Can you plz attach an image of the graph :O
It is a straight line in a graphical format.
Answer:
The proof is below
Step-by-step explanation:
Given a parallelogram ABCD. Diagonals AC and BD intersect at E. We have to prove that AE is congruent to CE and BE is congruent to DE i.e diagonals of a parallelogram bisect each other.
In ΔACD and ΔBEC
AD=BC (∵Opposite sides of a parallelogram are equal)
∠DAC=∠BCE (∵Alternate angles)
∠ADC=∠CBE (∵Alternate angles)
By ASA rule, ΔACD≅ΔBEC
By CPCT(Corresponding Parts of Congruent triangles)
AE=EC and DE=EB
Hence, AE is conruent to CE and BE is congruent to DE
Answer:
Step-by-step explanation:
its B
Answer:
1st one
Step-by-step explanation:
Since the x's for each one are going up by the same number, you are looking for a column of y's to either by going up by the same number or down by the same number.
The first table has a column of y's that look like this: 1/2 , 1 , 1 1/2, 2
This is actually going up for 1/2 each time.
This table shows a linear function.
