Answer: yes
Step-by-step explanation:
Do you have a typo?
Besides that, is this problem a multiple choice question
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

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

Using the linearity of the integral and Barrow's Theorem we have

As a result, the value of the double integral is -48-
When you bisect something, you cut it into two equally sized pieces. (from Latin: "bi" = two, "sect" = cut)
Bisecting an interval creates two smaller intervals each with half the length of the original interval. Some examples:
• bisecting [0, 2] gives the intervals [0, 1] and [1, 2]
• bisecting [-1, 1] gives the intervals [-1, 0] and [0, 1]
• bisecting an arbitrary interval
gives the intervals
and ![\left[\frac{a+b}2,b\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cfrac%7Ba%2Bb%7D2%2Cb%5Cright%5D)
The answer is B
you have to find the cubic roots of 27 and 729 to find the side lengths of the cubes. The first one has a side length is 3, and the second one has a side length of 9 which s 3x larger.
pls make me brainliest ! ;-)