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)
Formula is y = a(x-h)^2 + k
Where h is 1 and k is 1
f (x) = a(x-1)^2 + 1
-3 = a(0-1)^2 + 1
-3 = a(-1)^2 + 1
-3 = a(1) + 1
-3 - 1 = a
-4 = a
a = -4
A must be equal to -4
y = -4(x-1)^2 + 1
0 = -4(x-1)^2 + 1
4(x^2 - 2x + 1) - 1 = 0
4x^2 - 8x + 4 - 1 = 0
4x^2 - 8x + 3 = 0
4x^2 - 8x = -3
Divide fpr 4 each term of the equation....x^2 - 2x = -3/4
We must factor the perfect square ax^2 + bx + c which we don't have. We must follow the rule (b/2)^2 where b is -2....(-2/2)^2 =
(-1)^2 = 1 and we add up that to both sides
x^2 - 2x + 1 = -3/4 + 1
x^2 - 2x + 1 = 1/4
(x-1)^2 = 1/4
square root both sides x-1 = (+/-) 1/2
x1 = +1/2 + 1 = 3/2
x2 = -1/2 + 1 = 1/2
x-intercepts are 1/2 and 3/2, in form (3/2,0); (1/2,0)
a= 34 degrees
b= 28 degrees
c= 62 degrees
Step-by-step explanation:
First you know that b is 1/2 of 56 degrees or 28.
The triangle with the a in it is isoceles because the two sides are both radii.
In the triangle the top angle = 112 because it is a centeral angle to the 112 arc.
Angle a and opposite to a are equal and then have to be 34 degrees to equal 180.
We know two arc lengths are 112 and 56 and the one with angle a has to be 34x2 or 68.
a whole circle equals 360.
360-56-68-112 = 124
Angle c = 1/2 of 124, or 62 degrees
-9x+5<17 answer x >-4/3 how I do to it
subtract -5 from both side and you get 12. so it be -9x<12
than you divide -9 from both side because you want x by it self you get -4/3 but you have to flip the sign since is negavite and you will be left over with ×>-4/3
