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garri49 [273]
3 years ago
6

Help -------------------------------

Mathematics
2 answers:
NeTakaya3 years ago
7 0

Answer:

E + F = 2c² + 5c + 4

Step-by-step explanation:

E = 6c^2 - 2c - 1 \\\\F = -4c^2 + 7c +5\\\\E + F = (6c^2 - 2 c -1) + ( -4c^2 + 7c +5)

         =(6c^2 -4c^2) +(-2c + 7c) + (-1 + 5)      [\ arranging \ terms\ ]

         = 2c^2 + 5c + 4

8090 [49]3 years ago
6 0

___________________________________

<h2>Let's Solve it!</h2>

<h3>Given that:</h3>

\quad\quad\quad\quad\tt{E = 6 {c}^{2}  - 2c - 1}

\quad\quad\quad\quad\tt{ F =  - 4 {c}^{2}   + 7c  + 5}

<h3>Formula:</h3>

\quad\quad\quad\quad\tt{E + F = ?}

<h3>Solution:</h3>

\quad\quad\quad\tt{(6 {c}^{2}  - 2c - 1) + ( - 4 {c}^{2}  + 7c + 5)}

<h3>Step by step:</h3>

\quad\quad\quad\quad\tt{  ⟶(6 {c}^{2} )  +  ( - 4 {c}^{2}) =  \boxed{ \pink{2 {c}^{2}  }}}

\quad\quad\quad\quad\tt{ ⟶ ( - 2c)  +  ( 7c) =  \boxed{ \pink{5c  }}}

\quad\quad\quad\quad\tt{ ⟶ (  - 1)  +  ( 5) =  \boxed{ \pink{4 }}}

<h3>Now, it will look like this:</h3>

\quad\quad\quad\quad\tt{E + F =2 {c}^{2}   + 5c + 4 }

<h2>So, E + F is equal to:</h2>

\quad\quad\quad\quad\tt{\boxed{ \boxed{ \color{magenta}{ 2 {c}^{2}   + 5c + 4 }}}}

___________________________________

#CarryOnLearning

✍︎ C.Rose❀

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Initially 15 grams of salt are dissolved into 25 liters of water. Brine with concentration of salt 4 grams per liter is added at
Alla [95]

Answer:

a) dx/dt = 600 - 6x

b) x = 100 - 4.12((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

c) The mass of salt in the tank attains the value of 20 g at time, t = 0.227 min = 13.62 s

Explanation:

Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out

Let the concentration of salt in the tank at anytime be C

Let the concentration of salt entering the tank be Cᵢ

Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)

Let the flowrate in be represented by Fᵢ

Let the flowrate out = F₀ = F

Fᵢ = F₀ = F = 6 L/min

a) Then the component balance for the salt

Rate of accumulation = rate of flow into the tank - rate of flow out of the tank

dx/dt = Fᵢxᵢ - Fx

Fᵢ = 6 L/min, C = 4 g/L, F = 6 L/min

dC/dt = 24 - 6C

dx/dt = 25 (dC/dt), (dC/dt) = (1/25) (dx/dt) and C = x/25

(1/25)(dx/dt) = 24 - (6/25)x

dx/dt = 600 - 6x

b) dC/dt = 24 - 6C

dC/(24 - 6C) = dt

∫ dC/(24 - 6C) = ∫ dt

(-1/6) In (24 - C) = t + k (k = constant of integration)

In (24 - 6C) = -6t - 6k

-6k = K

In (24 - 6C) = K - 6t

At t = 0, C = 15 g/25 L = 0.6 g/L

In (24 - 6(0.6)) = K

In 20.4 = K

K = 3.02

So, the equation describing concentration of salt at anytime in the tank is

In (24 - 6C) = K - 6t

In (24 - 6C) = 3.02 - 6t

24 - 6C = e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾

6C = 24 - (⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)

C = 4 - ((e⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)/6

C = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

But C = x/25

x/25 = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

x = 100 - 4.12((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

c) when x = 20 g

20 = 100 - 4.12(e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

80 = (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

- (6t - 3.02) = In 80

- (6t - 3.02) = 4.382

(6t - 3.02) = -4.382

6t = -4.382 + 3.02

t = 1.362/6 = 0.227 min = 13.62 s

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