Question

Anyone knows how to do this

Answer 1

well, in 2005 there were 55000 folks, and then 3 years later it was 61600, since we know the increment is linear, then we can just call year 2005 year 0 and use a table for that

`\begin{array}{ccll} \stackrel{x}{year}&\stackrel{y}{population}\\ \cline{1-2} 0&55000\\ 3&61600 \end{array}\hspace{5em} (\stackrel{x_1}{0}~,~\stackrel{y_1}{55000})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{61600}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{61600}-\stackrel{y1}{55000}}}{\underset{\textit{\large run}} {\underset{x_2}{3}-\underset{x_1}{0}}} \implies \cfrac{ 6600 }{ 3 } \implies 2200`

`\begin{array}ll \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{55000}=\stackrel{m}{ 2200}(x-\stackrel{x_1}{0}) \\\\\\ y-55000=2200x\implies {\Large \begin{array}{llll} y=2200x+55000 \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{in 2010 from 2005 that's 5 years later, x = 5}}{y = 2200(5)+55000}\implies {\Large \begin{array}{llll} y=66000 \end{array}}`