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JessicaNicoleCraven Group Title

k^3(k7/5)^-5

  • 2 years ago
  • 2 years ago

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  1. angela210793 Group Title
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    |dw:1352048933|dw:1352048968202:dw|095:dw|

    • 2 years ago
  2. CliffSedge Group Title
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    ? Is \(\large k^3 (\frac{k^7}{5})^{-5}\) ?

    • 2 years ago
  3. angela210793 Group Title
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    :O where did my solution go :O

    • 2 years ago
  4. CliffSedge Group Title
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    ^ ? I dunno, looks like you were trying to post a drawing.

    • 2 years ago
  5. JessicaNicoleCraven Group Title
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    I am so confused. Its suppose to be k with an exponent of 3, times k(7/5) with an exponent of negative 5.

    • 2 years ago
  6. CliffSedge Group Title
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    Then is it \(\large k^3 \cdot (k^{7/5})^{-5} \) ?

    • 2 years ago
  7. JessicaNicoleCraven Group Title
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    Yeah.

    • 2 years ago
  8. CliffSedge Group Title
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    Ok. Order of operations, handle the exponent first. A power-to-a-power, multiply the exponents.

    • 2 years ago
  9. angela210793 Group Title
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    All u have to know is that x^a*x^b=x^(a+b) (x^a)^b=x^ab and x^-a=1/x^a so u'll have |dw:1352049321095:dw|

    • 2 years ago
  10. CliffSedge Group Title
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    \(\huge \rightarrow k^3 \cdot k^{(\frac{7}{5} \cdot -5)} \) \(\huge \rightarrow k^3 \cdot k^{-7}\)

    • 2 years ago
  11. JessicaNicoleCraven Group Title
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    Oh wow, thats a lot more simple than I was trying to make it. Thank you!

    • 2 years ago
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