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anonymous

  • 5 years ago

rationalizing denominators, the cube root of 2y squared divided by the cube root of 9x squared

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  1. anonymous
    • 5 years ago
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    \[\frac{\sqrt[3]{2y}}{\sqrt[3]{9x^2}} \times \frac{\sqrt[3]{9x^2}}{\sqrt[3]{9x^2}}\]\[\implies\frac{\sqrt[3]{2y * 9x^2}}{9x^2}\] \[\implies\frac{\sqrt[3]{18yx^2}}{9x^2}\]

  2. amistre64
    • 5 years ago
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    cbrt(9x^2)^2 does not equal 9x^2...

  3. amistre64
    • 5 years ago
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    cbrt((3^2 x^2) times (cbrt(3x)) = 3x

  4. anonymous
    • 5 years ago
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    Sorry, I was thinking of a regular square root, thanks for catching that amistre. :P

  5. amistre64
    • 5 years ago
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    no prob; happens to me all the time :)

  6. amistre64
    • 5 years ago
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    trig test in 20mins...yawn :)

  7. anonymous
    • 5 years ago
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    So the denominator should be \[\sqrt[3]{81x^4}\].

  8. amistre64
    • 5 years ago
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    denom will rationalize to 3x

  9. amistre64
    • 5 years ago
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    cbrt(3^2 x^2 3 x) = cbrt(3^3 x^3) = 3x

  10. anonymous
    • 5 years ago
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    \[thats what I got but the answer \in the back of the book says cube \root of6xy squred divided by 3x\]

  11. anonymous
    • 5 years ago
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    You're right. :P So the numerator should, in fact, be \[\sqrt[3]{6yx}\] and the denominator\[3x.\]

  12. amistre64
    • 5 years ago
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    top: cbrt(2y) cbrt(3x) = cbrt(6xy) yep; checks out on my "mr proffesor"

  13. radar
    • 5 years ago
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    Is it\[\sqrt[3]{(2y)2}\] or\[\sqrt[3]2y{2}\]

  14. amistre64
    • 5 years ago
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  15. anonymous
    • 5 years ago
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    how?

  16. amistre64
    • 5 years ago
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    Ciao :)

  17. radar
    • 5 years ago
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    The first expression is sort of confusing as it is written.

  18. anonymous
    • 5 years ago
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    @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).

  19. radar
    • 5 years ago
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    My first post was just seeking clarification.

  20. anonymous
    • 5 years ago
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    Ohhhhhhh, sorry, it is y^2...but that stays constant throughout, didn't see that before.

  21. radar
    • 5 years ago
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    I am talking about the original problem posted by abdon

  22. radar
    • 5 years ago
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    What is squared the 2y or just the y

  23. anonymous
    • 5 years ago
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    I believe it was \[\sqrt[3]{2y^2}/\sqrt[3]{9x^2}.\]

  24. radar
    • 5 years ago
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    OK. Thanks, now I will see if I get the same as you.

  25. anonymous
    • 5 years ago
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    Sorry for having such broken up responses, here's the condensed form of what I got:\[\frac{\sqrt[3]{6xy^2}}{3x}.\]

  26. anonymous
    • 5 years ago
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    \[got class ,thanks still confusing\]

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