Mathematics
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OpenStudy (anonymous):
Algebra I
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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. \[\sqrt{5} = 5^{\frac{ 1 }{ 2 }}\]
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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 = ... ?
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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
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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, \[x^{3/4}\]
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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 :)
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OpenStudy (anonymous):
Thanks for helping me out :)
hartnn (hartnn):
welcome ^_^