answer:Since f(x) = -f(x+1), it follows that f(x+2) = f(x), which means the function has a period of 2. Let x in [-1, 0], then x+2016 in [2015, 2016], Since when x in [2015, 2016], f(x) = x - 2017, we have f(x) = f(x+2016) = x - 1. Let x in [0, 1], then -x in [-1, 0], thus f(-x) = -x - 1, Since f(x) is an even function on mathbb{R}, we get f(x) = -x - 1 for x in [0, 1]. Since sin 1 > cos 1, it follows that f(sin 1) < f(cos 1). Therefore, the correct choice is boxed{text{D}}. By determining that the function has a period of 2 and f(x) = -x - 1 for x in [0, 1], we can conclude based on the monotonicity of the function. This problem tests the understanding of the monotonicity and periodicity of functions, as well as computational skills, and is considered a medium-level question.

answer:By shifting the graph of f(x) = cos omega x (omega > 0) to the right by frac{pi}{3} units, we obtain the graph of the function y = g(x) = cos omega left(x - frac{pi}{3}right). If y = g(x) is an odd function, then frac{pi}{3}omega = kpi + frac{pi}{2}, where k in mathbb{Z}. When k = 0, omega reaches its minimum value of frac{3}{2}. Therefore, the correct choice is: boxed{C}. This problem mainly examines the transformation rules of the graph of the function y = Asin(omega x + phi) and the even-odd properties of the cosine function, aiming to find the minimum value of omega. It is a basic question focusing on the transformation rules of the graph of the function y = Asin(omega x + phi) and the symmetry of the cosine function's graph.

answer:To solve this problem, we'll break it down into the two parts as requested. # Part 1: Average Number of Computers Infected in Each Round Let's denote x as the average number of computers one computer infects in each round. After two rounds of infection, the total number of infected computers is given by the initial computer plus the computers it infects in the first round, plus the computers those infected in the first round go on to infect in the second round. This can be represented as: [1 + x + x^2 = 81] This equation simplifies to: [(1 + x)^2 = 81] Taking the square root of both sides gives us two possible equations: [1 + x = 9 quad text{or} quad 1 + x = -9] Solving these for x gives: [x = 9 - 1 = 8 quad text{and} quad x = -9 - 1 = -10] We discard x = -10 since a negative number of infections doesn't make sense in this context. Therefore, we have: [x = 8] So, on average, each computer will infect boxed{8} computers in each round of infection. # Part 2: Number of Infected Computers After 3 Rounds To find out if the number of infected computers exceeds 700 after 3 rounds of infection, we use the formula for the total number of infections after n rounds, which, with our value of x = 8, becomes: [(1 + x)^3 = (1 + 8)^3] Calculating this gives: [9^3 = 729] Since 729 > 700, we can conclude that after 3 rounds of infection, the number of infected computers will indeed exceed 700. Therefore, the answer is boxed{text{Yes}}, the number of infected computers will exceed 700 after 3 rounds of infection.

answer:To find out how long the person traveled, we can use the formula: Time = Distance / Speed Given that the distance is 195 km and the speed is 65 km/hr, we can plug these values into the formula: Time = 195 km / 65 km/hr Time = 3 hours So, the person traveled for boxed{3} hours to reach the destination.