answer:Let's denote the profit share of A as ( P_A ), the profit share of B as ( P_B ), and the profit share of C as ( P_C ). We know that ( P_B = Rs. 1700 ). The profit share of each partner is proportional to their investment and the time period for which the capital was invested. Since the time period is the same for all, we can ignore it for this calculation. We can set up the following proportion for A and B: [ frac{P_A}{8000} = frac{P_B}{10000} ] We know ( P_B = Rs. 1700 ), so we can solve for ( P_A ): [ P_A = frac{8000}{10000} times 1700 ] [ P_A = frac{8}{10} times 1700 ] [ P_A = frac{4}{5} times 1700 ] [ P_A = 4 times 340 ] [ P_A = Rs. 1360 ] Now, we know the difference between the profit shares of A and C is Rs. 680: [ P_C - P_A = Rs. 680 ] [ P_C = P_A + Rs. 680 ] [ P_C = Rs. 1360 + Rs. 680 ] [ P_C = Rs. 2040 ] Now, let's denote the investment of C as ( I_C ). We can set up the following proportion for B and C: [ frac{P_B}{10000} = frac{P_C}{I_C} ] We know ( P_B = Rs. 1700 ) and ( P_C = Rs. 2040 ), so we can solve for ( I_C ): [ frac{1700}{10000} = frac{2040}{I_C} ] [ I_C = frac{2040 times 10000}{1700} ] [ I_C = frac{2040}{17} times 1000 ] [ I_C = 120 times 1000 ] [ I_C = Rs. 120000 ] Therefore, C invested Rs. boxed{120000} in the business.

answer:# Step-by-Step Solution Part (1): Discussing the Monotonicity of f(x) 1. First, we find the derivative of f(x): [ f'(x) = frac{a}{x} - x = frac{-x^2 + a}{x}. ] 2. For a leqslant 0, we have: [ f'(x) < 0 quad text{for all } x > 0, ] indicating that f(x) is monotonically decreasing on (0, +infty). 3. For a > 0, setting f'(x) = 0 gives: [ x = sqrt{a}. ] - For x in (0, sqrt{a}), we find f'(x) > 0, meaning f(x) is monotonically increasing. - For x in (sqrt{a}, +infty), we find f'(x) < 0, meaning f(x) is monotonically decreasing. Therefore, f(x) is monotonically increasing on (0, sqrt{a}) and monotonically decreasing on (sqrt{a}, +infty). Part (2)①: Finding the Range of Real Number a 1. From the monotonicity analysis, when a leqslant 0, f(x) cannot have two zeros as it is monotonically decreasing. 2. For a > 0, f(x) has a maximum at x = sqrt{a}, and we require: [ f(sqrt{a}) = alnsqrt{a} - frac{1}{2}a > 0, ] which simplifies to a > e. 3. Considering the behavior of f(x) as x rightarrow +infty and x rightarrow +0, we conclude that f(x) can have two zeros only if a > e. Thus, the range of a is boxed{(e, +infty)}. Part (2)②: Proving x_{1} + x_{2} > 2x_{0} 1. The tangent lines at (x_{1}, 0) and (x_{2}, 0) are given by: [ l_{1}: y = left(frac{a}{x_{1}} - x_{1}right)(x - x_{1}), quad l_{2}: y = left(frac{a}{x_{2}} - x_{2}right)(x - x_{2}). ] 2. Solving these equations simultaneously yields: [ x_{0} = frac{x_{1} + x_{2}}{frac{a}{x_{1}x_{2}} + 1}. ] 3. To prove x_{1} + x_{2} > 2x_{0}, it suffices to show: [ frac{x_{1} + x_{2}}{x_{0}} > 2 quad Leftrightarrow quad frac{a}{x_{1}x_{2}} > 1. ] 4. Given the zeros of f(x), we find: [ a = frac{frac{1}{2}(x_{1}^2 - x_{2}^2)}{ln{x_{1}} - ln{x_{2}}}. ] 5. Letting t = frac{x_{1}}{x_{2}} < 1, we analyze the function h(t) = ln t - frac{1}{2}left(t - frac{1}{t}right) for 0 < t < 1. 6. We find that h'(t) < 0 for 0 < t < 1, indicating h(t) is monotonically decreasing. 7. Since h(t) > h(1) = 0, we conclude: [ ln t > frac{1}{2}left(t - frac{1}{t}right) quad text{for } 0 < t < 1, ] which implies frac{a}{x_{1}x_{2}} > 1. Therefore, we have proven that x_{1} + x_{2} > 2x_{0}, encapsulating the final answer as boxed{x_{1} + x_{2} > 2x_{0}}.

answer:Let the side length of the equilateral triangle be ( s ), which is also the side length of the square. The area of the square is ( s^2 ). The area of the equilateral triangle is given by: [ text{Area of the triangle} = frac{sqrt{3}}{4}s^2 ] The total area of the hexagon, which is the sum of the areas of the square and the equilateral triangle, is: [ text{Total area} = s^2 + frac{sqrt{3}}{4}s^2 = left(1 + frac{sqrt{3}}{4}right)s^2 ] The fraction of the hexagon's area that is the area of the triangle is: [ frac{frac{sqrt{3}}{4}s^2}{left(1 + frac{sqrt{3}}{4}right)s^2} = frac{frac{sqrt{3}}{4}}{1 + frac{sqrt{3}}{4}} = frac{sqrt{3}}{4 + sqrt{3}} ] Simplifying further: [ frac{sqrt{3}}{4 + sqrt{3}} = frac{sqrt{3}}{4 + sqrt{3}} cdot frac{4 - sqrt{3}}{4 - sqrt{3}} = frac{4sqrt{3} - 3}{16 - 3} = frac{4sqrt{3} - 3}{13} ] Converting to percentage: [ left(frac{4sqrt{3} - 3}{13}right) times 100 approx 20.77% ] Thus, the percentage of the hexagon's area that is the area of the equilateral triangle is (boxed{20.77%}).

answer:The average of a and b is 30, so the sum of a and b is 2 times 30, which is 60. Therefore, a + b = 60. The average of b and c is 60, so the sum of b and c is 2 times 60, which is 120. Therefore, b + c = 120. To find the value of c - a, we can subtract the first equation from the second equation: (b + c) - (a + b) = 120 - 60 c - a = 60 Therefore, the value of c - a is boxed{60} .