answer:1. Given the points and lines, let's first establish that ( C I ) is the angle bisector of (angle P C Q): [ P I = I Q ] This implies that point ( I ) lies at the midpoint of the arc ( P Q ) on the circumcircle of triangle ( A I C ). 2. Define ( T ) as the intersection point of lines ( A P ) and ( C Q ). 3. Apply the Thales Theorem to the situation. The theorem suggests that the sought parallelism ( L S parallel P Q ) is equivalent to the proportionality condition: [ frac{T L}{T P} = frac{T S}{T Q} ] 4. Using the properties of the angle bisector and triangle similarity: - For triangle ( T P C ): [ frac{T L}{T P} = frac{T C}{P C} ] - For triangle ( T Q A ): [ frac{T S}{T Q} = frac{T A}{Q A} ] 5. Because triangles ( T P C ) and ( T Q A ) are similar by the Angle-Angle (AA) criterion (their corresponding angles are equal), the following relation holds: [ frac{T C}{P C} = frac{T A}{Q A} ] 6. Now, using these established proportions: - From the similarity ratios, we substitute back into the initial proportion: [ frac{T L}{T P} = frac{T C}{P C} quad text{and} quad frac{T S}{T Q} = frac{T A}{Q A} ] - Since ( frac{T C}{P C} = frac{T A}{Q A} ), this implies: [ frac{T L}{T P} = frac{T S}{T Q} ] which directly leads to the parallelism condition: [ L S parallel P Q ] # Conclusion: [ boxed{L S parallel P Q} ]

answer:[Analysis] This problem involves understanding the even/odd nature of a function and its monotonicity. First, by using the derivative, we analyze the monotonicity of the function in the interval left(0, frac{pi}{2}right). Then, utilizing the function's even/odd nature, we can transform and solve the inequality. [Solution] Given x in left[-frac{pi}{2}, frac{pi}{2}right], We have f(-x) = (-x)^2 - cos(-x) = x^2 - cos x = f(x). So, f(x) is an even function. When x in left(0, frac{pi}{2}right), f'(x) = 2x + sin x > 0, which means f(x) is strictly increasing. Therefore, f(x_0) > fleft(frac{pi}{3}right) is equivalent to |x_0| > frac{pi}{3}. Solving for x_0, we have x_0 < -frac{pi}{3} or x_0 > frac{pi}{3}. Considering x_0 in left[-frac{pi}{2}, frac{pi}{2}right], we have x_0 in left[-frac{pi}{2}, -frac{pi}{3}right) cup left(frac{pi}{3}, frac{pi}{2}right]. Thus, the answer is boxed{left[-frac{pi}{2}, -frac{pi}{3}right) cup left(frac{pi}{3}, frac{pi}{2}right]}.

answer:First, identify the boxes with at least 50,000. These are the boxes containing 50,000, 75,000, 100,000, 200,000, 300,000, 400,000, 500,000, 750,000, and 1,000,000. This totals 9 boxes. To have at least a half chance of holding one of these boxes, the participant must be left with at most 9 other boxes, as this provides the participant with a 9/18 = 1/2 chance. Hence, the total number of boxes remaining (including the box the participant is holding) should be 18. Since the game starts with 26 boxes, and one is held by the participant, the number of boxes that need to be eliminated is: [ 26 - 18 = boxed{8}. ] Conclusion: For a 50% chance to have at least 50,000, 8 boxes must be eliminated.

answer:Let's denote the amount Shaniqua makes for every haircut as H dollars. From the information given, we know that Shaniqua made 221 by giving 8 haircuts and 5 styles. We can write this as an equation: 8H (for the haircuts) + 5 * 25 (for the styles) = 221 We know that each style is 25, so 5 styles would be: 5 * 25 = 125 Now, we can subtract the total amount made from styles from the total amount Shaniqua made to find out how much she made from haircuts: 221 (total amount made) - 125 (amount made from styles) = 96 (amount made from haircuts) Now we have the equation: 8H = 96 To find the amount made per haircut (H), we divide both sides of the equation by 8: H = 96 / 8 H = 12 Shaniqua makes boxed{12} for every haircut.