answer:1. **Formulate the common quadratic equation**: From the given conditions, we have 4x_1^2 - kx_1 = c and 4x_2^2 - kx_2 = c. These can be rewritten as 4x_1^2 - kx_1 - c = 0 and 4x_2^2 - kx_2 - c = 0, implying both x_1 and x_2 are roots of: [ 4x^2 - kx - c = 0. ] 2. **Apply Vieta's formulas**: For a quadratic equation ax^2 + bx + c = 0, the sum of the roots (x_1 + x_2) is given by -frac{b}{a}. Here, a = 4, b = -k, so: [ x_1 + x_2 = -frac{-k}{4} = frac{k}{4}. ] 3. **Conclusion**: The sum of x_1 and x_2 is frac{k}{4}. Hence, the answer should be boxed as frac{k{4}}. The final answer is boxed{B}

answer:1. **Express the angle bisectors in terms of the sides of the triangle:** Given the lengths of the angle bisectors ( x, y, z ) of a triangle with sides ( a, b, c ), we have: [ x^2 = frac{4bcs(s-a)}{(b+c)^2}, quad y^2 = frac{4cas(s-b)}{(c+a)^2}, quad z^2 = frac{4abs(s-c)}{(a+b)^2} ] where ( s ) is the semi-perimeter of the triangle, ( s = frac{a+b+c}{2} ). 2. **Apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality:** By the AM-GM inequality, we know: [ (b+c)^2 geq 4bc, quad (c+a)^2 geq 4ca, quad (a+b)^2 geq 4ab ] This implies: [ frac{1}{x^2} = frac{(b+c)^2}{4bcs(s-a)} leq frac{1}{s(s-a)}, quad frac{1}{y^2} = frac{(c+a)^2}{4cas(s-b)} leq frac{1}{s(s-b)}, quad frac{1}{z^2} = frac{(a+b)^2}{4abs(s-c)} leq frac{1}{s(s-c)} ] 3. **Sum the inequalities:** Summing these inequalities, we get: [ frac{1}{x^2} + frac{1}{y^2} + frac{1}{z^2} geq frac{1}{s(s-a)} + frac{1}{s(s-b)} + frac{1}{s(s-c)} ] 4. **Apply the Arithmetic Mean-Harmonic Mean (AM-HM) inequality:** By the AM-HM inequality, we have: [ frac{1}{s-a} + frac{1}{s-b} + frac{1}{s-c} geq frac{9}{(s-a) + (s-b) + (s-c)} ] Since ( (s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s ), we get: [ frac{1}{s-a} + frac{1}{s-b} + frac{1}{s-c} geq frac{9}{s} ] 5. **Combine the results:** Therefore, [ frac{1}{x^2} + frac{1}{y^2} + frac{1}{z^2} geq frac{1}{s} left( frac{1}{s-a} + frac{1}{s-b} + frac{1}{s-c} right) geq frac{1}{s} cdot frac{9}{s} = frac{9}{s^2} ] Given that the perimeter of the triangle is 6, we have ( s = frac{6}{2} = 3 ). Thus, [ frac{9}{s^2} = frac{9}{3^2} = 1 ] Hence, [ frac{1}{x^2} + frac{1}{y^2} + frac{1}{z^2} geq 1 ] (blacksquare)

answer:Given that for (n geq 3), [b_n = frac{b_1 + b_2 + dots + b_{n-1}}{n-1},] it translates to [(n-1)b_n = b_1 + b_2 + dots + b_{n-1}.] Similarly, for (b_{n+1}), [n b_{n+1} = b_1 + b_2 + dots + b_{n-1} + b_n.] By subtraction, we get: [n b_{n+1} - (n-1)b_n = b_n,] implies [n b_{n+1} = n b_n,] so [b_{n+1} = b_n.] This infers that all terms from (b_3) onwards are equal to (b_3). Knowing (b_{12} = 150), it follows (b_3 = b_4 = dots = b_{12} = 150). Now, (b_3) as the mean of (b_1) and (b_2) gives: [frac{24 + b_2}{2} = 150.] Thus, [24 + b_2 = 300 quad Rightarrow quad b_2 = 300 - 24 = 276.] Therefore, (b_2 = boxed{276}.)

answer:1. **Understand the Problem:** - There are 16 progamers in a single-elimination tournament. - Each match, the losing progamer is eliminated. - Each winning progamer has a ceremonial celebration. - The first ceremony is 10 seconds, and each subsequent ceremony increases by 10 seconds. 2. **Determine the Rounds:** - First round: 16 progamers become 8 winners. - Second round: 8 winners become 4 winners. - Third round: 4 winners become 2 winners. - Final round: 2 winners become 1 winner. 3. **Calculate the Ceremony Times:** - **First Round:** Each of the 8 winners has a 10-second ceremony: [ 8 text{ progamers} times 10 text{ seconds} = 80 text{ seconds} ] - **Second Round:** Each of the 4 winners has a 20-second ceremony: [ 4 text{ progamers} times 20 text{ seconds} = 80 text{ seconds} ] - **Third Round:** Each of the 2 winners has a 30-second ceremony: [ 2 text{ progamers} times 30 text{ seconds} = 60 text{ seconds} ] - **Final Round:** The single winner has a 40-second ceremony: [ 1 text{ progamer} times 40 text{ seconds} = 40 text{ seconds} ] 4. **Add Up All Ceremonies:** - Sum the total duration of all ceremonies: [ 80 + 80 + 60 + 40 = 260 text{ seconds} ] # Conclusion: [ boxed{260} ]