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baby456

  • one year ago

(3x^2 y)^3/(6x^-2 y^5) please help medal + fan and please explain it.

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  1. hlambach
    • one year ago
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    Your missing a part there after 6x

  2. hlambach
    • one year ago
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    oops *You're haha

  3. baby456
    • one year ago
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    -2

  4. baby456
    • one year ago
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    is right there

  5. hlambach
    • one year ago
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    What power is 6x being raised to? 2y^5?

  6. baby456
    • one year ago
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    -2

  7. baby456
    • one year ago
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    let me draw it

  8. hlambach
    • one year ago
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    Okay.

  9. baby456
    • one year ago
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    |dw:1435273567323:dw|

  10. hlambach
    • one year ago
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    Got it. :) So you're simplifying it, right?

  11. baby456
    • one year ago
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    yes

  12. anonymous
    • one year ago
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    use your basic properties of exponents. \(\dfrac{a^M}{a^N}=a^{(M-N)}\\\) etc.

  13. baby456
    • one year ago
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    ok

  14. baby456
    • one year ago
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    |dw:1435273964905:dw|

  15. hlambach
    • one year ago
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    Tip: First simplify \[(3x ^{3}y)^{3}\]

  16. baby456
    • one year ago
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    9x^9y^3

  17. hlambach
    • one year ago
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    Good, now you can use the formulas Jenny gave you.

  18. hlambach
    • one year ago
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    Hold on, 3^3 is what?

  19. baby456
    • one year ago
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    |dw:1435274627143:dw|

  20. anonymous
    • one year ago
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    3^3=3*3*3 3^3=/=3*3

  21. hlambach
    • one year ago
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    Again, what Jenny said... I was wrong to say that you were right in your simplification.

  22. baby456
    • one year ago
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    oh i simpfieed it 27X^6y^3

  23. baby456
    • one year ago
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    |dw:1435274935189:dw|

  24. hlambach
    • one year ago
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    There you go.

  25. baby456
    • one year ago
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    ok so how about the bottom

  26. hlambach
    • one year ago
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    Well, now all you have to use is the Exponent Rule: \[\frac{ x ^{a} }{ x ^{b} } = x ^{a-b}\]

  27. baby456
    • one year ago
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    how about the 6

  28. baby456
    • one year ago
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    do you subtract 27-6

  29. hlambach
    • one year ago
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    Well, this is kind of hard to explain in text, but I'll try. :) You have (ignoring the y) 27x^9 ------ 6x^-2 To divide it into 27, make it 2(3x^-2) since 3 can be divided into 27. Then use the exponent rule.

  30. baby456
    • one year ago
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    how? again

  31. baby456
    • one year ago
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    |dw:1435275710920:dw|

  32. hlambach
    • one year ago
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    |dw:1435275705062:dw| SOrry for my terrible drawing skills haha. Now just divide out the 3x^-2

  33. baby456
    • one year ago
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    |dw:1435275773927:dw|

  34. baby456
    • one year ago
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    how

  35. hlambach
    • one year ago
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    Just divide 27x^9 ------ 3x^-2 You will leave the 2 at the bottom.

  36. baby456
    • one year ago
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    9x^7

  37. hlambach
    • one year ago
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    remember its x^a --- = x^a - b x^b

  38. hlambach
    • one year ago
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    9 - (-2) is what?

  39. baby456
    • one year ago
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    x to 6 not x to the 9

  40. anonymous
    • one year ago
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    split it up like this \(\dfrac{(3x^2 y)^3}{(6x^{-2} y^5) }=\dfrac{27x^6y^3}{6x^{-2}y^5}=\dfrac{27}{6}*\dfrac{x^6y^3}{x^{-2}y^5}\)

  41. baby456
    • one year ago
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    i am so confused

  42. anonymous
    • one year ago
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    read your textbook. she's explained the property several times over

  43. baby456
    • one year ago
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    i dont have a textbook

  44. baby456
    • one year ago
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    still confused @jenny1994 is confusing me

  45. hlambach
    • one year ago
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    Let's go with Jenny's way For now, focus on simplifying x^6 y^3 ------- x^-2y^5

  46. hlambach
    • one year ago
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    Using the rule that I am too lazy to write out. (Just look above)

  47. anonymous
    • one year ago
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    you can't follow a simple pattern? \(\large \dfrac{x^a}{x^b}=a^{(a-b)}\implies \dfrac{x^6}{x^{-2}}=?\)

  48. anonymous
    • one year ago
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    x^(a-b)*

  49. hlambach
    • one year ago
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    We can do this Jenny... haha Patience is a virtue. I wish I could put "OpenStudy Math Tutor on my transcript.

  50. anonymous
    • one year ago
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    Don't mention openstudy, but say you tutor. Resumes are all about embellishment

  51. hlambach
    • one year ago
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    Ha, I was joking, but that's a good idea!

  52. hlambach
    • one year ago
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    Sorry, back on subject.

