Answer:
put a closed for on -1 and go to the left with an arrow
B. Not moving, is the correct answer.
A is not correct because constant speed on a distance vs. time graph would vertically point to the right. C is not correct because acceleration on a distance vs. time graph would curve upwards. D is not correct because B applies to the graph.
Hope this helps :)
Answer:
yea
Step-by-step explanation:
cuz then there would be a graph and stuff and you would analyze which student spends more time
Answer:
The answer to this question is A >
Step-by-step explanation:
I got it right on edge 2020 :)
Answer:
Here we have the function:
S(t) = 500 - 400*t^(-1)
Then the rate of change at the value t, will be:
S'(t) = dS(t)/dt
This differentiation will be:
S'(t) = -400/t^2
Then:
a) the rate of change at t = 1 is:
S'(1) = -400/1^2 = -400
The rate of change after one year is -400
b) t = 10
S'(10) = -400/10^2 = -400/100 = -4
The rate of change after 10 years is -4, it reduced as the years passed, as expected.
