answer:# Problem: Given x_{i} > 0 and k geq 1, prove that: sum_{i=1}^{n} frac{1}{1+x_{i}} cdot sum_{i=1}^{n} x_{i} leq sum_{i=1}^{n} frac{x_{i}^{k+1}}{1+x_{i}} cdot sum_{i=1}^{n} frac{1}{x_{i}^{k}}. **Solution One:** We start with the inequality we want to prove: sum_{i=1}^{n} frac{1}{1+x_{i}} cdot sum_{i=1}^{n} x_{i} leq sum_{i=1}^{n} frac{x_{i}^{k+1}}{1+x_{i}} cdot sum_{i=1}^{n} frac{1}{x_{i}^{k}}. We subtract the left-hand side from the right-hand side and show that the result is non-negative. sum_{i=1}^{n} frac{x_{i}^{k+1}}{1+x_{i}} cdot sum_{i=1}^{n} frac{1}{x_{i}^{k}} - sum_{i=1}^{n} frac{1}{1+x_{i}} cdot sum_{i=1}^{n} x_{i} geq 0. Let us analyze the difference by expanding both summations: [ begin{aligned} &sum_{i neq j} frac{x_{i}^{k+1}}{1+x_{i}} cdot frac{1}{x_{j}^{k}} - sum_{i neq j} frac{x_{j}}{1+x_{i}} &= sum_{i neq j} left(frac{x_{i}^{k+1}}{(1+x_{i}) x_{j}^{k}} - frac{x_{j}}{(1+x_{i})}right) &= sum_{i neq j} frac{x_{i}^{k+1} - x_{j}^{k+1}}{(1+x_{i}) x_{j}^{k}} &= frac{1}{2} sum_{i neq j} left[frac{x_{i}^{k+1} - x_{j}^{k+1}}{(1+x_{i}) x_{j}^{k}} + frac{x_{j}^{k+1} - x_{i}^{k+1}}{(1+x_{j}) x_{i}^{k}}right] &= frac{1}{2} sum_{i neq j} (x_{i}^{k+1} - x_{j}^{k+1}) frac{(1+x_{j}) x_{i}^{k} - (1+x_{i}) x_{j}^{k}}{(1+x_{i})(1+x_{j}) x_{i}^{k} x_{j}^{k}} &= frac{1}{2} sum_{i neq j} (x_{i}^{k+1} - x_{j}^{k+1}) frac{(x_{i}^{k} - x_{j}^{k}) + x_{i} x_{j} (x_{i}^{k-1} - x_{j}^{k-1})}{(1+x_{i})(1+x_{j}) x_{i}^{k} x_{j}^{k}}. end{aligned} ] Since x_{i} > 0, x_{i}^{k+1} geq x_{j}^{k+1}, and x_{i}^{k} geq x_{j}^{k}, we have: sum_{i=1}^{n} frac{1}{1+x_{i}} cdot sum_{i=1}^{n} x_{i} leq sum_{i=1}^{n} frac{x_{i}^{k+1}}{1+x_{i}} cdot sum_{i=1}^{n} frac{1}{x_{i}^{k}}. **Solution Two:** Assume x_{1} geq x_{2} geq cdots geq x_{n} > 0. Then, we have the following inequalities: frac{1}{x_{1}^{k}} leq frac{1}{x_{2}^{k}} leq cdots leq frac{1}{x_{n}^{k}}, frac{x_{1}^{k}}{1+x_{1}} geq frac{x_{2}^{k}}{1+x_{2}} geq cdots geq frac{x_{n}^{k}}{1+x_{n}}. By applying Chebyshev's inequality, we get: [ begin{aligned} text{LHS} &= left( sum_{i=1}^{n} frac{1}{1+x_{i}}right) left( sum_{i=1}^{n} x_{i} right) &= left( sum_{i=1}^{n} frac{1}{x_{i}^{k}} cdot frac{x_{i}^{k}}{1+x_{i}} right) left( sum_{i=1}^{n} x_{i} right) &leq left( sum_{i=1}^{n} frac{1}{x_{i}^{k}}right) cdot left( sum_{i=1}^{n} frac{x_{i}^{k}}{1+x_{i}} right) left( frac{1}{n} right) cdot left( x_{1} + x_{2} + cdots + x_{n}right) &leq left( sum_{i=1}^{n} x_{i} cdot frac{x_{i}^{k}}{1+x_{i}} right) left( sum_{i=1}^{n} frac{1}{x_{i}^{k}} right) &= left( sum_{i=1}^{n} frac{x_{i}^{k+1}}{1+x_{i}} right) left( sum_{i=1}^{n} frac{1}{x_{i}^{k}} right). end{aligned} ] Therefore, we have shown that: sum_{i=1}^{n} frac{1}{1+x_{i}} cdot sum_{i=1}^{n} x_{i} leq sum_{i=1}^{n} frac{x_{i}^{k+1}}{1+x_{i}} cdot sum_{i=1}^{n} frac{1}{x_{i}^{k}}. blacksquare

