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# A certain piece of news is being broadcast to a potential audience of 200,000 people. Let f(t) be the number of people who have heard the news after t hours. Suppose that y=f(t) satisfies$y^{\prime}=.07(200,000-y), \quad y(0)=10 .$Describe this initial-value problem in words.

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Let $f(t)$ satisfies the initial-value problem

\begin{aligned} y'=.07(200,000-y), \quad y(0)=10, \end{aligned}

where $f(t)$ represents the number of people who have heard the some news after $t$ hours. Those news were being broadcast to a potential audience of $200,000$ people.

We see that there are $10$ people at the beginning who have heard the news. We can calculate here $y'(0)$ substituting $y(0)$ into the equation, to get

\begin{aligned} y'(0)=.07(200,000-10)=13,999.3>0, \end{aligned}

so we conclude that the number of people who have heard the news after $t$ hours is proportional to the difference between the number of a potential audience, $200,000$, and that number $y(t)$ and also that the obtained function is increasing.

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