Answer:
c
Step-by-step explanation:
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)
That answer will be 7.
Explanation:
=1-3+9
=10-3
=7
Answer:
the two expressions are equivalent.
Step-by-step explanation:
We know that f(x)= 3x-7 and g(x)= -2x+5, therefore:
f(x) - (-g(x)) = 3x-7 - ( +2x-5) = 3x - 7 - 2x + 5 = x -2
f(x) + g(x) = 3x-7 -2x + 5 = x - 2
Therefore, f(x) - (-g(x)) is equivalent to f(x) + g(x).
Another way to check that the two expressions are equivalent is by solving the parenthesis:
f(x) - (-g(x)) → f(x) + g(x)
Therefore, the two expressions are equivalent.
