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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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    \[\large (\frac{ 3n }{ 5m^{-3} })^2 (\frac{ 3m^4 }{ 2n^2 })^{-3}\]

  2. calculusxy
    • one year ago
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    @jim_thompson5910 @Nnesha @satellite73

  3. anonymous
    • one year ago
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    you want to work inside first or outside first?

  4. calculusxy
    • one year ago
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    I don't know... anything is fine as long as i get a correct solution

  5. anonymous
    • one year ago
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    ok lets take care of the outside exponents first, not that it matter

  6. anonymous
    • one year ago
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    \[\large (\frac{ 3n }{ 5m^{-3} })^2 (\frac{ 3m^4 }{ 2n^2 })^{-3}\]multiply each exponent in the first parentheses by 2, in the second by \(-3\)

  7. calculusxy
    • one year ago
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    \[(\frac{ a^2b }{ b^{-3}c^4 })^2 -> \frac{ a^6b^3 }{ b^{-9}c^{12} }\]

  8. calculusxy
    • one year ago
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    i meant to say (a^3)^2

  9. anonymous
    • one year ago
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    you lost me there

  10. anonymous
    • one year ago
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    your question has m and n in it, not a and b

  11. calculusxy
    • one year ago
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    oh god.. i wrote the wrong question.. wait a minute . (sorry)

  12. calculusxy
    • one year ago
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    \[\huge (\frac{ a^2b^3 }{ b^{-3}c^4 })(a^{-3}b)^{-2}\]

  13. anonymous
    • one year ago
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    oooh

  14. calculusxy
    • one year ago
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    The first parenthesis is raised to the third power.

  15. anonymous
    • one year ago
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    second one is \[\huge a^6b^{-2}=\frac{a^6}{b^2}\]

  16. calculusxy
    • one year ago
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    and the b doesn't have the third power...

  17. anonymous
    • one year ago
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    is that part ok? multiplied each exponent by \(-2\)

  18. calculusxy
    • one year ago
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    the whole fraction is raised to the power of three.

  19. anonymous
    • one year ago
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    can you post a screen shot?

  20. calculusxy
    • one year ago
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    \[\huge (\frac{ a^2b }{ b^{-3}c^4 })^3(a^{-3}b)^{-2}\]

  21. calculusxy
    • one year ago
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    @satellite73

  22. calculusxy
    • one year ago
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    @jim_thompson5910

  23. anonymous
    • one year ago
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    ooh k

  24. anonymous
    • one year ago
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    multiply each exponent in the first parentheses by 3

  25. calculusxy
    • one year ago
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  26. anonymous
    • one year ago
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    correct, minus the typo

  27. anonymous
    • one year ago
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    \[(\frac{ a^2b }{ b^{-3}c^4 })^3 \to \frac{ a^6b^3 }{ b^{-9}c^{12} }\]

  28. anonymous
    • one year ago
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    now before we mess with the second part, lets clean this up first

  29. anonymous
    • one year ago
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    the exponent of \(-9\) in the denominator brings the term up to the numerator, adding the exponetns makes this \[\large \frac{a^6b^{12}}{c^{12}}\]

  30. calculusxy
    • one year ago
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    okay

  31. anonymous
    • one year ago
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    so far so good?

  32. calculusxy
    • one year ago
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    yes

  33. anonymous
    • one year ago
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    next term is \[(a^{-3}b)^{-2}\] multiply each exponent by \(-2\) get \[\large a^6b^{-2}\]

  34. calculusxy
    • one year ago
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    \[\frac{ 1 }{ a^{-6}b^2 }\]

  35. anonymous
    • one year ago
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    on no don't introduce a negative exponent, leave it be

  36. anonymous
    • one year ago
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    we got \[\large \frac{a^6b^{12}}{c^{12}}\times a^6b^{-2}\] add up the exponents to finish it

  37. calculusxy
    • one year ago
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    but since the exponent is negative doesn't it let the base become it's reciprocal?

  38. anonymous
    • one year ago
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    ok lets back a second and look at \[a^6b^{-2}\]

  39. anonymous
    • one year ago
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    the exponent that is negative is attached to the \(b\) only

  40. calculusxy
    • one year ago
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    no it's the whole expression including the a^-3 and b

  41. anonymous
    • one year ago
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    if for some reason you want to write this with positive exponents only, it would be \[\large a^6b^{-2}=\frac{a^6}{b^2}\]

  42. calculusxy
    • one year ago
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    since the whole thing has a parenthesis around it and there is a -2 as the exponent.

  43. anonymous
    • one year ago
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    yes it was \[\large (a^{-3}b)^{-2}\] right?

  44. calculusxy
    • one year ago
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    yes

  45. anonymous
    • one year ago
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    ok so you COULD write this as \[\frac{1}{(a^{-3}b)^2}\] but you do not need to or want to

  46. calculusxy
    • one year ago
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    but my teacher preferred for us to go this way, and that's why i feel comfortable using it.

  47. anonymous
    • one year ago
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    what you want to do is multiply each exponent by \(-2\) to get \[\large a^6b^{-2}\]

  48. anonymous
    • one year ago
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    than multiplying that by the first part is easy add the exponents

  49. anonymous
    • one year ago
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    if you do it the way your teacher likes you will get the same thing because \[\frac{1}{(a^{-3}b)^2}=\frac{1}{a^{-6}b^2}=\frac{a^6}{b^2}\]

  50. calculusxy
    • one year ago
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    i just want to know if my final answer is correct or not: a^12c^12b^10

  51. anonymous
    • one year ago
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    the c is in the denominator right?

  52. calculusxy
    • one year ago
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    oh yes

  53. anonymous
    • one year ago
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    if you want to write it in one line it should be \[a^{12}b^{10}c^{-12}\]

  54. calculusxy
    • one year ago
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    i prefer the fraction way... but thanks for the "one line" thing

  55. calculusxy
    • one year ago
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    and i just want to make sure my answer for another question as well.

  56. anonymous
    • one year ago
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    yw

  57. anonymous
    • one year ago
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    kk

  58. calculusxy
    • one year ago
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    \[\large (5x^3y^{-3})^{-2}(2x^5y^{-4})^{-3} = \frac{ y^18 }{ 200x^{21} }\]

  59. calculusxy
    • one year ago
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    that was a y^{18}

  60. anonymous
    • one year ago
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    i knew that ~

  61. anonymous
    • one year ago
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    looks good to me!

  62. anonymous
    • one year ago
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    not to make light of this but it is really just bookkeping right? \[(-3)(-2)+(-4)(-3)=6+12=18\] for example

  63. calculusxy
    • one year ago
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    THANK YOU FOR YOUR HELP!

  64. anonymous
    • one year ago
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    yw

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