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TomTurner · Mar 29, 2007, 02:56 AM

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

Capuchin · Mar 29, 2007, 02:59 AM

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

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?

TomTurner · Mar 29, 2007, 07:19 AM

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

Capuchin · Mar 29, 2007, 07:24 AM

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?

asterisk_man · Mar 29, 2007, 08:00 AM

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

Capuchin · Mar 29, 2007, 09:48 AM

Well I didn't understand that bit :p