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wild

  • one year ago

plz help

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  1. LovingMyself
    • one year ago
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    hi

  2. LovingMyself
    • one year ago
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    whats the question

  3. wild
    • one year ago
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    \[(3x)^-1 \over 4\]

  4. wild
    • one year ago
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    write each expression so that it contains only positive exponents

  5. LovingMyself
    • one year ago
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    I dunno this 1 im so sorry

  6. satellite73
    • one year ago
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    \[\frac{(3x)^{-1}}{ 4}=\frac{1}{4(3x)}\]

  7. wild
    • one year ago
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    how did you do that

  8. satellite73
    • one year ago
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    \(b^{-1}=\frac{1}{b}\)

  9. satellite73
    • one year ago
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    in general, \[b^{-n}=\frac{1}{b^n}\]

  10. wild
    • one year ago
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    how did you get b

  11. satellite73
    • one year ago
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    i picked \(b\) as a variable you can use anyone you prefer

  12. wild
    • one year ago
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    oh ok

  13. wild
    • one year ago
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    satellite are you there:(

  14. Grazes
    • one year ago
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    I don't know what else there is to say. What don't you understand?

  15. wild
    • one year ago
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    idk i just dont know how to do it

  16. wild
    • one year ago
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    what happened to the -1

  17. Grazes
    • one year ago
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    This is what your problem looks like, right? \[\frac{ (3x)^{-1} }{ 4 }\]

  18. Grazes
    • one year ago
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    This is what your problem looks like, right? \[\frac{ (3x)^{-1} }{ 4 }\]

  19. wild
    • one year ago
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    yes

  20. Grazes
    • one year ago
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    Did you know that \[x^{-1}=\frac{ 1 }{ x }\]?

  21. wild
    • one year ago
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    no

  22. wild
    • one year ago
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    i thought x-1=1/x1

  23. Grazes
    • one year ago
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    Same thing. \[x^1=x\] just like \[3^{1}=3\]

  24. wild
    • one year ago
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    ok

  25. wild
    • one year ago
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    but isnt negative 1

  26. Grazes
    • one year ago
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    One thing first. Whenever you want to say that x has an exponential value of -1, you should type x^-1 or x to the power of -1 rather than x-1 because x-1 is just subtraction.

  27. wild
    • one year ago
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    oh ok sorry

  28. Grazes
    • one year ago
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    Is this what you're saying?\[x^{-1}=\frac{ 1 }{ x^{1} }\]

  29. wild
    • one year ago
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    yes

  30. Grazes
    • one year ago
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    Then you are right. But, as I said, \[x^{1}=x\] So you can just change the \[x^{1}\] to \[x\]

  31. wild
    • one year ago
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    oh ok

  32. Grazes
    • one year ago
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    This means that \[x^{−1}=\frac{ 1 }{ x }\]

  33. wild
    • one year ago
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    yeah i know

  34. Grazes
    • one year ago
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    Rewrite this so that there are only positive exponents:\[\frac{ 1 }{ x^{-1} }\]

  35. wild
    • one year ago
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    uh \[1\over x\]

  36. wild
    • one year ago
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    i think

  37. wild
    • one year ago
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    am i right

  38. Grazes
    • one year ago
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    No. General rule of thumb. If you see a negative number as an exponent, flip the term(with the exponent) over the fraction bar and make the exponent positive

  39. wild
    • one year ago
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    ?

  40. Grazes
    • one year ago
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    so I flip x^-1 over to the numerator and get \[x^{-1}\] then I take away the negative sign which gets me \[x^{1}\] or \[x\]

  41. wild
    • one year ago
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    so every time there an equation like this \[1 \over x ^-1\] you have to flip

  42. Grazes
    • one year ago
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    Yes. But ONLY if the exponent is negative. Next problem:\[\frac{ 1 }{ x^{-2} }\]

  43. wild
    • one year ago
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    so I fliped x^21 over to the numerator and get x−2 then I take away the negative sign which gets me x2

  44. wild
    • one year ago
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    -2

  45. Grazes
    • one year ago
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    x^2 is right. Next one:\[x ^{-3}\]

  46. wild
    • one year ago
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    \[1\over x^2\]

  47. Grazes
    • one year ago
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    Are you sure?

  48. Grazes
    • one year ago
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    I think it's a silly mistake, but it was x^3, not x^2

  49. Grazes
    • one year ago
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    Might be because the number was so small.

