answer:Since the domain of f(2^{x}) is [0,1], 0 leqslant x leqslant 1, 1 leqslant 2^{x} leqslant 2, Hence, in f(log_{2}x), let 1 leqslant log_{2}x leqslant 2, Solving it gives 2 leqslant x leqslant 4, Therefore, the answer is boxed{text{C}}. From the domain of f(2^{x}) being [0,1], we can derive that 1 leqslant 2^{x} leqslant 2, which leads to 1 leqslant log_{2}x leqslant 2 in f(log_{2}x), from which we can find the domain of f(log_{2}x). This problem tests the understanding of the domain of a function and how to find it, as well as the operations of exponential and logarithmic functions. It is a basic question.

answer:To compute the domain of f(x), we need to find when lfloor x^2 - 9x + 20 rfloor is not 0 because division by zero is undefined. The discriminant of the quadratic x^2 - 9x + 20 is computed as: [ 9^2 - 4(1)(20) = 81 - 80 = 1 > 0 ] This indicates that the quadratic has real roots. Now, solve x^2 - 9x + 20 = 0 by factoring: [ x^2 - 9x + 20 = (x-4)(x-5) = 0 ] Thus, x = 4 and x = 5 are roots. Next, find the interval where this quadratic is less than 1: [ x^2 - 9x + 20 < 1 quad rightarrow quad x^2 - 9x + 19 < 0 ] Factoring x^2 - 9x + 19: [ x^2 - 9x + 19 = (x - 4.5)^2 - 0.25 < 0 ] Take the square root: [ (x - 4.5)^2 < 0.25 quad rightarrow quad -0.5 < x - 4.5 < 0.5 ] [ 4 < x < 5 ] Thus, x^2 - 9x + 19 is negative between 4 and 5, meaning lfloor x^2-9x+20 rfloor = 0 in this interval. The domain of f(x) excludes (4,5), where the floor of the quadratic is 0. Therefore, the domain of f is: boxed{(-infty, 4] cup [5, infty)}

answer:Since complement_U A={0}, it means 0 in U and 0 notin A. This implies that the equation x^2-2x=0 holds true. Solving this equation, we get x_1=2 and x_2=0. When x_1=2, we have |2x-1|=3 in A, which is consistent with the problem statement. When x_2=0, we have |2x-1|=1. In this case, set A does not satisfy the condition of having distinct elements, so x_2=0 is discarded. Therefore, such a real number x exists, and it is boxed{x=2}.

answer:First, let's calculate the time it takes for Mobius to travel from Florence to Rome with a typical load at a top speed of 11 miles per hour. Time = Distance / Speed For the trip to Rome with a load: Time with load = 143 miles / 11 mph = 13 hours Now, let's calculate the time it takes for Mobius to return to Florence without a load at a top speed of 13 miles per hour. For the trip back to Florence without a load: Time without load = 143 miles / 13 mph = 11 hours The total time spent traveling at top speed is the sum of the time with load and the time without load: Total travel time = Time with load + Time without load Total travel time = 13 hours + 11 hours Total travel time = 24 hours The trip takes 26 hours in total, so the time spent on rest stops is the difference between the total trip time and the total travel time: Rest stop time = Total trip time - Total travel time Rest stop time = 26 hours - 24 hours Rest stop time = 2 hours Since the question asks for the rest stops during each half of the trip, we will divide the total rest stop time by 2: Rest stop time per half = Rest stop time / 2 Rest stop time per half = 2 hours / 2 Rest stop time per half = 1 hour Therefore, Mobius takes a boxed{1-hour} rest stop during each half of the trip.