2-5 x 1/2 = 0.2 (Length x Width)
(1/4)^-2 - (5^0 x 2) x 1^-1 =
(4/1)^2 - (1 x 2) x 1 = 16-2 = 14
If you raise something to the power of -2, swap numerator and denominator and remove the minus.
So (1/4)^-2 = 4^2 = 16
Also 1^-1 is just 1, not -1.
Answer:
How many of these passwords contain at least one occurrence of at least one of the five special characters?