2 years ago

Here's a graph of a linear function. Write the

1 answer:

8 0

I took 2 points from the graph and found my slope by using the slope formula, then I found my y-int which is the where the y crosses and that is -3. Hope this helps good luck !!

Answer: y=1/2x-3

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Now, we know that 90°< θ <180°, that simply means the angle θ is in the II quadrant, where sine is positive and cosine is negative.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{13}}\impliedby \textit{now let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
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c=hypotenuse\\
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$\bf \pm\sqrt{13^2-5^2}=a\implies \pm\sqrt{144}=a\implies \pm 12 =a \implies \stackrel{II~quadrant}{-12=a}
\\\\\\
therefore \qquad cos(\theta )=\cfrac{\stackrel{adjacent}{-12}}{\stackrel{hypotenuse}{13}}
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