So Marci's sore was 45 because every percent has to be out of 100 so to get it to 50 you divide by two so 90 / 2 = 45. And 45-39 = 6
Marci got 6 more correct then Donovan.
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:
L.C.M of 432 and 486: 3888
H.C.F of 432 and 486: 54
Step-by-step explanation:
If the parabola has y = -4 at both x = 2 and x = 3, then since a parabola is symmetric, its axis of symmetry must be between x = 2 and x = 3, or at x = 5/2. Our general equation can then be:
y = a(x - 5/2)^2 + k
Substitute (1, -2): -2 = a(-3/2)^2 + k
-2 = 9a/4 + k
Substitute (2, -4): -4 = a(-1/2)^2 + k
-4 = a/4 + k
Subtracting: 2 = 2a, so a = 1. Substituting back gives k = -17/4.
So the equation is y = (x - 5/2)^2 - 17/4
Expanding: y = x^2 - 5x + 25/4 - 17/4
y = x^2 - 5x + 2 (This is the standard form.)
