Question

z varies directly as x2 and inversely as y2. If z=128 when x=8 and y=9, find z if x=7 and y=4. (Round off your answer to the nearest hundredth.)

Answer 1

Answer:

z = 496.13

Explanation:

Let the statement be written mathematically as follows:

$z\propto x^2\propto\frac{1}{y^2}$

So that:

$z=\frac{kx^2}{y^2}$

Where k is constant.

For z = 128, x = 8, and y = 9, we have:

$\begin{gathered} 128=\frac{k\times8^2}{9^2} \\ \\ k=\frac{128\times9^2}{8^2^{}}=162 \end{gathered}$

Using this value of k, we write the formula as:

$z=\frac{162x^2}{y^2}$

Now for x = 7, y = 4, we need to find z.

$z=\frac{162\times7^2}{4^2}=496.13$