There are 4! = 24 poosible permutations of the four letters. Let the letters be A, B, C and D. Two permutations will have only letter A in the correct envelope, two more permutations will have only letter B in the correct envelope, two more will have only letter C in the correct envelope and two more will have only letter D in the correct envelope. Therefore 8 out of the 24 possible permutations will have only one letter with the correct address. The required probability is 8/24 = 1/3.
It all depends in what there exponents are
The answer is the last one
square tan theta
(tan^2) + 1 = sec^2
so sec^2 = 11/4
sec = - (sqrt 11)/2
cos= - 2/ (sqrt 11)
answer choice a if you simplify
