Consider two unique parllel lines. What aspects of these two lines are the same? What aspects of these two lines would have to be different? Explain your reasoning.

$9.41

step-by-step explanation:

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 ☕

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.