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Lena [83]
3 years ago
12

What is the slope of the line that passes through the points (-8, -1)(−8,−1) and (-6, -2) ?(−6,−2)? Write your answer in simples

t form.
Mathematics
1 answer:
ziro4ka [17]3 years ago
3 0

The slope of line passing through the given points is: -1/2

Step-by-step explanation:

Given points are:

(x1,y1) = (-8,-1)

(x2,y2) = (-6,-2)

Slope is defined as the steepness of a line.

It is denoted by m.

The formula for slope is:

m = \frac{y_2-y_1}{x_2-x_1}

Putting the values

m = \frac{-2-(-1)}{-6-(-8)}\\= \frac{-2+1}{-6+8}\\=\frac{-1}{2}

The slope of line passing through the given points is: -1/2

Keywords: Slope, steepness

Learn more about slope at:

  • brainly.com/question/899976
  • brainly.com/question/884169

#LearnwithBrainly

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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

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                   = - \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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Someone please help me i’m so llst
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