Answer:
The average acceleration of train is 
Step-by-step explanation:
We have,
A train that slows down from 15 m/s to 8.6 m/s in 1.2 s. It means that 15 m/s is its initial velocity and 8.6 m/s is its final velocity.
It is required to find the average acceleration of the train.

The average acceleration of train is
. Negative signs shows that the train is decelerating.
The 50 is the error because cents are represented with decimals so it has to be .50
So let the smaller integer be 2x+1 and the larger one be 2x+3. So 2x+1+2x+3 is 32. Simplifying, we get 4x+4=32. So subtracting 2 from both sides we get 4x+2=30. So dividing by 2 we get 2x+1=15. So 2x+3 is 17. So the numbers are 15 and 17
Circle A -- center(2, 0), radius 8 Circle A' -- center(-1, 5), radius 3
That equation has one solution