answer:1. We are given an inequality that holds for any real numbers ( x_1, x_2, ldots, x_n ): [ sum_{k=1}^{n} r_k (x_k - a_k) leq left( sum_{k=1}^{n} x_k^2 right)^{frac{1}{2}} - left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} ] 2. First, let ( x_1 = x_2 = cdots = x_n = 0 ). Substituting these values into the inequality: [ sum_{k=1}^{n} r_k (0 - a_k) leq left( sum_{k=1}^{n} 0^2 right)^{frac{1}{2}} - left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} ] Simplifying the left and right sides: [ -sum_{k=1}^{n} r_k a_k leq 0 - left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} ] Therefore, we get: [ sum_{k=1}^{n} r_k a_k geq left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} ] 3. Next, let us choose ( x_k = 2a_k ) for ( k = 1,2, ldots, n ). Substituting these values into the inequality: [ sum_{k=1}^{n} r_k (2a_k - a_k) leq left( sum_{k=1}^{n} (2a_k)^2 right)^{frac{1}{2}} - left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} ] Simplifying, we have: [ sum_{k=1}^{n} r_k a_k leq left( sum_{k=1}^{n} 4a_k^2 right)^{frac{1}{2}} - left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} ] [ sum_{k=1}^{n} r_k a_k leq 2 left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} - left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} ] [ sum_{k=1}^{n} r_k a_k leq left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} ] 4. Combining results from steps 2 and 3: [ sum_{k=1}^{n} r_k a_k = left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} ] 5. By Cauchy-Schwarz inequality: [ sum_{k=1}^{n} r_k a_k leq left( sum_{k=1}^{n} r_k^2 right)^{frac{1}{2}} cdot left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} ] Based on the previous equality: [ left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} leq left( sum_{k=1}^{n} r_k^2 right)^{frac{1}{2}} cdot left( sum_{k=1}^{n} a_k^2 right)^{frac{1}{2}} ] Simplifying: [ 1 leq left( sum_{k=1}^n r_k^2 right)^{frac{1}{2}} ] Squaring both sides: [ sum_{k=1}^{n} r_k^2 geq 1 ] 6. Consider ( x_k = r_k ) for ( k = 1,2, ldots, n ). Substituting these values into the inequality: [ sum_{k=1}^n r_k (r_k - a_k) leq left( sum_{k=1}^n r_k^2 right)^{frac{1}{2}} - left( sum_{k=1}^n a_k^2 right)^{frac{1}{2}} ] Simplifying: [ sum_{k=1}^n r_k^2 - sum_{k=1}^n r_k a_k leq left( sum_{k=1}^n r_k^2 right)^{frac{1}{2}} - left( sum_{k=1}^n a_k^2 right)^{frac{1}{2}} ] Using the previously found result: [ sum_{k=1}^n r_k a_k = left( sum_{k=1}^n a_k^2 right)^{frac{1}{2}} ] Thus: [ sum_{k=1}^n r_k^2 - left( sum_{k=1}^n a_k^2 right)^{frac{1}{2}} leq left( sum_{k=1}^n r_k^2 right)^{frac{1}{2}} - left( sum_{k=1}^n a_k^2 right)^{frac{1}{2}} ] [ sum_{k=1}^n r_k^2 leq left( sum_{k=1}^n r_k^2 right)^{frac{1}{2}} ] Squaring both sides: [ sum_{k=1}^n r_k^2 leq 1 ] 7. Combining the inequalities (sum_{k=1}^n r_k^2 geq 1) and (sum_{k=1}^n r_k^2 leq 1), we get: [ sum_{k=1}^n r_k^2 = 1 ] 8. With ( sum_{k=1}^n r_k^2 = 1 ) and ( sum_{k=1}^n r_k a_k = left( sum_{k=1}^n a_k^2 right)^{frac{1}{2}} ), according to the condition for equality in the Cauchy-Schwarz inequality, there exists a real number ( lambda ) such that: [ r_k = lambda a_k quad text{for } k = 1, 2, ldots, n ] 9. From (sum_{k=1}^n r_k^2 = 1): [ sum_{k=1}^n (lambda a_k)^2 = 1 ] [ lambda^2 sum_{k=1}^n a_k^2 = 1 ] [ lambda = left( sum_{k=1}^n a_k^2 right)^{-frac{1}{2}} ] 10. Thus, the values of ( r_k ) are: [ r_k = frac{a_k}{sqrt{sum_{k=1}^n a_k^2}} quad text{for } k = 1, 2, ldots, n ] # Conclusion: [ boxed{r_i = frac{a_i}{sqrt{sum_{k=1}^n a_i^2}}, quad i = 1, 2, ldots, n} ]

