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nataly862011 [7]
3 years ago
15

PLEASE HELPP Which equation is parallel to y = –3 x + 4 and passes through the point (–3 , 4)? y = –3 x – 5 y = 13 x + 3 y = –3

x + 15 y = 13 x + 5
Mathematics
1 answer:
wolverine [178]3 years ago
6 0

Answer:

A.(-3,4) is your answer.

Step-by-step explanation:

-3x+4

For A:

x=-3

y=4

Plug in 3 where x is and multiply -3 x -3, which equals 0.

Then you plug in 4 where y is, so then your equation is (4=0+4)

Then you add 0 +4, which sum is 4.

Final equation is 4=4, so yes your answer is A. (-3 , 4)

Can you mark me brainliest anwer.

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Let F⃗ =2(x+y)i⃗ +8sin(y)j⃗ .
Alik [6]

Answer:

-42

Step-by-step explanation:

The objective is to find the line integral of F around the perimeter of the rectangle with corners (4,0), (4,3), (−3,3), (−3,0), traversed in that order.

We will use <em>the Green's Theorem </em>to evaluate this integral. The rectangle is presented below.

We have that

           F(x,y) = 2(x+y)i + 8j \sin y = \langle 2(x+y), 8\sin y \rangle

Therefore,

                  P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y

Let's calculate the needed partial derivatives.

                              P_y = \frac{\partial P}{\partial y} (x,y) = (2(x+y))'_y = 2\\Q_x =\frac{\partial Q}{\partial x} (x,y) = (8\sin y)'_x = 0

Thus,

                                    Q_x -P_y = 0 -2 = - 2

Now, by the Green's theorem, we have

\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA = \int \limits_{-3}^{4} \int \limits_{0}^{3} (-2)\,dy\, dx \\ \\\phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-2y) \Big|_{0}^{3} \; dx\\ \phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-6)\; dx = -6x  \Big|_{-3}^{4} = -42

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