algebra word problem - help!!!?

Question

A bus leaves a station at 1 P.M., traveling west at an average
rate of 44 mi/h. One hour later a second bus leaves the same station, traveling
east at a rate of 48 mi/h. At what time will the two buses be 274 mi apart?

thank you to everyone who has helped me understand this better. i really do appreciate it.

Answer 1

274 = 44 mph * t + 48 mph * (t-1)

notice the first bus travels for t hours, while the second bus travels for (t -1) hour or one hour less because it left an hour later.

274 = 44t + 48t - 48
322 = 92 t
t = 3.5 hours

= so at 4:30 PM, aka 3.5 hours after 1 PM

Answer 2

Slightly modify the problem to indicate that when the second bus leaves, the first bus is 44 miles distant. Since the buses are moving apart at 92 mi/hr, if T is the time the SECOND BUS has traveled, 44+92*T=274. Solve for T, which is the additional time after 2PM.

Answer 3

After one hour (2:00), they are 44 mi apart, and they separate 92 mi/h every hour after that. So D, the distance between them can be written:

D = 44 + 92(t - 1) [t = hours after 1:00]
274 mi = 44 + 92(t - 1) >>
t = 3.5 hours
So they are 274 mi apart at 4:30 pm

Answer 4

after {1+ [274 -- 44]/[44 + 48]} hrs = 3.5 hrs
i.e. at (1 + 3.5) PM = 4.30 PM

Answer 5

If the scholars contemporary grade is a ninety% then we’re of direction assuming it’s out of a hundred% ninety/a hundred Now because of the fact the scholars grade more desirable 20% from the final grading era then you certainly upload that to the previous a hundred% providing you with ninety/ a hundred and twenty = .seventy 5 and because grades are in possibilities you multiply .seventy 5 via a hundred providing you with seventy 5% because of the fact the scholars previous grade