answer:Given the line equation -frac{3}{4} - 3kx = 7y and the point left(frac{1}{3}, -8right), we substitute x = frac{1}{3} and y = -8 into the equation: [ -frac{3}{4} - 3kleft(frac{1}{3}right) = 7(-8) ] Simplify the equation: [ -frac{3}{4} - k = -56 ] Solving for k: [ -k = -56 + frac{3}{4} = -56 + 0.75 = -55.25 ] [ k = boxed{55.25} ]

answer:Solution: Since the complex number z=1-i (where i is the imaginary unit), we have dot{z}=1+i. Thus, dot{z}+ frac {2i}{z}=1+i+ frac {2i}{1-i}=1+i+ frac {2i(1+i)}{(1-i)(1+i)}=1+i+i-1=2i. Therefore, the correct choice is boxed{D}. From the complex number z=1-i (where i is the imaginary unit), we find dot{z}=1+i. Then, by simplifying the operation dot{z}+ frac {2i}{z} using the algebraic form of division for complex numbers, the answer can be determined. This question examines the multiplication and division operations in the algebraic form of complex numbers, and explores the basic concepts of complex numbers. It is a fundamental question.

answer:According to the problem, let's assume the price needs to be reduced by x%. Let the original retail price be 1. We have 1 times (1 + 10%) = 1 times (1 + 25%) times (1 - x%), Solving this equation, we get x = 12. Therefore, the correct choice is boxed{text{B}}.

answer:1. **Understanding the Left-Hand Side (LHS):** The expression (binom{2n+1}{2k+1}) counts the number of ways to choose (2k+1) elements from a set of (2n+1) elements. 2. **Considering the Combinations in a Natural Order:** Consider a particular combination of these (2k+1) elements, ( A = {a_1, a_2, ldots, a_{2k+1}} ), in natural increasing order (where (a_1 < a_2 < ldots < a_{2k+1})). 3. **Identifying the Middle Element:** Focus on the element at the middle of this ordered selection, specifically the ((k+1))-th smallest element (a_{k+1}). 4. **Enumeration Based on the Middle Element:** Analyze how many different ways the remaining elements can be chosen, given that (a_{k+1} = m): - The elements before (a_{k+1}) (i.e., ({a_1, a_2, ldots, a_k})) must be selected from ([1, m-1]). - The elements after (a_{k+1}) (i.e., ({a_{k+2}, ldots, a_{2k+1}})) must be selected from ([m+1, 2n+1]). 5. **First Case: Middle Element is (k+1):** - Before (k+1): Choose (k) elements from (k) elements (which are ({1, 2, ldots, k})). [ binom{k}{k} ] - After (k+1): Choose (k) elements from (2n-k) elements (i.e., ({k+2, k+3, ldots, 2n+1})). [ binom{2n-k}{k} ] - Total ways for (a_{k+1} = k+1): [ binom{k}{k} binom{2n-k}{k} ] 6. **Second Case: Middle Element is (k+2):** - Before (k+2): Choose (k) elements from (k+1) elements (which are ({1, 2, ldots, k+1})). [ binom{k+1}{k} ] - After (k+2): Choose (k) elements from (2n-k-1) elements (i.e., ({k+3, k+4, ldots, 2n+1})). [ binom{2n-k-1}{k} ] - Total ways for (a_{k+1} = k+2): [ binom{k+1}{k} binom{2n-k-1}{k} ] 7. **General Case: Middle Element is (m+1):** - Before (m+1): Choose (k) elements from (m) elements (which are ({1, 2, ldots, m})). [ binom{m}{k} ] - After (m+1): Choose (k) elements from (2n-m) elements (which are ({m+2, m+3, ldots, 2n+1})). [ binom{2n-m}{k} ] - Total ways for (a_{k+1} = m+1): [ binom{m}{k} binom{2n-m}{k} ] 8. **Final Case: Middle Element is (2n+1-k):** - Before (2n+1-k): Choose (k) elements from (2n-k) elements. [ binom{2n-k}{k} ] - After (2n+1-k): Select remaining (k) elements from (k) elements. [ binom{k}{k} ] - Total ways for (a_{k+1} = 2n+1-k): [ binom{2n-k}{k} binom{k}{k} ] 9. **Summing Up All the Possibilities:** Summing over all possible middle elements from (k+1) to (2n+1 - k), we get: [ binom{2n+1}{2k+1} = sum_{m=k}^{2n-k} binom{m}{k} binom{2n-m}{k} ] # Conclusion: [ boxed{binom{2n+1}{2k+1}= binom{k}{k} binom{2n-k}{k} + binom{k+1}{k} binom{2n-k-1}{k} + cdots + binom{2n-k}{k} binom{k}{k}} ]