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pav-90 [236]
3 years ago
9

WOI

Mathematics
1 answer:
Alexxandr [17]3 years ago
3 0
He would run a 1=dx jsnsissnsnjss
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If the measure of arc RT = 80 degrees and the measure of arc US = 40 degrees, what is the m<2?
tester [92]
80+40=120
360-120=240
240/2=120
So answer is 120
6 0
3 years ago
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horrorfan [7]

Answer:

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Step-by-step explanation:

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2 years ago
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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

4 0
3 years ago
∆ ABC and ∆ CDE are similar right triangles. The coordinates of all the vertices are integers.
Eva8 [605]

Answer: Not sure but try D

Step-by-step explanation:

not sure

6 0
3 years ago
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