answer:# Problem: For a positive integer (n geq 4), if convex n-sides polygon (A_{1} A_{2} cdots A_{n}) satisfies for any (2 leq k leq n-2), A_{1} A_{k+1}=A_{2} A_{k+2}=cdots =A_{n-1} A_{k-1}=A_{n} A_{k}, it is called a "regular n-sides polygon". For instance, a rectangle and an isosceles trapezoid are regular quadrilaterals. (1) Please construct a regular hexagon that is not a perfect hexagon. (2) Proof: For (n geq 7), all regular n-sides polygons are perfect n-sides polygons. **Proof for (2):** 1. Start with the given properties: - We know that for any (2 leq k leq n-2), A_{1} A_{k+1}=A_{2} A_{k+2}=cdots =A_{n-1} A_{k-1}=A_{n} A_{k}. - Notice that this implies equal distances between non-adjacent vertices at a fixed step interval. Essentially, this implies that all side lengths and angles must be equal (`regular polygon` property). 2. Consider the properties of roots of unity in complex analysis: - A regular n-sides polygon can be represented by the vertices as roots of unity on the complex plane, given by the (n)th roots of unity: A_k = e^{2pi i k / n} quad text{for} quad k = 0, 1, 2, ldots, n-1. - This setup gives vertex coordinates that evenly distribute along the unit circle on the complex plane. 3. Establish a contradiction for (n geq 7) if a regular (n)-polygon was not a perfect polygon: - Assume there exists a regular n-sides polygon that is not a regular n-sides polygon. - This would mean that there are different side lengths and angles that satisfy the equal distance condition between non-adjacent vertices. However, this would result in a contradiction because having equal non-adjacent distances while the sides and angles are unequal is not geometrically possible for (n geq 7). 4. Thus, for (n geq 7), such conditions force the polygon to have equal side lengths and angles, making it a regular polygon: boxed{text{For } n geq 7, ; text{all regular } ntext{-sides polygons are perfect } ntext{-sides polygons.}} (blacksquare)

answer:1. **Objective**: Find two factors of 1950 whose difference is minimized. 2. **Calculate the approximate square root of 1950**: [ sqrt{1950} approx sqrt{4 times 487.5} approx 2 times sqrt{487.5} approx 2 times 22.09 approx 44.18 ] So, the factors are close to 44. 3. **Prime factorization of 1950**: [ 1950 = 2 times 3 times 5^2 times 13 ] 4. **Identify factors near sqrt{1950}**: - Factorize around 44: The prime factor 13 is closer to half of 44, and the remaining factors can compose another number close to it. Combine smaller primes to approach this target: [ 3 times 5^2 = 75 ] Choosing 39 and 50 for the two factors gives: [ 39 = 3 times 13, quad 50 = 2 times 5^2 ] 5. **Calculate the difference between these factors**: [ 50 - 39 = 11 ] 6. **Conclusion**: The smallest difference between two factors of 1950 given the conditions is 11. The correct answer is boxed{Btext{) }11}.

answer:To calculate the total cost the school will pay for the new seats, we proceed as follows: 1. First, we calculate the cost of ten seats without the discount. Since each seat costs 30, the cost for ten seats is: [30 times 10 = 300.] 2. Next, we calculate the discount offered for each group of ten seats. The discount is 10% of the cost of ten seats, which is: [300 times frac{10}{100} = 30.] 3. Therefore, the total cost for every ten seats after applying the discount is: [300 - 30 = 270.] 4. The school plans to add 5 rows of 8 seats each, totaling: [5 times 8 = 40 text{ seats}.] 5. This means the school will be purchasing: [frac{40}{10} = 4 text{ sets of 10 seats}.] 6. Finally, the total cost for the new seats, after the discount, is calculated by multiplying the cost per set of ten seats by the number of sets: [270 times 4 = 1080.] Therefore, the school will pay a total of boxed{1080} for the new seats.

answer:According to the problem, we have: x^2-4x-21>0, which simplifies to (x+3)(x-7)>0. Solving this inequality, we find x<-3 or x>7. Therefore, the domain of the function is (-infty, -3) cup (7, +infty). Hence, the answer is boxed{(-infty, -3) cup (7, +infty)}.