Go to this website called papa math it will help you with graphs
Which is the correct choice.
1 answer:
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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)
![\bf A_y=\cfrac{6^2}{2}\left[\cfrac{\pi\cdot 120 }{180}~-~sin(120^o ) \right] \\\\\\ A_y=18\left[\cfrac{2\pi }{3}~-~\cfrac{\sqrt{3}}{2} \right]\implies \boxed{A_y=12\pi -9\sqrt{3}}\\\\ -------------------------------\\\\ \textit{shaded area}\qquad \stackrel{A_x}{6\pi }~~+~~\stackrel{A_y}{12\pi -9\sqrt{3}}\implies 18\pi -9\sqrt{3}](https://tex.z-dn.net/?f=%5Cbf%20A_y%3D%5Ccfrac%7B6%5E2%7D%7B2%7D%5Cleft%5B%5Ccfrac%7B%5Cpi%5Ccdot%20120%20%7D%7B180%7D~-~sin%28120%5Eo%20%29%20%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C%0AA_y%3D18%5Cleft%5B%5Ccfrac%7B2%5Cpi%20%7D%7B3%7D~-~%5Ccfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%20%5Cright%5D%5Cimplies%20%5Cboxed%7BA_y%3D12%5Cpi%20-9%5Csqrt%7B3%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bshaded%20area%7D%5Cqquad%20%5Cstackrel%7BA_x%7D%7B6%5Cpi%20%7D~~%2B~~%5Cstackrel%7BA_y%7D%7B12%5Cpi%20-9%5Csqrt%7B3%7D%7D%5Cimplies%2018%5Cpi%20-9%5Csqrt%7B3%7D)
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°.
PART A
The given equation is
In order to find the maximum height, we write the function in the vertex form.
We factor -16 out of the first two terms to get,
We add and subtract
to get,
We again factor -16 out of the first two terms to get,
This implies that,
The quadratic trinomial above is a perfect square.
This finally simplifies to,
The vertex of this function is
The y-value of the vertex is the maximum value.
Therefore the maximum value is,
PART B
When the ball hits the ground,
This implies that,
We add -29 to both sides to get,
This implies that,
or
Since time cannot be negative, we discard the negative value and pick,
The given equation is
In order to find the maximum height, we write the function in the vertex form.
We factor -16 out of the first two terms to get,
We add and subtract
to get,
We again factor -16 out of the first two terms to get,
This implies that,
The quadratic trinomial above is a perfect square.
This finally simplifies to,
The vertex of this function is
The y-value of the vertex is the maximum value.
Therefore the maximum value is,
PART B
When the ball hits the ground,
This implies that,
We add -29 to both sides to get,
This implies that,
or
Since time cannot be negative, we discard the negative value and pick,