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:
7,042
Step-by-step explanation:
Just think of it as normal subtraction. They both have the same denominators all you need to do is subtract. 11/12-7/12 is 4/12 if you simplify which just means to find the number that both 4 and 12 can go into equally the simplified answer would be 1/3. You just need to divide 4 by both the top and the bottom (numerator and denominator).
Answer:
BC 1
AC 2
AABC 3
Step-by-step explanation:
BC=16
AC = 15
AABC= 15
Answer: f(-3) = -27 & f(3) = 27
Step-by-step explanation:
(-3)^3
-27
(3)^3
27
