X+y=57
y=x+15
x+x+15=57
2x=42
x=21
y=21+15
y=36
C=8πab
Divide both sides by 8πb
c/(8πb) = a
increase in x results in the decrease in y
Let's define variables:
s = original speed
s + 12 = faster speed
The time for the half of the route is:
60 / s
The time for the second half of the route is:
60 / (s + 12)
The equation for the time of the trip is:
60 / s + 60 / (s + 12) + 1/6 = 120 / s
Where,
1/6: held up for 10 minutes (in hours).
Rewriting the equation we have:
6s (60) + s (s + 12) = 60 * 6 (s + 12)
360s + s ^ 2 + 12s = 360s + 4320
s ^ 2 + 12s = 4320
s ^ 2 + 12s - 4320 = 0
We factor the equation:
(s + 72) (s-60) = 0
We take the positive root so that the problem makes physical sense.
s = 60 Km / h
Answer:
The original speed of the train before it was held up is:
s = 60 Km / h
No, the graphs of the lines are not parallel because if you simplify 27x-3y=-81, you get y=9x+27, and in order to have the lines parallel, both equations must have the same "m" in y=mx+b. 6 and 9 do not equal, so no they are not parallel.
