answer:Solution: (1) overrightarrow{a} - overrightarrow{b} = (3, -3-m); Since overrightarrow{a} perp (overrightarrow{a} - overrightarrow{b}), we have overrightarrow{a} cdot (overrightarrow{a} - overrightarrow{b}) = 3 + 3(3+m) = 0; Therefore, m = boxed{-4}; (2) overrightarrow{b} = (-2, -4), overrightarrow{a} = (1, -3); Therefore, costheta = frac{overrightarrow{a} cdot overrightarrow{b}}{|overrightarrow{a}||overrightarrow{b}|} = frac{10}{sqrt{20} cdot sqrt{10}} = frac{sqrt{2}}{2}; theta in [0, pi]; Therefore, theta = boxed{frac{pi}{4}}; That is, the angle theta between vectors overrightarrow{a} and overrightarrow{b} is frac{pi}{4}; (3) koverrightarrow{a} + overrightarrow{b} = (k-2, -3k-4), overrightarrow{a} - overrightarrow{b} = (3, 1); When koverrightarrow{a} + overrightarrow{b} is parallel to overrightarrow{a} - overrightarrow{b}, (k-2) cdot 1 - 3 cdot (-3k-4) = 0; Therefore, k = boxed{-1}.

answer:Since {a_n} is an increasing sequence, for n leq 5, we have (5-a) > 0 which implies a < 5. Similarly, since the sequence continues to increase for n > 5, a must be positive, hence a > 1. Next, we need to ensure the transition at n=5 is increasing, which means a_5 < a_6. Thus: (5-a) cdot 5 - 11 < a^{6-4}, Simplifying the inequality, we get: 5(5 - a) - 11 < a^2, Solving for a, we establish: 25 - 5a - 11 < a^2, 14 < 5a + a^2, 0 < a^2 + 5a - 14. Considering this as a quadratic equation ( a^2 + 5a - 14 = 0 ), by calculating its roots we find: a = -frac{5}{2} pm frac{sqrt{5^2 - 4(-14)}}{2} = -frac{5}{2} pm frac{sqrt{25 +56}}{2} = -frac{5}{2} pm frac{sqrt{81}}{2}, a = -frac{5}{2} + frac{9}{2} = 2 quad text{or} quad a = -frac{5}{2} - frac{9}{2} = -7. Since a > 1 and a < 5, we exclude the negative root, and we find that the valid range for a is 2 < a < 5. Therefore the correct answer is boxed{D}.

answer:1. **Isolate the Absolute Value**: Begin by isolating the absolute value term. Subtract 5 from both sides: [ |3x - 1| + 5 = 41 implies |3x - 1| = 41 - 5 = 36 ] 2. **Handle Two Cases**: The equation |3x - 1| = 36 splits into two scenarios: - (3x - 1 = 36) - (3x - 1 = -36) **First Case**: [ 3x - 1 = 36 implies 3x = 36 + 1 = 37 implies x = frac{37}{3} ] **Second Case**: [ 3x - 1 = -36 implies 3x = -36 + 1 = -35 implies x = frac{-35}{3} ] 3. **Sum the Solutions**: The sum of the solutions from both cases is: [ frac{37}{3} + frac{-35}{3} = frac{37 - 35}{3} = frac{2}{3} ] We thus find: [ boxed{frac{2}{3}} ]

answer:1. **Identify the Key Elements**: - Triangle ABC is inscribed in a circle with center O'. - Another circle with center O, which is the circumcenter of ABC, suggesting OA = OB = OC. 2. **Analyze the Triangle and Circle Properties**: - Since O is the circumcenter of triangle ABC, BO is the perpendicular bisector of side AC. - The triangle triangle O'BE is isosceles with O'B = O'E. 3. **Examine triangle ABE and triangle CBE**: - Since O'B = O'E, OB remains as a radius of the circle through O'. - AE = BE as angle AEB = angle BEA, since ABCE is cyclic (inscribed in a circle). - Similarly, CE = BE. 4. **Conclude**: - We have both AE = BE and CE = BE, so AE = CE = BE. Conclusion: The correct relationship among AE, CE, and BE is that they are all equal. The final answer is AE = CE = BE. The final answer is boxed{D}.