Ok so remember that a circle is 360 degrees and a line is 180 degrees
so
IKJ
ok so we see that we have a triangle
to find the internal angles (which add to 180)
we do, opposite angles are equal
JIK=59
also CJI+KJI=180 so 132+KJI=180, KJI=48
so
59+48+IKJ=180
73=IKJ
HIB
we can assume that AB is paralell to CD
so therefor IJK=HIB=48 (from previous question)
EIJ is adding to EIH to get 180
EIJ+59=180
EIJ=121
AIK=AIJ+JIK
JIK=59
AIJ=HIB=48
AIK=59+48=107
IKJ=73
HIB=48
EIJ=121
AIK=107
(x-8) ^ 2 = 121
(x-8) = + / - root (121)
x = 8 +/- root (121)
The solutions are:
x1 = 8 + root (121)
x2 = 8 - root (121)
2a ^ 2 = 8a-6
2a ^ 2-8a + 6 = 0
a ^ 2-4a + 3 = 0
(a-1) (a-3) = 0
The solutions are:
a1 = 1
a2 = 3
x ^ 2 + 12x + 36 = 4
x ^ 2 + 12x + 36-4 = 0
x ^ 2 + 12x + 32 = 0
(x + 4) (x + 8) = 0
The solutions are:
x1 = -8
x2 = -4
x ^ 2-x + 30 = 0
x = (- b +/- root (b ^ 2 - 4 * a * c)) / 2 * a
x = (- (- 1) +/- root ((- 1) ^ 2 - 4 * (1) * (30))) / 2 * (1)
x = (1 +/- root (1 - 120))) / 2
x = (1 +/- root (-119))) / 2
x = (1 +/- root (119) * i)) / 2
The solutions are:
x1 = (1 + root (119) * i)) / 2
x2 = (1 - root (119) * i)) / 2
Answer:
c
Step-by-step explanation:
With only the description, my answer is 20 students.
