1.<AFC
2.<AFD
3.<AFE
4.<AFB
#5, 6 and 7
m<BAE = 59
m<BAC + m<CAD + m<DAE = m<BAE
4x - 20 + x + 12 + x + 1 = 59
6x - 7 = 59
6x = 59 + 7
6x = 66
x = 11
m<BAC = 4x - 20 = 4(11) - 20 = 24
m<CAD = x + 12 = 11 + 12 = 23
m<BAE = x + 1 = 11 + 1 = 12
answer
5.
m<BAC = 24
6.
m<CAD = 23
7.
m<BAE = 12
#8, 9 and 10
m<BAE = 130
m<BAC + m<CAD + m<DAE = m<BAE
3x - 10 + 2x + 5 + x + 15 = 130
6x + 10 = 130
6x = 120
x = 20
m<BAC = 3x - 10 = 3(20) - 10 = 60 -10 = 50
m<CAD = 2x + 5 = 2(20) + 5 = 40 + 5 = 45
m<DAE = x + 5 = 20 + 15 = 35
answer
8.
m<BAC = 50
9.
m<CAD = 45
10.
m<DAE = 35
Yes they’re similar, but since length is 18 for the large triangle and 6 for the smaller one you’d need to take 6 times 3 to equal 18 so i think you need to divide 12 by 3 to get the length of BS which would end up being 4 if i’m right
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2f+6+4f=6+6f
6f+6-6=6-6+6f
6f=6f
This means f can equal any number which we call mathematically infinite solution or infinitely many solutions
