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OpenStudy (goformit100):
I think this guy is Pursuing Graduation degree. Maths question is of Honours Level
OpenStudy (mayaal):
hahaha
OpenStudy (anonymous):
haha what! i can't get this equation right i will try my best!! to type it out
OpenStudy (mayaal):
sure
OpenStudy (anonymous):
((1/y^2)-(3/y^4)) (y+5y^3)
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OpenStudy (anonymous):
i am sorry but my professor is just an retricebecause he never explains and just write off the book the first easy problem and just says follow the pattern :(
OpenStudy (anonymous):
anyone? yahaha
OpenStudy (shamim):
i hv an attschment
OpenStudy (anonymous):
where is the attachment?
OpenStudy (shamim):
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OpenStudy (shamim):
feel free to ask if u hv any confusion
OpenStudy (anonymous):
okay so you just multiplied both to make one term first?
OpenStudy (shamim):
ya i multiplied first
OpenStudy (anonymous):
the answer should be 5+(14/y^2)+(9/y^4)
OpenStudy (shamim):
r u able to get ur result frm my last line
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OpenStudy (shamim):
its easy
OpenStudy (anonymous):
yes it is but when you move the denominator to top all you do is rase the power to negative
OpenStudy (shamim):
ya
1/y^4=y^-4
OpenStudy (shamim):
1/y^3=y^-3
OpenStudy (anonymous):
the first 1/y^3? or 1/y^2
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OpenStudy (shamim):
differentiation of y^-3 is
-3y^(-3-1)
OpenStudy (anonymous):
i am just lost at the cancellation part or from denominator to numerator