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:
1/12 yards
Step-by-step explanation:
5×1/6 yards = 5/6 yards
5/6 yards - 3/4 yards = 10/12 yards - 9/12 yards
= 1/12 yards
10√3 (-7 - √5)
10√3 × -7 + 10√3 × -√5
-70√3 + 10√3 × -√5
-70√3 - 10√3√5
-70√3 -10√3 × 5
-70√3 - 10√15
hope that helps, God bless!
D.y = -3/5x - 2
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(f * g)(x) = 4x^2(x + 1)
4x^3 + 4x^2
