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Annette [7]
4 years ago
9

What is the simplified form of the square root of 48n^9

Mathematics
1 answer:
MrRa [10]4 years ago
3 0
\bf \sqrt{48n^9}\qquad\begin{cases}
48=2\cdot 2\cdot 2\cdot 2\cdot 3\\
\qquad 2^2\cdot 2^2\cdot 3\\
\qquad (2^2)^2\cdot 3\\
n^9=n^{8+1}\\
\qquad n^8\cdot  n^1\\
\qquad n^{4\cdot 2}\cdot n\\
\qquad (n^4)^2\cdot n
\end{cases}\implies \sqrt{(2^2)^2\cdot 3\cdot (n^4)^2\cdot n}
\\\\\\
2^2n^4\sqrt{3n}\implies 4n^4\sqrt{3n}
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Answer:

How to solve your problem

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7

2

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2

2

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3

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2

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5

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-7y^{2}-2y^{2}+y^{3}y-2y+5y^{3}-2y

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-7y^{2}-2y^{2}+{\color{#c92786}{y^{3}y}}-2y+5y^{3}-2y

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7

2

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2

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−

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3

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-7y^{2}-2y^{2}+{\color{#c92786}{y^{4}}}-2y+5y^{3}-2y

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+

4

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3

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{\color{#c92786}{-7y^{2}}}{\color{#c92786}{-2y^{2}}}+y^{4}-2y+5y^{3}-2y

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2

+

4

−

2

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5

3

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2

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-9y^{2}+y^{4}{\color{#c92786}{-2y}}+5y^{3}{\color{#c92786}{-2y}}

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Rearrange terms

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4

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{\color{#c92786}{-9y^{2}+y^{4}-4y+5y^{3}}}

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