answer:1. Let the initial number of balls be ( x ) (we can simply take ( x = 1 ) or any other value since we are working with percentages). 2. The first step involves giving ( 20% ) of the balls to Yi (乙). Thus, the remaining percentage of balls after this step is: [ 100% - 20% = 80%. ] Representing this mathematically, the balls left with Jia (甲) can be written as: [ 0.80 cdot x. ] 3. Next, Jia (甲) gives ( 10% ) of the remaining balls to Bing (丙). Thus, the remaining balls after this distribution is: [ 100% - 10% = 90% text{ of the previous amount}. ] Therefore, the balls left with Jia (甲) can be represented as: [ 0.90 cdot (0.80 cdot x) = 0.72 cdot x. ] 4. Finally, Jia (甲) gives ( 25% ) of the remaining balls to Ding (丁). Hence, the percentage of balls left after this final distribution is: [ 100% - 25% = 75% text{ of the previous amount}. ] So the balls left with Jia (甲) can be represented as: [ 0.75 cdot (0.72 cdot x) = 0.54 cdot x. ] 5. Converting the final amount to a percentage of the initial amount, we find that Jia (甲) retains: [ 54% text{ of the initial number of balls.} ] # Conclusion: [ boxed{E , 54%} ]

answer:To solve the inequality -x^2 + 4x + 5 < 0, we first rewrite it as x^2 - 4x - 5 > 0. This can be factored into (x - 5)(x + 1) > 0. Next, we determine the critical points where the product becomes zero, which are x = 5 and x = -1. To find where the inequality holds, we test values from the intervals defined by these critical points: 1. For x > 5, choose x = 6: (6 - 5)(6 + 1) = 1 cdot 7 > 0, so the inequality holds in this interval. 2. For -1 < x < 5, choose x = 0: (0 - 5)(0 + 1) = -5 cdot 1 < 0, so the inequality does not hold in this interval. 3. For x < -1, choose x = -2: (-2 - 5)(-2 + 1) = -7 cdot -1 > 0, the inequality holds in this interval. As a result, the solution set for the inequality is x > 5 or x < -1. Therefore, the solution set is: boxed{{x mid x > 5 text{ or } x < -1}}

answer:1. Let's understand the given angles in the problem: (110^{circ}, 75^{circ}, 65^{circ}, 15^{circ}). 2. We have two types of triangles: an acute-angled triangle and an obtuse-angled triangle. - An acute-angled triangle has all its angles less than (90^{circ}). - An obtuse-angled triangle has one angle greater than (90^{circ}) and the other two less than (90^{circ}). 3. Identify the angle that must belong to the obtuse-angled triangle. The angle (110^{circ}) is obtuse because it is greater than (90^{circ}), so it belongs to the obtuse-angled triangle. 4. The remaining angles are (75^{circ}, 65^{circ},) and (15^{circ}). - Among these angles, find the two angles that add up with (110^{circ}) to (180^{circ}): (75^{circ} + 65^{circ}): [ 110^{circ} + 75^{circ} + 65^{circ} = 250^{circ} ] This is more than (180^{circ}), thus they cannot form a triangle. Let's re-examine. 5. Assign the angles (75^{circ}) and (110^{circ}) to the obtuse triangle: [ 110^{circ} + 65^{circ} + 15^{circ} = 190^{circ} ] Again, it doesn't add to (180^{circ}). 6. Let's consider the angle sums: - (110^{circ} + 75^{circ} + x = 180^{circ} ); thus (x = 180^{circ} - 185^{circ} = 5^{circ}) - (110^{circ} + 65^{circ} + x = 180^{circ} ); thus (x = 180^{circ} - 150^{circ} = 30^{circ}) - (110^{circ} + 15^{circ} = 125^{circ}), then the remaining would be (55^{circ}) 7. We missed trying (75^{circ}, 65^{circ}) and (35^{circ}, etc. 8. Find (15^circ) - (75^{circ} + 65^{circ} + 40 = 180^{circ}) 9. (c): - The acute-angled triangle is (40 < 90) # Conclusion: The smallest angle in the acute-angled triangle is: ( boxed{15 )