Quadrilateral ABCD has vertices at A(3,−2), B(4,3), C(−2,0), and D(−3,−5). Quadrilateral A B C D in the plane with the given coordinates. © 2019 StrongMind. Created using GeoGebra. Which answers prove that quadrilateral ABCD is a parallelogram? Select all that apply. slope of AB¯¯¯¯¯¯¯¯=slope of DC¯¯¯¯¯¯¯¯; AD¯¯¯¯¯¯¯¯≅BC¯¯¯¯¯¯¯¯ AB¯¯¯¯¯¯¯¯∥DC¯¯¯¯¯¯¯¯ and AD¯¯¯¯¯¯¯¯∥BC¯¯¯¯¯¯¯¯ ∠A is supplementary to ∠B; ∠A≅∠C AD¯¯¯¯¯¯¯¯≅BC¯¯¯¯¯¯¯¯≅CD¯¯¯¯¯¯¯¯ slope of AD¯¯¯¯¯¯¯¯=slope of BC¯¯¯¯¯¯¯¯; AD¯¯¯¯¯¯¯¯≅BC¯¯¯¯¯¯¯¯

The quadrilateral ABCD is a parallelogram as, slope of AB = slope of DC = 5, and slope of AD = slope of BC = 0.5.

Step-by-step explanation:

Quadrilateral ABCD has vertices at A(3,−2), B(4,3), C(−2,0), and D(−3,−5).

So, the slope of AB =

Now, the slope of BC =

The slope of CD =

Finally, the slope of DA =

Therefore, the quadrilateral ABCD is a parallelogram as, slope of AB = slope of DC = 5, and slope of AD = slope of BC = 0.5. (Answer)

Note: A quadrilateral to be a parallelogram the slope of each pair of opposite sides must be the same.

understand that i don't know the answers

will someone answer his question

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.