Question

y varies inversely as the square of x. When y = $16$, x = $1$. When $x=8$, y equals:

• A. $2$
• B. $128$
• C. $64$
• D. $\frac{1}{4}$
• E. $1024$

Answer 1

The correct option is D $\frac{1}{4}$

$y=\frac{k}{x^2},16=\frac{k}{1},k=16;y=\frac{16}{x^2},y=\frac{16}{8.8}=\frac{1}{4}$
$\frac{y_1}{y_2}=\frac{x_1^2}{x_1^2},\frac{16}{y_2}=\frac{8^2}{1^2};y_2=\frac{1}{4}$.