Thread: Dont understand how this works

Suppose y varies directly as the square of x and inversly as the cube of w. We know that y=9 when x=2 and w=1. Find y when x=4 and w=2.

Originally Posted by brianfisher1208

Suppose y varies directly as the square of x and inversly as the cube of w. We know that y=9 when x=2 and w=1. Find y when x=4 and w=2.

The first sentence is telling you the following formula is correct (where k is the 'constant of variation'): $\displaystyle y=k\frac{x^2}{w^3}$

Use the information in the second sentence to solve for k. Once you know that, you can use the equation to get the answer to the third sentence.

Originally Posted by brianfisher1208

Suppose y varies directly as the square of x and inversly as the cube of w. We know that y=9 when x=2 and w=1. Find y when x=4 and w=2.

$\displaystyle y \propto x^2$, so $\displaystyle y = k_1x^2$.

Substitute $\displaystyle x$ and $\displaystyle y$ to find $\displaystyle k_1$.


$\displaystyle y \propto \frac{1}{w^3}$, so $\displaystyle y = \frac{k_2}{w^3}$.

Substitute $\displaystyle w$ and $\displaystyle y$ to find $\displaystyle k_2$.



Then you should be able to finish the question.