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sp2606 [1]
3 years ago
12

Section 1 - Topic 9 Parallel and Perpendicular Lines - Part 2 (please help)

Mathematics
1 answer:
RUDIKE [14]3 years ago
6 0

9514 1404 393

Answer:

  1. y = -1/7x
  2. y = 2x -7

Step-by-step explanation:

1. The slope of the given line is the x-coefficient, 7. The slope of the perpendicular line is the negative reciprocal of this: -1/7. Since the line goes through the origin, the y-intercept is zero.

  y = -1/7x

__

2. The first part of this problem is to find the solution to the given system of equations. We can substitute for x to do that:

  3(5 -y)-2y = 10

  15 -5y = 10 . . . simplify

  -3 +y = -2 . . . . divide by -2

  y = 1, x = 4 . . . add 3 to find y; subtract y from 5 to find x

__

The second part of this problem requires we note the slope of the parallel line. It is the x-coefficient, 2. The y-intercept can be found from ...

  b = y -mx = 1 -2(4) = -7

So, the desired line in y=mx+b form is ...

  y = 2x -7

_____

The first attachment shows the perpendicular lines of problem 1.

The second attachment shows the parallel lines of problem 2.

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Answer:

{- 2, - 4, - 6, - 8, - 10 }

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6 0
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If someone could do this exam for me I'll do anything else school related!
trapecia [35]
C. just shorten fraction and you'll only get 1/(y²z²)
you'll get (6xy²z²)/(6xy⁴z⁴)=1/(y²z²)
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