Answer:
a) ![(-2, 0)](https://tex.z-dn.net/?f=%28-2%2C%200%29)
Step-by-step explanation:
x + 2 = 3x + 6
-3x - 3x
___________
−2x + 2 = 6
- 2 - 2
_________
4 = −2x
_ ___
−2 −2
[Plug this back into both equations above to get the y-coordinate of 0]; ![0 = y](https://tex.z-dn.net/?f=0%20%3D%20y)
I am joyous to assist you anytime.
First, subtract px2 from both sides.
Now you have:
x3 - px2 = (1 - p) x1
Next, divide both sides by (1 - p)
So now you have
x3 - px2/(1 - p) = x1
...as your final answer
*You can decide if you want to leave the parenthesis in your final answer, I left them there so it could be visible where I put them. :)
5.67 (performed on a calculator)
-28 because using PEMDAS you would do -5x8 which =-40 plus 12 is -28
![{ \qquad\qquad\huge\underline{{\sf Answer}}}](https://tex.z-dn.net/?f=%7B%20%5Cqquad%5Cqquad%5Chuge%5Cunderline%7B%7B%5Csf%20Answer%7D%7D%7D%20)
Let's find the inverse of given function ~
![\qquad \sf \dashrightarrow \: f(x) = 2x - 1](https://tex.z-dn.net/?f=%5Cqquad%20%5Csf%20%20%5Cdashrightarrow%20%5C%3A%20f%28x%29%20%3D%202x%20-%201)
![\qquad \sf \dashrightarrow \: 2x = f(x) + 1](https://tex.z-dn.net/?f=%5Cqquad%20%5Csf%20%20%5Cdashrightarrow%20%5C%3A%202x%20%3D%20f%28x%29%20%2B%201)
![\qquad \sf \dashrightarrow \: x = \cfrac{f(x) + 1}{2}](https://tex.z-dn.net/?f=%5Cqquad%20%5Csf%20%20%5Cdashrightarrow%20%5C%3A%20x%20%3D%20%20%5Ccfrac%7Bf%28x%29%20%2B%201%7D%7B2%7D%20)
[ now, replace f(x) and x with
]
![\qquad \sf \dashrightarrow \: f {}^{ - 1} (x) = \cfrac{x + 1}{2}](https://tex.z-dn.net/?f=%5Cqquad%20%5Csf%20%20%5Cdashrightarrow%20%5C%3A%20f%20%7B%7D%5E%7B%20-%201%7D%20%28x%29%20%3D%20%20%5Ccfrac%7Bx%20%2B%201%7D%7B2%7D%20)
That's the required inverse function ~