  53. baby456
    • one year ago
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    who way should i follow

  54. hlambach
    • one year ago
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    Go with Jenny's Simplify x^6 y^3 ------- x^-2 y^5

  55. baby456
    • one year ago
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    you lost me at what to do about the bottom part please explain more

  56. hlambach
    • one year ago
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    Using the rule that has been mentioned several times, thus I won't write it out again, simplify. EXAMPLE x^4 --- = x^(4-2) = x^2 x^2 Get it? But to that do your formula

  57. hlambach
    • one year ago
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    *but do that to your formula.

  58. baby456
    • one year ago
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    so is it x4 since its 6-2

  59. hlambach
    • one year ago
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    I'm sorry, I typed it wrong, it's x^9 y^3 ------ x^-2 y^5 So it would be x^11 (since 9 - (-2) = 11) But you grasped the concept. Now what about the y?

  60. baby456
    • one year ago
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    where are you getting x^9 from

  61. hlambach
    • one year ago
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    From when you simplified (3x^3 y)^3 = 27x^9 y^3

  62. baby456
    • one year ago
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    (3x^2 y)^3= 27X^6y^3

  63. baby456
    • one year ago
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    2^3=6

  64. zepdrix
    • one year ago
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    Baby the subtraction comes from the rule, the bottom number also had a negative so don't let that extra negative get lost in the process! :)\[\Large\rm \frac{x^{\color{orangered}{6}}}{x^{\color{royalblue}{-2}}}=x^{\color{orangered}{6}-\color{royalblue}{-2}}\]You don't get x^4, see that?

  65. zepdrix
    • one year ago
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    2^3 does not equal 6. I think you meant 2*3 = 6 :)

  66. hlambach
    • one year ago
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    Oh, I'm sorry. But remember cubed mean 2 * 2 *2

  67. baby456
    • one year ago
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    OPPS SORRY

  68. baby456
    • one year ago
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    SO THEN X=8

  69. hlambach
    • one year ago
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    Hey, I gotta run. @zepdrix could you take over?

  70. zepdrix
    • one year ago
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    you mean for the final power on x? for the 6 - - 2? ya that looks better baby :)

  71. zepdrix
    • one year ago
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    ya you get outta here missy :P hehe

  72. hlambach
    • one year ago
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    haha, thanks, I;m already running late. :)

  73. zepdrix
    • one year ago
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    So you understood this first step, yes?\[\Large\rm \frac{(3x^2 y)^3}{6x^{-2} y^5}=\frac{27x^6 y^3}{6x^{-2} y^5}\]And then did you understand the part with the x's?\[\Large\rm \frac{27\color{orangered}{x^6} y^3}{6\color{orangered}{x^{-2}} y^5}=\frac{27\color{orangered}{x^8}y^3}{6y^5}\] I'm just trying to figure out where we're at :O This thread is so long and messy

  74. baby456
    • one year ago
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    YEP

  75. zepdrix
    • one year ago
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    \[\Large\rm \frac{27x^8\color{royalblue}{y^3}}{6\color{royalblue}{y^5}}=\frac{27x^8\color{royalblue}{y^?}}{6}\]What about the y's? :) What do you think will happen there?

  76. baby456
    • one year ago
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    3-5=-2

  77. zepdrix
    • one year ago
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    \[\Large\rm \frac{27x^8\color{royalblue}{y^{-2}}}{6}\]Mmmm good. As a final step, you want to notice that the 27 and 6 share some factor.

  78. baby456
    • one year ago
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    3

  79. zepdrix
    • one year ago
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    \[\Large\rm \frac{27x^8y^{-2}}{6}=\frac{9\cdot3x^8y^{-2}}{2\cdot3}\]Yah that sounds right

  80. zepdrix
    • one year ago
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    \[\Large\rm \frac{9\cdot\cancel3x^8y^{-2}}{2\cdot\cancel3}\]

  81. zepdrix
    • one year ago
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    Sometimes your teacher will ask for the final answer with NO NEGATIVE POWERS. Do you understand how to deal with the y if you have to get rid of the negative power? :o

  82. baby456
    • one year ago
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    NOPE

  83. zepdrix
    • one year ago
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    mm k, well our exponent rule tells us to flip it. if its in the numerator, it gets tossed down the denominator and the sign changes on the power. if it's in the denominator it goes up to the numerator and the sign changes on the power. examples:\[\Large\rm \frac{1}{2x^{-7}}=\frac{x^7}{2}\] \[\Large\rm \frac{x^{-4}}{3}=\frac{1}{3x^4}\]

  84. baby456
    • one year ago
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    OH I REMEMBER THIS YOU JUST SWITCH IT LIKE A RECIPROCAL

  85. zepdrix
    • one year ago
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    good good :) reciprocal, and change from negative to positive

  86. baby456
    • one year ago
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    OK THANKS I HAVE TO GO NOW

  87. baby456
    • one year ago
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    THANKS FOR ALL YOUR HELP

  88. baby456
    • one year ago
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    sorry about caplocks

  89. zepdrix
    • one year ago
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    \[\Large\rm \frac{9x^8y^{-2}}{2}=\frac{9x^8}{2y^{2}}\]so ya that will fix the y :)

  90. zepdrix
    • one year ago
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    np

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