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some.random.cool.kid

  • one year ago

What is the simplified form of left parenthesis 27 x to the power of negative six end power y to the power of 12 end power right parenthesis to the power of start fraction two over three end fraction end power

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  1. some.random.cool.kid
    • one year ago
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    @zepdrix

  2. some.random.cool.kid
    • one year ago
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    @triciaal

  3. some.random.cool.kid
    • one year ago
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  4. some.random.cool.kid
    • one year ago
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    @nincompoop @whpalmer4 @pooja195 @paki @ganeshie8

  5. some.random.cool.kid
    • one year ago
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    mind helping?

  6. zepdrix
    • one year ago
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    \[\large\rm \left[abc\right]^{n}=a^n b^n c^n\]When we have a bunch of stuff being raised to a power, we have to apply the exponent to each of them.

  7. zepdrix
    • one year ago
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    \[\large\rm \left[27x^{-6}y^{12}\right]^{2/3}=27^{2/3}(x^{-6})^{2/3}(y^{12})^{2/3}\]Ok with that step? :)

  8. some.random.cool.kid
    • one year ago
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    mm got it am i supposed to add across? for the powers part?

  9. some.random.cool.kid
    • one year ago
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    or exponents i should say?

  10. zepdrix
    • one year ago
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    Have to apply one of your exponent rules from that point:\[\large\rm \color{orangered}{(x^a)^b=x^{a\cdot b}}\]When you have a power and another power, you `multiply`

  11. some.random.cool.kid
    • one year ago
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    mmm ok ...

  12. zepdrix
    • one year ago
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    So for the exponent on x, we'll multiply -6 and 2/3.

  13. some.random.cool.kid
    • one year ago
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    -6 x 12? I belive. what each of those two thirds?

  14. some.random.cool.kid
    • one year ago
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    oh

  15. some.random.cool.kid
    • one year ago
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    I belive that is -4? correct...

  16. zepdrix
    • one year ago
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    \[\large\rm \left[27x^{-6}y^{12}\right]^{2/3}=27^{2/3}x^{-4}(y^{12})^{2/3}\]k looks good :)

  17. some.random.cool.kid
    • one year ago
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    mmm next step is ? -4 x 12 or -4 x the other 2/3rds?

  18. some.random.cool.kid
    • one year ago
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    or is it -4 x 12 and 2/3 x 2/3?

  19. some.random.cool.kid
    • one year ago
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    or is it just multiply across lol? i believe its what i said first though

  20. zepdrix
    • one year ago
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    Notice that these two things have `different bases`. One has a base of x, the other a base of y, that's letting us know that they won't interact with one another. \(\large\rm x^{a}\cdot x^{b}=x^{a+b}\) This rule that you're familiar with only works `when they have the same base`.

  21. zepdrix
    • one year ago
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    So the -4 on the x isn't going to interact in any way with the powers on y.

  22. some.random.cool.kid
    • one year ago
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    right so find the lcm?

  23. zepdrix
    • one year ago
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    You have \(\large\rm (y^{12})^{2/3}\). Our exponent rule tells us to `multiply` these powers, ya? 12 * 2/3?

  24. some.random.cool.kid
    • one year ago
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    8

  25. zepdrix
    • one year ago
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    \[\large\rm \left[27x^{-6}y^{12}\right]^{2/3}=\color{orangered}{27^{2/3}}x^{-4}y^{8}\]k good :) almost done

  26. some.random.cool.kid
    • one year ago
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    is this going to be 8 over 4?

  27. some.random.cool.kid
    • one year ago
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    no

  28. zepdrix
    • one year ago
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    Try to pay attention to what I've been saying. `They have different bases`, so they won't interact with one another in any way.

  29. some.random.cool.kid
    • one year ago
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    its not nvm i answered my own thing i belive you leave those alon now right so now i have 9 x^4y^8?

  30. zepdrix
    • one year ago
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    yay good job \c:/

  31. some.random.cool.kid
    • one year ago
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    :D mind helping me with a few more?

  32. zepdrix
    • one year ago
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    stay cool kid (⌐▨_▨) oh sure

  33. some.random.cool.kid
    • one year ago
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    Ok great I have another one this one is a bit different then the last a little bit ...

  34. some.random.cool.kid
    • one year ago
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    A student uses the quadratic formula to solve a quadratic equation and determines that one of the solutions is x equals negative five plus square root of negative 147 end square root. What are the values of a and b if this solution is written in the form x = a + bi, where a and b are real numbers?

