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calculusxy

  • one year ago

Help with exponents ... Question attached below.

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  1. calculusxy
    • one year ago
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    \[\huge \frac{ (2p m^{-1}q^0)^{-4} \times 2m^{-1}p^3 }{ 2pq^2}\]

  2. calculusxy
    • one year ago
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    @hartnn

  3. hartnn
    • one year ago
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    tried it ?

  4. hartnn
    • one year ago
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    just take one variable at a time assume there's only p

  5. hartnn
    • one year ago
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    \(\Large \dfrac{p^{-4}p^3}{p} = ..?\)

  6. calculusxy
    • one year ago
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    Actually what I did was this: \[\large (2p m^{-1}q^0)^4 = 2^{-4}p^{-4}m^{-5}q^{-4}\]

  7. calculusxy
    • one year ago
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    @hartnn

  8. hartnn
    • one year ago
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    for 2 and p thats correct. for others it isn't.

  9. hartnn
    • one year ago
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    \(\Large (m^{-1})^{-4} = m^{-1\times -4} = m^{???}\)

  10. calculusxy
    • one year ago
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    oh okay, so it's m^4

  11. hartnn
    • one year ago
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    yes

  12. hartnn
    • one year ago
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    \(\Large (q^{0})^{-4} = q^{0\times -4} = q^{???}\)

  13. calculusxy
    • one year ago
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    q^0

  14. calculusxy
    • one year ago
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    which is 1

  15. hartnn
    • one year ago
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    correct!

  16. calculusxy
    • one year ago
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    so we would now have: \[\large \frac{ 2^{-4}p^{-4}m^4q^0 \times 2m^{-1}p^3 }{ 2pq^2 }\]

  17. hartnn
    • one year ago
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    yes whats m^4 * m^{-1} = .. ?

  18. calculusxy
    • one year ago
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    Before that, is it possible to group the terms together according to the variables and exponents?

  19. hartnn
    • one year ago
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    according to variables, yes!

  20. calculusxy
    • one year ago
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    also what would we do with the 2^{-4}?

  21. hartnn
    • one year ago
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    combine like terms! \(\dfrac{2^{-4} \times 2}{2} = 2^{-4} = \dfrac{1}{2^4}\)

  22. calculusxy
    • one year ago
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    so the other two (without the variable), you took it out from 2m^{-1}?

  23. hartnn
    • one year ago
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    \(\Large m^4 \times m^{-1}= .. \) \(\Large \dfrac{p^{-4}p^3}{p} = ..?\)

  24. calculusxy
    • one year ago
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    what variable would the 2 have? \[2^1\] or \[2^0\]

  25. hartnn
    • one year ago
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    2 = 2^1

  26. calculusxy
    • one year ago
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    \[\frac{ 2^{-4} \times 2^1 }{ 2 } = 2^{-4}\]

  27. calculusxy
    • one year ago
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    am i correct?

  28. hartnn
    • one year ago
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    yes

  29. calculusxy
    • one year ago
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    ok let me go on..

  30. calculusxy
    • one year ago
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    now if i did the other variables: \[\frac{ p^{-4}p^3 }{ p^1 } = p^{-2}\] \[m^4 \times m^{-1} = m^3\] \[\frac{ q^0 }{ q^2 } = q^{-2}\]

  31. hartnn
    • one year ago
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    all correct! :)

  32. madhu.mukherjee.946
    • one year ago
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    good

  33. calculusxy
    • one year ago
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    okay so now I combine them?

  34. hartnn
    • one year ago
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    yes

  35. calculusxy
    • one year ago
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    \[\frac{ m^3 }{ 2^4q^2p^2 }\]

  36. hartnn
    • one year ago
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    \(\huge \checkmark \) *applause*

  37. calculusxy
    • one year ago
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    thank you! i have another question, can you help me on that as well?

  38. hartnn
    • one year ago
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    I can try :)

  39. calculusxy
    • one year ago
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    \[\huge \frac{ (2hj^2k^{-2} \times h^4h^{-1}k^4)^0 }{ 2h^{-3}j^{-4}k^{-2}}\]

  40. calculusxy
    • one year ago
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    @hartnn

  41. hartnn
    • one year ago
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    lol (anything except 0)^0 = 1 so there you have your numerator!

  42. calculusxy
    • one year ago
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    yeah that's what i was thinking.. but what what about solving it with the denominator?

  43. hartnn
    • one year ago
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    |dw:1442770780050:dw|

  44. hartnn
    • one year ago
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    1/h^(-3) = h^3

  45. calculusxy
    • one year ago
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    oh yeah! my teacher taught us that.. it just flushed out of my brain. so the answer is: \[\huge 2h^3j^4k^2\]

  46. hartnn
    • one year ago
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    (h^3 j^4 k^2)/2

  47. calculusxy
    • one year ago
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    i don't understand where that came from

  48. hartnn
    • one year ago
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    |dw:1442771044268:dw|

  49. calculusxy
    • one year ago
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    oh okay.. the positive 2 stayed on the bottom.

  50. calculusxy
    • one year ago
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    since it always was on the bottom...

  51. calculusxy
    • one year ago
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    okay so it's \[\frac{ h^3j^4k^2 }{ 2 }\]

  52. hartnn
    • one year ago
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    yes! correct :)

  53. calculusxy
    • one year ago
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    thank you!

  54. calculusxy
    • one year ago
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    sorry but i just wanted to check an answer with you..

  55. calculusxy
    • one year ago
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    \[\large x^4y^3 \times (2y^2)^0\] would the answer be \[x^4y^3\]

  56. hartnn
    • one year ago
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    yes correct :)

  57. calculusxy
    • one year ago
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    thank you tremendously!

  58. hartnn
    • one year ago
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    welcome tremendously! ^_^

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