The equation of line ab is y = 5x + 1. write an equation of a line parallel to line ab in slope-intercept form that contains point (4, 5). (1 point) y = 5x − 15 y = −5x + 15 y = 1 over 5 x + 21 over 5 y = 1 over 5 x − 29 over 5

y = 5x -15

Step-by-step explanation:

A parallel line has the same slope, hence the same x-coefficient in its equation. This makes it be ...

... y = 5x +b . . . . . . . for some value of b that will give the desired point

Putting the given point values into the equation, we can find the necessary y-intercept constant.

... 5 = 5·4 +b

... 5 - 20 = b = -15

The the desired equation is ...

... y = 5x -15

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.