SOLUTION: How do I get this formula to read with the r on the left hand side reading r = ??????? V = 1/3 3.142 r2 h I really would love some help. Thank you oh so much.

Algebra.Com
Question 94394: How do I get this formula to read with the r on the left hand side reading r = ???????
V = 1/3 3.142 r2 h
I really would love some help. Thank you oh so much.

Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
V = (1/3) (3.142) r^2 h
Multiply both sides by 3 to get:
(3.142)r^2h = 3V
Divide both sides by 3.142h to get:
r^2 = (3V)/(3.124h)
Take the square root of both sides to get:
r = sqrt[(3V)/(3.124h)]
==========
Cheers,
Stan H.

Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
How do I get this formula to read with the r on the left hand side reading r = ???????
V = 1/3 3.142 r2 h
:
This looks like the volume of a cone which is:
V = *
:
Multiply equation by 3:
3V =
:
Divide both sides by
= r^2
:
r =
:
Is this what you had in mind?

RELATED QUESTIONS

The volume of rubber in cubic centimeters is given by V = 4 / 3 3.14 * R ^ 3 - 4 / 3 *... (answered by Alan3354)
Please help me solve the formula for the indicated variable and evaluate the written... (answered by stanbon)
If the formula for Volume of a cone is v=1/3*pi*r^2*h how do I find for h or r? (answered by Earlsdon)
v=1/3 pi r2 h (make h the subject of the formula) ive got to v-1/3=pi r2 h whats (answered by scott8148)
Hi there! I have been stuck on this one logic problem for the longest time. I am a... (answered by jim_thompson5910)
V = 1/3 Pi r²h = 1/3Pi * 10² *24 = 240 Pi V = 753.98 m³ SA = Pi r² + Pi r√2r +... (answered by stanbon)
The volume of rubber(in cubic centimeters)in a hollow rubber ball used in raquetball is... (answered by tutor_paul)
Hello there Im strugling with this problem.Could you help? Thanks For all R =/ + and... (answered by nerdybill)
I saw that people can solve logic proofs on this site. I've looked everywhere for help.... (answered by Edwin McCravy)