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Anuta_ua [19.1K]
2 years ago
13

For the given equation, find the values of a, b, and c, determine the direction in which the parabola opens, and determine the y

-intercept. Decide which table best illustrates these values for the equation:
y = 3 x squared + 2 x minus 8

Mathematics
1 answer:
mariarad [96]2 years ago
7 0

Answer:

Table A

Step-by-step explanation:

y  = 3x^2 +2x -8

The quadratic is in the form

y = a x^2 +bx+c

a = 3  b = 2 c = -8

Since a > 0 it opens up

The y intercept is c so the y intercept is -8

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

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\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e
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Answer: 205 lbs

here are all the answers http://washburn.mpls.k12.mn.us/uploads/normal_calculations_practice_key.pdf

Step-by-step explanation:


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Step-by-step explanation:

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