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Mathematics

1 answer:

Fittoniya [83]3 years ago

8 0

Answer:

II and III are right

Step-by-step explanation:

The reason I is wrong is the zeros are (-2, 0) and (4, 0)

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The rate of change of the mass, A, of salt at time t is proportional to the square of the mass of salt present at time t. Initia

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Answer:

(a) -dA/dt = kA², A₀ = 10

(b) A =10/(1+ kt)

Step-by-step explanation:

(a) Find the IVP

A differential equation with an initial condition y₀ = f(x₀) is called an initial value problem.

The rate of decrease of A is proportional to A², and A₀ = 10, so the IVP is

-dA/dt = kA², A₀ = 10

(b) Solve the IVP

$\begin{array}{rcl}-\dfrac{\text{d}A}{\text{d}t} & = & kA^{2}\\\\\dfrac{\text{d}A}{A^{2}} & =&-k\text{d}t\\\\-\dfrac{1}{A} & = & -kt + C\\\\\dfrac{1}{A} & = & kt + C\\\\\end{array}$

Apply the initial condition: A₀ = 10 (when t = 0)

$\dfrac{1}{10} = C\\\\\dfrac{1}{A} = kt + \dfrac{1}{10}\\\\\dfrac{10}{A} = 10kt + 1\\\\\mathbf{A} = \mathbf{\dfrac{10}{1+ kt}}$

(c) Find the time when A(t) < 1

(i) Find the value of k (A₁₀ = 4)

$\begin{array}{rcl}A& =&\dfrac{10}{1+ kt}\\\\4 & =& \dfrac{10}{1 + 10k}\\\\4 + 40 k & = & 10\\\\40k & = & 6\\k & = & 0.15\\\end{array}\\\\\mathbf{A} =\mathbf{ \dfrac{10}{0.15t + 1}}$

(ii) Find t when A < 1

$\begin{array}{rcl}A(t) & < & 1\\\dfrac{10}{0.15t + 1} & < & 1\\\\10 & < & 0.15t + 1\\9 & < & 0.15t\\t & > & 60\\\\\end{array}\\\mathbf{A < 1} \textbf{ when }\mathbf{ t >60}$