume is approximately \( 7232.85 \) mm³.
### Question 5
**The volume of a container measures 3,200 in³ and contains a dozen tennis balls. Each tennis ball has a radius of 2.5 in. How much space in the container is empty? Use 3.14 as an approximation for pi. Round to the nearest whole number, if necessary.**
First, find the volume of one tennis ball:
\[
V = \frac{4}{3} \pi r^3
\]
Where \( r = 2.5 \) in:
\[
V = \frac{4}{3} \cdot 3.14 \cdot (2.5)^3 = \frac{4}{3} \cdot 3.14 \cdot 15.625 = \frac{4 \cdot 3.14 \cdot 15.625}{3} \approx \frac{197.29167}{3} \approx 65.76
\]
So, the volume of one tennis ball is approximately \( 65.76 \) in³. For a dozen (12 balls):
\[
\text{Total volume of tennis balls} = 12 \cdot 65.76 \approx 789.12 \text{ in³}
\]
Now, subtract this volume from the container's volume:
\[
\text{Empty space} = 3200 - 789.12 \approx 2410.88
\]
Rounded to the nearest whole number, the empty space is approximately \( 2411 \) in³.
### Summary of Answers
1. \( 972 \pi \) cm³
2. \( 113.04 \) m³
3. \( 288 \pi \) m³
4. \( 7232.85 \) mm³
5. \( 2411 \) in³