I assume the answer is 180 but u have not put a question
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:
B(x) is the better option
Step-by-step explanation:
Let's say that these functions represent the pay f(x) in hours, x. Function B would increase linearly on a greater scale because each X is being multiplied by 20.
You'll notice this trend for larger numbers, such as 10.
A(x) = 5(10) + 20 = 70
B(x) = 20(10) + 5 = 205
Therefore, your answer is B.
