well, we know it's a rectangle, so that means the sides JK = IL and JI = KL, so
![\stackrel{JK}{3x+21}~~ = ~~\stackrel{IL}{6y}\implies 3(x+7)=6y\implies x+7=\cfrac{6y}{3} \\\\\\ x+7=2y\implies \boxed{x=2y-7} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{JI}{6y-6}~~ = ~~\stackrel{KL}{2x+20}\implies 6(y-1)=2(x+10)\implies \cfrac{6(y-1)}{2}=x+10 \\\\\\ 3(y-1)=x+10\implies 3y-3=x+10\implies \stackrel{\textit{substituting from the 1st equation}}{3y-3=(2y-7)+10} \\\\\\ 3y-3=2y+3\implies y-3=3\implies \blacksquare~~ y=6 ~~\blacksquare ~\hfill \blacksquare~~ \stackrel{2(6)~~ - ~~7}{x=5} ~~\blacksquare](https://tex.z-dn.net/?f=%5Cstackrel%7BJK%7D%7B3x%2B21%7D~~%20%3D%20~~%5Cstackrel%7BIL%7D%7B6y%7D%5Cimplies%203%28x%2B7%29%3D6y%5Cimplies%20x%2B7%3D%5Ccfrac%7B6y%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20x%2B7%3D2y%5Cimplies%20%5Cboxed%7Bx%3D2y-7%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7BJI%7D%7B6y-6%7D~~%20%3D%20~~%5Cstackrel%7BKL%7D%7B2x%2B20%7D%5Cimplies%206%28y-1%29%3D2%28x%2B10%29%5Cimplies%20%5Ccfrac%7B6%28y-1%29%7D%7B2%7D%3Dx%2B10%20%5C%5C%5C%5C%5C%5C%203%28y-1%29%3Dx%2B10%5Cimplies%203y-3%3Dx%2B10%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20from%20the%201st%20equation%7D%7D%7B3y-3%3D%282y-7%29%2B10%7D%20%5C%5C%5C%5C%5C%5C%203y-3%3D2y%2B3%5Cimplies%20y-3%3D3%5Cimplies%20%5Cblacksquare~~%20y%3D6%20~~%5Cblacksquare%20~%5Chfill%20%5Cblacksquare~~%20%5Cstackrel%7B2%286%29~~%20-%20~~7%7D%7Bx%3D5%7D%20~~%5Cblacksquare)
Answer:
-15
Step-by-step explanation:
start from the inside and go out.
So first plug in -3 into g(x)
g(-3) = -3 - 7 = -10
then plug in -10 into f(x)
f(-10) = 2(-10) + 5 = -15
so f(g(x)) = -15
Answer:
The letter "x" is often used in algebra to mean a value that is not yet known.
It is called a "variable" or sometimes an "unknown".
But in some cases, x can be equal to 1 like example when working with exponents.
Step-by-step explanation:
Answer:
P(Y=1|X=3)=0.125
Step-by-step explanation:
Given :
p(1,1)=0
p(2,1)=0.1
p(3,1)=0.05
p(1,2)=0.05
p(2,2)=0.3
p(3,2)=0.1
p(1,3)=0.05
p(2,3)=0.1
p(3,3)=0.25
Now we are supposed to find the conditional mass function of Y given X=3 : P(Y=1|X=3)
P(X=3) = P(X=3,Y=1)+P(X=3,Y=2) +P(X=3,Y=3)
P(X=3)=p(3,1) +p(3,2) +p(3,3)
P(X=3)=0.05+0.1+0.25=0.4

Hence P(Y=1|X=3)=0.125
Answer:
that is actually 13/42
Step-by-step explanation:
cuz 13 cant be simplified by any number hope it help