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.

4 0

2 years ago

If (-3, y) lies on the graph of y = 3-x, then y = 1/27 -27 27

Vilka [71]

Answer:

y = 6

Step-by-step explanation:

[] We can plug in our x value of -3, from the coordinate point, and solve for y.

y = 3 - x

y = 3 - (-3)

y = 3 + 3

y = 6

[] We can also graph the equation, and find the point where x = -3

-> See attached

I do not see this option in the answers you have provided, so either I miss interpreted something, or something else got messed up.

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

6 0

2 years ago