F(x)=-4x-1 if you sub 3 into it you get 3(-4)-1 which is -13 and 2x+4=10
Answer:
16.7
Step-by-step explanation:
We can use the Pythagorean theorem, a^2 + b^2 = c^2.
a = 11
c = 20
11^2 + b^2 = 20^2
121 + b^2 = 400
b^2 = 279
b = sqrt(279) = 16.7
How many different combinations of “on” and “off” are possible with 8 “light bulbs”?
Answer: The total number of combinations of "on" and "off" that are possible with 8 "light bulbs" is 
Therefore, 256 different combinations of “on” and “off” are possible with 8 “light bulbs”
Ok here is a graph
Hope this helps and don’t forget to mark as brainliest if you thought it was most helpful :)
It’s c I know the answer because I took the test