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AngelaB97

  • one year ago

can someone please explain this to me thoroughly because i keep getting confused on how o simplify this expression

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  1. AngelaB97
    • one year ago
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    \[(x^2/y^-2)^-2 (2y^-3/x^2)^3\]

  2. AngelaB97
    • one year ago
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    |dw:1441669920633:dw|

  3. AngelaB97
    • one year ago
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    it is supposed to be 8/x^10y^13 bu i keep getting 8/x^10y^5

  4. Jhannybean
    • one year ago
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    \[\large \left(\frac{x^2}{y^{-2}}\right)^{-2}\left(\frac{2y^{-3}}{x^2}\right)^3\]Change the variables inside the parenthesis so they are all positive. \[\large ( x^2y^2)^{-2}\left(\frac{2}{x^2y^3}\right)^3\]Now that we've got all the variables INSIDE the parenthesis with positive powers, now we take care of the outer powers. Take the first one and put it over 1. \[\frac{1}{(x^2y^2)^2}\cdot \left(\frac{2}{x^2y^3}\right)^3\]

  5. Jhannybean
    • one year ago
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    Now let's distribute the powers through all variables inside the parenthesis. \[\frac{1}{(x^2)^2(y^2)^2} \cdot \frac{2^3}{(x^2)^3(y^3)^3}\]

  6. AngelaB97
    • one year ago
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    so since the exponent outside of the parentheses is negative aren't you supposed to flip it so it becomes y^2/x^2 ?

  7. AngelaB97
    • one year ago
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    r it that where i am making the mistake?

  8. Jhannybean
    • one year ago
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    Now let's distribute the powers through all variables inside the parenthesis. \[\frac{1}{(x^2)^2(y^2)^2} \cdot \frac{2^3}{(x^2)^3(y^3)^3}\]

  9. AngelaB97
    • one year ago
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    so since the exponent outside of the parentheses is negative aren't you supposed to flip it so it becomes y^2/x^2 ?

  10. Jhannybean
    • one year ago
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    Hmm... not exactly. We take care of each step individually.

  11. AngelaB97
    • one year ago
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    see, its (x^2/y^-2) isn't it supposed to be (y^2/x^2)^2

  12. AngelaB97
    • one year ago
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    @Jhannybean

  13. Jhannybean
    • one year ago
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    So we can write out \(\left(\dfrac{x^2}{y^{-2}}\right)^{-2}\) as \(\dfrac{1}{\left( \dfrac{x^2}{\dfrac{1}{y^2}}\right)^2}\)

  14. Jhannybean
    • one year ago
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    Do you see whats happening here?

  15. Jhannybean
    • one year ago
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    I don't think it's proper to say \(\left(\dfrac{x^2}{y^{-2}}\right)^{-2} \iff \left(\dfrac{y^2}{x^2}\right)^2\)

  16. AngelaB97
    • one year ago
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    oh i understand now

  17. AngelaB97
    • one year ago
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    can you also help me with another problem please?

  18. Jhannybean
    • one year ago
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    Are you finished with this one?

  19. AngelaB97
    • one year ago
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    yes i am

  20. Jhannybean
    • one year ago
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    Alright :)

  21. AngelaB97
    • one year ago
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    okay so it's (2y^-1z/z^2)^-1 (y/3z^2)^2

  22. Jhannybean
    • one year ago
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    This is a new problem?

  23. AngelaB97
    • one year ago
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    yes

  24. Jhannybean
    • one year ago
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    \[\left(\frac{2y^{-1}z}{z^2}\right)^{-1}\cdot \left(\frac{y}{3z^2}\right)^2\]

  25. Jhannybean
    • one year ago
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    So like our previous problem, what will you want to do first?

  26. AngelaB97
    • one year ago
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    flip the problem?

  27. Jhannybean
    • one year ago
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    No, what do we do to all the negative variables within the parenthesis?

