# Algebra II

Simplify
(2x^-2y^2/3xz^-2) times (4xzy^-3/9xyz)^-1

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asked by Lee
1. Flip the negative exponents from numerator to denominator and vice versa.

Therefore, we have:

2y^2(z^2)/3x^3

[4xz/9xy^4(z)]^-1 = 9xy^4(z)/4xz = 9y^4/4

Now, let's multiply the two fractions:

2y^2(z^2)/3x^3 * 9y^4/4 =

18y^6(z^2)/12x^3 =

3y^6(z^2)/2x^3

If I haven't missed anything, that should be it. Remember your rules for multiplying and dividing exponents and simplify from there.

I hope this will help.

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posted by MathGuru

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