Suppose p varies directly as the cube of x, and inversely with y. If p=5/8 when x=4 and y=10, find p when x=2 and y=16

How do you set up the steps?

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- May 17th 2009, 03:31 PMreuger_Rational Functions?
Suppose p varies directly as the cube of x, and inversely with y. If p=5/8 when x=4 and y=10, find p when x=2 and y=16

How do you set up the steps? - May 17th 2009, 03:43 PMJen
When P varies directly with x, it means, $\displaystyle p=k_1x$ for some constant k.

If P varies directly with the cube of x then, $\displaystyle P = k_1 x^3$

When P varies inversely with y it means, $\displaystyle P = \frac{k_2}{y}$

Now if you use the initial conditions provided you should be able to solve for the constants $\displaystyle k_1 \mbox{ and } k_2$ - May 17th 2009, 03:51 PMreuger_thanks
- May 17th 2009, 04:02 PMJen
No Problem. :D