since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.
we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-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-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-7%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B-5%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-2-%28-4%29%5D%5E2%2B%5B-5-%28-7%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-2%2B4%29%5E2%2B%28-5%2B7%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B2%5E2%2B2%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B2%5Ccdot%202%5E2%7D%5Cimplies%20d%3D2%5Csqrt%7B2%7D~%5Chfill%20%5Cstackrel%7B~%5Chfill%20radius%7D%7B%5Ccfrac%7B2%5Csqrt%7B2%7D%7D%7B2%7D%5Cimplies%5Cboxed%7B%20%5Csqrt%7B2%7D%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-7%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B-5%7D%29%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%20%5Ccfrac%7B-2-4%7D%7B2%7D~~%2C~~%5Ccfrac%7B-5-7%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B-6%7D%7B2%7D~%2C~%5Ccfrac%7B-12%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cstackrel%7Bcenter%7D%7B%5Cboxed%7B%28-3%2C-6%29%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2](https://tex.z-dn.net/?f=%5Cbf%20%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-3%7D%7B%20h%7D%2C%5Cstackrel%7B-6%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20radius%3D%5Cstackrel%7B%5Csqrt%7B2%7D%7D%7B%20r%7D%20%5C%5C%5B2em%5D%20%5Bx-%28-3%29%5D%5E2%2B%5By-%28-6%29%5D%5E2%3D%28%5Csqrt%7B2%7D%29%5E2%5Cimplies%20%28x%2B3%29%5E2%2B%28y%2B6%29%5E2%3D2)
Answer:
All of them
Step-by-step explanation:
none of the graphs repeat any numbers on the x
Answer:
you needed to mark this question for the science section
Step-by-step explanation:
Answer:
-14
Step-by-step explanation:
f(x) = 3x-5
g(x) = -x^2 +1
g(2) = - (2^2) +1
g(2) =-4+1 = -3
Then stick the -3 in f(x)
f(-3) = 3(-3)-5 = -9 -5 = -14
f(g(2) = -14
Answer:
y=5/6x+8
Step-by-step explanation
This is definitely parallel
