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chakshu Group Title

homogeneous equations....how to find degree???

  • one year ago
  • one year ago

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  1. chakshu Group Title
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    can u explain by example??

    • one year ago
  2. chakshu Group Title
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    do we have to convert equation to y/x form everytime??

    • one year ago
  3. chakshu Group Title
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    yes

    • one year ago
  4. chakshu Group Title
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    what i know is that if we have equation like|dw:1354030150259:dw|

    • one year ago
  5. chakshu Group Title
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    we find degree by |dw:1354030245650:dw|

    • one year ago
  6. chakshu Group Title
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    its a function

    • one year ago
  7. chakshu Group Title
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    say f(x,y)

    • one year ago
  8. chakshu Group Title
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    the denominator x cancels numerator x n what we get degree as 2

    • one year ago
  9. chakshu Group Title
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    since x has power 2

    • one year ago
  10. chakshu Group Title
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    u catchin me?

    • one year ago
  11. chakshu Group Title
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    but in another question

    • one year ago
  12. chakshu Group Title
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    \[x^n f1(y/x)+y^-n f2(x/y)\]

    • one year ago
  13. chakshu Group Title
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    degree is considered both + and - n

    • one year ago
  14. chakshu Group Title
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    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
  15. across Group Title
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    A function \(f\) is said to be homogeneous of degree \(k\) if \(f(\alpha\mathbf{x})=\alpha^kf(\mathbf{x})\). In your case,\[f(\alpha x,\alpha y)=\frac{\alpha^3x^3+\alpha^3y^3}{\alpha x-\alpha y}=\alpha^2\frac{x^3+y^3}{x-y}=\alpha^2f(x,y).\]

    • one year ago
  16. chakshu Group Title
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    @across thanx for ya help aneways

    • one year ago
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