Ok im not sure the answer yet so ima work on it at the same time while im explaining it. (The answer will probably be near the end)
We can use the elimination method to eliminate y out. To do that we multiply the first equation by 3.
6x+3y=-12
Now just subtract it from the other equation.
6x+3y=-12
5x+3y=-6
***x=-6***
Usually after doing the elimination method you will have to solve for x but in this case its already solved for you. If you want to find y now you just take the first equation and fill x with -6 and solve for y.
2(-6)+y=-4
-12+y=-4
y=8
Brainliest my answer if it helps you out?
Answer:
since angle B and angle C contains the same arc AD. set the angles equal
11x-3=8x+15
x=6
so the angle has a measure of 63 degrees
the central angle is the angle of the arc
and it is always 2 times the angle that we just that contains the arc but does touches the end of the circle.
so its 63 *2 =126 degrees.
Answer:
m∠GDE
alternate interior angles
Answer:
A.
Step-by-step explanation:
![\bf f(x)=\cfrac{2x-3}{x+1}~\hspace{10em}g(x)=\cfrac{x+3}{2-x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ f(~~g(x)~~)\implies \cfrac{2[g(x)]-3}{[g(x)]+1}\implies \cfrac{2\left( \frac{x+3}{2-x} \right)-3}{\left( \frac{x+3}{2-x} \right)+1}\implies \cfrac{\frac{2x+6}{2-x}-3}{\frac{x+3}{2-x}+1} \\\\\\ \cfrac{\frac{2x+6-6+3x}{2-x}}{\frac{x+3+2-x}{2-x}}\implies \cfrac{2x+6-6+3x}{2-x}\cdot \cfrac{2-x}{x+3+2-x}\implies \cfrac{5x}{5}\implies x](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3D%5Ccfrac%7B2x-3%7D%7Bx%2B1%7D~%5Chspace%7B10em%7Dg%28x%29%3D%5Ccfrac%7Bx%2B3%7D%7B2-x%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0Af%28~~g%28x%29~~%29%5Cimplies%20%5Ccfrac%7B2%5Bg%28x%29%5D-3%7D%7B%5Bg%28x%29%5D%2B1%7D%5Cimplies%20%5Ccfrac%7B2%5Cleft%28%20%5Cfrac%7Bx%2B3%7D%7B2-x%7D%20%5Cright%29-3%7D%7B%5Cleft%28%20%5Cfrac%7Bx%2B3%7D%7B2-x%7D%20%5Cright%29%2B1%7D%5Cimplies%0A%5Ccfrac%7B%5Cfrac%7B2x%2B6%7D%7B2-x%7D-3%7D%7B%5Cfrac%7Bx%2B3%7D%7B2-x%7D%2B1%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5Cfrac%7B2x%2B6-6%2B3x%7D%7B2-x%7D%7D%7B%5Cfrac%7Bx%2B3%2B2-x%7D%7B2-x%7D%7D%5Cimplies%20%5Ccfrac%7B2x%2B6-6%2B3x%7D%7B2-x%7D%5Ccdot%20%5Ccfrac%7B2-x%7D%7Bx%2B3%2B2-x%7D%5Cimplies%20%5Ccfrac%7B5x%7D%7B5%7D%5Cimplies%20x)
![\bf \rule{34em}{0.25pt}\\\\ g(~~f(x)~~)\implies \cfrac{[f(x)]+3}{2-[f(x)]}\implies \cfrac{\frac{2x-3}{x+1}+3}{2-\frac{2x-3}{x+1}}\implies \cfrac{\frac{2x-3+3x+3}{x+1}}{\frac{2x+2-(2x-3)}{x+1}} \\\\\\ \cfrac{2x-3+3x+3}{x+1}\cdot \cfrac{x+1}{2x+2-(2x-3)}\implies \cfrac{2x-3+3x+3}{x+1}\cdot \cfrac{x+1}{2x+2-2x+3} \\\\\\ \cfrac{5x}{5}\implies x](https://tex.z-dn.net/?f=%5Cbf%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0Ag%28~~f%28x%29~~%29%5Cimplies%20%5Ccfrac%7B%5Bf%28x%29%5D%2B3%7D%7B2-%5Bf%28x%29%5D%7D%5Cimplies%20%5Ccfrac%7B%5Cfrac%7B2x-3%7D%7Bx%2B1%7D%2B3%7D%7B2-%5Cfrac%7B2x-3%7D%7Bx%2B1%7D%7D%5Cimplies%20%5Ccfrac%7B%5Cfrac%7B2x-3%2B3x%2B3%7D%7Bx%2B1%7D%7D%7B%5Cfrac%7B2x%2B2-%282x-3%29%7D%7Bx%2B1%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2x-3%2B3x%2B3%7D%7Bx%2B1%7D%5Ccdot%20%5Ccfrac%7Bx%2B1%7D%7B2x%2B2-%282x-3%29%7D%5Cimplies%20%5Ccfrac%7B2x-3%2B3x%2B3%7D%7Bx%2B1%7D%5Ccdot%20%5Ccfrac%7Bx%2B1%7D%7B2x%2B2-2x%2B3%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B5x%7D%7B5%7D%5Cimplies%20x)
and in case you recall your inverses, when f( g(x) ) = x, or g( f(x) ) = x, simply means, they're inverse of each other.
