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angelwings996

  • one year ago

Simplest form help ?

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  1. hba
    • one year ago
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    @angelwings996 Um,Where is the question ?

  2. angelwings996
    • one year ago
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    Sorry, Im trying to put it into the equation

  3. hba
    • one year ago
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    I gotta go sleep. Someone will be here to help you soon :)

  4. angelwings996
    • one year ago
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    \[\frac{ \sqrt[3]{x ^{3}} }{ \sqrt[5]{x ^{2}} }\]

  5. angelwings996
    • one year ago
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    This is the equation I need to put in simplest form

  6. blondie16
    • one year ago
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    have no fear amparo is here LOL

  7. angelwings996
    • one year ago
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    Haha Okaay

  8. hba
    • one year ago
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    \[[\huge\ \frac{ \sqrt[3]{x ^{3}} }{ \sqrt[5]{x ^{2}} }] =(x^3)^{1/3}/(x^2)^{1/5}\]

  9. artofspeed
    • one year ago
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    @angelwings996 , when meeting stuff like this, put the number at LEFT of the square root sign as DENOMINATOR

  10. artofspeed
    • one year ago
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    so \[\sqrt[3]{x^3} = x^\frac{3}{3}\]

  11. artofspeed
    • one year ago
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    \[\sqrt[3]{x^2}=x^\frac{2}{3}\]

  12. angelwings996
    • one year ago
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    Wouldn't it be 5 instead of 3 for the second one ?

  13. artofspeed
    • one year ago
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    \[\sqrt[100]{x^7}=x^\frac{7}{100}\]

  14. artofspeed
    • one year ago
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    in your question, yes

  15. angelwings996
    • one year ago
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    How did you get 100 and 7 ?

  16. angelwings996
    • one year ago
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    or is it an example ?

  17. artofspeed
    • one year ago
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    that's just an example to show the use of the trick

  18. angelwings996
    • one year ago
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    Okay, so what would I do now ?

  19. artofspeed
    • one year ago
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    if they numbers have the same base, u do: exponent on top - exponent below

  20. hba
    • one year ago
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    \[\sqrt{x}=x^{1/2}\] \[\sqrt[3]{x}=x^{1/3}\] similarly \[x^a=x^{1/a}\]

  21. artofspeed
    • one year ago
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    so it's 3/3-2/5 = 3/5 answer should be x^(3/5)

  22. angelwings996
    • one year ago
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    Oh I see!

  23. angelwings996
    • one year ago
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    Thank you so much for your guys' help (:

  24. artofspeed
    • one year ago
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    Np :)

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