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anonymous
 5 years ago
rationalizing denominators, the cube root of 2y squared divided by the cube root of 9x squared
anonymous
 5 years ago
rationalizing denominators, the cube root of 2y squared divided by the cube root of 9x squared

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\sqrt[3]{2y}}{\sqrt[3]{9x^2}} \times \frac{\sqrt[3]{9x^2}}{\sqrt[3]{9x^2}}\]\[\implies\frac{\sqrt[3]{2y * 9x^2}}{9x^2}\] \[\implies\frac{\sqrt[3]{18yx^2}}{9x^2}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cbrt(9x^2)^2 does not equal 9x^2...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cbrt((3^2 x^2) times (cbrt(3x)) = 3x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, I was thinking of a regular square root, thanks for catching that amistre. :P

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0no prob; happens to me all the time :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0trig test in 20mins...yawn :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the denominator should be \[\sqrt[3]{81x^4}\].

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0denom will rationalize to 3x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cbrt(3^2 x^2 3 x) = cbrt(3^3 x^3) = 3x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[thats what I got but the answer \in the back of the book says cube \root of6xy squred divided by 3x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're right. :P So the numerator should, in fact, be \[\sqrt[3]{6yx}\] and the denominator\[3x.\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0top: cbrt(2y) cbrt(3x) = cbrt(6xy) yep; checks out on my "mr proffesor"

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Is it\[\sqrt[3]{(2y)2}\] or\[\sqrt[3]2y{2}\]

radar
 5 years ago
Best ResponseYou've already chosen the best response.0The first expression is sort of confusing as it is written.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@abdon: If you multiply the denominator by cbrt(3x) you get cbrt(27*x^3) which simplifies to 3x. Now, because you multiplied the denominator by that factor, you have to multiply the numerator by the same factor, and you get cbrt(6xy).

radar
 5 years ago
Best ResponseYou've already chosen the best response.0My first post was just seeking clarification.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ohhhhhhh, sorry, it is y^2...but that stays constant throughout, didn't see that before.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0I am talking about the original problem posted by abdon

radar
 5 years ago
Best ResponseYou've already chosen the best response.0What is squared the 2y or just the y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I believe it was \[\sqrt[3]{2y^2}/\sqrt[3]{9x^2}.\]

radar
 5 years ago
Best ResponseYou've already chosen the best response.0OK. Thanks, now I will see if I get the same as you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry for having such broken up responses, here's the condensed form of what I got:\[\frac{\sqrt[3]{6xy^2}}{3x}.\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[got class ,thanks still confusing\]
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