SOLUTION: I'm not sure if this is Linear Algebra, but i have a good feeling. So: Suppose Y varies directly with the square of W and the cube of X. When W=4 and X=2, Y=72. Find Y when W=1 an

Question 66175: I'm not sure if this is Linear Algebra, but i have a good feeling. So:
Suppose Y varies directly with the square of W and the cube of X. When W=4 and X=2, Y=72. Find Y when W=1 and X=5.
I'm having problems setting the problem up and I don't know what it means when it says it varies directly with the (square, cube, etc..) of X (or whatever).
Much appricated thanks!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I'm not sure if this is Linear Algebra, but i have a good feeling. So:
Suppose Y varies directly with the square of W and the cube of X. When W=4 and X=2, Y=72. Find Y when W=1 and X=5.
I'm having problems setting the problem up and I don't know what it means when it says it varies directly with the (square, cube, etc..) of X (or whatever).
------------------
If y=3x, y varies directly as x; that means "as x increases, y increases".
But if y=3/x, y varies indirectly as x: that means "as x increases, y decreases".
----------------
So for your problem:
y = k(w^2X^3)
We need to find "k" as follows:
72=k(4^2*2^3}
72=k(128)
k=72/128=9/16
------------
Cheers,
Stan H.
EQUATION:
y=(9/16)w^2x^3
--------------
Then if w=1 and x=5,
y=(9/16)(1^2*5^3)
y=(9/16)(125)
y=70.3125

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