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anonymous
 5 years ago
Can someone explain how to solve \[3^{1/3}\times27^{1/2}\] Thanks! :)
anonymous
 5 years ago
Can someone explain how to solve \[3^{1/3}\times27^{1/2}\] Thanks! :)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[3^{1/3}\times3^{3(1/2)}\] \[3^{1/3+(3/2)}\] \[3^{1/33/2}\] \[3^{(29)/6}\] \[3^{11/6}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you simplify that further? :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[27=3^{3}\] then solve the fraction...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[a ^{m} \times a ^{n}=a ^{m+n}\] the base must equal...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mean, how do you simplify \[3^{11/6}\] further? because actually, i got that answer too. but the answer isn't in the choices given. thanks!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt[6]{3}\] \[\sqrt[6]{3}/9\] \[3\sqrt[6]{3}/2\] and \[3\sqrt[6]{3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[3^{\frac{11}{6}} =3^{2}.3^{1/6}={\sqrt[6]{3} \over 9}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is \[\sqrt[6]{3}/9\] \[3^{11/6}=\] \[\sqrt[6]{3^{11}}\] \[\sqrt[6]{3^{1}/3^{12}}\] \[\sqrt[6]{3^{1}}/\sqrt[6]{12}\] \[\sqrt[6]{3}/3^{12(1/6)}\] \[\sqrt[6]{3}/3^{2}\] \[\sqrt[6]{3}/9\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Anwara, how is 3−116=3−2.31/6 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow,anwar.how do u do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i meant, how did you get to the 2nd equation? and thanks, mohsen!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because \(2+1/6= \frac{12+1}{6}=11/6\), which is the same exponent :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0same question with phoebeca :) bt u answered it :) thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm a teacher but still impress your answer,anwar :)
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