anonymous
  • anonymous
homogeneous equations....how to find degree???
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
can u explain by example??
anonymous
  • anonymous
do we have to convert equation to y/x form everytime??
anonymous
  • anonymous
yes

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
what i know is that if we have equation like|dw:1354030150259:dw|
anonymous
  • anonymous
we find degree by |dw:1354030245650:dw|
anonymous
  • anonymous
its a function
anonymous
  • anonymous
say f(x,y)
anonymous
  • anonymous
the denominator x cancels numerator x n what we get degree as 2
anonymous
  • anonymous
since x has power 2
anonymous
  • anonymous
u catchin me?
anonymous
  • anonymous
but in another question
anonymous
  • anonymous
\[x^n f1(y/x)+y^-n f2(x/y)\]
anonymous
  • anonymous
degree is considered both + and - n
anonymous
  • anonymous
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
across
  • across
A function \(f\) is said to be homogeneous of degree \(k\) if \(f(\alpha\mathbf{x})=\alpha^kf(\mathbf{x})\). In your case,\[f(\alpha x,\alpha y)=\frac{\alpha^3x^3+\alpha^3y^3}{\alpha x-\alpha y}=\alpha^2\frac{x^3+y^3}{x-y}=\alpha^2f(x,y).\]
anonymous
  • anonymous
@across thanx for ya help aneways

Looking for something else?

Not the answer you are looking for? Search for more explanations.