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What exactly are you required to do for this problem?
Im sorry only the m ^2. Solve by adding.
What class are you taking?
I believe the answer is -1+m / n^3. The m wasnt ^2 it was the n. sorry im confusing. Prealgebra
I'm confused how you got that jamie. I'm gonna write the problem out again. It is -1/n + m/n^2
I have a prgram that does the problem for me if i type it in but let me check
yeah thats what i got.
Wow then I totally do not know what im doing. dang it
hahah download the program its really nice. its called algerbrator i think its 30$
I will look into that
if you're still confused, multiply -1/n by n/n (=1) then you'll have common denominators (n^2)
INT jamie said her answer is right and mine is wrong so i have been sitting here confused trying to figure it out
yeah, jamie is correct (-1/n)*(n/n)=(-n)/(n^2) [(-n)/(n^2)]+[(m^2)/(n^2)]=(m^2-n)/(n^2)
you can do this because n/n is just 1
Thanks so much