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anonymous

  • 5 years ago

how do i solve 3/ -4k^2- 5sqrt k^4?

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  1. amistre64
    • 5 years ago
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    is the equation: (3/-4k^2) - 5 sqrt(k^4) ?

  2. anonymous
    • 5 years ago
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    no its 3 over the entire -4k^2- 5sqrt k^4.

  3. amistre64
    • 5 years ago
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    3/ (4k^2 -5 sqrt(k^4)) then?

  4. amistre64
    • 5 years ago
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    with the (-)

  5. anonymous
    • 5 years ago
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    yes

  6. amistre64
    • 5 years ago
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    lets work on that "sqrt(k^4)" and see if we cant "undo" it..ok? what do you know of square roots and exponents?

  7. anonymous
    • 5 years ago
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    they r opposites of each other. you can use one to undo the other.

  8. amistre64
    • 5 years ago
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    good... how can we re-write sqrt(k^4) so that we can "undo" it? sqrt(k^2 k^2) is a start; and everything with a "^2" can be taken out right? So, kk can come out, which = k^2 do you agree?

  9. anonymous
    • 5 years ago
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    yes

  10. amistre64
    • 5 years ago
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    Then lets see how this helps us with the rest of the denominator: 4k^2 -5k^2 what can we do with this?

  11. amistre64
    • 5 years ago
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    i keep forgetting the (-). -4k^2 -5k^2

  12. anonymous
    • 5 years ago
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    its cool. we can take out k^2 rite?

  13. amistre64
    • 5 years ago
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    yes, you are right... k^2(-4-5) = -9k^2 right?

  14. anonymous
    • 5 years ago
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    yea

  15. amistre64
    • 5 years ago
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    now lets look how this helps us with the rest of it: 3/(-9k^2) do you see anything else we can do to this?

  16. amistre64
    • 5 years ago
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    3 ------ -9 k^2

  17. anonymous
    • 5 years ago
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    -1 ----- 3k^s ?

  18. amistre64
    • 5 years ago
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    yes, exactly :) even tho the ^s is a typo.... I will let it slide :)

  19. anonymous
    • 5 years ago
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    thank you so much!

  20. amistre64
    • 5 years ago
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    youre welcome...

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