Umm can you be more specific on your question so I can help you need to explain what you need help on
Answer:
1/2
((-2)^2-(4*2)^1/3)/abs(-2*2)
-2^2 = 4
4*2 = 8
-2*2 = -4 and abs of that is 4
4-(8)^1/3/4
8^1/3 = 2
4-2 = 2
2/4 = 1/2
Step-by-step explanation:
Answer:
2
Step-by-step explana
(6x-y+4)*2= 12x -2y+8= (3x+9x)-(2y-8)
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:
12
Step-by-step explanation:
h(n)=3n
h(4)
1. Subsitute 4 for n
h(4)=3*4
h(4)=12
