Answer:
66
Step-by-step explanation:
Answer:
Q 1 is option B and q 2 is option D
Step-by-step explanation:
Answer: {-16,4,92}. 2) {-16,10 ,42}. 3) {0,10,42}. 4) {0,4,92}. 5 The function f(x) = 2x. 2 + 6x - 12 has a domain.
Step-by-step explanation:
Answer:
the equation of the line in slope-intercept form would be y=5/2x-5
Step-by-step explanation:
hope this helped!! :D
Answer:
-60
Step-by-step explanation:
The objective is to state whether or not the following limit exists
.
First, we simplify the expression in the numerator of the fraction.

Now, we obtain

and the fraction is transformed into

Therefore, the following limit is

You can plug in
in the equation, hence
