Given: circle k(O), DC∥AB, AC ∩ DB=O, m AD=108°

Find: m∠A and m∠AOB

m<A = 54 degrees

m<AOB = 72 degrees

Step-by-step explanation:

First, if arcDA is 108 degrees, than m<DOA is 108 degrees

m<DOA and m<AOB are supplementary, so m<AOB=180-108=72 degrees

Next m<A

If this double triangle kind of helix is in the center of an ellipse, it means it is congruent. If it this happens in a circle it meant the triangles are both congruent and isosceles.

Hence, triAOB is isosceles.

SO, the angle of A and B are congruent and the equation is half of the supplement of AOB, in which case is DOA---

m<A=m<B=m<DOA/2

m<A=108/2

m<A=54 degrees

I also went to RSM,

so good luck for the rest of Eighth Grade Geometry.

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5