Could ΔABC be congruent to ΔADC by SSS? Explain. Yes, but only if AB ≅ DC. Yes, but only if BC ≅ DC. No, because AB is not congruent to AC. No, because AB ≅ DA.

Yes, but only if BC ≅ DC; Option B

Step-by-step explanation:

There are 4 possible ways to determine whether two triangles are congruent, and they are the following ;

ASA ( Angle - Side - Angle ),

SAS ( Side - Angle - Side ),

AAS ( Angle - Angle - Side ), and

SSS ( Side - Side - Side )

It is known here that we must prove these triangles congruence through SSS. A triangle can thus be made possible through the congruence of corresponding sides;

Now if we were to create these triangles it would be that BC and DC are, if the triangles were to coincide with one another, the apparent same length if it were that these two Δs really are ≅. Thus;

Solution; Yes, but only if BC ≅ DC

The Answer is B.

Step-by-step explanation:

step-by-step explanation:

when the line is solid:

when the line is dashed:

this > inequality sign means:

this < inequality sign means:

this > inequality sign means:

this < inequality sign means:

about the problem:

srry i'm not done ! i'll

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.