Answer:For every right triangle, the circumcenter is always the midpoint of the hypotenuse. All four points of concurrency can themselves be concurrent! Only with an equilateral triangle will the centroid, circumcenter, incenter and orthocenter always be the same point!
Step-by-step explanation:
It’s the first one plug it in
Answer: 2 quarters, a nickel, and 2 pennies
Step-by-step explanation:
Answer:
The reasons are given below.
Step-by-step explanation:
In triangle ΔAXC and ΔBXC, we are given that angles 3 and 4 are right angles and AX = BX. we have to match the reasons in the given proof of congruency of triangles △AXC ≅ △BXC
In ΔAXC and ΔBXC,
AX=BX (Given)
∠3 = ∠4 = 90° (both right angles)
CX=CX (Common i.e reflexive property of equality)
Hence by SAS similarity theorem ΔAXC ≅ ΔBXC
hence, the above are the reasons of the statements in given proof.
Answer: Hi!
The only thing we can do to simplify the equation is combine like terms.
5x^2 + 4x^2 = 9x^2
-2y - 7y = -9y
3 - 1 + 7 = 9
Our equation now looks like this:
9x^2 - 9y + 9
We have nothing left to simplify, so we're done!
Hope this helps!
