# If y varies inversely as the cube of x and x=2.5 when y=0.015, how do you find y when x=5?

$y = .001875$ when $x = 5$
y prop 1/x^3 or y = k* 1/x^3 ; x =2.5 when $y = 0.015 \therefore 0.015 = k \cdot \frac{1}{2.5} ^ 3 \mathmr{and} k = {2.5}^{3} \cdot 0.015 = 0.234375$ So the variation equation stands $y = 0.234375 \cdot \frac{1}{x} ^ 3$ Now putting $x = 5$ we get $y = 0.234375 \cdot \frac{1}{5} ^ 3 = .001875$[Ans]