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
Answer:
The last option
Step-by-step explanation:
The last option
Answer:
Oh hi lauryn
Step-by-step explanation:
Fancy seeing you here
Answer:
26/5
Step-by-step explanation:
(2)(2)(4)/5 −8+10
(4)(4)/5−8+10
=16/5−8+10
=−24/5+10
=26/5
Alternate form
5.2
Answer:
Step-by-step explanation:
Given the system of equations:
2x + 3y = 15
X + 3y = 0
What is x?
2x + 3y = 15
X + 3y = 0 subtract the bottom equation fron the top one
2x - x + 3y - 3y = 15 - 0 see how the 3y - 3y equates to zero
2x - x = 15
x = 15 Now for extra credit what is y?
X + 3y = 0
15 + 3y = 0
15 - 15 + 3y = 0 - 15
3y = -15
3y/3 = -15/3
y = -5
-15
-5
0
5
15
