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svet-max [94.6K]
3 years ago
14

Are the graphs of the lines in the pair parallel? Explain. Find the slope of a line parallel to 5x+2y=6

Mathematics
1 answer:
Sonja [21]3 years ago
8 0

Answer:

Yes the graphs of the lines are in pair parallel.

Step-by-step explanation:

They are in pair parallel because they do not touch each other. To find the slope of the line that is parallel to 5x+2y=6...

The first thing to do is subtract 5x since its a positive so it would look like this:

5x+2y=6

<u> -5x      =-5x</u>

Next you would divide by 2 on each side so the y can be isolated:

5x+2y=6

<u>-5x     =-5x</u>

      <u>2y</u> = <u>-5x + 6</u>

      2          2

Lastly the 2 divided by 2 gets crossed out because they cancel each other out and divide -5x by 2 and 6 by 2 then you would be stuck with this:

y= -2.5x+3

So your parallel slope would be -2.5x.

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Step-by-step explanation:

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Prove that sinA-sin3A+sin5A-sin7A/cosA-cos3A-cos5A+cos7A= cot2A
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Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.

We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)

N = -(sin7A - sinA) + sin5A - sin3A
    = -2cos4A*sin3A + 2cos4A*sinA
    = 2cos4A(sinA - sin3A)
    = 2cos4A*2cos(2A)sin(-A)
    = -4cos4A*cos2A*sinA

D = cos7A + cosA - (cos5A + cos3A)
   = 2cos4A*cos3A - 2cos4A*cosA
   = 2cos4A(cos3A - cosA)
   = 2cos4A*(-2)sin2A*sinA
   = -4cos4A*sin2A*sinA

Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
       = cos2A/sin2A
      = cot2A

This verifies the identity.
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3 years ago
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