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)
Well i have to say is that PEMDAS exponents first. Hope this helps.
8 and 1/2 would be probably the answer
Answer:
100 gallon for every 50 mile
Step-by-step explanation:
if you look at it the 50 meet up with 100 and 50 says it's mile and 100 is gallon so you got 100 gallon for every 50 mile