  50. wild
    • one year ago
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    oh yeah it was a little as i looked closer it was a 3

  51. wild
    • one year ago
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    \[1 \over x^3\]

  52. Grazes
    • one year ago
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    Next one(and the important one to help you with your problem):\[\frac{ 4 }{ x^{-2} }\]

  53. wild
    • one year ago
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    idk 4/x^2

  54. Grazes
    • one year ago
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    Why don't you bring x^-2 to the numerator and take away the negative sign?

  55. wild
    • one year ago
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    so it would be 4x^2

  56. Grazes
    • one year ago
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    Exactly. Now can you do this?\[\frac{ (3x)^{-1} }{ 4 }\]

  57. Grazes
    • one year ago
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    If you really don't understand, no is a good answer too.

  58. wild
    • one year ago
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    im still thinking

  59. wild
    • one year ago
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    3x/4

  60. wild
    • one year ago
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    i got it wrong

  61. Grazes
    • one year ago
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    Then change it :P

  62. wild
    • one year ago
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    change what

  63. Grazes
    • one year ago
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    remember what we did with \[x^{-2}\]

  64. wild
    • one year ago
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    1/x^2

  65. Grazes
    • one year ago
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    We flipped it into the denominator and took away the negative sign to get \[\frac{ 1 }{ x^{2} }\]

  66. Grazes
    • one year ago
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    Can you apply this to your problem or should I just explain?

  67. wild
    • one year ago
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    but 3 is a numerator too it can turn to 1

  68. wild
    • one year ago
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    yeah you should explain i get everything you taught me i just can put it all together

  69. Grazes
    • one year ago
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    Do you agree that \[(3x)^{-1}=\frac{ 1 }{ (3x)^{1} }\]or \[\frac{ 1 }{ (3x) }\]

  70. wild
    • one year ago
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    \[1\over (3x)^1\]

  71. Grazes
    • one year ago
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    Errr I mean that it equals both but the lower one without the exponent of 1 is simplified

  72. Grazes
    • one year ago
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    But you agree with it, right?

  73. wild
    • one year ago
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    yes

  74. wild
    • one year ago
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    i should've looked at it as a whole

  75. Grazes
    • one year ago
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    Now do you agree that \[(3x)^{−1}\times \frac{ 1 }{ 4 }=\frac{ (3x)^{−1} }{ 4 }\]

  76. wild
    • one year ago
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    yes

  77. Grazes
    • one year ago
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    so if you substitute the (3x)^-1 with \[\frac{ 1 }{ 3x}\] in the expression, you get \[\frac{ 1 }{ 4(3x) }\]

  78. Grazes
    • one year ago
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    and that simplifies to\[\frac{ 1 }{ 12x }\]

  79. Grazes
    • one year ago
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    But if you think about it, all I actually did was flip the (3x)^-1 over to the denominator and take away the negative sign|dw:1363833108473:dw|

  80. wild
    • one year ago
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    is it the same thing as multiplying 1/3x and 1/4

  81. Grazes
    • one year ago
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    Yes, but the point is that I just flipped it over and that it's incredibly simple.

  82. Grazes
    • one year ago
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    Everyone else does it like:|dw:1363833370136:dw|

  83. wild
    • one year ago
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    you know what your right its really easy well thank you very much i truly get thank you very much you should be a teacher and again thank you very much 100x:)

  84. Grazes
    • one year ago
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    lolwut. Just understand it. and REMEMBER IT.

  85. wild
    • one year ago
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    yes sir

  86. wild
    • one year ago
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    lol

  87. wild
    • one year ago
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    hahahadont make me take out my shoe

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