answer:1. **Define the areas of the triangles:** Let (left[ cdot right]) denote the area. We define the areas of the triangles formed by extending the sides of the pentagon: [ left[ triangle A_3 B_1 A_4 right] = left[ triangle A_4 B_2 A_5 right] = left[ triangle A_5 B_3 A_1 right] = left[ triangle A_1 B_4 A_2 right] = left[ triangle A_2 B_5 A_3 right] = T. ] 2. **Express the area of quadrilateral (A_2A_5B_2B_5):** [ left[ A_2 A_5 B_2 B_5 right] = left[ A_1 A_2 A_3 A_4 A_5 right] - left[ triangle A_1 A_2 A_5 right] + 2T. ] 3. **Calculate the area of the regular pentagon (A_1A_2A_3A_4A_5):** Since (A_1A_2A_3A_4A_5) is a regular pentagon with side length 1, we need to find its apothem (a). The area of the pentagon is given by: [ left[ A_1 A_2 A_3 A_4 A_5 right] = frac{1}{2} a cdot P, ] where (P = 5) is the perimeter of the pentagon. To find (a), we use the tangent of the central angle: [ tan(36^circ) = frac{frac{1}{2}}{a} implies a = frac{cot(36^circ)}{2}. ] Hence, [ left[ A_1 A_2 A_3 A_4 A_5 right] = frac{5}{4} cdot cot(36^circ). ] 4. **Calculate the area of (triangle A_1 A_2 A_5):** [ left[ triangle A_1 A_2 A_5 right] = frac{1}{2} cdot A_1 A_2 cdot A_1 A_5 cdot sin angle A_2 A_1 A_5 = frac{1}{2} cdot 1^2 cdot sin(108^circ). ] 5. **Calculate the area (T):** To find (T), we need the height (h) of one of the outer triangles, say (triangle B_1 A_3 A_4). The height (h) is given by: [ tan(72^circ) = frac{h}{frac{1}{2}} implies h = frac{tan(72^circ)}{2}. ] Thus, [ T = frac{1}{2} cdot h cdot 1 = frac{tan(72^circ)}{4}. ] 6. **Combine the areas:** [ left[ A_2 A_5 B_2 B_5 right] = frac{5}{4} cdot cot(36^circ) - frac{sin(108^circ)}{2} + frac{tan(72^circ)}{2}. ] 7. **Simplify the trigonometric expressions:** Using the known values: [ sin(108^circ) = cos(18^circ), quad tan(72^circ) = cot(18^circ), quad cot(36^circ) = tan(54^circ). ] We find: [ left[ A_2 A_5 B_2 B_5 right] = frac{5}{4} cdot tan(54^circ) - frac{cos(18^circ)}{2} + frac{cot(18^circ)}{2}. ] 8. **Calculate the area of the 10-sided polygon:** [ left[ text{10-sided polygon} right] = left[ A_1 A_2 A_3 A_4 A_5 right] + 5T = frac{5}{4} cdot cot(36^circ) + 5T. ] 9. **Simplify the final ratio:** [ frac{left[ A_2 A_5 B_2 B_5 right]}{left[ text{10-sided polygon} right]} = frac{frac{5}{4} cdot tan(54^circ) - frac{cos(18^circ)}{2} + frac{cot(18^circ)}{2}}{frac{5}{4} cdot tan(54^circ) + frac{5}{4} cdot tan(72^circ)}. ] Simplifying further, we find: [ boxed{frac{1}{2}}. ]

answer:The graph of the function f(x) = 2x^2 + 4x - 1 is a parabola that opens upwards and has its axis of symmetry along the line x = -1. Since the parabola is symmetrical about x = -1 and opens upwards, the function is decreasing on the interval [-2, -1] and increasing on the interval [-1, 2]. We evaluate the function at the endpoints of the interval [-2, 2]: - At x = -2, f(-2) = 2(-2)^2 + 4(-2) - 1 = 8 - 8 - 1 = -1. - At x = 2, f(2) = 2(2)^2 + 4(2) - 1 = 16 + 8 - 1 = 23. Additionally, we examine the vertex of the parabola, which lies at the axis of symmetry x = -1: - At x = -1, f(-1) = 2(-1)^2 + 4(-1) - 1 = 2 - 4 - 1 = -3. Since the function is continuous and differentiable on the interval [-2, 2], and the parabola opens upwards, the maximum value occurs at one of the endpoints of the interval. Comparing the calculated values, f(-2) = -1, f(2) = 23, and f(-1) = -3, we conclude that the maximum value of f(x) on the interval [-2, 2] is boxed{23}.

answer:Let's denote the number of yellow marbles as Y and the number of blue marbles as B. According to the information given, we have two equations: 1) B = Y - 2 (since there are 2 fewer blue marbles than yellow marbles) 2) Y + B = 240 (since the sum of the yellow and blue marbles is 240) Now, we can substitute the first equation into the second equation to solve for Y: Y + (Y - 2) = 240 2Y - 2 = 240 2Y = 240 + 2 2Y = 242 Y = 242 / 2 Y = 121 So, there are boxed{121} yellow marbles.

answer:**Analysis** This question mainly examines the understanding of the graph of a quadratic function. For a quadratic inequality, it is necessary to pay attention to its direction of opening, axis of symmetry, and discriminant. When m leqslant 0, it is obviously not valid; when m > 0, since f(0)=1 > 0, it is only necessary to discuss the axis of symmetry. **Solution** Solution: When m=0, f(x)=-8x+1 > 0 does not always hold, g(x)=0, this condition is not met; When m < 0, g(x)=mx is always negative for x > 0, and the graph of f(x)=2mx^2-2(4-m)x+1 opens downward, So for any x > 0, it is obviously not always positive; When m > 0, g(x)=mx is always positive for x > 0, so it is only necessary that f(x)=2mx^2-2(4-m)x+1, is always positive for x leqslant 0, if -dfrac{b}{2a} =dfrac{4−m}{2m} geqslant 0, that is 0 < m leqslant 4, the conclusion is obviously valid, If -dfrac{b}{2a} =dfrac{4−m}{2m} < 0, that is m > 4, at this time, as long as Delta=4(4-m)^2-8m < 0 is satisfied, we get 4 < m < 8． In summary, the range of m is 0 < m < 8, Therefore, the correct choice is boxed{text{B}}.