Question

If y varies inversely as the square of x, and y=7/4/ when x=1, find y when x=3

Answer 1

Y varies inversely as the square of x. So we can write the relation between x and y as:

$y= \frac{k}{ x^{2} }$

where k is a constant of proportionality. It is given that y = 7/4 when x = 1.
Using these values, we can write:

$\frac{7}{4}= \frac{k}{1} \\ \\ k= \frac{7}{4}$

So, now the equation in terms of x and y can be written as:

$y= \frac{7}{4 x^{2} }$

We are to find y when x = 3. So we can write:

$y= \frac{7}{4(9)}=\frac{7}{36}$

Thus, for x= 3 value of y will be 7/36