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Nataly_w [17]
3 years ago
7

I ONLY NEED HELP ON NUMBER 2

Mathematics
2 answers:
Readme [11.4K]3 years ago
5 0
The radius of the circle
Multiply that by two to get the diameter
That is basically the height of the center of the rectangle
multiply that by the base
The radius is two
which means that the diameter is 4
Multiply the height times the base to get the area of the whole rectangle
4*8 = 32 That is the area of the whole rectangle you need only the shaded side
this means you have to find the area circle and subtract it from the area of the whole rectangle
Area of the circle is 
A = pi r^2
plug in what you know
A = pi (2)^2
A = pi 4
That is <span>12.56

Subtract it from the area of the rectangle
32 minus </span><span>12.56.
The area of the shaded region is </span><span>19.4.

Hope this helps :)</span>
REY [17]3 years ago
4 0
So, the area of a circle is a=pi r^2
so do that and get a=pi(4)
and since the circle touches the edge of the rectangle, the rectangle is 4 cm tall, so 4*8 is 32 and pi(4) is about 12.56, so 32-12.56 is 19.44, and I believe that is the answer. (19.4)
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\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ X(\stackrel{x_1}{-6}~,~\stackrel{y_1}{2})\qquad Z(\stackrel{x_2}{5}~,~\stackrel{y_2}{8})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ XZ=\sqrt{[5-(-6)]^2+[8-2]^2}\implies XZ=\sqrt{(5+6)^2+(8-2)^2} \\\\\\ XZ=\sqrt{121+36}\implies \boxed{XZ=\sqrt{157}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ Z(\stackrel{x_2}{5}~,~\stackrel{y_2}{8})\qquad Y(\stackrel{x_2}{8}~,~\stackrel{y_2}{2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ ZY=\sqrt{(8-5)^2+(2-8)^2}\implies ZY=\sqrt{9+36}\implies \boxed{ZY=\sqrt{45}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{perimeter}{14+\sqrt{157}+\sqrt{45}}\qquad \approx \qquad 33.2

6 0
3 years ago
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