  28. AngelaB97
    • one year ago
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    we witch them dont we?

  29. Jhannybean
    • one year ago
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    By switch you mean... make them positive? :)

  30. AngelaB97
    • one year ago
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    yes

  31. Jhannybean
    • one year ago
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    You are correct. Therefore we will have \[\left(\frac{2z}{yz^2}\right)^{-1}\cdot \left(\frac{y}{3z^2}\right)^2\]

  32. AngelaB97
    • one year ago
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    yes so then it becomes (yz^2/2z)^1 ?

  33. AngelaB97
    • one year ago
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    multiplied by (y/3z^2)^2

  34. AngelaB97
    • one year ago
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    so first you make the numbers inside the parentheses positive, then you flip it whenever there's a negative exponent over the parentheses?

  35. Jhannybean
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @AngelaB97 so first you make the numbers inside the parentheses positive, `then you flip it whenever there's a negative exponent over the parentheses?` \(\color{#0cbb34}{\text{End of Quote}}\) Im not understanding what you mean by that highlighted portion.

  36. AngelaB97
    • one year ago
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    as in whenever you have a negative exponent both outside and inside the parentheses, you have to deal with the numbers inside the parentheses?

  37. Jhannybean
    • one year ago
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    Yes, you deal with the numbers inside the ( ) first, making them + , then we take care of the outside ( ) by putting over 1. [i.e 1/ ( ) ]

  38. Jhannybean
    • one year ago
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    Does that make sense?

  39. AngelaB97
    • one year ago
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    yes exactly

  40. Jhannybean
    • one year ago
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    Cool! \(\checkmark\) Glad you're following along.

  41. Jhannybean
    • one year ago
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    Now we have: \[\left(\frac{2z}{yz^2}\right)^{-1}\cdot \left(\frac{y}{3z^2}\right)^2\] We're going to put the first one over 1 to make it +. \[\frac{1}{\left(\dfrac{2z}{yz^2}\right)} \cdot \left(\frac{y}{3z^2}\right)^2\]

  42. Jhannybean
    • one year ago
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    SOrry, bear with me, Im lagging a lot right now.

  43. AngelaB97
    • one year ago
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    don't worry you're helping me alot

  44. Jhannybean
    • one year ago
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    This becomes: \[\frac{yz^2}{2z} \cdot \left(\frac{y}{3z^2}\right)^2\]

  45. Jhannybean
    • one year ago
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    Now we can reduce our fraction. Remember, \(\dfrac{x^m}{x^n} = x^{m-n} \therefore \dfrac{z^2}{z} = z^{2-1} = z\) Now we can rewrite this as: \[\frac{yz}{2} \cdot \left(\frac{y}{3z^2}\right)^2\]

  46. AngelaB97
    • one year ago
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    then yz/2 * y^2/9z^4

  47. Jhannybean
    • one year ago
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    exactly.

  48. Jhannybean
    • one year ago
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    We get \[=\frac{y^3z}{18z^4}\] \[\frac{z}{z^4} =z^{1-4} =z^{-3}\] \[=\frac{y^3}{18z^{-3}} \iff \boxed{\frac{y^3z^3}{18}}\]

  49. Jhannybean
    • one year ago
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    Whoop whoop wait a minute.

  50. Jhannybean
    • one year ago
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    Typo.

  51. Jhannybean
    • one year ago
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    we got \(z^{-3}\) therefore it would be: \[\frac{y^3z^{-3}}{18} \iff \boxed{\frac{y^3}{18z^3}} \]

  52. AngelaB97
    • one year ago
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    gotcha, thanks sosososo much!!! :) <3

  53. AngelaB97
    • one year ago
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    you helped me out so much today

  54. Jhannybean
    • one year ago
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    No problem :) I've got to head off now. SO good luck on the rest!

  55. Jhannybean
    • one year ago
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    Im glad you participated instead of blatantly asking for the answer.

  56. AngelaB97
    • one year ago
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    thank you!!

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