# If y varies inversely as the cube root of x, and y = 12 when x = 8, how do you find y when x = 1?

Jul 3, 2016

$24$

#### Explanation:

Write the english as maths: $y \propto \frac{1}{\sqrt[3]{x}}$

Change the variation to an equation by multiplying by a constant.

$y = \frac{k}{\sqrt[3]{x}} \text{ we need to find a value for } k$

$k = \sqrt[3]{x} \times y = \sqrt[3]{8} \times 12 = 2 \times 12 = 24$

Replace $k$ with $24$ to obtain the equation

$y = \frac{24}{\sqrt[3]{x}}$

Now we can calculate y for any given x.

$y = \frac{24}{\sqrt[3]{1}} = \frac{24}{1} = 24$