SOLUTION: if y varies directly as x and inversely as z^2, and y=3 when x=2 and z=4 what is the value of x when y=9 and z=4?
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Question 246151: if y varies directly as x and inversely as z^2, and y=3 when x=2 and z=4 what is the value of x when y=9 and z=4? Answer by dabanfield(803) (Show Source):
You can put this solution on YOUR website! if y varies directly as x and inversely as z^2, and y=3 when x=2 and z=4 what is the value of x when y=9 and z=4?
Restated the above says y = k*(x/(z^2)) where k is a constant of proportionality yet to be determined.
As stated above y=3 when x=2 and z=4 so substituting in our formula:
3 = k*(2/(4^2))
3 = k*(2/16)= k/8
k=24
So our equation becomes y=24*(x/(z^2))
To solve the problem we need to substitute y = 9 and z = 4 and then solve for x in the equation above.
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