Section 1 - Topic 9 Parallel and Perpendicular Lines - Part 2 (please help) 1. Write the equation of the line that it is perpendicular to y = 7x - 3 and passes through the origin.

2. Write the equation of the line that it is parallel to y = 2x + 1 and passes through the solution of the following system of equations. (3x - 2y = 10 x + y = 5

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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

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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

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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

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The first attachment shows the perpendicular lines of problem 1.

The second attachment shows the parallel lines of problem 2.

answer: e 9, 5.75

step-by-step explanation:

step 1

find the variation ab in the x coordinate

delta x=(x2-x1)=10-> 8 units

find the variation ab  in the y coordinate 

delta y=(y2-y1)=6-> 2 units

step 2

find the coordinates point c

  x coordinate 

cx=ax+delta x*2/> cx=2+8*(2/> cx=3

  y coordinate 

cy=ay+delta y*2/> cy=4+2*(2/> cy=4.25

the coordinates of point c are (3,4.25)

step 3

find the coordinates point d

  x  coordinate 

dx=ax+delta x*4/> dx=2+8*(4/> dx=4

  y  coordinate 

dy=ay+delta y*4/> dy=4+2*(4/> dy=4.5

the coordinates of point d are (4, 4.5)

step 4

find the coordinates point e

  x  coordinate 

ex=ax+delta x*4/> ex=2+8*(6/> ex=5

  y  coordinate 

ey=ay+delta y*4/> ey=4+2*(6/> ey=4.75

the coordinates of point e are (5, 4.75)

step 5

find the coordinates point f

  x  coordinate 

fx=ax+delta x*4/> fx=2+8*(8/> fx=6

  y  coordinate 

fy=ay+delta y*4/> fy=4+2*(8/> fy=5

the coordinates of point f are (6,5)

step 6

find the coordinates point g

  x  coordinate 

gx=ax+delta x*4/> gx=2+8*(10/> gx=7

  y  coordinate 

gy=ay+delta y*4/> gy=4+2*(10/> gy=5.25

the coordinates of point g are (7,5.25)

step 7

find the coordinates point h

  x  coordinate 

hx=ax+delta x*4/> hx=2+8*(12/> hx=8

  y  coordinate 

hy=ay+delta y*4/> hy=4+2*(12/> hy=5.5

the coordinates of point h are (8,5.5)

step 8

find the coordinates point i

  x  coordinate 

ix=ax+delta x*4/> ix=2+8*(14/> ix=9

  y  coordinate 

iy=ay+delta y*4/> iy=4+2*(14/> iy=5.75

the coordinates of point i are (9,5.75)

- r3ktforgood ☕

for this case we have that by definition, the surface area of a cylinder is given by:

where:

r: it is the radius of the cylinder

h: it is the height of the cylinder

we can also rewrite it as:

where:

la: it's the lateral area

ba: area of the base

option a