Answer:
As per theorem Segment joining midpoints of two sides of a triangle is parallel to the third side.
Step-by-step explanation:
Given : In a triangle ABC, D and E are the midpoint of sides AB and AC.
To Proof : DE║BC
Construction: A line CF parallel to BA is drawn and De is extended to point F.
Proof : In a triangle ADE and EFC
Sides AD║CF and DF is transverse.Then ∠ADE = ∠CFE (Alternate angles)
Similarly AD║CF and AC is transverse then ∠DAE = ∠FCA (Alternate angles)
Therefore ΔADE≅ΔCEF (AAS property)
From this property side AD = Side CF (corresponding sides of the congruent triangles)
Therefore BD = CF
So sides BD and CF are equal and parallel to each other which represents DBCF is a parallelogram.
Therefore DE║BC
Hence proved.
We factor the denominators
Factor x^2 - 16
x^2 - 4^2
We use a^2 - b^2 = (a+b)(a-b)
so x^2 - 4^2 = (x+4)(x-4)
Replace it in the given equation
Excluded values are the values that makes the denominator 0
we have (x-4) and (x+4) in the denominator
We set the denominator =0 and solve for x
x-4 =0
Add 4 on both sides
x= 4
x+4=0
subtract 4 onboth side
so x= -4
Excluded values are x=-4 and x=4