When 9^2/3 is written in simplest radicsl form, which value remains under the radical? 3 6 9 27

2 answers:

7 0

$\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 9^{\frac{2}{3}}\implies (3^2)^{\frac{2}{3}}\implies 3^{2\cdot \frac{2}{3}}\implies 3^{\frac{4}{3}}\implies \sqrt[3]{3^4}\implies \sqrt[3]{3^3\cdot 3^1}\implies 3\sqrt[3]{\stackrel{\textit{this one}}{3}}$

faust18 [17]3 years ago

3 0

Answer:

\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 9^{\frac{2}{3}}\implies (3^2)^{\frac{2}{3}}\implies 3^{2\cdot \frac{2}{3}}\implies 3^{\frac{4}{3}}\implies \sqrt[3]{3^4}\implies \sqrt[3]{3^3\cdot 3^1}\implies 3\sqrt[3]{\stackrel{\textit{this one}}{3}}

Step-by-step explanation:

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Ket [755]

Answer:

B. Step 2 uses the associative property, and step 3 uses the commutative property.

Step-by-step explanation:

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The commutative property lets you change the order of any pair of terms involved in addition. It appears that in Step 3, the order of the terms within the group has been swapped.

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Comment on associative and commutative properties

What applies to terms in addition applies to factors in multiplication.

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3 years ago

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a_sh-v [17]

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