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:
what do you mean
Step-by-step explanation:
what do you mean
I cant answer if i dont know what math would you mind telling me what math
X/2+1/3=x/3+1/2
x/2=x/3+1/2-1/3
x/2 - x/3 = 3/6 - 2/6
x/2 - x/3 = 1/6
3x/6 - 2x/6 = 1/6
x/6 = 1/6
x = 1/6 * 6
x = 1
7/12-9/8(2x-1)-5/6(3x+2)=0
(7*2-9(2x-1)*3-5(3x+2)*4)/24=0
(14-9(2x-1)*3-5(3x+2)*4)/24=0
(14-27(2x-1)-20(3x+2)/24=0
(14-54x+27-60x-40)/24=0
(14+(-54x-60x)+27-40)/24=0
(14-114x+27-40)/24
(-144x+1)/24=0
-114x+1=0
-114x=-1
x=1/144
