H(1)=-6 h(2)= 20 h(n)= h(n-2)+h(n-1) What is the value of h(3)?
1 answer:
You might be interested in
Answer:
The time duration of the two pipes restricted flow before the flow became normal is 4.5 minutes
Step-by-step explanation:
The given information are;
The time duration for the volume, V, of the basin to be filled by one of the pipe, A, = 12 minutes
The time duration for the volume, V, of the basin to be filled by the other pipe, B, = 16 minutes
Therefore, the flow rate of pipe A = V/12
The flow rate of pipe B = V/16
Due to the restriction, we have;
The proportion of its carrying capacity the first pipe, A, carries = 7/8 of the carrying capacity
The proportion of its carrying capacity the second pipe, B, carries = 5/6 of the carrying capacity
Whereby the tank is filled 3 minutes after the restriction is removed, we have;
Simplifying gives;
2·t + 7 = 16
t = (16 - 7)/2 = 4.5 minutes
Therefore, it took 4.5 seconds of the restricted flow before the the flow of water in the two pipes became normal