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So, I have an equation V=1/3h^(3r-h). The "^" signifies "squared" so it would be h-squared. Please solve for r and explain each step please. Thanks!

2007-09-24 09:24:34 · 4 answers · asked by =(Disullusioned_Dreams=( 1 in Science & Mathematics Mathematics

Sorries. I forgot that after the 1/3 and before the h^ there is the Pi sign (3.14).

2007-09-24 09:26:52 · update #1

4 answers

V=1/3πh²(3r-h)
3V/πh²=3r-h
(3V/πh² + h)/3 = r

2007-09-24 09:32:00 · answer #1 · answered by chasrmck 6 · 0 0

V= 1/3 pi*h^2 ( 3r-h)

Distribute:
pi*r*h^2 - 1/3*pi*h^3 =V

Add the second term to both sides, effectively canceling on the left side:
pi*r*h^2 = 1/3*pi*h^3 + V

Divide by pi*h^2 on both sides, effectively canceling it on the left side:
r = (1/3*pi*h^3 +V) / (pi*h^2)

2007-09-24 16:37:36 · answer #2 · answered by sayamiam 6 · 0 0

so divide both sides by pi h^2 and multiply by 3.
Now you have 3V/pi h^2 = 3r - h
add h
(3V/pi h^2) +h = r
and divide by 3
(V/pi h^2) + h/3 = r

2007-09-24 16:33:45 · answer #3 · answered by ccw 4 · 0 0

V=pi/3h^2(3r-h)
3V = pi*h^2*(3r-h)
3V/(pi*h^2)= 3r-h
3V/(pi*h^2) +h =3r
[ 3V/(pi*h^2) +h]/3 = r

2007-09-24 16:45:54 · answer #4 · answered by ironduke8159 7 · 0 0

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