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Mekhanik [1.2K]
3 years ago
9

Which equation represents the line that passes through points (1, –5) and (3, –17)? y = negative 6 x + 1 y = 6 x + 1 y = negativ

e 6 x minus 1 y = 6 x minus 1 im on a time limit pls hurry
Mathematics
1 answer:
Paraphin [41]3 years ago
6 0

Answer:

Using y = mx + c

slope = -17+5/3-1 = -12/2 = -6

Using point (1 , - 5)

y + 5 = -6(x - 1)

y + 5 = - 6x + 6

y = - 6x + 6 - 5

y = - 6x + 1

Hope this helps.

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~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{-26}~,~\stackrel{y_2}{120})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{(~~-26 - 0~~)^2 + (~~120 - 0~~)^2} \implies r=\sqrt{(-26)^2 + (120 )^2} \\\\\\ r=\sqrt{( -26 )^2 + ( 120 )^2} \implies r=\sqrt{ 676 + 14400 } \implies r=\sqrt{ 15076 } \\\\[-0.35em] ~\dotfill

\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-26}{h}~~,~~\underset{120}{k})}\qquad \stackrel{radius}{\underset{\sqrt{15076}}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-26) ~~ )^2 ~~ + ~~ ( ~~ y-120 ~~ )^2~~ = ~~(\sqrt{15076})^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (x+26)^2+(y-120)^2 = 15076~\hfill

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