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

Question

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

Answer 1

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 $k$:

$$10=\frac{k}{{10}^3}⟹10=\frac{k}{1000}$$

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:

$$\boxed{{\displaystyle 5}}$$