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shusha [124]
3 years ago
12

I need help like NOW pronto im confused HELP

Mathematics
2 answers:
Igoryamba3 years ago
7 0

Answer:

A) A dilation by a scale factor of 2, centered at the origin, followed by a 180° clockwise rotation about the origin.

Step-by-step explanation:

a dilation by a scale factor of 2, centered at the origin will double its size and move point A from (-2,2) to (-4,4). Then the clockwise rotation of 180° degrees will rotate the figure so that it's point A (now called point D) will be at (4,-4).

Brilliant_brown [7]3 years ago
3 0

Answer:

a pex

Step-by-step explanation:

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3 years ago
Find the area of the shaded region. Round your answer to the nearest tenth.
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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.

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\bf \textit{area of a sector of a circle}\\\\&#10;A_x=\cfrac{\theta \pi r^2}{360}\quad &#10;\begin{cases}&#10;r=radius\\&#10;\theta =angle~in\\&#10;\qquad degrees\\&#10;------\\&#10;r=6\\&#10;\theta =60&#10;\end{cases}\implies A_x=\cfrac{60\cdot \pi \cdot 6^2}{360}\implies \boxed{A_x=6\pi} \\\\&#10;-------------------------------\\\\

\bf \textit{area of a segment of a circle}\\\\&#10;A_y=\cfrac{r^2}{2}\left[\cfrac{\pi \theta }{180}~-~sin(\theta )  \right]&#10;\begin{cases}&#10;r=radius\\&#10;\theta =angle~in\\&#10;\qquad degrees\\&#10;------\\&#10;r=6\\&#10;\theta =120&#10;\end{cases}

\bf A_y=\cfrac{6^2}{2}\left[\cfrac{\pi\cdot 120 }{180}~-~sin(120^o )  \right]&#10;\\\\\\&#10;A_y=18\left[\cfrac{2\pi }{3}~-~\cfrac{\sqrt{3}}{2} \right]\implies \boxed{A_y=12\pi -9\sqrt{3}}\\\\&#10;-------------------------------\\\\&#10;\textit{shaded area}\qquad \stackrel{A_x}{6\pi }~~+~~\stackrel{A_y}{12\pi -9\sqrt{3}}\implies 18\pi -9\sqrt{3}

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3 years ago
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