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

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

4 0

3 years ago

What is the value of x in the diagram?

Molodets [167]

The answer would be 42 degrees because the total of a triangle is 180 degrees. So you would subtract 180 and 42 (180-42) and that gives you 138 and since the two angles that are touching are the same

6 0

3 years ago

Write a real world problem for: f + 5 = 9

Triss [41]

9-5=4 Therefore f=4
Hope this helps

4 0

3 years ago

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If ON is a midsegment of KLM, find LM. A.2 B.3 C.4 D.5

Aleksandr-060686 [28]

NO is 2 times LM

2(x+1) = 4x

2x + 2 = 4x

2 = 2x

1 = x

Plug that in for 4x

4(1)

4

4 0

3 years ago

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