Mathematics, 26.03.2021 05:40 rstredway7693

A tank contains 350 liters of fluid in which 40 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate. Let the A(t) be the number of grams of salt in the tank at time t. Find the rate at which the number of grams of salt in the tank is changing at time t.

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Answer from: brettriley4402

$\frac{dA}{A} = -\frac{1}{70} dt$

A(t) = $40 e^{-\frac{t}{70}}$

Step-by-step explanation:

Given - A tank contains 350 liters of fluid in which 40 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate. Let the A(t) be the number of grams of salt in the tank at time t.

To find - Find the rate at which the number of grams of salt in the tank is changing at time t.

Find the number A(t) of grams of salt in the tank at time t.

Proof -

Given that,

Total fluid in the tank = 350 litres

Total salt = 40 grams

Rate in flow = 5 L/min

Rate out flow = 5 L/min

Concentration in = 0

Concentration out = (Amount of salt at time t )/ (Solution in tank at time t)

Now,

Total solution in the tank at time t = Total fluid in the tank + (Rate in flow - Rate out flow )t

= 350 + (5 - 5)t

= 350

⇒Solution in the tank at time t = 350

⇒Concentration out = $\frac{A(t)}{350}$

Now,

Rate of change in the tank at time t = $\frac{dA}{dt}$

= (Rate in flow )( Concentration In) - (Rate out flow )( Concentration out)

= 5(0) - 5( $\frac{A(t)}{350}$ )

= 0 - $\frac{A(t)}{70}$

= $-\frac{A(t)}{70}$

⇒ $\frac{dA}{dt}$ = $-\frac{A(t)}{70}$

⇒ $\frac{dA}{A} = -\frac{1}{70} dt$

Integrate both sides , we get

ln(A) = $-\frac{1}{70}t + C$

⇒A(t) = $e^{-\frac{t}{70}+ C}$

= $e^{-\frac{t}{70}}*e^{C}$

= $C_{1} e^{-\frac{t}{70}}$

⇒A(t) = $C_{1} e^{-\frac{t}{70}}$

Now,

Given that,

Initially, the tank has 40 grams of salt

⇒A(0) = 40

⇒C₁ = 40

⇒A(t) = $40 e^{-\frac{t}{70}}$

Answer from: Quest

Idont knowgfdhdhfhggg.

Answer from: Quest

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in $\delta bea$ and $\delta dea$ ;

$ba=da$ , as these are radii of our given circle.

$ea=ea$

therefore, $\delta bea\cong \delta dea$ by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

$3x-24=x+10$

$3x-x=24+10$

$2x=34$

$x=\frac{34}{2}$

$x=17$

now let us substitute x=17 in the expression for the length of be.

$\text{length of be}=3x-24$

$\text{length of be}=3\cdot 17-24$

$\text{length of be}=51-24$

$\text{length of be}=27$

therefore, length of be will be 27 units and option c is the correct choice.

In circle a, ∠bae ≅ ∠dae. what is the length of be? 14 units 17 units 27 units 34 units

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