The answer to this question is 23. How you get this is by adding all the sides like this: 2n+6+135+151+140+63=540. Then you just solve the equation.
Answer:
2 units is perpendicular bisector of line segment
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:
# of terms: 2
inConstant(s): might be wrong but i think its 5
Coefficient(s): might be y itself but i dont know
Highest degree: 5
I did what I could but this number is not factorable with rational numbers
Hey,
The lower quartile is the median of the lower half of the data set.
The median is the middle most number or the average of the two middle most numbers depending on how many values there are.
Cheers,
Izzy
