answer:Since in the arithmetic sequence {a_{n}}, s_{6} > s_{7} > s_{5}, It follows that a_{1} > 0, d < 0, hence statement ① is incorrect; Since s_{6} > s_{7} > s_{5}, it follows that a_{6}=S_{6}-S_{5} > 0, a_{7}=S_{7}-S_{6} < 0, S_{11}=11a_{1}+55d=11(a_{1}+5d)=11a_{6} > 0, hence statement ② is correct; Since s_{6} > s_{7} > s_{5}, it follows that a_{6}+a_{7}=S_{7}-S_{5} > 0, Therefore, S_{12}=12a_{1}+66d=12(a_{1}+a_{12})=12(a_{6}+a_{7}) > 0, hence statement ③ is incorrect; Therefore, a_{1}+6d < 0, a_{1}+5d > 0, therefore S_{6} is the maximum, hence statement ④ is incorrect; Since a_{6}=S_{6}-S_{5} > 0, a_{7}=S_{7}-S_{6} < 0, a_{6}+a_{7}=S_{7}-S_{5} > 0, Therefore, |a_{5}| > |a_{7}|, hence statement ⑤ is correct. Therefore, the answer is: boxed{A}. This problem is solved directly using the formula for the general term and the sum of the first n terms of an arithmetic sequence. It examines the ability to judge the truth of statements, the basic knowledge of the sum formula and the general term formula of an arithmetic sequence, reasoning and computational skills, and the idea of transformation and simplification. It is a medium-difficulty problem.

answer:1. **Listing the Numbers**: Let's consider the given problem by writing the five n-digit numbers vertically, one under the other. We will label these digits as D_{i,j} where i is the index of the number (ranging from 1 to 5) and j is the digit position (ranging from 1 to n). 2. **Calculating Pairs**: In each column of digits, we can form binom{5}{2} = 10 unordered pairs. Since there are n columns, we will have a total of 10n pairs across all columns. 3. **Identifying Matching Pairs**: A pair (D_{i,j}, D_{k,j}) is considered a matching pair if D_{i,j} = D_{k,j}. There are only two possible choices for each digit (1 or 2). Therefore, in each column, the number of matching pairs (i.e., pairs where both digits are the same) must be either 4 or 6. - If a column contains three 1s and two 2s, then the matching pairs will be [ binom{3}{2} + binom{2}{2} = 3 + 1 = 4 ] pairs. - If a column contains four 1s and one 2, or vice versa, then the matching pairs will be [ binom{4}{2} + binom{1}{2} = 6 + 0 = 6 ] pairs. 4. **Counting All Matching Pairs**: Summing over all n columns, the total number of matching pairs is bounded by the number of columns multiplied by the minimum and maximum possible matching pairs per column: [ 4n leq text{Total matching pairs} leq 6n ] 5. **Alternative Calculation from Conditions**: According to the problem, each number should match in exactly m positions with each of the other four numbers. Therefore, for each number pair there are m matching positions. Given that there are binom{5}{2} = 10 pairs of numbers: [ text{Total matching pairs} = 10m ] 6. **Setting Up the Inequality**: Equating the two expressions for the total number of matching pairs, we derive: [ 4n leq 10m leq 6n ] 7. **Analyzing the Ratios**: Dividing through by 10n to normalize we get: [ frac{4n}{10n} leq frac{10m}{10n} leq frac{6n}{10n} implies frac{2}{5} leq frac{m}{n} leq frac{3}{5} ] # Conclusion: Thus, we have proven that the ratio frac{m}{n} must lie between frac{2}{5} and frac{3}{5}, inclusive. [ boxed{frac{2}{5} leq frac{m}{n} leq frac{3}{5}} ]

answer:# Step-by-Step Solution Part (1): Find the measure of angle C Given: ccosB+frac{{sqrt{3}}}{3}bsinC-a=0 1. Rewrite the equation using trigonometric identities: [ ccos B + frac{sqrt{3}}{3}bsin C - a = 0 ] becomes [ sin Ccos B + frac{sqrt{3}}{3}sin Bsin C - sin A = 0 ] 2. Use the identity sin(A + B + C) = sin(pi) = 0 to express sin A as sin(B + C): [ sin Ccos B + frac{sqrt{3}}{3}sin Bsin C - sin(B + C) = 0 ] 3. Expand sin(B + C) using the sum of angles identity: [ sin Ccos B + frac{sqrt{3}}{3}sin Bsin C - (sin Bcos C + cos Bsin C) = 0 ] 4. Simplify to isolate terms involving sin C and cos C: [ frac{sqrt{3}}{3}sin Bsin C - sin Bcos C = 0 ] 5. Factor out sin B: [ sin Bleft(frac{sqrt{3}}{3}sin C - cos Cright) = 0 ] 6. Solve for tan C: [ frac{sqrt{3}}{3}sin C - cos C = 0 implies tan C = sqrt{3} ] 7. Since C in (0, pi), we find: [ C = frac{pi}{3} ] Final answer for part (1): [ boxed{C = frac{pi}{3}} ] Part (2): Find the value of a + b Given: c = 3 and the area of triangle ABC is frac{3sqrt{3}}{4} 1. Use the area formula: [ S_{triangle ABC} = frac{1}{2}absin C = frac{3sqrt{3}}{4} ] Since C = frac{pi}{3}, sin C = frac{sqrt{3}}{2}, we get: [ frac{1}{2}ableft(frac{sqrt{3}}{2}right) = frac{3sqrt{3}}{4} implies ab = 3 ] 2. Apply the cosine rule: [ c^2 = a^2 + b^2 - ab implies 9 = a^2 + b^2 - 3 ] Simplify to find: [ a^2 + b^2 = 12 ] 3. Calculate (a + b)^2: [ (a + b)^2 = a^2 + b^2 + 2ab = 12 + 2(3) = 18 ] Therefore: [ a + b = sqrt{18} = 3sqrt{2} ] Final answer for part (2): [ boxed{a + b = 3sqrt{2}} ] **Analysis:** The solution involves applying trigonometric identities, the sine rule, and the cosine rule to solve for the specific measures and relationships within the triangle. This approach demonstrates a methodical application of trigonometry and algebra to geometric problems, highlighting the interconnectedness of these mathematical domains.

answer:When frac{pi}{6} leq x leq frac{pi}{2}, we have frac{1}{2} leq sin x leq 1, Therefore, the value range of the function y=sin x (frac{pi}{6} leq x leq frac{pi}{2}) is left[frac{1}{2}, 1right]. Hence, the correct option is: boxed{text{D}}. This is determined by the graph and properties of the sine function, which directly lead to the corresponding result. This question tests the application of the graph and properties of the sine function and is considered a basic question.