  35. some.random.cool.kid
    • one year ago
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    hold on i need to attach the numbers part this is too messy

  36. zepdrix
    • one year ago
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    \[\large\rm x=-5+\sqrt{-147}\]

  37. some.random.cool.kid
    • one year ago
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  38. some.random.cool.kid
    • one year ago
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    there we go :D

  39. zepdrix
    • one year ago
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    Well, recall that we define the imaginary unit i, to be \(\large\rm \sqrt{-1}\)

  40. zepdrix
    • one year ago
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    So we can do something clever with the square root in our problem.

  41. zepdrix
    • one year ago
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    \[\large\rm \sqrt{-147}=\quad\sqrt{-1\cdot147}=\quad\sqrt{-1}\cdot\sqrt{147}\]I broke it up in a clever way to pull the sqrt(-1) outside like that. Do you see the i?

  42. some.random.cool.kid
    • one year ago
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    yep I have used this method before...

  43. zepdrix
    • one year ago
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    \[\large\rm x=-5+\sqrt{-147}\]\[\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{147}\]

  44. some.random.cool.kid
    • one year ago
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    first before you continue

  45. some.random.cool.kid
    • one year ago
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    shouldnt I be trying to find the sqrt?

  46. some.random.cool.kid
    • one year ago
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    and trying to figure out whetther its irrational or not?

  47. some.random.cool.kid
    • one year ago
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    because I believe that is 12.124355653

  48. zepdrix
    • one year ago
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    Yes, try to simplify the square root if you're able to. In doesn't matter which order you do these steps in though :) No no, we don't want an ugly decimal approximation.

  49. some.random.cool.kid
    • one year ago
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    which is irrational so now you may preceed

  50. some.random.cool.kid
    • one year ago
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    eh 12.12

  51. zepdrix
    • one year ago
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    no...

  52. some.random.cool.kid
    • one year ago
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    ...

  53. zepdrix
    • one year ago
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    You don't use your calculator for this step.

  54. some.random.cool.kid
    • one year ago
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    no i rounded lol

  55. zepdrix
    • one year ago
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    But you first used a calculator.. that's not what we're doing here.

  56. zepdrix
    • one year ago
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    You try to simplify the root by pulling out a `perfect square`. Example: \(\large\rm \sqrt{72}=\sqrt{2\cdot36}=\sqrt{2}\cdot\sqrt{36}=\sqrt{2}\cdot 6\)

  57. some.random.cool.kid
    • one year ago
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    oooo nvm its not 12.12

  58. zepdrix
    • one year ago
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    So in my example here, I noticed that 72 was a multiple of 36, 36 can be taken out of the root since it's a perfect square. it comes out as a 6

  59. some.random.cool.kid
    • one year ago
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    its either 21 or 7sqrtof 3 I believe correct? not sure which one yet tough

  60. some.random.cool.kid
    • one year ago
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    though*

  61. some.random.cool.kid
    • one year ago
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    im not sure 6 is simplest of forms though?

  62. zepdrix
    • one year ago
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    6sqrt2 is the simplified form of sqrt(72). That was just an example though.

  63. zepdrix
    • one year ago
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    Break down 147, what are some factors that we can use? One little math trick is to add up the digits, 1 + 4 + 7 = 12. 12 is a multiple of 3, so this number is divisible by 3. So 147 = 3 * _

  64. some.random.cool.kid
    • one year ago
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    /uh

  65. some.random.cool.kid
    • one year ago
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    not sure

  66. zepdrix
    • one year ago
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    Just use a calculator then :U 147 / 3

  67. some.random.cool.kid
    • one year ago
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    49

  68. zepdrix
    • one year ago
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    ok\[\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{147}\]\[\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{3\cdot49}\]\[\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{3}\cdot\sqrt{49}\]

  69. zepdrix
    • one year ago
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    \[\large\rm x=-5+\color{orangered}{\mathcal i}\cdot\sqrt{3}\cdot\sqrt{49}\]What does sqrt(49) simplify to?

  70. some.random.cool.kid
    • one year ago
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    ok

  71. some.random.cool.kid
    • one year ago
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    7

  72. zepdrix
    • one year ago
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    Good good good. Let's write the i in the back, so it looks a lil bit cleaner.\[\large\rm x=-5+7\sqrt3~\mathcal i\]

  73. some.random.cool.kid
    • one year ago
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    ok

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