A and B are both 5 units from it
Check the picture below on the left-side.
we know the central angle of the "empty" area is 120°, however the legs coming from the center of the circle, namely the radius, are always 6, therefore the legs stemming from the 120° angle, are both 6, making that triangle an isosceles.
now, using the "inscribed angle" theorem, check the picture on the right-side, we know that the inscribed angle there, in red, is 30°, that means the intercepted arc is twice as much, thus 60°, and since arcs get their angle measurement from the central angle they're in, the central angle making up that arc is also 60°, as in the picture.
so, the shaded area is really just the area of that circle's "sector" with 60°, PLUS the area of the circle's "segment" with 120°.

![\bf \textit{area of a segment of a circle}\\\\ A_y=\cfrac{r^2}{2}\left[\cfrac{\pi \theta }{180}~-~sin(\theta ) \right] \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ ------\\ r=6\\ \theta =120 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20segment%20of%20a%20circle%7D%5C%5C%5C%5C%0AA_y%3D%5Ccfrac%7Br%5E2%7D%7B2%7D%5Cleft%5B%5Ccfrac%7B%5Cpi%20%5Ctheta%20%7D%7B180%7D~-~sin%28%5Ctheta%20%29%20%20%5Cright%5D%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0A%5Ctheta%20%3Dangle~in%5C%5C%0A%5Cqquad%20degrees%5C%5C%0A------%5C%5C%0Ar%3D6%5C%5C%0A%5Ctheta%20%3D120%0A%5Cend%7Bcases%7D)
Answer:
If we have a set of N elements:
{x₁, x₂, ..., xₙ}
The mean value (also called the average value) Is calculated as:
Average = (x₁ + x₂ + ... + xₙ)/n
So if x is the average of 13, -16 and 6 ( a total of 3 values)
x will be equal to:
x = (13 + (-16) + 6)/3 = (19 - 16)/3 = 3/3 = 1
x = 1
And we know that:
y = ∛8
Remember that:
2*2 = 4
and
4*2 = 8
then
2*2*2 = 2^3 = 8
then ∛8 = 2.
So we have:
y = ∛8 = 2
Now we can replace these values in the equation:
x^2 + y^3
replacing:
x = 1
y = 2
we get:
1^2 + 2^3 = 1 + 8 = 9