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anonymous
 one year ago
Simplify square root of 2 multiplied by the cube root of 2.
anonymous
 one year ago
Simplify square root of 2 multiplied by the cube root of 2.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{2}*\sqrt[3]{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Change each radical to rational expression. \[\sqrt{2}>2^{\frac{ 1 }{ 2 }}\] \[\sqrt[3]{2}>2^\frac{ 1 }{ 3 }\] Then use LCD in both fractions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have no idea how to do it. I think its 2 2/3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and yes it looks like that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is the least common multiple of 2 and 3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02: 2, 4, 6, 8, 10 3: 3, 6, 9, 12, 15

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes! So, dw:1438119775405:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Good! Than multiply 3 to the numerator.dw:1438119897874:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry lost connection

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now do it to the other one. dw:1438120062129:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then multiply it by the numerator of 1/3.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438120283484:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438120302016:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes. Do you want to turn it into radical?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no thankyou it doesnt ask to do that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thankyou for helping me understand!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay! You're welcome. :)
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