SOLUTION: I was not sure what category to pick so I choose test. Sorry. My Algebra II book calls this type of problem variation. Anyway here is my problem: For a constant temperature,

Question 21997: I was not sure what category to pick so I choose test. Sorry. My Algebra II book calls this type of problem variation. Anyway here is my problem:
For a constant temperature, the pressure (P) of gas varies inversely as the volume (V). If the pressure is 25 lb/in squared when the volume is 400 ft to the third, find the pressure when the volume is 150 ft to the third.
So i know the formula is y=k/x and then you substitue P=k/x. then substitue 25 squared = k/400 third. I think at this point I want to get rid of the denominator of 400 to the third by multiplying both sides by 400 to the third, but I am not sure. Any help would be much appreciated. Thank you for your time.

Answer by stanbon!(97) (Show Source):
You can put this solution on YOUR website!
I think you are getting confused in reading your problem.
The pressuse is 25 lb per sq. in. not (25 lb/in)^2.
The volume is 400 cubic ft. not (400 ft.)^3.
So you have in place of P = k/V the following:
25 = k/400
k = 10,000
The equation you want is P= 10,000/V
Now, find the pressure when the volume is 150 cu. ft.
P = 10,000/150= 66 2/3 lb./sq.in
Cheers,
Stan H.