When considering similar triangles, we need congruent angles and proportional sides.
Hence
"Angles B and B' are congruent, and angles C and C' are congruent." is sufficient to prove similarity of two triangles.
"Segments AC and A'C' are congruent, and segments BC and B'C' are congruent." does not prove anything because we know nothing about the angles.
"Angle C=C', angle B=B', and segments BC and B'C' are congruent." would prove ABC is congruent to A'B'C' if and only if AB is congruent to A'B' (not just proportional).
"<span>Segment BC=B'C', segment AC=A'C', and angles B and B' are congruent</span>" is not sufficient to prove similarity nor congruence because SSA is not generally sufficient.
To conclude, the first option is sufficient to prove similarity (AAA)
Answer:B.) Bisector Of An Angle
Step-by-step explanation:
Perhaps the most concise way to factor is by "completing the square" which is how the quadratic formula is derived...
x^2+6x+8=0 move constant to other side, subtract 8 from both sides
x^2+6x=-8, halve the linear coefficient, square it, then add that to both sides, in this case (6/2)^2=3^2=9
x^2+6x+9=1 now the left side is a perfect square of the form
(x+3)^2=1 take the square root of both sides
x+3=±√1 subtract 3 from both sides
x=-3±√1
x=-3±1
x=-4 and -2
Since the zeros occur when x=-4 and -2 the factors of the equation are:
(x+2)(x+4)
For this case we have the following polynomial:
7x2 + 68xy - 20y2
Factoring we have:
(7x-2y) (x + 10y)
We verify the factorization:
7x2 + 70xy - 2xy - 20y2
Rewriting we have:
7x2 + 68xy - 20y2
Therefore, the factorization is correct.
Answer:
A) (7x - 2y) (x + 10y)
Answer:
1. 4.22
2. 19.415
3. nine and thirty- five
4. Three and four hundred twelve
Step-by-step explanation
