It looks like the integral is

where <em>C</em> is the circle of radius 2 centered at the origin.
You can compute the line integral directly by parameterizing <em>C</em>. Let <em>x</em> = 2 cos(<em>t</em> ) and <em>y</em> = 2 sin(<em>t</em> ), with 0 ≤ <em>t</em> ≤ 2<em>π</em>. Then

Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on <em>C</em> nor in the region bounded by <em>C</em>, so

where <em>D</em> is the interior of <em>C</em>, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result:
.
Answer:
Do you think you can zoom into the pic?
Step-by-step explanation:
Answer:
∠ABD=150°
Step-by-step explanation:
6c+12+51-c=178
5c=178-12-51
5c=178-63=115
c=115÷5=23
∠ABD=6c+12=6×23+12=138+12=150°
The quadrilateral that best describes the picture is a square
I don’t really know the answer but hopefully you got it