Find three consecutive odd integers such that the largest decreased by three times the second is 47 less than the smallest

X = smallest y = second z = biggest

x = x
y = x + 2
z = x + 4

(x+4) - 3(x+2) = x - 47
x+4 - 3x - 6 = x - 47
-2x - 2 = x - 47
-3x - 2 = -47
-3x = -45
x = 15

x = 15
y = 15 + 2
z = 15 + 4

x = 15
y = 17
z = 19

The answer for you question is b

The answer is -pie over 2