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riadik2000 [5.3K]
3 years ago
10

How many solutions are there to the following system of equations?

Mathematics
2 answers:
Bas_tet [7]3 years ago
7 0
-4x-8y=0\\
\underline{-3x+8y=-7}\\
-7x=-7\\
x=1 \Rightarrow B
sweet [91]3 years ago
6 0
4x-8y=0 \\
\underline{-3x+8y=-7} \\
4x-3x=0-7 \\
x=-7 \\ \\
4x-8y=0 \\
4 \times (-7)-8y=0 \\
-28-8y=0 \\
-8y=28 \\
y=\frac{28}{-8} \\
y=-\frac{7}{2} \\ \\
 \left \{ {{x=-7} \atop {y=-\frac{7}{2}}} \right.

There is one solution to this system of equations. The answer is B.
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Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p
Juli2301 [7.4K]
I'm reading this as

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy

with \nabla f=(2xe^{-y},2y-x^2e^{-y}).

The value of the integral will be independent of the path if we can find a function f(x,y) that satisfies the gradient equation above.

You have

\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}

Integrate \dfrac{\partial f}{\partial x} with respect to x. You get

\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx
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Differentiate with respect to y. You get

\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]
2y-x^2e^{-y}=-x^2e^{-y}+g'(y)
2y=g'(y)

Integrate both sides with respect to y to arrive at

\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy
y^2=g(y)+C
g(y)=y^2+C

So you have

f(x,y)=x^2e^{-y}+y^2+C

The gradient is continuous for all x,y, so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e
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timofeeve [1]
Hello,
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