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calculusxy

  • one year ago

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  1. calculusxy
    • one year ago
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    @hartnn @Nnesha

  2. calculusxy
    • one year ago
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    \[\huge \frac{ (2m^2)^{-1} }{ m^2 }\]

  3. calculusxy
    • one year ago
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    \[\large \frac{ 1 }{ 2m^1 }\]

  4. hartnn
    • one year ago
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    \(\dfrac{m^{-2}}{m^2 } = \dfrac{1}{m^2 \times m^2 }\)

  5. anonymous
    • one year ago
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    \[x ^{-1}=\frac{ 1 }{ x^1 }\]

  6. hartnn
    • one year ago
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    how did you get m^1 ?

  7. anonymous
    • one year ago
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    Hartnn can you help me with my problem after her ?

  8. hartnn
    • one year ago
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    I'll try rosie :)

  9. calculusxy
    • one year ago
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    \[\large (2m^2)^{-1} = 2^{-1}m^{2-(-1)} = 2^{-1}m^3\]

  10. hartnn
    • one year ago
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    \(\Huge (a^b)^c = a^{bc}\)

  11. hartnn
    • one year ago
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    \(\Large (m^2)^{-1} = m^{-2}\)

  12. calculusxy
    • one year ago
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    oh yes sorry my mistake once again...

  13. hartnn
    • one year ago
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    np :) m^-2 = 1/m^2

  14. calculusxy
    • one year ago
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    would 2^{-1} be 1/2^1

  15. hartnn
    • one year ago
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    yesss

  16. calculusxy
    • one year ago
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    and then m^{-2 - 2} = m^{-4} = 1/m^4

  17. calculusxy
    • one year ago
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    would the answer be like \[\large \frac{ 1 }{ 2^1m^4 }\]

  18. hartnn
    • one year ago
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    write 2^2 as 2 only :) and yes

  19. hartnn
    • one year ago
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    i mean 2^1 =2

  20. calculusxy
    • one year ago
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    ok thank you

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spraguer (Moderator)
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