b. V=4∏r2L :L
.......5

it is r squared not 2 L
I need L as the subject

and 4 over 5

Please and thanks

V = \frac{4}{5}\pi r^2L

Is this what you mean? Now you need L as the subject and \frac{4}{5} as the subject? Just making sure I understand, then I can help you :)

What does the :L mean?

Yes that is what I meant , the :L is what needs to be the subject, it is how it was written by my tutor but just L I don't need the 4 over 5 as the subject

Okay, Let me explain it to you this way: You always do the same operation to both sides.

So to get L as the subject, we can either get get rid of everythign on its side of the equation, or we can move L to the side with the least stuff and then get rid of everything else on that side.

The latter is the best option here, because V is the only thing on the other side:

V = \frac{4}{5}\pi r^2L

So, step by step we:

divide both sides by L:

\frac{V}{L}=\frac{4}{5}\pi r^2

divide both sides by V:

\frac{1}{L}=\frac{\frac{4}{5}\pi r^2}{V}

take the inverse of both sides (Raising both sides to the power of -1):

L = \frac{V}{\frac{4}{5}\pi r^2}

simplify:

L = \frac{5V}{4\pi r^2}

and there we go.

Now can you do the same steps to make 4/5 the subject?

Cap, he doesn't need 4/5 the subject, he was just telling you that it was 4/5 in the original formula. :)

Well I didn't understand that bit :p