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OpenStudy (anonymous):

Algebra I

OpenStudy (anonymous):

\[Simplify \left( \sqrt{5} )(\sqrt[3]{5}\right)\]

OpenStudy (anonymous):

(A) 5^(5/6) (B) 5^(1/6) (C) 5^(2/3) (D) 5^(7/6)

OpenStudy (anonymous):

I think the answer is C.

OpenStudy (anonymous):

@ganeshie8 @hartnn @texaschic101

OpenStudy (anonymous):

It's A. \[\sqrt{5} = 5^{\frac{ 1 }{ 2 }}\]

hartnn (hartnn):

so C isn't correct..

OpenStudy (anonymous):

Wait how?

hartnn (hartnn):

ok \(\sqrt x = x^{1/2} \\ \sqrt[3]x = x^{1/3}\)

hartnn (hartnn):

and \(\Large x^m x^n = x^{m+n}\)

hartnn (hartnn):

so, 1/2 +1/3 = ... ?

OpenStudy (anonymous):

Oh yeah thnx @Sadworld and @hartnn I understand it now! :)

OpenStudy (anonymous):

I probably had done it a more complicated way. That's why

hartnn (hartnn):

can you do a similar other question on your own ?

OpenStudy (anonymous):

\[\sqrt{5} = 5^{\frac{ 1 }{ 2 }} ; \sqrt[3]{5} = 5^{\frac{ 1 }{ 3 }}\]

OpenStudy (anonymous):

Yeah. Give me a question plz

hartnn (hartnn):

\(\Large \sqrt[4]{x^3}\times \sqrt[4]x = ... ?\)

OpenStudy (anonymous):

Give me some time tho

hartnn (hartnn):

sure take your time :)

hartnn (hartnn):

hint : \(\Large \sqrt[p]{x^q} = x^{q/p}\)

OpenStudy (anonymous):

Okay so, \[x^{3/4}\]

OpenStudy (anonymous):

Then, \[x^\frac{ 1 }{ 4 }\]

hartnn (hartnn):

\(\Large x^{3/4} \times x^{1/4} = ... ?\)

hartnn (hartnn):

\(\Large x^{3/4} \times x^{1/4} = x^{3/4+1/4}=... ?\)

OpenStudy (anonymous):

x

hartnn (hartnn):

excellent :)

OpenStudy (anonymous):

Thanks for helping me out :)

hartnn (hartnn):

welcome ^_^

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