Q:

How do I solve rate of change problems? (With picture) thanks!**please help me understand the 3 problems by explaining

Accepted Solution

A:
5)[tex]\bf \begin{array}{|cc|cccc|ll} \cline{1-6} sodas&x&\underline{24}&28&\underline{32}&36\\ \cline{1-6} cost&y&\underline{18}&21&\underline{24}&27\\ \cline{1-6} \end{array}~\hspace{9em} (\stackrel{x_1}{24}~,~\stackrel{y_1}{18})\qquad (\stackrel{x_2}{32}~,~\stackrel{y_2}{24}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{24-18}{32-24}\implies \cfrac{6}{8}\implies \cfrac{3}{4}[/tex]6)[tex]\bf \begin{array}{|cc|cc|ll} \cline{1-4} year&x&0&12\\ \cline{1-4} \$&y&720&1080\\ \cline{1-4} \end{array}~\hspace{10em} (\stackrel{x_1}{0}~,~\stackrel{y_1}{720})\qquad (\stackrel{x_2}{12}~,~\stackrel{y_2}{1080}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1080-720}{12-0}\implies \cfrac{360}{12}\implies \cfrac{30}{1}\implies 30[/tex]7)slope as you should already know is rise/run, or how much something moves in relation something else, namely how much the y-axis go up as the x-axis moves sideways, one moves, the other follows, but the increments will be different, sometimes the same, but usually different.the y-intercept means, when the graph of the equation touches or intercepts the y-axis, and when that happens x = 0, or the horizontal distance is at bay.for the slope on  6), 30 or 30/1 means, for every 1 year(x) passed, the worth(y) increased by 30, or jumped by 30 units, so as the x-axis moved 1, the y-axis moved 30.  After 12 years 30 * 12 = 360, and we add the initial 720 and we end up with 1080.the y-intercept, well, as aforementioned is when x = 0, is year 0.