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 .

- anonymous

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- anonymous

@DaBest21

- Michele_Laino

here we have to use this formula:
\[\Large V = \frac{{\pi {r^2}h}}{3}\]

- Michele_Laino

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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## More answers

- anonymous

\[3\frac{ 183.26}{ 3.14*5*5 }\]is the formula @Aliypop

- anonymous

so do 3.14*5*5 first @Aliypop

- Michele_Laino

from that formula above we get:
\[h = \frac{{3V}}{{\pi {r^2}}} = \frac{{3 \times 183.26}}{{3.14 \times 5 \times 5}} = ...feet\]

- anonymous

78.5

- anonymous

yep now do 183.26/78.5

- anonymous

2.33452293

- Michele_Laino

hint:
\[h = \frac{{3V}}{{\pi {r^2}}} = \frac{{3 \times 183.26}}{{3.14 \times 5 \times 5}} = 3 \times 2.335 = ...feet\]

- anonymous

2.3335

- anonymous

what's the next step ?

- Michele_Laino

as I wrote before, what is:
3*2.335=...?

- anonymous

7.005

- Michele_Laino

that's right!

- anonymous

so7.005 is the answer?

- Michele_Laino

yes!

- anonymous

no

- anonymous

u have to round

- anonymous

okay :D

- anonymous

@Aliypop round

- Michele_Laino

yes! sorry @Aliypop you have to round off to the hearest unit of foot

- Michele_Laino

nearest*

- anonymous

so what does 0 tell u to do to 7 Hint less than 5 means stay the same

- anonymous

so would it be 8.00

- anonymous

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

- anonymous

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 .

- anonymous

it appeared wrong again

- anonymous

we did this already didnt we? @Aliypop

- anonymous

@welshfella

- anonymous

@welshfella @welshfella

- anonymous

I got 211 now as the answer

- Michele_Laino

hint:
the volume V of a sphere is given by the subsequent equation (pi=3.14):
\[\begin{gathered}
V = \frac{{4 \times 3.14}}{3}{r^3} = \frac{{12.56}}{3} \times {r^3} = \hfill \\
\hfill \\
= \frac{{12.56}}{3} \times {4^3} = \frac{{12.56}}{3} \times 64 = ...? \hfill \\
\end{gathered} \]

- anonymous

2,411.25

- Michele_Laino

I got this:
\[\begin{gathered}
V = \frac{{4 \times 3.14}}{3}{r^3} = \frac{{12.56}}{3} \times {r^3} = \hfill \\
\hfill \\
= \frac{{12.56}}{3} \times {4^3} = \frac{{12.56}}{3} \times 64 = 267.85c{m^3} \hfill \\
\end{gathered} \]

- anonymous

okay

- anonymous

so 267.9

- Michele_Laino

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:
\[V = 100 \times 267.9466 = 26794.66 \cong 26794.7c{m^3}\]

- anonymous

okay

- anonymous

so rounded to the nearest foot it is 267.9

- Michele_Laino

267.9 is the volume of a single volleyball

- Michele_Laino

the requested volume of needed white chocolate is:
\[V \cong 26794.7c{m^3}\]

- anonymous

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.

- Michele_Laino

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|

- anonymous

okay

- Michele_Laino

substituting your data, we have:
\[\begin{gathered}
V = \frac{1}{3}L \times L \times H = \hfill \\
\hfill \\
= \frac{1}{3}8 \times 8 \times 10 = \frac{{640}}{3} = ...inche{s^3} \hfill \\
\end{gathered} \]

- anonymous

213.333333

- Michele_Laino

correct!

- Michele_Laino

Now, we have 250 of such pyramids, so the requested volume of needed glass, is:
\[{V_{TOTAL}} = 213.3333 \times 250 = \]

- anonymous

53,33.333333

- Michele_Laino

that's right! Now we have to round off that result to the nearest unit, so we get:
\[{V_{TOTAL}} = 213.3333 \times 250 = {\text{53333}}{\text{.325}} \cong {\text{53333}}inche{s^3}\]

- anonymous

ok

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