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Free_Kalibri [48]
3 years ago
12

What is the equation of the following line? Be sure to scroll down first to see all answer options.

Mathematics
1 answer:
Annette [7]3 years ago
8 0
From the figure the given line passes through the points (0, 0) and (-4, 8).

Recall that the equation of a straight line is given by
\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}

Thus, The equation of the given figure is given by
\frac{y-0}{x-0} = \frac{8-0}{-4-0}= \frac{8}{-4}  \\  \\ -4y=8x \\  \\ y=-2x
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\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}

 

 

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