Answer:
y = -3/2 x +13
Step-by-step explanation:
We want our line to be perpendicular to
y = 2/3 x -1
The slope of this line is 2/3 (since it is written in the form y = mx+b and m is the slope)
Perpendicular lines have negative reciprocal slopes
m = -(3/2)
The slope of our new line is -3/2
We can use point slope form of the equation
y-y1 - m (x-x1)
y - 7 = -3/2 (x-4)
Distribute
y-7 = -3/2x +6
Add 7 to each side
y-7+7 = -3/2 x +6+7
y = -3/2 x +13
Answer:
Aviara his more boys
AOMS 400 students / 5 = 80 X 3 = 240 girls
VMS 360 students / 8 = 45 X 5 = 225 girls
Answer:
C. UPR
Step-by-step explanation:
Opposite from TPQ
Step-by-step explanation:
53-5 is 48
48/6 is 8
X=8
therefore it's 8
if the endpoints are there, that means that segment with those endpoints is the diameter of the circle, and that also means that the midpoint of that diameter is the center of the circle.
it also means that the distance from the midpoint to either endpoint, is the radius of the circle.
![\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{8+2}{2}~~,~~\cfrac{-1-5}{2} \right)\implies (5,-3)\impliedby center \\\\[-0.35em] \rule{34em}{0.25pt}](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%7B2%7D~%2C~%5Cstackrel%7By_1%7D%7B-5%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B-1%7D%29%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%7B8%2B2%7D%7B2%7D~~%2C~~%5Ccfrac%7B-1-5%7D%7B2%7D%20%5Cright%29%5Cimplies%20%285%2C-3%29%5Cimpliedby%20center%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{midpoint}{(\stackrel{x_1}{5}~,~\stackrel{y_1}{-3})}\qquad \stackrel{endpoint}{(\stackrel{x_2}{8}~,~\stackrel{y_2}{-1})}\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[8-5]^2+[-1-(-3)]^2}\implies r=\sqrt{(8-5)^2+(-1+3)^2} \\\\\\ r=\sqrt{3^2+2^2}\implies r=\sqrt{13} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%5Cstackrel%7Bmidpoint%7D%7B%28%5Cstackrel%7Bx_1%7D%7B5%7D~%2C~%5Cstackrel%7By_1%7D%7B-3%7D%29%7D%5Cqquad%20%5Cstackrel%7Bendpoint%7D%7B%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B-1%7D%29%7D%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%7Bradius%7D%7Br%7D%3D%5Csqrt%7B%5B8-5%5D%5E2%2B%5B-1-%28-3%29%5D%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B%288-5%29%5E2%2B%28-1%2B3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20r%3D%5Csqrt%7B3%5E2%2B2%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B13%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{5}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\sqrt{13}}{ r} \\[2em] [x-5]^2+[y-(-3)]^2=(\sqrt{13})^2\implies \boxed{(x-5)^2+(y+3)^2=13}](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%7B5%7D%7B%20h%7D%2C%5Cstackrel%7B-3%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20radius%3D%5Cstackrel%7B%5Csqrt%7B13%7D%7D%7B%20r%7D%20%5C%5C%5B2em%5D%20%5Bx-5%5D%5E2%2B%5By-%28-3%29%5D%5E2%3D%28%5Csqrt%7B13%7D%29%5E2%5Cimplies%20%5Cboxed%7B%28x-5%29%5E2%2B%28y%2B3%29%5E2%3D13%7D)
