How do you simplify x^5 /( x^2 y^-2)?

Apr 24, 2018

${x}^{3} / {y}^{-} 2$

Explanation:

Let's look at the $x$ terms first. We have

${x}^{5} / {x}^{2}$

which would evaluate to ${x}^{3}$, because of the property ${x}^{a} / {x}^{b} = {x}^{a - b}$.

There is no other $y$ terms except in the denominator, so it'll stay there. We have

${x}^{3} / {y}^{-} 2$

Hope this helps!

Apr 24, 2018

${y}^{2} \cdot {x}^{3}$

Explanation:

So, I think the important thing to talk about here is what ${y}^{-} 2$ means.

${y}^{-} 2$ can also be rewritten as $\frac{1}{y} ^ 2$ because of the negative exponent. For example, ${1}^{-} 5$ is also $\frac{1}{5}$. This is simply the rule of negative exponents. It's an important one! I'd remember it.

So after knowing this, we can arrive at:
${x}^{5} / {x}^{2} / {y}^{2}$

Then we cancel terms to achieve
${x}^{3} / \frac{1}{y} ^ 2$

Then just invert and multiply to obtain:
${y}^{2} \cdot {x}^{3}$