Answer:
A composite figure is made up of simple geometric shapes. To find the area of a composite figure or other irregular-shaped figure, divide it into simple, non overlapping figures. Find the area of each simpler figure, and then add the areas together to find the total area of the composite figure.
Step-by-step explanation:
Okay. So I find it easiest to calculate the full area as if the shape was a rectangle, then subtracting open space. We already know that length is 100, so now we just need to find width. 50 + 25 + 50 + 25 + 50 = 200. 200 x 100 =20,000. So now we just need to find the areas of open space and subtract. Both spaces have a length of 75 and a width of 25. 25 x 75 = 1875. There were 2 spaces, and 1875 x 2 = 3,750. 20,000 - 3,750 = 16,250.
The area of the figure is 16,250 feet.
Answer:
A and C
Step-by-step explanation:
you will see a double sign inequality when your circle on the number line is shaded in. aka a closed circle
Travis got 43 dollars
Robin collected 35 percent more so we can model this with
Travis + (Travis x .35)
43 + 15.05 =58.05
Jessica colelcted 20 percent less so
Travis - (Travis x .20)
43 - 8.6 = 34.3
34.4 + 58.05 + 43 = 135.45
C is the closest
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r}\\\\ -------------------------------\\\\ (x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~ \begin{cases} \stackrel{center}{(-1,0)}\\ \stackrel{radius}{6} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B%7D%7B%20r%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%28x%2B1%29%5E2%2By%5E2%3D36%5Cimplies%20%5Bx-%28%5Cstackrel%7Bh%7D%7B-1%7D%29%5D%5E2%2B%5By-%5Cstackrel%7Bk%7D%7B0%7D%5D%5E2%3D%5Cstackrel%7Br%7D%7B6%5E2%7D~~~~%0A%5Cbegin%7Bcases%7D%0A%5Cstackrel%7Bcenter%7D%7B%28-1%2C0%29%7D%5C%5C%0A%5Cstackrel%7Bradius%7D%7B6%7D%0A%5Cend%7Bcases%7D)
so, that's the equation of the circle, and that's its center, any point "ON" the circle, namely on its circumference, will have a distance to the center of 6 units, since that's the radius.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2} \\\\\\ d=\sqrt{0+1}\implies d=1](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%0AA%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%5Cqquad%20%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdistance%7D%7Bd%7D%3D%5Csqrt%7B%5B-1-%28-1%29%5D%5E2%2B%281-0%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-1%2B1%29%5E2%2B1%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B0%2B1%7D%5Cimplies%20d%3D1)
well, the distance from the center to A is 1, namely is "inside the circle".
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{6})\\\\\\ \stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(6-0)^2}\implies d=\sqrt{(-1+1)^2+6^2} \\\\\\ d=\sqrt{0+36}\implies d=6](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%0AB%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B6%7D%29%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdistance%7D%7Bd%7D%3D%5Csqrt%7B%5B-1-%28-1%29%5D%5E2%2B%286-0%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-1%2B1%29%5E2%2B6%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B0%2B36%7D%5Cimplies%20d%3D6)
notice, the distance to B is exactly 6, and you know what that means.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-8}) \\\\\\ \stackrel{distance}{d}=\sqrt{[4-(-1)]^2+[-8-0]^2}\implies d=\sqrt{(4+1)^2+(-8)^2} \\\\\\ d=\sqrt{25+64}\implies d=\sqrt{89}\implies d\approx 9.43398](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%0AC%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B-8%7D%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdistance%7D%7Bd%7D%3D%5Csqrt%7B%5B4-%28-1%29%5D%5E2%2B%5B-8-0%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%284%2B1%29%5E2%2B%28-8%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B25%2B64%7D%5Cimplies%20d%3D%5Csqrt%7B89%7D%5Cimplies%20d%5Capprox%209.43398)
notice, C is farther than the radius 6, meaning is outside the circle, hiking about on the plane.
