LCM:
10: 0,10,20,30,40,50
4: 0,4,8,12,16,20
ANSWER IS 20
LCM:
10: see above numbers
6: 0,6,12,18,24,30
ANSWER IS 30
So, 10 and 8 would be 40
Answer:
{8,-4,-10}
Step-by-step explanation:
re arrange relation, im going to assume x is the domain input and y is the range.
-3x+2=y
plug in every element of the set x to get the range set y
-3(-2)+2=y
y=8
-3(2)+2=y
y=-4
-3(4)+2=y
y=-10
so the range is {8,-4,-10}
Y + 4 = -3/4x --------------------> y = -3/4x - 4
Answer:
A D 3 3
B E 3 3
C F 3 3
A G 5 5
B H 5 5
C I 5 5
A J 6.4 6.4
B K 6.4 6.4
C L 6.4 6.4
Step-by-step explanation:
no idea, just copied from the site lol
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)
