Which of the following are not sufficient to prove that a quadrilateral is a parallelogram?

i. two pairs of opposite angles congruent.
ii. a pair of adjacent angles are supplementary.
iii. both pairs of opposite sides are congruent.
iv. a pair of opposite angles congruent and a pair of opposite sides congruent.
v. both pairs of opposite sides are parallel.
vi. a pair of opposite sides parallel and the other pair of opposite sides congruent.
vii. one pair of opposite sides are both parallel and congruent.
viii. the diagonals bisect each other.

Answer with explanation:

We can prove that a quadrilateral is a parallelogram , if

I. Two pairs of opposite angles congruent.

Consider, a Quadrilateral ABCD, in which , ∠A=∠C,and ∠B=∠D.

Now, ∠A+∠B+∠C+∠D=360°

which gives, ∠A+∠B=180 and, ∠A+∠D=180°

Showing, AB║CD, and AD║BC.

Join ,either, AC or BD.

So, By AAS,we can prove that ,the two triangles are congruent.

So, Quadrilateral ABCD is a Parallelogram.

III.Both pairs of opposite sides are congruent.→Axiom

V.Both pairs of opposite sides are parallel. →→Proved in I

VI.One pair of opposite sides are both parallel and congruent.

In Quadrilateral ,ABCD

AB║CD,AB=CD

Join AC

In Δ ABC and ΔADC

∠BAC=∠DCA→Alternate interior  Angles

AC is common.

AB=CD→[given]

Δ ABC ≅ ΔADC→→[SAS]

VIII.The diagonals bisect each other.

In ΔAOB and ΔCOD

AO=OC

BO=OD

∠AOB=∠COD→[Vertically opposite angles]

→∠AOB≅∠COD[SAS]

which gives, AB=CD.By taking other two triangles, ∠AOD=∠COB, we can prove, AD=BC.

So, when opposite sides are equal in a Quadrilateral it is a parallelogram.

The Condition which are not Sufficient to prove that a quadrilateral is a parallelogram:

II: A pair of adjacent angles are supplementary.

IV: A pair of opposite angles congruent and a pair of opposite sides congruent.

VI: A pair of opposite sides parallel and the other pair of opposite sides congruent.

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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