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anonymous

  • one year ago

I need help will fan+medal?!?!?!?!

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  1. anonymous
    • one year ago
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    simplify \[\sqrt[3]{7}/\sqrt[5]{7}\]

  2. anonymous
    • one year ago
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    A. 7 to the power of 1 over 5 B. 7 to the power of 8 over 15 C. 7 to the power of 5 over 3 D. 7 to the power of 2 over 15

  3. anonymous
    • one year ago
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    @Nasa7 CAN U HELP PLZ

  4. anonymous
    • one year ago
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    @rock_mit182 do u know how to simplify this?

  5. Anguyennn
    • one year ago
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    first question would be 1 because they are both the same and when you divide them together it would be 1

  6. anonymous
    • one year ago
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    i need it simplified though not the answer

  7. rock_mit182
    • one year ago
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    |dw:1442284199542:dw|

  8. rock_mit182
    • one year ago
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    @Queli what did you get ?

  9. anonymous
    • one year ago
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    Im not sure..

  10. rock_mit182
    • one year ago
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    keep in mind in our case: a = 7

  11. rock_mit182
    • one year ago
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    ill help you out with the numerator

  12. anonymous
    • one year ago
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    \[7^{2/3}\] and \[7^{2/5}\]

  13. rock_mit182
    • one year ago
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    |dw:1442284606506:dw|

  14. rock_mit182
    • one year ago
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    you were close enough, try again. keep in mind what i told you about : \[7 = 7^{1}\]

  15. anonymous
    • one year ago
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    \[7^{1/3} and 7^{1/5}\] so the answer would be d right

  16. rock_mit182
    • one year ago
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    \[\frac{ 7^{\frac{ 1 }{ 3 }} }{ 7^{\frac{ 1 }{ 5 }} } = 7 ^{\frac{ 1 }{ 3 }}* 7\frac{- 1 }{ 5 }\]

  17. rock_mit182
    • one year ago
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    WHisch is the same as: \[=7^{\frac{ 1 }{ 3 }-\frac{ 1 }{ 5 }}\]

  18. rock_mit182
    • one year ago
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    can you solve this by your own now ?

  19. rock_mit182
    • one year ago
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    1/3 - 1/5 ?

  20. rock_mit182
    • one year ago
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    7 to the power of .... ?

  21. anonymous
    • one year ago
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    the common denominator would be 15 so 5/15-3/15=2/15

  22. rock_mit182
    • one year ago
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    yeah :)

  23. rock_mit182
    • one year ago
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    nice work

  24. rock_mit182
    • one year ago
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    that's all my friend

  25. anonymous
    • one year ago
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    thanks been stuck on that problem for hrs : )

  26. rock_mit182
    • one year ago
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    bu you're smart enough to figure out anything. you're welcome

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