answer:First, set up the equation f(f(x)) = f(x): f(f(x)) = (f(x))^2 - 5f(x) + 1, text{ where } f(x) = x^2 - 5x + 1. Substituting f(x) into the expression for f(f(x)), we get: (x^2 - 5x + 1)^2 - 5(x^2 - 5x + 1) + 1 = x^2 - 5x + 1. Simplify the equation: (x^2 - 5x + 1)^2 - 5x^2 + 25x - 5 + 1 = x^2 - 5x + 1. Subtract x^2 - 5x + 1 from both sides: (x^2 - 5x + 1)^2 - 5x^2 + 25x - 5 + 1 - (x^2 - 5x + 1) = 0. (x^2 - 5x + 1)^2 - 6x^2 + 30x - 5 = 0. Factoring out x^2 - 5x + 1: (x^2 - 5x + 1)((x^2 - 5x + 1) - 6x + 4) = 0. Let's factor the quadratic: (x^2 - 5x + 1)(x^2 - 5x + 1 - 6x + 4) = 0, (x^2 - 5x + 1)(x^2 - 11x + 5) = 0. x^2 - 5x + 1 = 0 text{ and } x^2 - 11x + 5 = 0. Solving these quadratics: x = frac{5 pm sqrt{25 - 4}}{2} = frac{5 pm sqrt{21}}{2}, x = frac{11 pm sqrt{121 - 20}}{2} = frac{11 pm sqrt{101}}{2}. Thus, the values of x are boxed{frac{5 pm sqrt{21}}{2}, frac{11 pm sqrt{101}}{2}}.

answer:First, we need to calculate the volume of one box. The volume of a box is found by multiplying its length by its width by its height. Volume of one box = Length × Width × Height Volume of one box = 15 inches × 12 inches × 10 inches Volume of one box = 1800 cubic inches Now, we know that the total volume occupied by the boxes is 1.08 million cubic inches. To find out how many boxes there are, we divide the total volume by the volume of one box. Number of boxes = Total volume / Volume of one box Number of boxes = 1,080,000 cubic inches / 1800 cubic inches Number of boxes = 600 Now that we know there are 600 boxes, we can calculate the total amount the company pays each month for record storage. Total amount paid per month = Number of boxes × Price per box per month Total amount paid per month = 600 boxes × 0.5 per box per month Total amount paid per month = 300 Therefore, the company pays boxed{300} each month for record storage.

answer:1. **Define Variables:** Let y be the population in 1997, and let k be the constant of proportionality. 2. **Setup Equations Based on Given Information:** - For n = 1995, the population difference between 1997 and 1995 is y - 45, and this is proportional to the population in 1996. Therefore, the equation is: [ y - 45 = k cdot 70 ] - For n = 1996, the population difference between 1998 and 1996 is 145 - 70 = 75, and this is proportional to the population in 1997. Therefore, the equation is: [ 75 = k cdot y ] 3. **Express k in Terms of y:** From 75 = k cdot y, solve for k: [ k = frac{75}{y} ] 4. **Substitute k Into the First Equation:** Substitute k = frac{75}{y} into y - 45 = k cdot 70: [ y - 45 = frac{75 cdot 70}{y} ] Simplify this to: [ y - 45 = frac{5250}{y} ] 5. **Form a Quadratic Equation:** Multiply both sides by y: [ y^2 - 45y - 5250 = 0 ] 6. **Factorize the Quadratic Equation:** Factorizing: [ y^2 - 45y - 5250 = (y - 105)(y + 50) = 0 ] 7. **Solve for y:** Set each factor equal to zero: [ y - 105 = 0 quad text{or} quad y + 50 = 0 ] Solving these, we find: [ y = 105 quad text{and} quad y = -50 ] Since the population cannot be negative, we have y = 105. 8. **Conclusion:** Therefore, the population in 1997 was 105. The final answer is boxed{C) 105}