Simplify Completely. (9x^-4y^2)^-1/(y^-3z^4) x (3y^-1)^2/(xz^5)

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Simplify Completely. (9x^-4y^2)^-1/(y^-3z^4) x (3y^-1)^2/(xz^5)

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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what is (x^2)^-1?
(1/x) = (x)^-1
Sorry this is a question dividing exponents and I thought I could use the ^ symbol to represent that, is that wrong? (9x^-4y^2)^-1/(y^-3z^4) x (3y^-1)^2/(xz^5) means 9x to the -4 power times y to the 2nd power divided by y to the -3 power times z to the 4th power, all in parentheses to the -1st power. then times that by 3y to the -1st power divided by xz to the 5th power in parentheses, all to the 2nd power

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that's not wrong, i meant 1/x equal to x^-1 so, (9x^-4y^2)^-1 = 1/(9x^-4y^2)

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