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valina [46]
3 years ago
13

Help with geometry please

Mathematics
2 answers:
Georgia [21]3 years ago
7 0

Step-by-step explanation:

Working off what the person below did, one way to figure it out is by taking the lines (A straight line is 180*)

So to figure it out you have the 121*, and that little segment next to it.

So 180*-121*= 59*. And since that little segment/corner is a vertical angle, You move the 59* Into the little corner in the triangle.

Same with 144*, it has a little segment next to it.

So 180*-144*=36*. Move that in.

A triangle is 180* on the inside. So 180*-36*-59*=85*. That is the last corner INSIDE the triangle.

Move that OUT.

it's in the little segment/corner next to <em>n</em> and it's on a line, So 180*-85*=95*

n=95*.

eimsori [14]3 years ago
4 0

Answer:

The answer to your question is: n = 95°

Step-by-step explanation:

The sum of the suplementary angles equals 180°

Then

           121 + x = 180

           x = 180 - 121

           x = 59°

 

           144 + y = 180

            y = 180 - 144

            y = 36°

The sum of the internal angles in a triangle equals 180°

   

           x + y + z = 180

          59 + 36 + z = 180

          z = 180 - 59 - 36

          z = 85

Finally

          n + 85 = 180

          n = 180 - 85

          n = 95°

           

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Answer:

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Step-by-step explanation:

Graph is shown in the attached sheet

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x     0   4.5    3                              x       0      2.5     3

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3 years ago
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y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{5}}x-1\qquad \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} \\\\[-0.35em] ~\dotfill

\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-2}{5}} ~\hfill \stackrel{reciprocal}{\cfrac{5}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{5}{-2}\implies \cfrac{5}{2}}}

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(\stackrel{x_1}{-2}~,~\stackrel{y_1}{14})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{5}{2} \\\\\\ \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}{14}=\stackrel{m}{ \cfrac{5}{2}}(x-\stackrel{x_1}{(-2)}) \implies y -14= \cfrac{5}{2} (x +2) \\\\\\ y-14=\cfrac{5}{2}x+5\implies {\Large \begin{array}{llll} y=\cfrac{5}{2}x+19 \end{array}}

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