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Can someone help me simplify this radical expression, please? √45n^5 Also can you tell me how you did it?

Pre-Algebra
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i take it is \[\sqrt{45n^5}\] right?
first step is to factor \(45=9\times 5\) and this helps because you know what the square root of 9 is, namely 3 you can write \[\sqrt{45n^5}=\sqrt{9\times 5\times n^4\times n}=\sqrt{9}\sqrt{n^4}\sqrt{5n}=3n^2\sqrt{5n}\]
we can take care of \(\sqrt{n^5}\) in your head two goes in to 5 twice, with a remainder of 1, so \(n^2\) comes out of the radical and one \(n\) stays in, making \(\sqrt{n^5}=n^2\sqrt{n}\)

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Other answers:

Okay hold on let me write this down
i don't understand
@satellite73 how did you get rid of the 5?
in the last step
He got rid of the 5 by putting 2 into 5 twice, and it had a remainder of 1, so it came out as n2
I still don't get it
can I plz get some help
LOL

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