answer:To analyze the given problem, we start by setting x_{1}=1-x_{2} in the given inequality. This substitution leads us to: [ frac{f(1-x_{2})}{f(x_{2})} + frac{f(x_{2})}{f(1-x_{2})} leq 2 ] Given that for any x in (0,1), f(x) > 0, we can apply the Cauchy-Schwarz inequality, which gives us: [ frac{f(1-x_{2})}{f(x_{2})} + frac{f(x_{2})}{f(1-x_{2})} geq 2sqrt{frac{f(1-x_{2})}{f(x_{2})} cdot frac{f(x_{2})}{f(1-x_{2})}} = 2 ] From the inequality and the application of the Cauchy-Schwarz inequality, we can conclude that: [ frac{f(1-x_{2})}{f(x_{2})} + frac{f(x_{2})}{f(1-x_{2})} = 2 ] This equality holds only when frac{f(1-x_{2})}{f(x_{2})} = frac{f(x_{2})}{f(1-x_{2})}, which simplifies to f(x_{2}) = f(1-x_{2}). This implies that the graph of f(x) is symmetric about the line x = frac{1}{2}, making statement ③ correct. For statement ①, since f(x) = f(1-x), it contradicts the claim that f(x) > f(1-x) for any x in (0,1). Therefore, statement ① is incorrect. Considering the inequality frac{f(x_{1})}{f(x_{2})} + frac{f(x_{1})}{f(x_{2})} leq 2 and knowing that f(x) > 0 for any x in (0,1), we deduce that f(x_{1}) leq f(x_{2}). This implies that f(x) is a constant function, say f(x) = k where k > 0. For statement ②, the function y = frac{f(x)}{x} + x = frac{k}{x} + x is analyzed. This function is monotonically decreasing on (0, sqrt{k}) and increasing on (sqrt{k}, +infty). Therefore, f(x) being monotonically decreasing on (0,1) only holds if k geq 1, which is not necessarily true given the information. Hence, statement ② is incorrect. In conclusion, only statement ③ is correct, which means the correct answer is: [ boxed{B} ]

answer:Start by rewriting the given equation of the circle to facilitate completing the square: [x^2 - 6x + y^2 - 18y = 9.] Next, complete the square for x: [x^2 - 6x implies x^2 - 6x + left(frac{-6}{2}right)^2 = x^2 - 6x + 9 = (x-3)^2.] Complete the square for y: [y^2 - 18y implies y^2 - 18y + left(frac{-18}{2}right)^2 = y^2 - 18y + 81 = (y-9)^2.] Substituting these back into the equation: [(x-3)^2 + (y-9)^2 = 99.] This tells us that the circle has a center at (3, 9). Therefore, h+k = 3+9. Conclusion: The value of h+k, where (h, k) is the center of the circle, is boxed{12}.

answer:To solve for (n), we first determine (-2839) modulo 10: [ -2839 div 10 = -283.9 ] Thus, (-2839) rounded to the nearest lower whole number is (-284), and (-284 times 10 = -2840). Therefore, (-2839 - (-2840) = 1). So, we have [ -2839 equiv 1 pmod{10} ] Thus, the integer (n) is (n = boxed{1}).

answer:To form a three-digit number using the digits 1, 2, and 3 without repetition, we need to consider all possible permutations of these three digits. 1. The first digit (hundreds place) can be any one of the three numbers. This gives us 3 options. 2. The second digit (tens place) can be any one of the remaining two numbers, since we cannot repeat the number chosen for the hundreds place. This gives us 2 options. 3. The third digit (ones place) can only be the remaining number, as the other two have already been used. This leaves us with 1 option. To find the total number of permutations, we multiply the number of options for each digit's place: 3 times 2 times 1 = 6 So, there are 6 different three-digit numbers that can be formed without repeating any digits using 1, 2, and 3. The answer is boxed{6}.