Answer:
whats the question lol ?
Step-by-step explanation:
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
f(x) + g(x) = 3x³ + 8x - 24
that is 3x³ + 7x - 26 + x + 2 = 3x³ + 8x - 24
Answer:
(14400x^4)^8
Step-by-step explanation:
sqrt10 to the 4th is 10^2 or 100, and xsqrt12 to the 4th is 144x^4. Multiply these to get (14400x^4), and take the 8th to get (14400x^4)^8.
2 116.666/275
hope it helps!
