Suppose x varies inversely with the cube of y. If x=10

when y=10
, what is y when x is 80?

1 answer

If \( x \) varies inversely with the cube of \( y \), we can express this relationship mathematically as:

\[ x = \frac{k}{y^3} \]

where \( k \) is a constant.

Given that \( x = 10 \) when \( y = 10 \), we can substitute these values into the equation to find \( k \):

\[ 10 = \frac{k}{10^3} \implies 10 = \frac{k}{1000} \]

Multiplying both sides by 1000:

\[ k = 10 \times 1000 = 10000 \]

Now we have the equation:

\[ x = \frac{10000}{y^3} \]

Next, we need to find \( y \) when \( x = 80 \):

\[ 80 = \frac{10000}{y^3} \]

To solve for \( y^3 \), we first multiply both sides by \( y^3 \):

\[ 80y^3 = 10000 \]

Now divide both sides by 80:

\[ y^3 = \frac{10000}{80} = 125 \]

Taking the cube root of both sides gives us:

\[ y = \sqrt[3]{125} = 5 \]

Thus, when \( x = 80 \), the corresponding value of \( y \) is:

\[ \boxed{5} \]