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anonymous

  • 5 years ago

Can someone please Help me with college algebra 2 ?

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  1. anonymous
    • 5 years ago
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    ok

  2. anonymous
    • 5 years ago
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    \[\sqrt[4]{32y^13}\]

  3. anonymous
    • 5 years ago
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    32 = 2 x 2 x 2 x 2

  4. anonymous
    • 5 years ago
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    i left out one 2

  5. anonymous
    • 5 years ago
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    its y raised to the 13th

  6. anonymous
    • 5 years ago
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    im guessing im factoring it?

  7. anonymous
    • 5 years ago
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    are you trying to solve for y?

  8. anonymous
    • 5 years ago
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    no i think you are trying to simplify the expression

  9. anonymous
    • 5 years ago
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    i was using prime factorization

  10. anonymous
    • 5 years ago
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    right simplify

  11. anonymous
    • 5 years ago
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    so you have 32 factored to 2^5, so take away 16 from inside the radical and leave in one 2 as for y, you have y^3 outside the radical and one y inside, since (y^3)^4 * y = y^13

  12. anonymous
    • 5 years ago
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    im pretty sure it is 2y^3*radical4(2y)

  13. anonymous
    • 5 years ago
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    so its 32^(1/4)*y^(13/4)*3^(1/4)

  14. anonymous
    • 5 years ago
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    its a 4 outside the radical and 32y^13 on the inside

  15. anonymous
    • 5 years ago
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    Factor completely: 6b^2+4a-8b-3ab

  16. anonymous
    • 5 years ago
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    wait, ok your previous expression was messed up, can you re type it?

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