Asked by Lee
Simplify
(2x^-2y^2/3xz^-2) times (4xzy^-3/9xyz)^-1
(2x^-2y^2/3xz^-2) times (4xzy^-3/9xyz)^-1
Answers
Answered by
MathGuru
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.
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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