(f+g)(x)=f(x)+g(x)
(f+g)(x)=3x+4+8x+1
(f+g)(x)=11x+5
(f+g)(0)=11(0)+5
(f+g)(0)=0+5
(f+g)(0)=5
You would take the answer of the square root problem and find the number that you can multiply by its self that would make that answer correct.
S+m+l=28
4s+2m+l=58
6s+5m+4l=135
Eliminate the variable l from the first 2 equations
s+m+l=28
-4s-2m-l=-58
-3s-m=-30
Elminiate the variable l from the last 2 equations
6s+5m+4l=135
-16s-8m-4l=-232
-10s-3m=-97
Now solve for s and m using the 2 equations without l
-3s-m=-30
-10s-3m=-97
9s+3m=90
-10s-3m=-97
-s=-7
s=7
Then plug in s into one of the equations without l
-3(7)-m=-30
-21-m=-30
-m=-9
m=9
Now plus in s and m into one of the original 3 equations
(7)+(9)+l=28
16+l=28
l=12
Final answer:
Small=$7
Medium=$9
Large=$12
I know it only asks for large but I wanted to show you how to find them all for future reference. :)
Answer:
There are two rational roots for f(x)
Step-by-step explanation:
We are given a function
To find the number of rational roots for f(x).
Let us use remainder theorem that when
f(a) =0, (x-a) is a factor of f(x) or x=a is one solution.
Substitute 1 for x
f(1) = 1-2-5+6=0
Hence x=1 is one solution.
Let us try x=-1
f(-1) = 1-2-5+6 =0
So x =-1 is also a solution and x+1 is a factor
We can write f(x) by trial and error as
We find that 
factor gives two irrational solutions as
±√3.
Hence number of rational roots are 2.
