Given: ABCD is a parallelogram.
Diagonals AC, BD intersect at E.
Prove: AE = CE and BE DE
Angles Segments Triangles Statements Reasons
ABCD is a parallelogram
ZABE and LODE
are alt. interior angles
ZBCE and ZDAE
are alt. interior angles
LCBE and ZADE
are alt. interior angles
Statements
Reasons
D
Assemble the proof by dragging tiles to
the Statements and Reasons columns.

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.

what’s the question?

i dont know

step-by-step explanation:

the reason i dont know is because i have not taken this class and for that reason i haven't learned this stuff

hope i could