Thato walks uphill at a rate of 2.5 mi/h from the base of a 6-mi trail. at the same time, edwin walks downhill at a rate of 3.5 mi/h from the top of the trail. how many hours pass before they meet on the trail?

1 hour

Explanation:

speed of Thato (St) = 2.5 mi/h

speed of Edwin (Se) = 3.5 mi/h

distance (d) = 6 mi

How many hours pass before they meet on the trail?

lets assume that they both meet when the time = T hoursAt time = T hours, the  distance covered by Thato = speed x time

        = 2.5 x T = 2.5T

At time = T hours, the  distance covered by Edwin = speed x time

        = 3.5 x T = 3.5T

At the time they meet (T hours) the addition of the distance covered by both Thato and Edwin would be 6 mi (the total distance covered by both of them would be 6 mi)2.5T + 3.5T = 6

      6T = 6

       T = 1 hour

The masses of the objects and their densities

Try b? sorry if i’m incorrect