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f(x, y)=\(\frac{x^3+y^3}{x^{99}+y^{98}x+y^{99}}\) find the value of fy at (x,y) = (0,1).

Posted: Thu Jul 14, 2022 9:29 am
by answerhappygod
a) 101
b) -96
c) 210
d) 0