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1. the answer is c.

   step-by-step explanation:

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2. child does not get an education so she can use the molla on junk

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3. [tex]c(t) = 0.044\cdot e ^{-0.066\cdot t}[/tex]

   Step-by-step explanation:

   The quantity of salt inside the tank is modelled after the Principle of Mass Conservation:

   Salt

   [tex]\dot m_{in, salt} - \dot m_{out, salt} = \frac{dm_{tank,salt}}{dt}[/tex]

   Water

   [tex]\dot m_{in,water} - \dot m_{out,water} = \frac{dm_{tank,water}}{dt}[/tex]

   Given that water is an incompressible fluid, the expression can be simplified into the following expression:

   [tex]\dot V_{in, water} - \dot V_{out,water} = \frac{dV_{tank, water}}{dt}[/tex]

   Both flows have the same rate and tank can be modelled as a steady state system.

   [tex]\dot V_{in, water} - \dot V_{out,water} = 0[/tex]

   The expression for salt concentration in the tank is:

   -[tex]-\dot V_{tank}\cdot c = V_{tank} \cdot \frac{dc}{dt}[/tex]

   After some handling, the following homogeneous first-order linear differential equation is found:

   [tex]\frac{V_{tank}}{\dot V_{tank}} \cdot \frac{dc}{dt} + c = 0[/tex]

   Where [tex]c (0) = 0.044\,\frac{lbm}{gal}[/tex]. The solution is obtained by using Laplace transforms:

   [tex]\frac{V_{tank}}{\dot V_{tank}} \cdot \left[s\cdot C(s) - c(0)\right] + C(s) = 0[/tex]

   [tex]\left(\frac{V_{tank}}{\dot V_{tank}}\cdot s + 1\right)\cdot C(s) = \frac{V_{tank}}{\dot V_{tank}}\cdot c(0)[/tex]

   [tex]C(s) = \frac{\frac{V_{tank}}{\dot V_{tank}}\cdot c(0) }{\left(\frac{V_{tank}}{\dot V_{tank}} \right)\cdot \left(s + \frac{\dot V_{tank}}{V_{tank}} \right)}[/tex]

   [tex]c(t) = c(0) \cdot e^{-\frac{\dot V_{tank}}{V_{tank}}\cdot t }[/tex]

   Where [tex]\frac{\dot V_{tank}}{V_{tank}} = 0.066\,min^{-1}[/tex].

   The formula for the concentration of salt in the tank is:

   [tex]c(t) = 0.044\cdot e ^{-0.066\cdot t}[/tex]

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