The volume of a cone is 183.26 cubic feet and its radius is 5 feet. What is the cone's height? Round to the nearest foot. Use 3.14 for \pi .

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The volume of a cone is 183.26 cubic feet and its radius is 5 feet. What is the cone's height? Round to the nearest foot. Use 3.14 for \pi .

Mathematics
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here we have to use this formula: \[\Large V = \frac{{\pi {r^2}h}}{3}\]
where V is the volume of the cone, r is the radius of its base, and h is the height of the cone

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Other answers:

\[3\frac{ 183.26}{ 3.14*5*5 }\]is the formula @Aliypop
so do 3.14*5*5 first @Aliypop
from that formula above we get: \[h = \frac{{3V}}{{\pi {r^2}}} = \frac{{3 \times 183.26}}{{3.14 \times 5 \times 5}} = ...feet\]
78.5
yep now do 183.26/78.5
2.33452293
hint: \[h = \frac{{3V}}{{\pi {r^2}}} = \frac{{3 \times 183.26}}{{3.14 \times 5 \times 5}} = 3 \times 2.335 = ...feet\]
2.3335
what's the next step ?
as I wrote before, what is: 3*2.335=...?
7.005
that's right!
so7.005 is the answer?
yes!
no
u have to round
okay :D
@Aliypop round
yes! sorry @Aliypop you have to round off to the hearest unit of foot
nearest*
so what does 0 tell u to do to 7 Hint less than 5 means stay the same
so would it be 8.00
no it would be 7 Don't forget to medal and fan if you already haven't. Plus send me a message if you need more help @Aliypop
Carson's Bakery and Chocolate Shop is experimenting with making chocolate in the shape of different types of sports equipment. Bayside High School's Volleyball team has ordered 100 solid white chocolate volleyballs. What volume of white chocolate is needed to fill this order if the radius of each volleyball is 4 cm? Rounded to the nearest tenth of a cubic centimeter. Use 3.14 for \pi .
it appeared wrong again
we did this already didnt we? @Aliypop
I got 211 now as the answer
hint: the volume V of a sphere is given by the subsequent equation (pi=3.14): \[\begin{gathered} V = \frac{{4 \times 3.14}}{3}{r^3} = \frac{{12.56}}{3} \times {r^3} = \hfill \\ \hfill \\ = \frac{{12.56}}{3} \times {4^3} = \frac{{12.56}}{3} \times 64 = ...? \hfill \\ \end{gathered} \]
2,411.25
I got this: \[\begin{gathered} V = \frac{{4 \times 3.14}}{3}{r^3} = \frac{{12.56}}{3} \times {r^3} = \hfill \\ \hfill \\ = \frac{{12.56}}{3} \times {4^3} = \frac{{12.56}}{3} \times 64 = 267.85c{m^3} \hfill \\ \end{gathered} \]
okay
so 267.9
sorry I have made an error, here is the right result: \[\begin{gathered} V = \frac{{4 \times 3.14}}{3}{r^3} = \frac{{12.56}}{3} \times {r^3} = \hfill \\ \hfill \\ = \frac{{12.56}}{3} \times {4^3} = \frac{{12.56}}{3} \times 64 = 267.9466c{m^3} \hfill \\ \end{gathered} \] so the total volume of the requested chocolate is: \[V = 100 \times 267.9466 = 26794.66 \cong 26794.7c{m^3}\]
okay
so rounded to the nearest foot it is 267.9
267.9 is the volume of a single volleyball
the requested volume of needed white chocolate is: \[V \cong 26794.7c{m^3}\]
The new business building uptown is building a wall made entirely of glass pyramids. Each pyramid will have a square base with length of 8 inches and height of 10 inches. If the wall requires 250 of these pyramids, what volume of glass will be used? Round to the nearest cubic inch.
we have to compute the volume V of a single pyramid. So, we can apply this formula: \[\Large V = \frac{1}{3}L \times L \times H\] where L is the length of the square base, and H is the height of the pyramid: |dw:1438014203141:dw|
okay
substituting your data, we have: \[\begin{gathered} V = \frac{1}{3}L \times L \times H = \hfill \\ \hfill \\ = \frac{1}{3}8 \times 8 \times 10 = \frac{{640}}{3} = ...inche{s^3} \hfill \\ \end{gathered} \]
213.333333
correct!
Now, we have 250 of such pyramids, so the requested volume of needed glass, is: \[{V_{TOTAL}} = 213.3333 \times 250 = \]
53,33.333333
that's right! Now we have to round off that result to the nearest unit, so we get: \[{V_{TOTAL}} = 213.3333 \times 250 = {\text{53333}}{\text{.325}} \cong {\text{53333}}inche{s^3}\]
ok

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