A water tank is box shaped with a square base of side length 5 ft. The tank is fed by a pipe above and drains from its base. In order to have a well mixed solution, both the feeder pipe and the drain must be opened. The tank is being filled at a rate of 25 ft3/s when the height of the liquid in the tank is 10 ft. If the drains are pulling 30 ft3/s, what is the rate of change in the depth of the solution in the tank

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

whole number data type

step-by-step explanation:

as whole number type represents whole number

like 1, 2 ,3, 4

in other words the whole numbers are those numbers which cannot be written infraction.

since we can take days as a whole number i.e 1 day 2 day and we cannot write 1.2 day hence we will use whole number data type

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.