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

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Answer 1 · 2016-07-03T09:18:58

$24$ $24$

Write the english as maths: $y \propto \frac{1}{\sqrt[3]{x}}$ $y prop1/root(3)x$

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

$y = \frac{k}{\sqrt[3]{x}} we need to find a value for k$ $y = k/root(3)x" we need to find a value for "k$

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

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

$y = \frac{24}{\sqrt[3]{x}}$ $y = 24/root(3)x$

Now we can calculate y for any given x.

$y = \frac{24}{\sqrt[3]{1}} = \frac{24}{1} = 24$ $y = 24/root(3)1 =24/1 =24$