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JessicaNicoleCraven

  • 3 years ago

k^3(k7/5)^-5

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  1. angela210793
    • 3 years ago
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    |dw:1352048933|dw:1352048968202:dw|095:dw|

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

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

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

  5. JessicaNicoleCraven
    • 3 years ago
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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.

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

  7. JessicaNicoleCraven
    • 3 years ago
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    Yeah.

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

  9. angela210793
    • 3 years ago
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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|

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

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

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