Solve V for h: Rewrite Formula V=1/3πr^2h

  • Thread starter Thread starter GLprincess02
  • Start date Start date
  • Tags Tags
    Formulas
AI Thread Summary
To rewrite the formula V=1/3πr^2h for h, first multiply both sides by 3 to eliminate the fraction, resulting in 3V=πr^2h. Then, isolate h by dividing both sides by πr^2, leading to h=3V/(πr^2). The user confirmed their understanding of the steps and expressed gratitude for the assistance. The discussion highlights the importance of maintaining balance in equations when manipulating them.
GLprincess02
Messages
50
Reaction score
0
On my homework, I was given the equation V=1/3 x pi x r squared x h. I need to rewrite the formula so that it is solved for h.

Any help will be greatly appreciated!
 
Physics news on Phys.org
:rolleyes: You probably should have posted this in the homework section instead, and if you can't find the topic here later, it was probably moved there.

But, I'm not mean. :-p

I'm not going to solve it for you, but think: If you do something to one side of the equals, the same thing has to happen on the other side.

Use the 1/3 as an example, and cancel it by multiplying by 3. Then, just make everything else go to the other side by multiplying it over. ^:biggrin:
 
I'm sorry about not posting this in the right forum...I just signed up today, so I'm still getting used to everything here!

And about your reply...I tried what you said, and I got 3V over pi x r squared. Is this right??
 
if you meant \frac{3V}{\pi r^2} then yes
 
Lol sorry...once again, I'm new here!

Thank you both very much for your help. :biggrin:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

Similar threads

Back
Top