Select all the correct answers:
1) Yes
2) No
x=8→h(8)=2(8)^2+5(8)+2=2(64)+40+2=128+40+2→h(8)=170
x=8→f(8)=3^8+2=6,561+2→f(8)=5,563>170=h(8)
3) Yes
4) No
5) Yes
rg=[g(3)-g(2)]/(3-2)=[g(3)-g(2)]/1→rg=g(3)-g(2)
g(3)=20(3)+4=60+4→g(3)=64
g(2)=20(2)+4=40+4→g(2)=44
rg=64-44→rg=20
rf=f(3)-f(2)
f(3)=3^3+2=27+2→f(3)=29
f(2)=3^2+2=9+2→f(2)=11
rf=29-11→rf=18
rh=h(3)-h(2)
h(3)=2(3)^2+5(3)+2=2(9)+15+2=18+15+2→h(3)=35
h(2)=2(2)^2+5(2)+2=2(4)+10+2=8+10+2→h(2)=20
rh=35-20→rh=15
rg=20>18=rf
rg=20>15=rh
6) No
x=4→g(4)=20(4)+4=80+4→g(4)=84
x=4→h(4)=2(4)^2+5(4)+2=2(16)+20+2=32+20+2→h(4)=54
x=4→f(4)=3^4+2=81+2→f(4)=83>54=h(4)
f(4)=83<84=g(4)
An equation is:
2(x+3)-7=9
Distribute the 2
2(x)+2(3) -7=9
2x+6-7=9
Combine like terms
2x-1=9
Add one to both sides
2x-1+1=9+1
2x=10
Divide both sides by two to isolate the variable.
2x/2=10/2
x=m
Step-by-step explanation:
To solve a system of equations, we can add the two equations and solve for one of the remaining variables -- let's try to eliminate the 
variable when we add the two equations together.
Right now, there's a 
term in the first equation, and a 
term in the second equation, so if we add those together, we'll be able to eliminate the variable altogether and solve for 
.
However, when we also have a 
term in the first equation and 
term in the second equation, so adding these together will also eliminate the term, leaving a 
on the left-hand side of the equation.
If we add the two numbers on the right side of the equation, we get 
, which does not equal , meaning there are no solutions to this system of equations.
1.5-6-12n. 1-12n
2.-4+40x-35. -39+40x
3.-36x+6+6. 12-36x
4.40n-32+8n. 48n-32
5.7n-28-6. 34-7n
6.-16+20x+3x 23x-16
7. -2n-16+8. -2n-8
8.25b-25-1. 25b-26
9.-4n+15-6n. -10n+15
-1-3x+6. -3x+5
