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3 years ago

A system of linear equations is shown below.

Mathematics

1 answer:

yulyashka [42]3 years ago

3 0

Answer: No solution

Step-by-step explanation:

2m+4p=22

6m+12p=66

First, to choose one variable to simplified, I choose to find 1st equations in m form.

2m+4p=22

2m=22-4p [Subtract 4p from both side]

m=22/2 - 4/2p [Divide 2 on both side to find m]

m= 11 - 2p

6m+12p=66

6(11-2p)+12p=66 [Plug in the 11-2p in to m,according to the 1st equation]

66-12p+12p=66

The -12p has cancel with 12p, so it means that there is no solution

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A brine solution of salt flows at a constant rate of 4 L/min into a large tank that initially held 100 L of pure water. The solu

frutty [35]

Answer:

a. $m(t) = 26.67 - 26.67e^{-0.03t}$ b. 7.44 s

Step-by-step explanation:

a. If the concentration of salt in the brine entering the tank is 0.2 kg/L, determine the mass of salt in the tank after t min.

Let m(t) be the mass of salt in the tank at any time, t.

Now, since a brine solution flows in at a rate of 4 L/min and has a concentration of 0.2 kg/L, the mass flowing in per minute is m' = 4 L/min × 0.2 kg/L = 0.8 kg/min

Now, the concentration in the tank of volume 100 L at any time, t is m(t)/100 L. Since water flows out at a rate of 3 L/min, the mass flowing out per minute is

m(t)/100 × 3 L/min = 3m(t)/100 kg/min

Now the net rate of change of mass of salt in the tank per minute dm/dt = mass flowing in -mass flowing out

dm/dt = 0.8 kg/min - 3m(t)/100 kg/min

So, dm/dt = 0.8 - 0.03m(t)

The initial mass of salt entering m(0) = 0 kg

dm/dt = 0.8 - 0.03m(t)

separating the variables, we have

dm/[0.8 - 0.03m(t)] = dt

Integrating, we have

∫dm/[0.8 - 0.03m(t)] = ∫dt

-0.03/-0.03 × ∫dm/[0.8 - 0.03m(t)] = ∫dt

1/(-0.03)∫-0.03dm/[0.8 - 0.03m(t)] = ∫dt

-1/0.03㏑[0.8 - 0.03m(t)] = t + C

㏑[0.8 - 0.03m(t)] = -0.03t - 0.03C

㏑[0.8 - 0.03m(t)] = -0.03t + C' (C'= -0.03C)

taking exponents of both sides, we have

$0.8 - 0.03m(t) = e^{-0.03t + C'} \\0.8 - 0.03m(t) = e^{-0.03t}e^{C'}\\0.8 - 0.03m(t) = Ae^{-0.03t} A = e^{C'}\\0.03m(t) = 0.8 - Ae^{-0.03t}\\m(t) = 26.67 - \frac{A}{0.03} e^{-0.03t}\\when t = 0 \\m(0) = 0\\m(0) = 26.67 - \frac{A}{0.03} e^{-0.03(0)}\\\\0 = 26.67 - \frac{A}{0.03} e^{0}\\26.67 = \frac{A}{0.03} \\\frac{A}{0.03} = 26.67\\\frac{A}{0.03} = 6.67\\A = 26.67 X 0.03\\A = 0.8\\m(t) = 26.67 - \frac{A}{0.03} e^{-0.03t}\\\\m(t) = 26.67 - \frac{0.8}{0.03} e^{-0.03t}\\$

So, the mass of the salt after t min is

$m(t) = 26.67 - 26.67e^{-0.03t}$

b. When will the concentration of salt in the tank reach 0.1 kg/L?

When the concentration of the salt reaches 0.1 kg/L, m(t) = 0.1 kg/L

Solving the equation for t,

$m(t) = 26.67 - 6.67e^{-0.03t}\\0.1 = 26.67 - 26.67e^{-0.03t}\\26.67e^{-0.03t} = 26.67 - 0.1\\26.67e^{-0.03t} = 26.57\\e^{-0.03t} = 26.56/26.67\\e^{-0.03t} = 0.9963\\$

taking natural logarithm of both sides, we have

-0.03t = ㏑0.9963

-0.03t = -0.0038

t = -0.0038/-0.03

t = 0.124 min

t = 0.124 × 60 s

t = 7.44 s

5 0

3 years ago