(02.04 MC)

1 answer:

6 0

$(\stackrel{x_1}{3}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{18}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{18}-\stackrel{y1}{6}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{3}}} \implies \cfrac{ 12 }{ 2 } \implies 6$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{6}=\stackrel{m}{ 6}(x-\stackrel{x_1}{3}) \\\\\\ y-6=6x-18\implies {\Large \begin{array}{llll} y=6x-12 \end{array}}$

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Ellen "The Wonderer" continues to walk in the path of an inscribed polygon. She starts at W, walks south until the boundary poin

mart [117]

Answer:

110°

Step-by-step explanation:

The path in the inscribed polygon walked by Ellen is shown in the diagram attached. Let the degrees in which Ellen must turn to get back to her starting point be T. Therefore to calculate T we use the theorem "An inscribed quadrilateral whose vertices all lie on a circle, the opposite angles of such a quadrilateral are in fact supplements for each other i.e their sum is 180 degrees."

Therefore: T + 70° = 180°

T = 180° - 70°

T = 110°

To get back to the starting point, Ellen must turn 110°