Usually I would use my head, but you can use whatever you want.
Since g(6)=6, and both functions are continuous, we have:
![\lim_{x \to 6} [3f(x)+f(x)g(x)] = 45\\\\\lim_{x \to 6} [3f(x)+6f(x)] = 45\\\\lim_{x \to 6} [9f(x)] = 45\\\\9\cdot lim_{x \to 6} f(x) = 45\\\\lim_{x \to 6} f(x)=5](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%206%7D%20%5B3f%28x%29%2Bf%28x%29g%28x%29%5D%20%3D%2045%5C%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20%5B3f%28x%29%2B6f%28x%29%5D%20%3D%2045%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20%5B9f%28x%29%5D%20%3D%2045%5C%5C%5C%5C9%5Ccdot%20lim_%7Bx%20%5Cto%206%7D%20f%28x%29%20%3D%2045%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20f%28x%29%3D5)
if a function is continuous at a point c, then

,
that is, in a c ∈ a continuous interval, f(c) and the limit of f as x approaches c are the same.
Thus, since

, f(6) = 5
Answer: 5
Answer:
12
Step-by-step explanation:
-2f-3=7
+2f-3=7+2
f-3=9
f+3=9+3
f=12
i think
Answer:
2
Step-by-step explanation:
slope=rise/run=20/10=2/1=2.
Answer:
t = 1 second
Step-by-step explanation:
The height of a golf ball, in meters, x seconds after it has been hit is given by the expression.

We need to find the time when the ball will hit the ground.
When it hits the ground, h=0

It is a quadratic equation. The solution of the above equation is :
x = 1 seconds, x = -0.02 seconds
Hence, after 1 second the ball will hit the ground.