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Sloan [31]
3 years ago
9

Write an equation of the line passing through point p that is parallel to the given line.

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
8 0

Answer:

Step-by-step explanation:

y=-3x+13

y = mx + b, where m is the slope and b is the y-intercept.

A line parallel to this line will have the same slope, -3.  We can start by writing:

y = -3x + b

No matter what is chosen for the value of b, this line will be parallel.  We do want the line to go through point (6,10), however.  The way that can be done is to find a value of b that would shift the line to go through (6,10).

Find b by entering the point (6,10) into the equation, and then solve for b:

y = -3x + b

10 = -3(6) + b

10 = -18 + b

b = 28

The line becomes y = 3x + 28

See the attachment.

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What is the perimeter of a polygon with vertices at (−2, 1) , ​ (−2, 4) ​, (2, 7) , ​ (6, 4) ​, and (6, 1) ​?
alekssr [168]
First we need the distances of the sides of the polygon, because perimeter = sum of all sides.
d =  \sqrt{{(y2 - y1)}^{2} + {(x2 - x1)}^{2} }
d1 =  \sqrt{{(1 - 4)}^{2} + {( - 2 -  - 2)}^{2} } \\  =  \sqrt{{(- 3)}^{2} + {(0)}^{2} } =  \sqrt{9}  \: = 3
d2 = \sqrt{{(4 - 7)}^{2} + {( - 2 - 2)}^{2} } \\  = \sqrt{{(- 3)}^{2} + {( -4)}^{2} } =  \sqrt{9 + 16}  \\  =  \sqrt{25}  \: = 5
d3 \: = \sqrt{{(7 - 4)}^{2} + {(2 - 6)}^{2} } \\  =  \sqrt{{(3)}^{2} +  {( - 4)}^{2} } =  \sqrt{9 + 16}  \\  =  \sqrt{25} \:  = 5
d4 = \sqrt{{(4 - 1)}^{2} + {(6 - 6)}^{2} } \\  =  \sqrt{ {(3)}^{2} +  {(0)}^{2}  }  =  \sqrt{9} \:  = 3
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p = d1 + d2 + d3 + d4 + d5
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     1

-----------

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    2- -2                 4                    1

---------------  =  --------------- = ---------------

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                                 (simplifyed)

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Given

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\Rightarrow \text{Number of Nickels = }8d-7

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