1) f(x)=2x+6
f(2)=2(2)+6
=4+6
=10
2)f(x)=3x
f(a+1)=3(a+1)
=3a+3
3)f(x)=3x-1 and g(x)=5x+3
f(2)=3(2)-1
f(2)=6-1
=5
g(3)=5x+3
=5(3)+32
=15+3
=18
f(2)+f(3)=5+18
=23.
I hope someone solves your problem :)
Answer:
- vertical asymptote: x = 7
- slant asymptote: y = x+9
Step-by-step explanation:
The vertical asymptotes are found where a denominator factor is zero (and there is no corresponding numerator factor to cancel it). Here, that is at x = 7.
There is no horizontal asymptote because the numerator is of higher degree than the denominator.
When you divide the numerator by the denominator, you get ...
y = (x +9) +60/(x -7)
Then when x gets large, the behavior is governed by the terms not having a denominator: y = x +9. This is the equation of the slant asymptote.
Answer:
16.9
Step-by-step explanation:
We can use a proportion also in this case
51.6 : 23 = 38 : NO
NO = (38 x 23)/51.6 = 16.9
Answer:
180 is the answer you multiple 6 times 30