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:
(x, y) → (x - 3, y + 5 )
Step-by-step explanation:
A shift of 3 units to the left means subtract 3 from the original x- coordinate.
A shift of 5 units up means add 5 to the original y- coordinate.
Thus the rule representing the translation is
(x, y ) → (x - 3, y + 5 )
Answer:
k = -6/35
Step-by-step explanation:
To make the function continuous
kx^2 = x+k
These must be equal where the function is defined for two different intervals
This is at the point x=-6 so let x=-6
k(-6)^2 = -6+k
36k = -6+k
Subtract k from each side
36k-k = -6+k-k
35k = -6
Divide by 35
35k/35 = -6/35
k = -6/35
Answer:
Yes
Step-by-step explanation:
If you do 7x3 it equals 21.