Find the average :8.4+7.5+12.9+10.6 divided by four
3/4 - 1/5
0.75 - 0.2
0.55
Answer: 0.55 or 11/20
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.
Answer:
1. x + 4 = 9
Hint: the word 'sum' generally refers to addition.
2. 10<em>a</em> = 70
3. 
= 15
4. 
- 4 = 4
