You have two angles congruent, plus a side that's NOT between them.
I guess you'd call that situation " AAS " for "angle-angle-side".
That's what you have, and it's NOT enough to prove the triangles
congruent. There can be many many different pairs of triangles
that have AAS = AAS.
So there's no congruence postulate to cover this case, because they're
not necessarily.
Y’’(x)= 6x + 1
y’(x)= 3x^2 + x + 2
y(x)= x^3 + 1/2x^2 + 2x + 5
Answer:
7/18
Step-by-step explanation:
4 2/3=14/3
(14/3)/12=14/36
reduces to
7/18
Answer:
<h3>Sv is angle bisector of rst.</h3>
