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IgorLugansk [536]
3 years ago
6

P and q are the zeroes of the polynomial 2x^2+5x-1, find the value of p+q+pq

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

Answer:

  1. Sorry I didn't get your text
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What expression are equivalent to 4(4x + 9)
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16x+36

Step-by-step explanation:

4(4x + 9)

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8 0
3 years ago
A tank contains 100 L of water. A solution with a salt con- centration of 0.4 kg/L is added at a rate of 5 L/min. The solution i
Fantom [35]

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!!!

8 0
3 years ago
Help plsls I’ll brainlist i promise this one’s abit confusing tho.
slega [8]

Answer:

25

Step-by-step explanation:

Please refer to the picture;

4 0
2 years ago
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