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marin [14]
3 years ago
14

I need help on how to find the slope of the first two question

Mathematics
2 answers:
nikitadnepr [17]3 years ago
7 0
Remember the equation for slope is y=mx+b {m=slope, and b=y-intercept.} {Slope= rise/run}
1.) y= 2x+0
2.) y= -3/4x+0.5

P.S.: hope I helped :)
xeze [42]3 years ago
7 0
For both questions, all you really have to do is choose 2 exact points on each graph, and use those to find the slope. Thankfully, it already has 2 plots on there for us, so all we have to do is plug in the values in the equation below :)

Slope equation : y₂-y₁ / x₂-x₁

1)  (1,2), (-2,-4)

-4 -2 / -2 -2

-6 / -4

Simplify.

The slope of number one is 3/2

2) (-2,2), (6,-4)

-4 -2 / 6 + 2

-6 / 8

Simplify.

The slope for question 2 is -3/4

~Hope I helped!~
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Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).
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keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

\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 y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5

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\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \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}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}

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