We have to assume that the speed before being stuck was sufficient to get to the destination on time had there been no delay. Call that speed "s" in km/h.
Since 200 km is "halfway", the total distance must be 400 km.
time = distance / speed
total time = (time for first half) + (delay) + (time for second half)
400/s = 200/s + 1 + 200/(s+10) . . . .times are in hours, distances in km
200/s = 1 + 200/(s+10) . . . . . . . . . . subtract 200/s
200(s+10) = s(s+10) +200s . . . . . . .multiply by s(s+10)
0 = s² +10s - 2000 . . . . . . . . . . . . . .subtract the left side
(s+50)(s-40) = 0
Solutions are s = -50, s = 40
The speed of the bus before the traffic holdup was 40 km/h.
It would be the 2nd one down and 3rd one up is the correct answer just look a little closer and you will see what i mean there ages are and and they have a 3o sec head start and they ran 60 yards in 30 sec.
Answer:
ok
Step-by-step explanation: i explain wait for sometime
In this case, you just multiply the expression under the root that gave the greatest integer (as you pull the square root of that number). See how I did: 4 multiplied by 31, which gives 124 then pull out the square root of 4, because 31 can not be, because it is a prime number and root would be an irrational number.
