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alexandr402 [8]
3 years ago
14

I need help Please hurry

Mathematics
2 answers:
Montano1993 [528]3 years ago
7 0

Answer:

81 Degrees

Step-by-step explanation:

These are the same number which means all you have to do is remember that triangles are 180 degrees total and subtract out the 18 and you get 162 which you then divide by 2.

Gelneren [198K]3 years ago
3 0

Answer:

x=81

Step-by-step explanation:

all triangles = 180 degrees

180 - 18 = 162

162/2 = 81

81 is your answer

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Answer with step-by-step explanation:

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\frac{\delta f}{\delta x}=2y(x^2+y^2)^2+8x^2y(x^2+y^2)=(x^2+y^2)(2x^2y+2y^3+8x^2y)=2(5x^2y+y^3)(x^2+y^2)

Differentiate again w.r.t x

\frac{\delta^2f}{\delta x^2}=2(10xy)(x^2+y^2)+4x(5x^2y+y^3)=20x^3y+20xy^3+20x^3y+4xy^3=40x^3y+24xy^3

Differentiate function w.r.t y

\frac{\delta f}{\delta y}=2x(x^2+y^2)^2+2xy\times 2(x^2+y^2)\times 2y

\frac{\delta f}{\delta y}=(x^2+y^2)(2x^3+2xy^2+8xy^2)=2(x^2+y^2)(x^3+5xy^2)

Again differentiate w.r.t y

\frac{\delta^2f}{\delta x^2}=2(2y)(x^3+5xy^2)+20xy(x^2+y^2)=4x^3y+20xy^3+20x^3y+20xy^3=24x^3y+40xy^3

Differentiate partially w.r.t y

\frac{\delta^2f}{\delta y\delta x}=2(2y(5x^2y+y^3)+(x^2+y^2)(5x^2+3y^2))=10x^4+36x^2y^2+10y^4

\frac{\delta^2f}{\delta y\delta x}=10x^4+36x^2y^2+10y^4\frac{\delta^2f}{\delta x\delat y}=2(2x(x^3+5xy^2)+(3x^2+5y^2)(x^2+y^2))=10x^4+36x^2y^2+10y^4

\frac{\delta^2f}{\delta x\delat y}=10x^4+36x^2y^2+10y^4

Hence, if f(x,y) is of class C^2 (is twice continuously differentiable), then the mixed partial derivatives are equal.

i.e\frac{\delta^2f}{\delta y\delta x}=\frac{\delta^2f}{\delta x\delta y}

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