answer:Let the original price of the jacket be P. The final price is 75% of the original price, mathematically expressed as: [ 0.75P ] Let x% be the percentage increase and decrease. The price after the increase and decrease is: [ Pleft(1 + frac{x}{100}right)left(1 - frac{x}{100}right) = Pleft(1 - left(frac{x}{100}right)^2right) ] Applying an additional 10% discount results in: [ Pleft(1 - left(frac{x}{100}right)^2right) times 0.9 ] Setting this equal to 0.75P, we get: [ Pleft(1 - left(frac{x}{100}right)^2right) times 0.9 = 0.75P ] Dividing both sides by P and isolating the terms: [ left(1 - left(frac{x}{100}right)^2right) times 0.9 = 0.75 ] [ 1 - left(frac{x}{100}right)^2 = frac{0.75}{0.9} ] [ 1 - left(frac{x}{100}right)^2 = frac{5}{6} ] [ left(frac{x}{100}right)^2 = frac{1}{6} ] [ frac{x}{100} = sqrt{frac{1}{6}} ] [ x = 100 times sqrt{frac{1}{6}} approx 40.82 ] Thus, the percentage by which the price was increased and then decreased is 40.82. The final answer is boxed{textbf{(C)} 40.82}.

answer:Let's denote the distance covered by the van as ( D ) kilometers. The van initially takes 6 hours to cover this distance. Let's denote the initial speed of the van as ( V ) km/h. So, we can write the relationship between distance, speed, and time as: [ D = V times 6 ] Now, we are told that to cover the same distance in ( frac{3}{2} ) of the previous time, the speed should be maintained at 32 km/h. Since ( frac{3}{2} ) of 6 hours is ( frac{3}{2} times 6 = 9 ) hours, we can write the new relationship as: [ D = 32 times 9 ] Now we can solve for ( D ) using the second equation: [ D = 32 times 9 ] [ D = 288 ] So, the distance covered by the van is boxed{288} kilometers.

answer:Let's denote the number of hours David mowed each day as ( h ). Since David worked for a week, he worked for 7 days. Therefore, the total number of hours he worked in a week is ( 7h ). The total amount of money he made from mowing the lawn is ( 14 times 7h = 98h ) dollars. He spent half of this money on a pair of shoes, so he was left with ( frac{98h}{2} = 49h ) dollars. He then gave half of the remaining money to his mom, so he was left with ( frac{49h}{2} ) dollars. According to the problem, after giving half of the remaining money to his mom, he had 49 left. Therefore, we can set up the equation: [ frac{49h}{2} = 49 ] To find ( h ), we multiply both sides of the equation by 2: [ 49h = 49 times 2 ] [ 49h = 98 ] Now, we divide both sides by 49 to solve for ( h ): [ h = frac{98}{49} ] [ h = 2 ] So, David mowed for boxed{2} hours each day.

answer:1. **Initial Information and Setup** Given the quadrilateral ABCD with AB = AD and CB = CD, and angle ABC = 90^circ. Points E and F are on line segments AB and AD respectively. Points P and Q are on line segment EF such that P is between E and Q and frac{AE}{EP} = frac{AF}{FQ}. Points X and Y are on line segments CP and CQ respectively such that BX perp CP and DY perp CQ. We need to prove that the points X, P, Q, and Y are concyclic. 2. **Triangles and Medians** From the problem setup, triangles ABC and ADC are right triangles with respect to AC: [ angle ABC = angle ADC = 90^circ. ] 3. **Intersection Point K** Let AC and EF intersect at point K. Since AK bisects angle EAF: [ Rightarrow frac{AE}{AF} = frac{KE}{KF}. ] 4. **Ratio Conditions** Given frac{AE}{AF} = frac{EP}{FQ}, it follows that: [ frac{KE}{KF} = frac{EP}{FQ} Rightarrow frac{KE}{EP} = frac{KF}{FQ}. ] 5. **Extending Lines CP and CQ** Extend CP to meet AE at S and extend CQ to meet AF at T. 6. **Menelaus' Theorem Application** For triangle KPC and the transversal ESA, Menelaus' theorem gives: [ frac{KE}{EP} cdot frac{PS}{SC} cdot frac{CA}{AK} = 1. ] 7. **Similar Argument for triangle KFQ** Similarly, applying Menelaus' theorem for triangle KFQ with transversal TCA: [ frac{KF}{FQ} cdot frac{QT}{TC} cdot frac{CA}{AK} = 1. ] 8. **Relating Ratios to CP and CQ** Therefore: [ frac{CS}{PS} = frac{KE}{EP} cdot frac{CA}{AK} = frac{KF}{FQ} cdot frac{CA}{AK} = frac{CT}{QT}, ] which implies: [ frac{CP}{CQ} = frac{CS}{CT}. ] 9. **Orthogonal Conditions** Considering BX perp CP and DY perp CQ, and since: [ angle ABC = angle ADC = 90^circ, ] we have: [ frac{CX cdot CP}{CY cdot CQ} = frac{CX cdot CS}{CY cdot CT} = left(frac{CB}{CD}right)^2 = 1. ] 10. **Conclusion** Therefore, the product of the powers of C with respect to the circles omega_X and omega_Y equals 1. This implies that the points X, P, Q, and Y are concyclic (lie on a common circle). [ boxed{X ,P, Q, Y text{ are concyclic}} ]