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.

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.

a

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a)   |-1|, |3/2|, |-2|, |-3|, |3.5|