well, since we know the endpoints for the diameter, its midpoint will be where the center of the circle is located, so
and if we get the distance between those endpoints, and take half of that, that'd be the radius of it.
![~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{11})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-4 - (-10)]^2 + [11 - 5]^2}\implies d=\sqrt{(-4+10)^2+6^2} \\\\\\ d=\sqrt{6^2+6^2}\implies d=\sqrt{72}~\hfill \stackrel{radius=half~that}{\cfrac{\sqrt{72}}{2}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-10%7D~%2C~%5Cstackrel%7By_1%7D%7B5%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-4%7D~%2C~%5Cstackrel%7By_2%7D%7B11%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%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B-4%20-%20%28-10%29%5D%5E2%20%2B%20%5B11%20-%205%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-4%2B10%29%5E2%2B6%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B6%5E2%2B6%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B72%7D~%5Chfill%20%5Cstackrel%7Bradius%3Dhalf~that%7D%7B%5Ccfrac%7B%5Csqrt%7B72%7D%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-7}{ h},\stackrel{8}{ k})\qquad \qquad radius=\stackrel{\frac{\sqrt{72}}{2}}{ r} \\\\\\\ [x-(-7)]^2~~ + ~~[y-8]^2~~ = ~~\left( \frac{\sqrt{72}}{2} \right)^2\implies (x+7)^2~~ + ~~(y-8)^2~~ = ~~18](https://tex.z-dn.net/?f=%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%20%5Cqquad%20center~~%28%5Cstackrel%7B-7%7D%7B%20h%7D%2C%5Cstackrel%7B8%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20radius%3D%5Cstackrel%7B%5Cfrac%7B%5Csqrt%7B72%7D%7D%7B2%7D%7D%7B%20r%7D%20%5C%5C%5C%5C%5C%5C%5C%20%5Bx-%28-7%29%5D%5E2~~%20%2B%20~~%5By-8%5D%5E2~~%20%3D%20~~%5Cleft%28%20%5Cfrac%7B%5Csqrt%7B72%7D%7D%7B2%7D%20%5Cright%29%5E2%5Cimplies%20%28x%2B7%29%5E2~~%20%2B%20~~%28y-8%29%5E2~~%20%3D%20~~18)
Answer:
IQR: is the length of the middle 50% of that interval of space
Step-by-step explanation:
In order to find IQR, subtract the third interquartile range - the first interquartile range
Answer:
x = 11
Step-by-step explanation:
Complementary means they add up to 90°
So add the angles up and set them equal to 90°
3x - 4 + 5x + 6 = 90
ADD the x's
8x - 4 + 6 = 90
COMBINE the -4 and 6
8x +2 = 90
SUBTRACT 2 from both sides.
8x = 88
DIVIDE by 8 on both sides.
8x/8 = 88/8
x = 11
Answer:
60%
Step-by-step explanation:
I changed 3/5 to 30/50, then multiplied them both by 2 to get 60/100, and since a percentage is out of a hundred, its 60%
I hope this helps! ^^
