Question

Starting from 130 feet away, a person on bicycle rides towards a checkpoint and then passes it. The rider is traveling at a constant rate of 25 feet per second. The distance between the bicycle and the checkpoint is given by the equationd equals absolute value of the quantity 130 minus 25 t. At what times is the bike 15 feet away from the checkpoint?
(1 point)
Responses

4.6 sec and 9.2 sec
4.6 sec and 9.2 sec

2.9 sec and 5.8 sec
2.9 sec and 5.8 sec

4.6 sec and 5.8 sec
4.6 sec and 5.8 sec

2.9 sec and 3.3 sec

Answer 1

suppose R={1,3,5,7,9,11,13,15,17} and D={3,6,9,12,15,18,21,24,27}, what is R◠D?

Answer 2

To find the times when the bike is 15 feet away from the checkpoint, we need to solve the equation:

|130 - 25t| = 15

Since we are taking the absolute value, we can have two cases:

Case 1: 130 - 25t = 15
Solving this equation, we get: t = 4.6 seconds

Case 2: 130 - 25t = -15
Solving this equation, we get: t = 5.8 seconds

Therefore, the correct answer is: 4.6 seconds and 5.8 seconds.