$\bf \begin{array}{|cc|ll} \cline{1-2} x&y\\ \cline{1-2} 40&32\\ 28&16\\ 16&12\\ \cline{1-2} \end{array}~\hfill \stackrel{\textit{average rate of change}}{slope} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{40}~,~\stackrel{y_1}{32})\qquad (\stackrel{x_2}{28}~,~\stackrel{y_2}{16}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{16-32}{28-40}\implies \cfrac{-16}{-12}\implies \boxed{\cfrac{4}{3}} \\\\[-0.35em] ~\dotfill$

$\bf (\stackrel{x_1}{28}~,~\stackrel{y_1}{16})\qquad (\stackrel{x_2}{16}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-16}{16-28}\implies \cfrac{-4}{-12}\implies \boxed{\cfrac{1}{3}}$

8 0

3 years ago

What is the slope-intercept equation of the line below? A.y=2x+3 B.y=-2x-3 C.y=2x-3 D.y=-2x+3

PtichkaEL [24]

Answer:

Step-by-step explanation:

First, lets find the y-intercept. The y-intercept where the line passes through the y-axis. In this case, it is -3.

Now, to find the slope, lets find the next non-decimal point(that way it is easier on us). That would be (1,-1). Now, to get to this point from the y-intercept, we have to go left 1 unit, and up 2 units. So, the change in y over change in x = 2/1 or 2. Now, using this information, we can form the equation:

y = 2x + -3. or y = 2x - 3.

7 0

3 years ago