answer:The derivative of the function is f′(x)=(2ax+b)e^{x}+(ax^{2}+bx+c)e^{x}=[ax^{2}+(2a+b)x+b+c]e^{x}. Let g(x)=ax^{2}+(2a+b)x+b+c. Since e^{x} > 0, the zeros of y=f′(x) are the same as the zeros of g(x)=ax^{2}+(2a+b)x+b+c, and f′(x) and g(x) have the same sign. Given that a > 0, when x < -3 or x > 0, g(x) > 0, i.e., f′(x) > 0; when -3 < x < 0, g(x) < 0, i.e., f′(x) < 0. Therefore, the monotonically increasing intervals of f(x) are (-infty,-3) and (0,+infty), and the monotonically decreasing interval is (-3,0). x=0 is the minimum point of f(x), so we have the following system of equations: begin{cases} & c=-1, & b+c=0, & 9a-3(2a+b)+b+c=0, end{cases} Solving the system, we get a=1, b=1, and c=-1. Therefore, boxed{a+b+c=1}.

answer:# Step-by-Step Solution: Part 1: Writing the Function Expression of Annual Profit S **Case 1:** When 0 < x leq 20, The sales revenue R is given by R = 500 - 2x. The cost includes a fixed R&D cost of 1.5 million yuan and a variable cost of 380 yuan per unit. Since x is in thousands of units, the total cost is 380x + 150 (in million yuan). Therefore, the profit S can be calculated as: [ begin{align*} S &= xR - 380x - 150 &= x(500 - 2x) - 380x - 150 &= 500x - 2x^2 - 380x - 150 &= -2x^2 + 120x - 150 end{align*} ] **Case 2:** When x > 20, The sales revenue R changes to R = 370 + frac{2140}{x} - frac{6250}{x^2}. Using the same approach for calculating profit S: [ begin{align*} S &= xR - 380x - 150 &= xleft(370 + frac{2140}{x} - frac{6250}{x^2}right) - 380x - 150 &= 370x + 2140 - frac{6250}{x} - 380x - 150 &= -10x - frac{6250}{x} + 1990 end{align*} ] Thus, the function expression of the annual profit S (in million yuan) in terms of the annual production quantity x (in thousand units) is: [ S = left{ begin{array}{ll} -2x^2 + 120x - 150, & 0 < x leq 20 -10x - frac{6250}{x} + 1990, & x > 20 end{array} right. ] Part 2: Finding the Maximum Profit and Corresponding Production Quantity **Case 1:** For 0 < x leq 20, Rewriting S as a completed square gives us: [ S = -2(x^2 - 60x + 900) + 1650 = -2(x - 30)^2 + 1650 ] The function S is increasing on (0, 20] because the vertex of the parabola (x = 30) is outside this interval. Thus, at x = 20, S reaches its maximum for this interval: [ S = -2(20^2) + 120 cdot 20 - 150 = 1450 ] **Case 2:** For x > 20, Using AM-GM inequality on x + frac{625}{x}: [ S = -10left(x + frac{625}{x}right) + 1990 leq -10 cdot 2sqrt{x cdot frac{625}{x}} + 1990 = 1490 ] The equality holds when x = frac{625}{x}, which implies x = 25. Therefore, at x = 25, S reaches its maximum for this interval: [ S = 1490 ] Comparing the maximum values from both cases, 1450 < 1490. Therefore, the company achieves maximum profit when the production quantity is 25 thousand units, and the maximum profit is 1,490 million yuan. [ boxed{text{Maximum profit is achieved at } x = 25 text{ thousand units, with a profit of } 1,490 text{ million yuan.}} ]

answer:To find Reeya's average score, we add up all her scores and then divide by the number of subjects. Her total score is: 55 + 67 + 76 + 82 + 55 = 335 The number of subjects is 5. Now we divide the total score by the number of subjects to find the average: Average score = Total score / Number of subjects Average score = 335 / 5 Average score = 67 Reeya's average score is boxed{67} out of 100.

answer:The given quadratic equation is (x^2 + 18x + 81 = 0). We attempt to factorize the quadratic: [ x^2 + 18x + 81 = (x + 9)^2 ] This gives us: [ (x + 9)^2 = 0 ] [ x + 9 = 0 ] [ x = -9 ] There is only one real solution, (x = -9). However, we were asked for the solution with the greatest absolute value, which here is (|,-9,| = 9). Therefore, the real number solution with the greatest absolute value is (boxed{9}).