Answer:
yes
Step-by-step explanation:
im not bad human but really nice so
Answer:
![\huge\underline\color{pink}{Answer ☘}](https://tex.z-dn.net/?f=%5Chuge%5Cunderline%5Ccolor%7Bpink%7D%7BAnswer%20%E2%98%98%7D)
<h3>the second one will be...</h3>
![\frac{40 \times 50}{4.0 \times 5.0 } \\ \\ = > \frac{2000}{20} = 100](https://tex.z-dn.net/?f=%20%5Cfrac%7B40%20%5Ctimes%2050%7D%7B4.0%20%5Ctimes%205.0%20%7D%20%20%5C%5C%20%5C%5C%20%20%3D%20%20%3E%20%20%5Cfrac%7B2000%7D%7B20%7D%20%20%3D%20100)
hope helpful~
Let ????C be the positively oriented square with vertices (0,0)(0,0), (2,0)(2,0), (2,2)(2,2), (0,2)(0,2). Use Green's Theorem to
bonufazy [111]
Answer:
-48
Step-by-step explanation:
Lets call L(x,y) = 10y²x, M(x,y) = 4x²y. Green's Theorem stays that the line integral over C can be calculed by computing the double integral over the inner square of Mx - Ly. In other words
![\int\limits_C {L(x,y)} \, dx + M(x,y) \, dy = \int\limits_0^2\int\limits_0^2 (M_x - L_y ) \, dx \, dy](https://tex.z-dn.net/?f=%5Cint%5Climits_C%20%7BL%28x%2Cy%29%7D%20%5C%2C%20dx%20%2B%20M%28x%2Cy%29%20%5C%2C%20dy%20%3D%20%20%5Cint%5Climits_0%5E2%5Cint%5Climits_0%5E2%20%28M_x%20-%20L_y%20%29%20%5C%2C%20dx%20%5C%2C%20dy)
Where Mx and Ly are the partial derivates of M and L with respect to the x variable and the y variable respectively. In other words, Mx is obtained from M by derivating over the variable x treating y as constant, and Ly is obtaining derivating L over y by treateing x as constant. Hence,
- M(x,y) = 4x²y
- Mx(x,y) = 8xy
- L(x,y) = 10y²x
- Ly(x,y) = 20xy
- Mx - Ly = -12xy
Therefore, the line integral can be computed as follows
![\int\limits_C {10y^2x} \, dx + {4x^2y} \,dy = \int\limits_0^2\int\limits_0^2 -12xy \, dx \, dy](https://tex.z-dn.net/?f=%5Cint%5Climits_C%20%7B10y%5E2x%7D%20%5C%2C%20dx%20%2B%20%7B4x%5E2y%7D%20%5C%2Cdy%20%3D%20%5Cint%5Climits_0%5E2%5Cint%5Climits_0%5E2%20-12xy%20%5C%2C%20dx%20%5C%2C%20dy)
Using the linearity of the integral and Barrow's Theorem we have
![\int\limits_0^2\int\limits_0^2 -12xy \, dx \, dy = -12 \int\limits_0^2\int\limits_0^2 xy \, dx \, dy = -12 \int\limits_0^2\frac{x^2y}{2} |_{x = 0}^{x=2} \, dy = -12 \int\limits_0^22y \, dy \\= -24 ( \frac{y^2}{2} |_0^2) = -24*2 = -48](https://tex.z-dn.net/?f=%5Cint%5Climits_0%5E2%5Cint%5Climits_0%5E2%20-12xy%20%5C%2C%20dx%20%5C%2C%20dy%20%3D%20-12%20%5Cint%5Climits_0%5E2%5Cint%5Climits_0%5E2%20xy%20%5C%2C%20dx%20%5C%2C%20dy%20%3D%20-12%20%5Cint%5Climits_0%5E2%5Cfrac%7Bx%5E2y%7D%7B2%7D%20%7C_%7Bx%20%3D%200%7D%5E%7Bx%3D2%7D%20%5C%2C%20dy%20%3D%20-12%20%5Cint%5Climits_0%5E22y%20%5C%2C%20dy%20%5C%5C%3D%20-24%20%28%20%5Cfrac%7By%5E2%7D%7B2%7D%20%7C_0%5E2%29%20%3D%20-24%2A2%20%3D%20-48)
As a result, the value of the double integral is -48-
The answer as a mixed number is 7
Answer:
11/12
Explanation:
1/3 + 3/2 = 11/6
11/6 x 1/2 (average is halfway) = 11/12