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3 years ago

what is the slope intercept form of the linear equation with a graph that passes through (-2,2) and is perpendicular to the grap

h of y=1/3x+9

Mathematics

1 answer:

gtnhenbr [62]3 years ago

6 0

$\text{The slope-intercept form}\ y=mx+b.\\\\Let\\k:y=m_1x+b_1\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\\------------------------\\\text{Let}\ k:y=\dfrac{1}{3}x+9\ \text{and}\ l:y=mx+b\\\\l\ \perp\ k\iff\dfrac{1}{3}m=-1\qquad|\cdot3\to m=-3\\\\l:y=-3x+b\\\\\text{Substitute the coordinates of the point (-2, 2) to the equation}\\\text{of the line}\ l:\\\\2=-3(-2)+b\\2=6+b\qquad|-6\\-4=b\to b=-4\\\\\boxed{Answer:\ y=-3x-4}$

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Cos^2x+cos^2(120°+x)+cos^2(120°-x)<br>i need this asap. pls help me

o-na [289]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

= - cos60cosx - sin60sinx

= - $\frac{1}{2}$ cosx - $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos² (120 + x)

= $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

= -cos60cosx + sin60sinx

= - $\frac{1}{2}$ cosx + $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos²(120 - x)

= $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------------

Putting it all together

cos²x + $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x + $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

= cos²x + $\frac{1}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ (cos²x + sin²x) = $\frac{3}{2}$

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