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chakshu Group TitleBest ResponseYou've already chosen the best response.0
can u explain by example??
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
do we have to convert equation to y/x form everytime??
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
what i know is that if we have equation likedw:1354030150259:dw
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
we find degree by dw:1354030245650:dw
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
its a function
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
say f(x,y)
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
the denominator x cancels numerator x n what we get degree as 2
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
since x has power 2
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
u catchin me?
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
but in another question
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
\[x^n f1(y/x)+y^n f2(x/y)\]
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
degree is considered both + and  n
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
i want to know in what form of fraction do we have to exactly convert an equation to find degree?? x/y or y/x?? hope u get what m askin
 one year ago

across Group TitleBest ResponseYou've already chosen the best response.1
A function \(f\) is said to be homogeneous of degree \(k\) if \(f(\alpha\mathbf{x})=\alpha^kf(\mathbf{x})\). In your case,\[f(\alpha x,\alpha y)=\frac{\alpha^3x^3+\alpha^3y^3}{\alpha x\alpha y}=\alpha^2\frac{x^3+y^3}{xy}=\alpha^2f(x,y).\]
 one year ago

chakshu Group TitleBest ResponseYou've already chosen the best response.0
@across thanx for ya help aneways
 one year ago
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