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Mathematics 15 Online
OpenStudy (anonymous):

HELP ME????

OpenStudy (anonymous):

The formula for finding the volume of a cylinder is V=(pie) * r^2 *h where r is the radius of the base, and h is the height of the cylinder. a) Solve the equation for h. b) Use your answer from part a) to find the height, h, of a cylinder with volume of 2240 cubic centimeters and radius of 5 centimeters. Show your work.

OpenStudy (anonymous):

just rearrange the equation of h is one side and the rest is on the other side.

OpenStudy (anonymous):

then for part (b), substitute in the numbers :)

OpenStudy (anonymous):

a little more help on A please

OpenStudy (anonymous):

V=h * r^2 * (pie) ????????

zepdrix (zepdrix):

\[\huge V= \pi r^2 h\]Divide both sides by pi.\[\huge \frac{V}{\pi}=r^2h\]Divide both sides by r^2.\[\huge \frac{V}{\pi r^2}=h\] Is this what you were trying to do ms katie? :o

zepdrix (zepdrix):

And also, there's no E in Pi, it's the number Pi, not an apple pie silly! c:

OpenStudy (anonymous):

ok sooo b would be.. \[\sqrt{2240\div (\pi*h)}=5\]

zepdrix (zepdrix):

Hmm it looks like you have your h and r in the wrong spot. \[\huge \frac{V}{\pi r^2}=h\] Becomes...\[\huge \frac{2240}{\pi (5)^2}=h\]

zepdrix (zepdrix):

I mean, yes your equation is correct.. but you didn't want to solve for R, you're giving yourself more work :O

OpenStudy (anonymous):

so the answer is ... \[h=\frac{ 448 }{ 5 }\]

zepdrix (zepdrix):

Hmm, where did your pi go? :\

OpenStudy (anonymous):

pi is next to 5

OpenStudy (anonymous):

it didnt show up

zepdrix (zepdrix):

k looks good c: With a problem like this, you probably want to simplify to a decimal.

zepdrix (zepdrix):

Not necessarily though, depends on what your teacher prefers c:

OpenStudy (anonymous):

which would be?

zepdrix (zepdrix):

If you don't have a Pi button on your calculator then just use 3.14. Punch it in! :D

OpenStudy (anonymous):

Thank you(:

OpenStudy (anonymous):

If your teacher doesn't specify I would leave it exact: \[h= \frac{448}{5 \pi}\]

OpenStudy (anonymous):

ok

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