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Your answer is
What is 4 yards 1 foot and 3 inches minus 2 yards 1 foot and 9 inches?
1 answer:
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Answer:
The total cost is $3,504.63
Step-by-step explanation:
Length of fence = 90 feet
Breadth of fence = 175 feet
Height of fence = 6 feet, 6 inches
1 inch = 0.0833333 feet
6 inches = 6*0.0833333 =0.4999998 feet
So, Height of fence = 6 feet + 0.49 feet =6.49 feet ≈ 6.5 feet
Perimeter =
Cost of erecting the fence is $1.25 per linear foot
So, Cost of erecting the fence in 530 feet =
Total surface area =
The cost of materials is $0.825 per square foot of fence
Cost of material for 3445 sq.feet =
So, total cost = 2842.125+662.5 = 3504.63
Hence The total cost is $3,504.63
Answer:
a) (dy/dt) = 2 - [3y/(100 + 2t)]
b) The solved differential equation gives
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
c) Concentration of salt in the tank after 20 minutes = 0.2275 kg/L
Step-by-step explanation:
First of, we take the overall balance for the system,
Let V = volume of solution in the tank at any time
The rate of change of the volume of solution in the tank = (Rate of flow into the tank) - (Rate of flow out of the tank)
The rate of change of the volume of solution = dV/dt
Rate of flow into the tank = Fᵢ = 5 L/min
Rate of flow out of the tank = F = 3 L/min
(dV/dt) = Fᵢ - F
(dV/dt) = (Fᵢ - F)
dV = (Fᵢ - F) dt
∫ dV = ∫ (Fᵢ - F) dt
Integrating the left hand side from 100 litres (initial volume) to V and the right hand side from 0 to t
V - 100 = (Fᵢ - F)t
V = 100 + (5 - 3)t
V = 100 + (2) t
V = (100 + 2t) L
Component balance for the amount of salt in the tank.
Let the initial amount of salt in the tank be y₀ = 0 kg
Let the rate of flow of the amount of salt coming into the tank = yᵢ = 0.4 kg/L × 5 L/min = 2 kg/min
Amount of salt in the tank, at any time = y kg
Concentration of salt in the tank at any time = (y/V) kg/L
Recall that V is the volume of water in the tank. V = 100 + 2t
Rate at which that amount of salt is leaving the tank = 3 L/min × (y/V) kg/L = (3y/V) kg/min
Rate of Change in the amount of salt in the tank = (Rate of flow of salt into the tank) - (Rate of flow of salt out of the tank)
(dy/dt) = 2 - (3y/V)
(dy/dt) = 2 - [3y/(100 + 2t)]
To solve this differential equation, it is done in the attached image to this question.
The solution of the differential equation is
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
c) Concentration after 20 minutes.
After 20 minutes, volume of water in tank will be
V(t) = 100 + 2t
V(20) = 100 + 2(20) = 140 L
Amount of salt in the tank after 20 minutes gives
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
y(20) = 0.4 [100 + 2(20)] - 40000 [100 + 2(20)]⁻¹•⁵
y(20) = 0.4 [100 + 40] - 40000 [100 + 40]⁻¹•⁵
y(20) = 0.4 [140] - 40000 [140]⁻¹•⁵
y(20) = 56 - 24.15 = 31.85 kg
Amount of salt in the tank after 20 minutes = 31.85 kg
Volume of water in the tank after 20 minutes = 140 L
Concentration of salt in the tank after 20 minutes = (31.85/140) = 0.2275 kg/L
Hope this Helps!!!
