Write (4,0) , (-2,1) as an equation in slope-intercept form

Mathematics

1 answer:

borishaifa [10]3 years ago

8 0

Answer:

y=-1/6x+2/3

Step-by-step explanation:

(1-0)/(-2-4)=-1/6 slope

y-1=-1/6(x+2)

y-1 = - 1/6x-1/3

y=-1/6x+2/3

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Suppose theta is an angle in the standard position whose terminal side is in quadrant 4 and cot theta = -6/7. find the exact val

zimovet [89]

First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

$\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill$

$\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}$

$\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}$