16+90+2y =180 since it is a straight line
106 +2y = 180
subtract 106 from each side
2y = 74
divide by 2 on each side
y=37
90+2y+O = 180 since it is a straight line
90 + 2(37) +O = 180
90+74 +O = 180
164+0 =180
subtract 164 from each side
O = 16
O +4x+90=180 sine it is a straight line
16 + 4x +90 = 180
106 +4x = 180
subtract 106 from each side
4x = 74
divide by 4
x=18.5
The circumference of the circle is 168.7 in
[8] The hypotenuses must be congruent, DE and IJ
HL is hypotenuse-leg, or where the hypotenuse and one leg are congruent.
Since we see a leg is congruent, to prove they are congruent with HL we will need to know that the hypotenuses are congruent.
[9] Another leg, AC and YX
LL is leg-leg. Since we have one leg, we will need the other leg to prove the triangles are congruent by LL.
[10] A leg, DE and E? or FE and E?
LA means leg-angle. The angles by point E are vertical angles, so they will be congruent. To finish prooving the congruence using LA we will also need a leg.
-> The question marks are because the letters are too blurry to read.
I hope this helps you
if in 350 students are 14 students were absent
in 100 students are ?
?.350=100.4
?=400/360
?=1.14%
