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faust18 [17]
4 years ago
10

The expression can be simplified and written without negative exponents as...

Mathematics
1 answer:
Ksju [112]4 years ago
6 0

\dfrac{(27y^{-2})^\frac{1}{3}}{y^{-\frac{2}{3}}}=(27)^\frac{1}{3}(y^{-2})^\frac{1}{3}:\dfrac{1}{y^\frac{2}{3}}=\sqrt[3]{27}y^{-2\cdot\frac{1}{3}}\cdot y^\frac{2}{3}\\\\=3y^{-\frac{2}{3}}\cdot y^\frac{2}{3}=3y^{-\frac{2}{3}+\frac{2}{3}}=3y^0=3\\\\\text{used:}\\\\(a\cdot b)^n=a^n\cdot b^n\\\\a^\frac{1}{n}=\sqrt[n]{a}\\\\a^{-n}=\dfrac{1}{a^n}\\\\(a^n)^m=a^{n\cdot m}\\\\a^n\cdot a^m=a^{n+m}\\\\a^0=1,\ a\neq0

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