Atrain traveling 50 mph left a station 30 minutes before a second train running at 55 mph. how soon did the second train overtake the first? if 50x represents the distance the slower train travels, then the faster train travels55(x - 0.5)55(x + 0.5)50 + 30x

Disatnce=speed times time

x is time in hours
the other train will travel with half an hour less
so the faster train is 55(x-0.5)
when will they travel the same distance?

50x=55(x-0.5)
50x=55x-27.5
minus 50x both sides
0=5x-27.5
add 27.5 both sides
27.5=5x
divide both sides by 5
5.5=x

after 5.5 hours or 5 hours and 30 mins

and the faster train is 55(x-0.5)

1st train DATA:
rate = 50 mph ; time = x hrs ; distance = rt = 50x miles

2nd train DATA:
rate = 55 mph ; time = (x-(1/2)) hrs ; distance = rt = 55(x-(1/2)) miles

Equation:
distance = distance 
50x = 55(x-(1/2)
50x = 55x - (55/2)
-5x = -55/2
x = 11/2
x = 5 1/2 hrs (time at which the 2nd train overtakes the 1st train)
Hope this helps!

360 - 120 - 150 = 90

Interior Angle 1 = 180 - 120 = 60°

Interior Angle 2 = 180 - 150 = 30°

Interior Angle 3 = 180 - 90 = 90°