1. I can no see your graph but will assume that velocity remains 4 m/s from t = 3 to t = 8 seconds. In that case, the object moves 4 m/s *5 s = 20 m farther during that interval. That would put it at 28 m from the "point from which measurements are made", at t = 8 s.
2. y(x) is a monotonically decreasing function over that interval.
0.90^0 = 1
0.90^6 = 0.531441
y will be at its highest value at x = 0 and at its lowest value at x = 6.
How do you do this question?
1. A speed versus time graph shows the segment from (3, 4) to (8,4). The units for the vertical axis are m/s. If the object whose speed was being graphed is 8 m away from the point from which measurements are made at t= 3s, how far away is the object at t = 8s?
2. Determine the maximum and minimum values for y = 250(0.90)^x for 0<x<6
1 answer
1 answer