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Answer :
To determine the radius of the conical water reservoir, we'll use the formula for the volume of a cone:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
where:
- [tex]\( V \)[/tex] is the volume of the reservoir,
- [tex]\( r \)[/tex] is the radius of the base,
- [tex]\( h \)[/tex] is the height,
- [tex]\( \pi \)[/tex] is a constant approximately equal to 3.14.
We are given:
- The volume [tex]\( V = 225 \)[/tex] cubic feet,
- The height [tex]\( h = 8.5 \)[/tex] feet.
We need to rearrange the volume formula to solve for the radius [tex]\( r \)[/tex]:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
To isolate [tex]\( r^2 \)[/tex], multiply both sides by 3 and divide by [tex]\(\pi h\)[/tex]:
[tex]\[ 3V = \pi r^2 h \][/tex]
[tex]\[ r^2 = \frac{3V}{\pi h} \][/tex]
Take the square root to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{3V}{\pi h}} \][/tex]
Substitute the known values into the equation:
[tex]\[ r = \sqrt{\frac{3 \times 225}{3.14 \times 8.5}} \][/tex]
Calculate the value inside the square root:
[tex]\[ r \approx \sqrt{\frac{675}{26.69}} \][/tex]
[tex]\[ r \approx \sqrt{25.29} \][/tex]
Now, find the square root:
[tex]\[ r \approx 5.03 \][/tex]
Therefore, the radius of the water reservoir, rounded to the nearest hundredth, is approximately [tex]\( 5.03 \)[/tex] feet. The correct formula from the given options is:
[tex]\[ r = \sqrt{\frac{3V}{3.14 h}}, \, r = 5.03 \text{ feet} \][/tex]
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
where:
- [tex]\( V \)[/tex] is the volume of the reservoir,
- [tex]\( r \)[/tex] is the radius of the base,
- [tex]\( h \)[/tex] is the height,
- [tex]\( \pi \)[/tex] is a constant approximately equal to 3.14.
We are given:
- The volume [tex]\( V = 225 \)[/tex] cubic feet,
- The height [tex]\( h = 8.5 \)[/tex] feet.
We need to rearrange the volume formula to solve for the radius [tex]\( r \)[/tex]:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
To isolate [tex]\( r^2 \)[/tex], multiply both sides by 3 and divide by [tex]\(\pi h\)[/tex]:
[tex]\[ 3V = \pi r^2 h \][/tex]
[tex]\[ r^2 = \frac{3V}{\pi h} \][/tex]
Take the square root to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{3V}{\pi h}} \][/tex]
Substitute the known values into the equation:
[tex]\[ r = \sqrt{\frac{3 \times 225}{3.14 \times 8.5}} \][/tex]
Calculate the value inside the square root:
[tex]\[ r \approx \sqrt{\frac{675}{26.69}} \][/tex]
[tex]\[ r \approx \sqrt{25.29} \][/tex]
Now, find the square root:
[tex]\[ r \approx 5.03 \][/tex]
Therefore, the radius of the water reservoir, rounded to the nearest hundredth, is approximately [tex]\( 5.03 \)[/tex] feet. The correct formula from the given options is:
[tex]\[ r = \sqrt{\frac{3V}{3.14 h}}, \, r = 5.03 \text{ feet} \][/tex]
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