Answer:
g(x) = ![\frac{1}{3}(x - 1) = \frac{x}{3} - \frac{1}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%28x%20-%201%29%20%3D%20%5Cfrac%7Bx%7D%7B3%7D%20-%20%5Cfrac%7B1%7D%7B3%7D)
Step-by-step explanation:
f(x) = 3x + 5
f[g(x)] = 3[g(x)] + 5
⇒ 3[g(x)] + 5 = x + 4
⇒ 3[g(x)] = x + 4 - 5
⇒ 3[g(x)] = x - 1
⇒ g(x) = ![\frac{1}{3}(x - 1) = \frac{x}{3} - \frac{1}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%28x%20-%201%29%20%3D%20%5Cfrac%7Bx%7D%7B3%7D%20-%20%5Cfrac%7B1%7D%7B3%7D)
Answer:
Step-by-step explanation:
-5+-15= -20
4+15= 19
-20+19
-1
4(9+7-9)
the nines cancel
4*7 =28
Given
![f(x) = x^2 + 6x + 9, g(x) = x^2 - 9](https://tex.z-dn.net/?f=f%28x%29%20%3D%20x%5E2%20%2B%206x%20%2B%209%2C%20g%28x%29%20%3D%20x%5E2%20-%209)
:
![(f+g)(x) = (x^2 + 6x+9)+(x^2-9) =2x^2+6x=2x(x+3)](https://tex.z-dn.net/?f=%28f%2Bg%29%28x%29%20%3D%20%28x%5E2%20%2B%206x%2B9%29%2B%28x%5E2-9%29%20%3D2x%5E2%2B6x%3D2x%28x%2B3%29)
, which is the second expression
![(f-g)(x) = (x^2 + 6x+9)-(x^2-9) =6x+18=6(x+3)](https://tex.z-dn.net/?f=%28f-g%29%28x%29%20%3D%20%28x%5E2%20%2B%206x%2B9%29-%28x%5E2-9%29%20%3D6x%2B18%3D6%28x%2B3%29)
, which is the first second expression
![(fxg)(x) = (x^2+6x+9)(x^2-9) = (x^4-9x^2)+(6x^3-54x)+(9x^2-81)=x^4+6x^3-54x-81](https://tex.z-dn.net/?f=%28fxg%29%28x%29%20%3D%20%28x%5E2%2B6x%2B9%29%28x%5E2-9%29%20%3D%20%28x%5E4-9x%5E2%29%2B%286x%5E3-54x%29%2B%289x%5E2-81%29%3Dx%5E4%2B6x%5E3-54x-81)
, which is the fourth expression
![(f/g)(x) = (x^2+6x+9)/(x^2-9) = \frac{(x+3)^2}{(x+3)(x-3)} = \frac{x+3}{x-3}](https://tex.z-dn.net/?f=%28f%2Fg%29%28x%29%20%3D%20%28x%5E2%2B6x%2B9%29%2F%28x%5E2-9%29%20%3D%20%5Cfrac%7B%28x%2B3%29%5E2%7D%7B%28x%2B3%29%28x-3%29%7D%20%3D%20%5Cfrac%7Bx%2B3%7D%7Bx-3%7D)
, which is the third expression
And always remember this applies to you too like honestly you’re amazing for doing this