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February 1, 2015

February 1, 2015

Posted by **ken** on Monday, November 1, 2010 at 6:52pm.

i think [4(pi)r(cubed)]/3=volume of hemisphere

and h (pi) r squared volume of the cylinder.

so

[4(pi) r (cubed)/3] + [45 (pi) r squared= volume of bin

so

[4 r (cubed)/3] + [45 (pi) r squared=volume

how do i solve for r?

sorry, i'm stuck

- algebra -
**bobpursley**, Monday, November 1, 2010 at 6:56pmthe volume of a sphere is 4/3 PI r^3, so a hemisphere would be half of that.

If you have an equation such as this,

a r^3 + b r^2 + c=0

it is a third degree equation. I think I would graph

f(r)= a r^3+b r^2+c and see where it crosses the axis, that is a solution.

There are ways to solve cubic equations, but for this, I would graph it.

- algebra -
**ken**, Monday, November 1, 2010 at 7:09pmthanks Mr. Pursley,

graphing it now.

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