answer:Since the foci of the ellipse are on the x-axis, we have m^{2} > 2+m, which simplifies to m^{2}-2-m > 0. Solving this inequality gives m > 2 or m < -1. Additionally, since 2+m > 0, we have m > -2. Therefore, the range of values for m is: m > 2 or -2 < m < -1. Hence, the correct option is boxed{text{D}}. First, based on the condition that the foci of the ellipse are on the x-axis, we establish m^{2} > 2+m. Also, considering 2+m > 0, the intersection of these two ranges gives the answer. This question mainly examines the standard equation of an ellipse. That is, for the standard equation of an ellipse frac {x^{2}}{a^{2}}+ frac {y^{2}}{b^{2}}=1, when the foci are on the x-axis, a > b; when the foci are on the y-axis, a < b.

answer:1. **Understanding the Problem:** We need to determine the positive integers ( n ) such that any triangle ( triangle ABC ) can be decomposed into ( n ) subtriangles, each similar to ( triangle ABC ). 2. **Initial Consideration:** Let's consider a triangle ( triangle ABC ) with side length 1. For a number ( n ) to be decomposable, we need to find a way to divide ( triangle ABC ) into ( n ) smaller triangles, each similar to ( triangle ABC ). 3. **Basic Decomposition:** - For ( n = 1 ), the triangle is itself, so it is trivially decomposable. - For ( n = 2 ), it is not possible to decompose a triangle into 2 smaller triangles that are similar to the original triangle. 4. **Decomposition for ( n = 4 ):** - We can decompose ( triangle ABC ) into 4 smaller triangles, each similar to ( triangle ABC ). This can be done by drawing lines parallel to the sides of ( triangle ABC ) at one-third of the height from each vertex, creating 4 smaller triangles. 5. **Generalizing the Decomposition:** - From ( n = 4 ), we can generate more decomposable numbers by subdividing one of the smaller triangles into 4 even smaller triangles. This gives us ( n = 4, 7, 10, 13, ldots ) (i.e., ( n = 4 + 3k ) for ( k geq 0 )). - Similarly, starting from ( n = 6 ), we can generate ( n = 6, 9, 12, 15, ldots ) (i.e., ( n = 6 + 3k ) for ( k geq 0 )). - Starting from ( n = 8 ), we get ( n = 8, 11, 14, 17, ldots ) (i.e., ( n = 8 + 3k ) for ( k geq 0 )). 6. **Non-decomposable Numbers:** - The only missing values are ( n = 2, 3, 5 ). It is easy to see that these numbers are not feasible because there is no way to decompose a triangle into 2, 3, or 5 smaller triangles that are similar to the original triangle. Conclusion: The positive integers that are decomposable numbers for every triangle are those of the form ( n = 1 ) or ( n = 3k + 1 ) for ( k geq 1 ). The final answer is ( boxed{ n = 1 } ) or ( n = 3k + 1 ) for ( k geq 1 ).

answer:Let's break down the solution step by step. 1. First, we can distribute the integral over the sum, resulting in two separate integrals: int_{- pi}^{pi} dx + int_{- pi}^{pi} sin x dx. 2. The first integral int_{- pi}^{pi} dx is simply the integral of 1 with respect to x from -pi to pi. This is equal to x evaluated from -pi to pi, which gives us pi - (-pi) = 2pi. 3. The second integral int_{- pi}^{pi} sin x dx is the integral of sin x with respect to x from -pi to pi. This is equal to -cos x evaluated from -pi to pi. Since cos is an even function, cos(-pi) = cos(pi), and so the second integral evaluates to 0. 4. Therefore, the sum of the two integrals is 2pi + 0 = 2pi. So, the answer is boxed{2pi}. This problem tests the application of definite integrals and computational skills.

answer:# Problem Solution: Part 1: Maximizing Daily Sales Profit Given: - Cost price per unit: 50 yuan - Daily sales quantity y related to selling price x: y = -x + 80 - 50 < x < 80 We need to find the selling price x to maximize daily sales profit w. **Step 1:** Express daily sales profit w in terms of x. [w = (x - 50) cdot y] Substitute y = -x + 80 into the equation: [w = (x - 50) cdot (-x + 80)] [w = -x^2 + 130x - 4000] **Step 2:** Rewrite the profit function to complete the square. [w = -(x^2 - 130x + 4000)] [w = -(x^2 - 130x + 4225 - 4225 + 4000)] [w = -(x - 65)^2 + 225] **Step 3:** Determine the maximum profit. Since the coefficient of (x - 65)^2 is negative, the parabola opens downwards, indicating a maximum point at x = 65. At x = 65, w reaches its maximum value: [w = 225] Therefore, the selling price should be set at boxed{65} yuan to maximize the daily sales profit, and the maximum profit is boxed{225} yuan. Part 2: Setting Selling Price for a Target Profit of 200 Yuan Given: - Target daily sales profit w = 200 yuan - Selling price should not exceed 68 yuan **Step 1:** Set up the equation for w = 200. [w = -x^2 + 130x - 4000 = 200] Simplify: [-x^2 + 130x - 4200 = 0] **Step 2:** Solve the quadratic equation. [x^2 - 130x + 4200 = 0] Using the quadratic formula, where a = 1, b = -130, and c = 4200: [x = frac{-(-130) pm sqrt{(-130)^2 - 4 cdot 1 cdot 4200}}{2 cdot 1}] [x_1 = 60, quad x_2 = 70] **Step 3:** Choose the valid solution. Since x_2 = 70 does not meet the requirement of being less than or equal to 68 yuan, we discard it. Therefore, the selling price per unit should be set at boxed{60} yuan to achieve a daily sales profit of 200 yuan.