I hope this helps you
Area=height.base/2
Area =2. (x+1)/2
Area =x+1
Answer:
D
Step-by-step explanation:
well, the assumption is that is a rectangle, namely it has two equal pairs, so we can just find the length of one of the pairs to get the dimensions.
hmmmm let's say let's get the length of the segment at (-1,-3), (1,3) for its length
and
the length of the segment at (-1, -3), (-4, -2) for its width
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{length}{L}=\sqrt{[1-(-1)]^2+[3-(-3)]^2}\implies L=\sqrt{(1+1)^2+(3+3)^2} \\\\\\ L=\sqrt{4+36}\implies L=\sqrt{40} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B-3%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B1%7D~%2C~%5Cstackrel%7By_2%7D%7B3%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Blength%7D%7BL%7D%3D%5Csqrt%7B%5B1-%28-1%29%5D%5E2%2B%5B3-%28-3%29%5D%5E2%7D%5Cimplies%20L%3D%5Csqrt%7B%281%2B1%29%5E2%2B%283%2B3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20L%3D%5Csqrt%7B4%2B36%7D%5Cimplies%20L%3D%5Csqrt%7B40%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{width}{w}=\sqrt{[-4-(-1)]^2+[-2-(-3)]^2}\implies w=\sqrt{(-4+1)^2+(-2+3)^2} \\\\\\ w=\sqrt{9+1}\implies w=\sqrt{10} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the rectangle}}{A=Lw}\implies \sqrt{40}\cdot \sqrt{10}\implies \sqrt{400}\implies \boxed{20}](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B-3%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-4%7D~%2C~%5Cstackrel%7By_2%7D%7B-2%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bwidth%7D%7Bw%7D%3D%5Csqrt%7B%5B-4-%28-1%29%5D%5E2%2B%5B-2-%28-3%29%5D%5E2%7D%5Cimplies%20w%3D%5Csqrt%7B%28-4%2B1%29%5E2%2B%28-2%2B3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20w%3D%5Csqrt%7B9%2B1%7D%5Cimplies%20w%3D%5Csqrt%7B10%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20rectangle%7D%7D%7BA%3DLw%7D%5Cimplies%20%5Csqrt%7B40%7D%5Ccdot%20%5Csqrt%7B10%7D%5Cimplies%20%5Csqrt%7B400%7D%5Cimplies%20%5Cboxed%7B20%7D)
Answer:
Diameter = 19.33
Step-by-step explanation:
Imaging a radius line from O to the endpoints of the 7. call this line R.
Label the part of the vertical line from O to the 90 degree intersection y.
Now you have a right triangle.
Using the Pythagorean theorem:
R² = 7² + y²
also
y = R - 3
substitute for y:
R² = 49 + (R-3)²
R² = 49 + R² - 6R + 9
simplify:
0 = 58 - 6R
6R = 58
R = 9.6667
Diameter = 2(9.6667) = 19.33
Answer:
$150
61 mph is 11 mph over 50 mph; see where the 11 mph mark is on the graph, and go up on the line to see what the fine is ($150).
Hope this helps :)
