answer:Let's denote the population of the town two years ago as P. In the first year, the population increased by 10%, so the population at the end of the first year would be P + 0.10P = 1.10P. In the second year, the population increased by 20%, so the population at the end of the second year would be 1.10P + 0.20(1.10P) = 1.10P * 1.20. We are given that the population at the end of the second year is 1320. So we can set up the equation: 1.10P * 1.20 = 1320 Now we can solve for P: 1.32P = 1320 P = 1320 / 1.32 P = 1000 So the population of the town two years ago was boxed{1000} .

answer:1. **Introduce Given Polynomials:** Let (P_1(x), P_2(x), P_3(x), ldots, P_{2019}(x)) be the given polynomials. Define their sum (H(x)) as follows: [ H(x) = P_1(x) + P_2(x) + P_3(x) + ldots + P_{2019}(x) ] Note that (H(x)) is a polynomial of degree at most 2018. 2. **Analyze the Roots:** Assume (x_1) is a root of (P_1(x)). According to the given condition, (x_1) is not a root of any other polynomial (P_2(x), P_3(x), ldots, P_{2019}(x)). However, (x_1) must be a root of the sum of the other polynomials: [ x_1 text{ is a root of } P_2(x) + P_3(x) + ldots + P_{2019}(x) ] Therefore, we can write this condition as: [ P_1(x_1) + (P_2(x_1) + P_3(x_1) + ldots + P_{2019}(x_1)) = 0 ] Since (x_1) is a root of (P_1(x)), it means (P_1(x_1) = 0). Thus: [ 0 + (P_2(x_1) + P_3(x_1) + ldots + P_{2019}(x_1)) = 0 implies P_2(x_1) + P_3(x_1) + ldots + P_{2019}(x_1) = 0 ] Consequently: [ H(x_1) = P_1(x_1) + P_2(x_1) + P_3(x_1) + ldots + P_{2019}(x_1) = 0 ] 3. **Repeat the Argument:** Similarly, let (x_2) be a root of (P_2(x)). This (x_2) is not a root of any other polynomial except for (P_2(x)) itself. By the condition, (x_2) must be a root of the sum of the other polynomials as well: [ x_2 text{ is a root of } P_1(x) + P_3(x) + ldots + P_{2019}(x) ] Hence: [ P_2(x_2) + (P_1(x_2) + P_3(x_2) + ldots + P_{2019}(x_2)) = 0 ] Since (P_2(x_2) = 0), we have: [ 0 + (P_1(x_2) + P_3(x_2) + ldots + P_{2019}(x_2)) = 0 implies P_1(x_2) + P_3(x_2) + ldots + P_{2019}(x_2) = 0 ] Therefore: [ H(x_2) = P_1(x_2) + P_2(x_2) + P_3(x_2) + ldots + P_{2019}(x_2) = 0 ] 4. **Generalize for All Roots:** By repeating this argument for every polynomial (P_i(x)) and their respective roots (x_1, x_2, ldots, x_{2019}), we see: [ H(x_i) = 0 quad text{for all } i = 1, 2, ldots, 2019 text{ and } x_1, x_2, ldots, x_{2019} text{ are distinct} ] 5. **Conclusion:** Since a polynomial (H(x)) of degree at most 2018 has 2019 distinct roots, it must be identically zero. By the Fundamental Theorem of Algebra, a polynomial of degree (d) can have at most (d) roots unless it is the zero polynomial. Therefore, (H(x) equiv 0): [ H(x) equiv 0 ] (boxed{0})

answer:To determine the interval where the function is monotonic increasing, we examine when the inside of the logarithm is positive. The inequality -x^2+2x>0 holds true when 0<x<2. Let t = -x^2 + 2x, then y = log_{frac{1}{2}}t. We need to evaluate the monotonicity of both the inner function t and the outer logarithmic function y: 1. The inner function t = -x^2 + 2x can be rewritten as t = -(x^2 - 2x). Completing the square gives t = -(x-1)^2 + 1. From this form, it is evident that t achieves a maximum when x = 1 and decreases on either side of this point. Therefore, t is a decreasing function on the interval (1,2). 2. The outer function y = log_{frac{1}{2}}t is a decreasing function because the base of the logarithm frac{1}{2} is between 0 and 1. In other words, as t increases, y decreases. By combining these two observations, we note that due to the "opposite" monotonic characteristics of the inner and outer functions (the inner decreases while the outer also decreases), the composite function f(x) will be increasing. Hence, the interval of monotonic increase for f(x) = log_{frac{1}{2}}(-x^2+2x) is (1,2). So the answer is: boxed{(1,2)}.

answer:(1) The sampling fraction is frac{400}{8000} = frac{1}{20}. - For freshmen (1st year), we sample 1600 times frac{1}{20} = 80 students. - For sophomores (2nd year), we sample 3200 times frac{1}{20} = 160 students. - For juniors (3rd year), we sample 2000 times frac{1}{20} = 100 students. - For seniors (4th year), we sample 1200 times frac{1}{20} = 60 students. Thus, the number of students sampled from each grade is boxed{80, 160, 100, text{ and } 60} respectively. (2) For sampling faculty and staff, we would use systematic sampling: - **Step 1**: Assign numbers to the 505 faculty and staff members, from 001 to 505. - **Step 2**: Use simple random sampling to exclude 5 individuals (the excluding method can be done using a random number table), and renumber the remaining 500 individuals from 001 to 500. - **Step 3**: Calculate the sampling interval, k = frac{500}{50} = 10, which means the population is divided into 50 segments with a gap of 10. - **Step 4**: Randomly select a starting number from the first segment, for example, 008. - **Step 5**: Starting from 008, select every 10th number. Therefore, we obtain 008, 018, 028, ..., up to 498. The corresponding faculty and staff for these numbers form a sample size of 50. So, the sample size for the faculty and staff will be boxed{50}, selected through systematic sampling.