$\bf \begin{cases}
(f\circ g)(x)=x^2-6x+8\\\\
(f\circ g)(x)=f(~~g(x)~~)\\\\
g(x)=x-3\\
\end{cases}
\\\\\\
\textit{now, one probability is }x^2-1
\\\\\\
f(x)=x^2-1\implies f(~~g(x)~~)=(x-3)^2-1
\\\\\\
f(~~g(x)~~)=(x^2-6x+9)-1\implies f(~~g(x)~~)=x^2-6x+8$

5 0

3 years ago

Please answer the 3rd Question please answer it fast............I WILL MARK YOU AS THE BRANLIEST.................

mojhsa [17]

since LM=LN there values are same which is given as 5.5 cm and MN =7cm

now draw a line LM which is 5.5 cm long. From one point of this line construct an arc 5.5 cm in upward direction.Then from the opposite end of the same line LM construct an arc 7 cm long in upward direction. Let it meet the the first arc at any point. The arcs will meet for sure at any angle. Join the two ends of line LN to this point where they meet. We get a triangle!

Remember to mark LM , LN and MN as soon as u draw them so as to avoid confusion.

<em>IF U WANT I'LL DO IT AND SEND A PHOTO</em>

8 0

3 years ago

katrin2010 [14]

Point slope and it would be y=2x-5

4 0

3 years ago

What is the graph of the function f(x) =x^2 minus 9x+20 over x-4

exis [7]

The answer to this problem is =x

5 0

3 years ago