View an Example Mulder and Scully are driving to the same town. Mulder leaves the office at :30 a. m. averaging mph. Scully leaves at :00 a. m., following the same path and averaging mph. At what time will Scully catch up with Mulder?

Complete question :

Mulder and Scully are driving to the same town. Mulder leaves the office at 9:30 a.m. averaging 51 mph. Scully leaves at 10:00 a.m., following the same path and averaging 57 mph. At what time will Scully catch up with Mulder?

2:15 p.m

Step-by-step explanation:

Given the following :

Mulder :

Speed = 51mph

Time he left office = 9:30 a.m

Scully :

Speed = 57mph

Time she left office = 10:00 a.m

Time difference = (10:00 - 9:30)a.m = 30 minutes = half an hour = 0.5 hour

Let Scully's travel time when she catches up = s

Therefore, Mulder's time at the same moment = (s + 0.5)

Recall:

distance = speed × time

Hence,

Scully's distance : 57 * s = 57s

Mulder's distance : 51 * (s + 0.5)

Equating both :

57s = 51(s + 0.5)

57s = 51s + 25.5

57s - 51s = 25.5

6s = 25.5

s = 25.5 / 6

s = 4.25 hours

Hence scully will catch up with Mulder after 4.25 hours

10:00am + 4.25 hours = 2:15p.m

ab will be equal to

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph