At 3:30 PM, Berto‘s train was 34 miles past the egg farm, traveling in an average speed of 85 mph. At the same time on a nearby track, Edwardos waters train was traveling at an average speed of 110 mph and had 12 miles to go before it reached the egg farm. To the nearest hundredth of an hour, after how much time will the trains meet up?

1.84 hours

Step-by-step explanation:

Let our egg farm be position 0 miles.

Since Berto's train is at 34 miles away from the egg farm at 3:30 PM and moves at an average speed of 85 mph, and his distance is increasing by 85t in time, t. So, his distance, d from the egg farm at time, t is d = 34 + 85t.

Also, Edwardos waters train is approaching the egg farm and is a distance of 12 miles from the egg farm moving at an average speed of 110 mph towards the egg farm, and his distance is increasing by 110t in time, t. So, his distance d' from the egg farm at time, t is d' = -12 + 110t. There is a negative in front of the 12 since it is a position before the egg farm

Now, when the trains meet, their distances are equal. So,

d = d'

34 + 85t = -12 + 110t

collecting like terms, we have

34 + 12 = 110t - 85t

46 = 25t

t = 1.84 h

So, the trains will meet after 1.84 hours

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents