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

2 Answers
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#

as our final answer.

Hope this helps!

#y^2*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 #1/y^2# because of the negative exponent. For example, #1^-5# is also #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^3/1/y^2#

Then just invert and multiply to obtain:
#y^2*x^3#