$\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{57})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{91}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{91}-\stackrel{y1}{57}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{4}}}\implies \cfrac{34}{2}\implies 17 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{57}=\stackrel{m}{17}(x-\stackrel{x_1}{4})\implies y-57=17x-68$

$\bf y=17x-11\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad \begin{cases} \stackrel{slope}{17}\\\\ \stackrel{y-intercept}{(0,-11)} \end{cases}$

4 0

3 years ago

Find the circumference and area <br> r = 5 m

stira [4]

Step-by-step explanation:

Let's get our relevant base equations listed out:

d = 2r

area = pi*r²

c = πd

Thus, d=2*5m = 10m, c=π*10m=10π m, and area = 25m²*π

Feel free to ask further questions!

4 0

3 years ago