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iplayffxiv

Find the second derivative of the function. s=(t^6+3t+3)/ t^2

  • 11 months ago
  • 11 months ago

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  1. Mertsj
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    It might be easier to write it like this: \[s=t^4+3t ^{-1}+3t ^{-2}\]

    • 11 months ago
  2. sasogeek
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    @iplayffxiv can you continue from the help @Mertsj has provided?

    • 11 months ago
  3. iplayffxiv
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    i cant

    • 11 months ago
  4. iplayffxiv
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    i dont understand that

    • 11 months ago
  5. sasogeek
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    He simplified the equation first. now it will be easier to find the derivatives :) still lost? here's a simple way to find a derivative: if \(\large y=ax^n \) the derivative \(\large dy \), is equal to \(\large dy=nax^{n-1} \)

    • 11 months ago
  6. sasogeek
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    \(\large a \) is a constant, if \(\large a \) is 1, it is not written

    • 11 months ago
  7. iplayffxiv
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    does not make sense to me. it is multiple choice. can you choose the right answer?

    • 11 months ago
  8. sasogeek
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    first derivative: \(\large s=4t^3-3t^{-2}-6t^{-3} \) apply the same trick to see if you get ur answer.... take ur time, shouldn't be that difficult :)

    • 11 months ago
  9. iplayffxiv
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    A) d2s/dt2 = 4t^3 -3/t2 -6/t3 B) d2s/dt2 = 12t^4 +6/t +18/t2 C) d2s/dt2 = 4t^2 -3/t3 -6/t4 D) d2s/dt2 = 12t^2 +6/t^3 +18/t4

    • 11 months ago
  10. iplayffxiv
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    you see?

    • 11 months ago
  11. sasogeek
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    find the derivative of \(\large 4t^3 \), your answer will be obvious :) remember, the derivative of \(\large ax^n =nax^{n-1} \)

    • 11 months ago
  12. iplayffxiv
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    i dont know what to do

    • 11 months ago
  13. sasogeek
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    do well to read on how to find derivatives, but the answer you're looking for is the last option

    • 11 months ago
  14. iplayffxiv
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    someone give me answer

    • 11 months ago
  15. Mertsj
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    Do you want to learn to do this or do you just want an answer?

    • 11 months ago
  16. iplayffxiv
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    i would like to have the answer

    • 11 months ago
  17. Mertsj
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    Then we are all leaving because OS is not an answering service. Goodbye and good luck. All the time that has been spent on this problem was ample for you to learn to do i had you so chosen.

    • 11 months ago
  18. Loser66
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    good luck bye!!!

    • 11 months ago
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