Answer:
dy/dt = - (1/5) ft/s = - 0.2 ft/s
Step-by-step explanation:
Given
L = 5 ft
Qin = 25 ft³/s
Qout = 30 ft³/s
h = 10 ft
dy/dt = ?
We can apply the relation
ΔQ = Qint - Qout = 25 ft³/s - 30 ft³/s
⇒ ΔQ = - 5 ft³/s
Then we use the formula
Q = v*A
where Q = ΔQ, A = L² is the area of square base and v = dy/dt is the rate of change in the depth of the solution in the tank
⇒ ΔQ = (dy/dt)*L²
⇒ dy/dt = ΔQ/L²
⇒ dy/dt = (- 5 ft³/s)/(5 ft)²
⇒ dy/dt = - (1/5) ft/s = - 0.2 ft/s
Answer:
-2, 4. -2 is x and 4 is y
Step-by-step explanation:
hope this helps!
Answer:
465371295673
Step-by-step explanation:
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