answer:From the Law of Sines, we know that frac{a}{sin A} = frac{b}{sin B} = 2R, Since sin A > sin B, it follows that a > b, which implies A > B. Conversely, since A > B, it implies a > b, Given a = 2Rsin A and b = 2Rsin B, it follows that sin A > sin B. Therefore, "A > B" is a necessary and sufficient condition for "sinA > sinB." Thus, the correct answer is boxed{C}. By the Law of Sines, we know that frac{a}{sin A} = frac{b}{sin B}, hence sin A > sin B Leftrightarrow a > b Leftrightarrow A > B, which leads to the conclusion. This problem uses a triangle as a medium to examine the meaning and judgment method of necessary and sufficient conditions. The key to solving the problem is the correct application of the Law of Sines and the properties of triangles, making it a basic question.

answer:Start by simplifying the expression frac{1}{sqrt{5} + 2}. Multiply the numerator and the denominator by the conjugate of the denominator: frac{1}{sqrt{5} + 2} = frac{1}{sqrt{5}+2} times frac{sqrt{5}-2}{sqrt{5}-2} = frac{sqrt{5}-2}{5-4} = frac{sqrt{5}-2}{1} = sqrt{5}-2. Next, substitute this back into the original expression: frac{1}{2+ frac{1}{sqrt{5}+2}} = frac{1}{2 + (sqrt{5} - 2)} = frac{1}{sqrt{5}}. Now, multiply the numerator and denominator by the conjugate of the denominator: frac{1}{sqrt{5}} = frac{1}{sqrt{5}} times frac{sqrt{5}}{sqrt{5}} = frac{sqrt{5}}{5}. Therefore, the final simplified and rationalized form of the expression is boxed{frac{sqrt{5}}{5}}.

answer:To solve this, we first acknowledge that n! denotes the product of all positive integers up to n. This means for any integer k such that 1 leq k leq n, k divides n!. Now we analyze the numbers in the range n! < a < n!+2n: - Any integer a = n! + k for 1 leq k leq n is divisible by k, and hence composite since k divides n! and thus n! + k. - For integers a = n! + k where n < k < 2n, k does not necessarily divide n. However, all such k are greater than n and less than 2n. It's necessary to check for primality individually for each k > n. Since n! + k with 1 leq k leq n are all composite, we focus on k from n+1 to 2n-1. The challenge is to determine if any of these k values, which are added to n!, yield a prime number. It turns out the question largely depends on the individual values of k in the extended range. Generally: - The difficulty of proving primality grows as n increases. - We know that for k leq n, n!+k is composite. Hence, we are interested in the intervals [n+1, 2n-1]. These terms are candidates for primality, but this requires individual checks because whether n!+k is prime is not trivially deduced purely by its form. Thus, instinctively, it's less likely to encounter many primes in this range, but it is not impossible. Each value k from n+1 to 2n-1 added to n! would need to be checked for primality. Conclusion: There are boxed{0} guaranteed primes unless specific k values in the set [n+1, 2n-1] are known to be prime, and n!+k might be prime, which is not ascertainable without specific primality tests for each such k.

answer:(I) Solution: According to the question, set A={1, 2, …, 1000} can be set. Then according to the question, it can be determined that aᵢ=i, i∈N∗ (i=1, 2, …, 1000). Therefore, for any m+n∈A (m, n=1, 2, …, 1000), there is aᵐ+aₙ=m+n∈A. Therefore, set A={1, 2, …, 1000} has property P. Therefore, set A={1, 2, …, 1000}. (II) Proof (by contradiction): Suppose there exists aᵏ∈A, such that aᵏ=2000. Then obviously k≤1000, According to the characteristics of property P, let i=j=1000, Therefore i+j=1000+1000=2000∈A, Then there is aᵢ+aₖ=a₁₀₀₀+a₁₀₀₀=2a₁₀₀₀∈A, but a₁₀₀₀ is the maximum value of the elements in set A, so 2a₁₀₀₀∉A, which is contradictory, and the assumption is not established. Therefore, it is not true that there exists aᵏ∈A, such that aᵏ=2000. (III) Solution: Let k be the largest positive integer such that aᵏ<2000, then a₁₀₀₀>a₉₉₉>…>aₖ₊₁≥2001. If aᵏ>1000, then there exists a positive integer i such that aᵏ=1000+i∈A, so a₁₀₀₀+aᵢ∈A. Similar to (II), a₁₀₀₀+aᵢ>a₁₀₀₀ cannot belong to set A. Therefore, aᵢ≤1000 (i=1, 2, …, k), and it is known from the question that aᵢ≥i (i=1, 2, …, 1000), So, aᵏ≥k, and there are at most 19 elements in set A that are greater than 2000, so k≥1000-19=981. Next, prove that aᵏ>k cannot be true. Assume aᵏ>k, then there exists a positive integer i such that aᵏ=k+i∈A, obviously i≤19, So there exists a positive integer m such that aₘ=aᵏ+aᵢ<2aᵏ<2000. While 2000>aₘ=aᵢ+aᵏ>aᵏ and k is the largest positive integer such that aᵏ<2000, which is contradictory, so aᵏ>k cannot be true. That is, aᵏ=k is true. When aᵏ=k, for any 1≤i, j≤k≤1000 that satisfy i+j∈A, obviously aᵢ+aₖ∈A is true. If aᵢ≥2001, then k<i+j≤2000, that is, i+j∉A, So, A={1, 2, …k, aₖ₊₁, aₖ₊₂, …, a₁₀₀₀}, where aᵢ>2000 (i=k+1, k+2, …, 1000) are all sets that meet the requirements. And k can take the values 981, 982, …, 1000, so the number of sets that meet the conditions is C_{ 19 }^{ 0 }+ C_{ 19 }^{ 1 }+… C_{ 19 }^{ 18 }+ C_{ 19 }^{ 19 }=2^{19}. Therefore, the number of sets A that meet the conditions is boxed{2^{19}}.