View Full Version : Rearanging formula
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