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MaRussiya [10]
3 years ago
8

Find the lcd. Add or subtract the fractions. Simplify if needed. (- 3/16) + (- 5/2)

Mathematics
1 answer:
Solnce55 [7]3 years ago
8 0

Answer:

\boxed{-\frac{43}{16}}

.

Step-by-step explanation:

-\frac{3}{16}  + (-\frac{5}{2} )

= -\frac{3}{16}  - \frac{5}{2}

= \frac{-3 - 5(8)}{16}

= \frac{-3 - 40}{16}

= -\frac{43}{16}

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Step-by-step explanation:

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Answer:

Option b is the correct answer

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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)]
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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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3 years ago
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