Match the following reasons with the statements given. complete the proof for the theorem 4-18 mentioned above. 1. do = ob, ao = oc if two sides = and ||, then a parallelogram. 2. doc = aob vertical angles are equal. 3. triangle cod congruent to triangle aob cpcte 4. 1 = 2, ab = dc given 5. ab||dc if alternate interior angles =, then lines parallel. 6. abcd is a parallelogram sas

Please see attachment.

Solution:

We need to match the column using column proof. Please see the attachment for matching column.

In ΔDOC and ΔBOA

      DO=BO               (Given)

 ∠DOC=∠BOA          (Vertically Opposite angle)

      OC=OA               (Given)

∴   ΔDOC ≅ ΔBOA   by  SAS congruence property

     ∠1=∠2  and  AB=DC     By CPCTE

Thus, AB||DC   (∠1 and ∠2 are alternate angles equal then lines parallel)

ABCD is a parallelogram. ( If two sides equal and parallel then a parallelogram.

Below is matched table.

DO = OB, AO = OC ⇒ Given      

∠ DOC =∠ AOB      ⇒ Vertical angles are equal

∆COD ≅ ∆AOB⇒  SAS             CPCTE

∠1 = ∠2,  AB = DC⇒  CPCTE          

AB||DC                 ⇒  If alternate interior angles =, then lines parallel

ABCD is a parallelogram ⇒ If two sides = and ||, then a parallelogram

Please see attachment for figure and matching.      

chad should not have included the values (1, 2) and (3, 6) because that would mean he only had half a healing card.

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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