The initial equation is
. The subsequent equation is
. By setting both equations equal, we can derive the following:
. Next, simplify:
. Then factor:
. Put x=0 or x=1 into the second equation to yield:
. Alternatively,
. Thus, the solutions are:
.
Answer:
Let the train's speed be denoted as x km/h.
Scenario 1:
Distance = 288 km
Speed = x km/h
Time = Distance divided by Speed
= 288/x hours
Scenario 2:
Distance = 288 km
Speed = (x + 4) km/h
Time = 288/(x + 4) hours
As 288/x is greater than 288/(x + 4)
288/x - 288/(x + 4) = 1
288[1/x - 1/(x + 4)] = 1
[x + 4 - x] / [x(x + 4)] = 1/288
[4 / (x^2 + 4x)] = 1/288
x² + 4x = 1152
x² + 4x - 1152 = 0
x² + 36x - 32x - 1152 = 0
x(x + 36) - 32(x + 36) = 0
(x + 36)(x - 32) = 0
x + 36 = 0, x - 32 = 0
x = -36, x = 32
x = -36 is not valid as speed cannot be negative.
Conclusively, the train's speed = 32 km/h
Answer:
23(6,8)(3,3)(13-3)(-3-)
Step-by-step explanation:
3-6=8-3
+13-3=
23
