Given that the heights and the volumes of the two solids are equivalent, find the diameter of the cylinder. Round your answer to the nearest hundredth

- anonymous

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- anonymous

##### 1 Attachment

- anonymous

A) 2.13mm
B) 2.31 mm
C) 4.17mm
D) 4.27mm

- triciaal

|dw:1437511958025:dw|

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## More answers

- anonymous

\[\large \sf Volume~of~rectangular~prism\]\[\large \sf V~=~l \times w \times h\] \[\large \sf Volume~of~cylinder\]\[\large \sf V~=~\pi r^{2} h\]

- triciaal

diameter is twice the radius, r

- anonymous

idk the radius

- anonymous

Find the volume of the rectangular prism. Put that volume into the cylinder's formula and solve for the radius.

- anonymous

i got 121.635

- anonymous

And that's the volume. So now solve for the radius.

- anonymous

how do i get the radius

- anonymous

\[\large \sf V~=~ \pi r^{2} h\\\large \sf 121.635~=~ \pi r^{2} 8.5\] get r alone

- anonymous

I don't know how to get R alone

- anonymous

Do you know how to get x alone in \[\large \sf 5x~=~25\]?

- anonymous

yes

- anonymous

What's different about getting r alone?

- anonymous

its harder i dont know how to put it into pie

- anonymous

\[\large \sf 121.635~=~r^{2} \times 8.5 \pi\]

- anonymous

How about that? Can you get the \(r^{2}\) alone?

- anonymous

−∞,∞)
{r|r∈R}

- anonymous

What?

- anonymous

i plugged that into the calculator and thats what i got

- anonymous

But that's completely unrelated...

- anonymous

i have no idea

- anonymous

Well what did you do to get x alone in \[\large \sf 5x=25\]?

- anonymous

x=5

- anonymous

I know that, but what did you have to do to get that answer?

- anonymous

5 goes into 25 5 times

- anonymous

this test is timed and i dont ahve that much longer i need to stick to this question

- anonymous

\[\large \sf 5x=25 \Rightarrow \frac{5x}{5}=\frac{25}{5} \Rightarrow x=5\]

- anonymous

Since the 5 and x are being multipled together, we divide both sides by 5 to get them apart. This is simple algebra

- anonymous

i need to help with THIS question not x and 25

- anonymous

Well you clearly need some help going over the basics since you don't even know how to solve the current problem.

- anonymous

pls leave if you arent gonna help me solve this question i dont have time for that

- anonymous

Hey buddy, I've been helping this whole time. To be 100% honest it's really irritating trying to help people like you since you clearly haven't been studying since the problem you're asking for is past the things you know how to do. So I'l ask you, do you want my help or not?

- anonymous

if you arent gonna help me get the answer then no i don't

- anonymous

Ok, then I won't ¯\_(ツ)_/¯

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