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anonymous

  • 5 years ago

-1/n + (m/n)^2?

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  1. anonymous
    • 5 years ago
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    What exactly are you required to do for this problem?

  2. anonymous
    • 5 years ago
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    Im sorry only the m ^2. Solve by adding.

  3. anonymous
    • 5 years ago
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    What class are you taking?

  4. anonymous
    • 5 years ago
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    I believe the answer is -1+m / n^3. The m wasnt ^2 it was the n. sorry im confusing. Prealgebra

  5. anonymous
    • 5 years ago
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    OH okay...

  6. anonymous
    • 5 years ago
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    -n+m^2 /n^2

  7. anonymous
    • 5 years ago
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    I'm confused how you got that jamie. I'm gonna write the problem out again. It is -1/n + m/n^2

  8. anonymous
    • 5 years ago
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    I have a prgram that does the problem for me if i type it in but let me check

  9. anonymous
    • 5 years ago
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    yeah thats what i got.

  10. anonymous
    • 5 years ago
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    Wow then I totally do not know what im doing. dang it

  11. anonymous
    • 5 years ago
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    hahah download the program its really nice. its called algerbrator i think its 30$

  12. anonymous
    • 5 years ago
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    I will look into that

  13. anonymous
    • 5 years ago
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    lol

  14. anonymous
    • 5 years ago
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    Thanks

  15. anonymous
    • 5 years ago
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    if you're still confused, multiply -1/n by n/n (=1) then you'll have common denominators (n^2)

  16. anonymous
    • 5 years ago
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    INT jamie said her answer is right and mine is wrong so i have been sitting here confused trying to figure it out

  17. anonymous
    • 5 years ago
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    yeah, jamie is correct (-1/n)*(n/n)=(-n)/(n^2) [(-n)/(n^2)]+[(m^2)/(n^2)]=(m^2-n)/(n^2)

  18. anonymous
    • 5 years ago
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    you can do this because n/n is just 1

  19. anonymous
    • 5 years ago
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    Thanks so much

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spraguer (Moderator)
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