# make r the subject of v=4/3 pie r^3

make r the subject of v=4/3 pie r^3 need some help guys
Simply rearrange to get r = ....
Original post by Thrug
Simply rearrange to get r = ....

How do i do it
Divide both sides by Pie, then multiply both sides by 3/4. finally cube root.
What are you stuck on?

Rearrange to get all terms in r (i.e: the r^3 term, in this case) on one side, and everything else on the other side.

Given $r^3 = \cdots$, then taking the cube root of both sides will give you an expression for r.
Original post by james.h
What are you stuck on?

Rearrange to get all terms in r (i.e: the r^3 term, in this case) on one side, and everything else on the other side.

Given $r^3 = \cdots$, then taking the cube root of both sides will give you an expression for r.

can you show me how
Original post by daviesblue
Divide both sides by Pie, then multiply both sides by 3/4. finally cube root.

dont you mean times both sides by 3
Original post by gorila7
can you show me how

I'll work through a different example to show you the ideas involved.

The volume of a cone is $V = \frac{1}{3} \pi r^2 h$

Multiply both sides by 3 to get rid of the 1/3: $3V = \pi r^2 h$

Divide both sides by pi: $\frac{3V}{\pi} = r^2 h$

Divide both sides by h: $\frac{3V}{\pi h} = r^2$

Take the square root of both sides: $\sqrt{\frac{3V}{\pi h}} = r$

Your example is very similar.
Original post by james.h
I'll work through a different example to show you the ideas involved.

The volume of a cone is $V = \frac{1}{3} \pi r^2 h$

Multiply both sides by 3 to get rid of the 1/3: $3V = \pi r^2 h$

Divide both sides by pi: $\frac{3V}{\pi} = r^2 h$

Divide both sides by h: $\frac{3V}{\pi h} = r^2$

Take the square root of both sides: $\sqrt{\frac{3V}{\pi h}} = r$

Your example is very similar.

is my answer the cubed root of 3v/4 pie
Original post by gorila7
is my answer the cubed root of 3v/4 pie

$r = \sqrt[3]{\frac{3V}{4\pi}}$
Original post by james.h
$r = \sqrt[3]{\frac{3V}{4\pi}}$

Thank you very much sir. I rep you
There is an answer which shows "R= Cube Root of (6V/PI) / 2" in an assessment book. Does anyone know how the author went to this final answer instead of what was given by James H. above?

Any idea?
Original post by rykellim
There is an answer which shows "R= Cube Root of (6V/PI) / 2" in an assessment book. Does anyone know how the author went to this final answer instead of what was given by James H. above?

Any idea?

3V/4 = 6V/8 so they've just taken the cube root of 8 (i.e. 2) outside the root sign
Show working pls
Original post by Grace Billy
Show working pls

Please don't resurrect 10 year old threads