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anonymous

  • one year ago

Please Help meeeeeeeee I'm meltingggggg!!!

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  1. anonymous
    • one year ago
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    \[(3^{-1}-2^{-1})^{-1}\]

  2. DecentNabeel
    • one year ago
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    1/(1/3 -1/2) \[\left(-2^{-1}+3^{-1}\right)^{-1}=-6\]

  3. anonymous
    • one year ago
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    ya, i did it on my calc and its -6

  4. anonymous
    • one year ago
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    |dw:1437259855827:dw|

  5. anonymous
    • one year ago
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    is that the way to do it? I know there are other ways to do it but is this viable?

  6. anonymous
    • one year ago
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    \[\left( 3^{-1}-2^{-1} \right)^{-1}=\left( \frac{ 1 }{ 3^1 }-\frac{ 1 }{ 2^1 } \right)^{-1}\] \[=\left( \frac{ 2-3 }{ 6 } \right)^{-1}=\left( \frac{ -1 }{ 6 } \right)^{-1}=\frac{ \left( -1 \right)^{-1} }{ \left( 6 \right)^{-1} }=\frac{ 6^1 }{ \left( -1 \right) }\]=-6

  7. anonymous
    • one year ago
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    how about the way i did it? is that okay too?

  8. anonymous
    • one year ago
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    ????

  9. anonymous
    • one year ago
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    no

  10. anonymous
    • one year ago
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    o...

  11. anonymous
    • one year ago
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    \[x^a=\frac{ 1 }{ x ^{-a} }~and~viceverse\]

  12. DecentNabeel
    • one year ago
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    \[\mathrm{Apply\:exponent\:rule}:\quad \:a^{-1}=\frac{1}{a}\] \[=\left(-\frac{1}{2}+\frac{1}{3}\right)^{-1}\]

  13. DecentNabeel
    • one year ago
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    \[\left(-\frac{1}{2}+\frac{1}{3}\right)^{-1}=\frac{1}{-\frac{1}{2}+\frac{1}{3}}\] \[\mathrm{Combine\:the\:fractions\:using\:the\:LCD}:\quad -\frac{1}{2}+\frac{1}{3}=\frac{2-3}{6}\]

  14. DecentNabeel
    • one year ago
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    \[=\frac{1}{-\frac{1}{6}}\] =-6

  15. DecentNabeel
    • one year ago
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    are you understand @yomamabf

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spraguer (Moderator)
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