you will do as you said yes yes i is fine with me but i’m going on the right now so bye bye u it was correct that
Answer:
![\large\boxed{-\bigg[(x-3)^2+2x\bigg]+1}\\\\\boxed{-x^2+4x-8}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B-%5Cbigg%5B%28x-3%29%5E2%2B2x%5Cbigg%5D%2B1%7D%5C%5C%5C%5C%5Cboxed%7B-x%5E2%2B4x-8%7D)
Step-by-step explanation:
![f(x)=(x-3)^2+2x\\\\g(x)=-x+1\\\\g\ \circ\ f\to\text{put f(x) instead of x in the function g(x)}:\\\\(g\ \circ\ f)(x)=-\bigg[\underbrace{(x-3)^2+2x}_{x}\bigg]+1](https://tex.z-dn.net/?f=f%28x%29%3D%28x-3%29%5E2%2B2x%5C%5C%5C%5Cg%28x%29%3D-x%2B1%5C%5C%5C%5Cg%5C%20%5Ccirc%5C%20f%5Cto%5Ctext%7Bput%20f%28x%29%20instead%20of%20x%20in%20the%20function%20g%28x%29%7D%3A%5C%5C%5C%5C%28g%5C%20%5Ccirc%5C%20f%29%28x%29%3D-%5Cbigg%5B%5Cunderbrace%7B%28x-3%29%5E2%2B2x%7D_%7Bx%7D%5Cbigg%5D%2B1)
-----------------------------------------
![-\bigg[(x-3)^2+2x\bigg]+1=-(x-3)^2-2x+1\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=-(x^2-(2)(x)(3)+3^2)-2x+1=-(x^2-6x+9)-2x+1\\\\=-x^2-(-6x)-9-2x+1=-x^2+6x-9-2x+1\\\\\text{combine like terms}\\\\=-x^2+(6x-2x)+(-9+1)=-x^2+4x-8](https://tex.z-dn.net/?f=-%5Cbigg%5B%28x-3%29%5E2%2B2x%5Cbigg%5D%2B1%3D-%28x-3%29%5E2-2x%2B1%5C%5C%5C%5C%5Ctext%7Buse%7D%5C%20%28a-b%29%5E2%3Da%5E2-2ab%2Bb%5E2%5C%5C%5C%5C%3D-%28x%5E2-%282%29%28x%29%283%29%2B3%5E2%29-2x%2B1%3D-%28x%5E2-6x%2B9%29-2x%2B1%5C%5C%5C%5C%3D-x%5E2-%28-6x%29-9-2x%2B1%3D-x%5E2%2B6x-9-2x%2B1%5C%5C%5C%5C%5Ctext%7Bcombine%20like%20terms%7D%5C%5C%5C%5C%3D-x%5E2%2B%286x-2x%29%2B%28-9%2B1%29%3D-x%5E2%2B4x-8)
-----------------------------------
To tell if an equation has infinite solutions, the equation will be equal to each other.
For example,
x = x
x + 1 = x + 1
x - y = x - y
And so on...
And in special cases,
0x = 0
0x = 0(y + 1)
They have infinite solutions because there's no constant to determine the variable.
Answer:
72.25663
Step-by-step explanation:
C=2πr=2·π·11.5≈72.25663
Answer:
-3
Step-by-step explanation:
y = mx + b
b is the y-intercept
