Answer:
B. 4
Step-by-step explanation:
We are looking for the coefficient of the term x⁵. When we see it in the polynomial as 4x⁵, our coefficient and answer would then be 4.
2(x+7) + x=20
(2)(x) + (2)(7) + x=20 Distribute
2x+14 + x =20
(2x+x) + (14) =20 Combine Like Terms
3x+14=20
- 14 -14 Subtract 14 from both sides
3x = 6
3x/3 6/3 Divide Both Sides by 3
x = 2
Let me know if you still don't understand
The formula is a permutation in which n represents the sample points in the set, while r represents the number of sample points in each permutation.
Answer:
-42
Step-by-step explanation:
The objective is to find the line integral of
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](https://tex.z-dn.net/?f=F%28x%2Cy%29%20%3D%202%28x%2By%29i%20%2B%208j%20%5Csin%20y%20%3D%20%5Clangle%202%28x%2By%29%2C%208%5Csin%20y%20%5Crangle)
Therefore,
![P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y](https://tex.z-dn.net/?f=P%28x%2Cy%29%20%3D%202%28x%2By%29%20%5Cquad%20%5Cwedge%20%5Cquad%20Q%28x%2Cy%29%20%3D%208%5Csin%20y)
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](https://tex.z-dn.net/?f=P_y%20%3D%20%5Cfrac%7B%5Cpartial%20P%7D%7B%5Cpartial%20y%7D%20%28x%2Cy%29%20%3D%20%282%28x%2By%29%29%27_y%20%3D%202%5C%5CQ_x%20%3D%5Cfrac%7B%5Cpartial%20Q%7D%7B%5Cpartial%20x%7D%20%28x%2Cy%29%20%3D%20%288%5Csin%20y%29%27_x%20%3D%200)
Thus,
![Q_x -P_y = 0 -2 = - 2](https://tex.z-dn.net/?f=Q_x%20-P_y%20%3D%200%20-2%20%3D%20-%202)
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](https://tex.z-dn.net/?f=%5Coint_C%20F%20%5C%2Cdr%20%3D%20%5Ciint_D%20%28Q_x-P_y%29%5C%2CdA%20%3D%20%5Cint%20%5Climits_%7B-3%7D%5E%7B4%7D%20%5Cint%20%5Climits_%7B0%7D%5E%7B3%7D%20%28-2%29%5C%2Cdy%5C%2C%20dx%20%5C%5C%20%5C%5C%5Cphantom%7B%5Coint_C%20F%20%5C%2Cdr%20%3D%20%5Ciint_D%20%28Q_x-P_y%29%5C%2CdA%7D%3D%20%5Cint%20%5Climits_%7B-3%7D%5E%7B4%7D%20%28-2y%29%20%5CBig%7C_%7B0%7D%5E%7B3%7D%20%5C%3B%20dx%5C%5C%20%5Cphantom%7B%5Coint_C%20F%20%5C%2Cdr%20%3D%20%5Ciint_D%20%28Q_x-P_y%29%5C%2CdA%7D%3D%20%5Cint%20%5Climits_%7B-3%7D%5E%7B4%7D%20%28-6%29%5C%3B%20dx%20%3D%20-6x%20%20%5CBig%7C_%7B-3%7D%5E%7B4%7D%20%3D%20-42)
Could you show us a picture of the question