HELP! BRAINLIEST! Parallelogram JKLM is shown on the coordinate plane below:

If parallelogram JKLM is rotated 270° clockwise around the origin, what are the coordinates of the endpoints of the side congruent to side JM in the image parallelogram?

J′(−2, −6); M′(1, −5)
J′(6, 2); M′(−5, 1)
J′(2, 6); M′(−1, 5)
J′(6, −2); M′(5, 1)

Given that rotating 90 degrees clockwise around the origin switches the x andy values and makes the new y value negative, we can, for example, switch (2, 1) to (1, -2). 180 degrees clockwise simply makes both values negative (-2, -1), and 270 degrees clockwise switches them and makes the new y value negative  (-1, 2), we can plug those in to our JM endpoints to turn (-5, 1) into (-1, -5) and (-6, 2) into (-2, -6)

Step-by-step explanation:

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.