The graph of a linear equation passes through (0, 5) and (-10, 0). What is the equation?

1 answer:

8 0

$(\stackrel{x_1}{0}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-10}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{-10}-\underset{x_1}{0}}}\implies \cfrac{-5}{-10}\cfrac{1}{2}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{0}) \\\\\\ y-5-\cfrac{1}{2}x\implies y=\cfrac{1}{2}x+5$

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<u>Step-by-step explanation:</u>

To prove:

$\cos 3x=\cos^3x-3\sin^2 x\cos x$

Identities used:

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$\sin 2A=2\sin A\cos A$ ........(2)

$\cos 2A=\cos^2A-\sin^2 A$ .......(3)

Taking the LHS:

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Using identity 1:

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Using identities 2 and 3:

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