answer:To calculate the total percent decrease over the two years, we need to apply each year's percentage decrease to the value of the card at the start of that year. Let's assume the original value of the card is 100 (you can use any starting value, the percentage decrease will be the same). First year decrease of 20%: 100 - (20% of 100) = 100 - 20 = 80 The value of the card after the first year is 80. Second year decrease of 10%: 80 - (10% of 80) = 80 - 8 = 72 The value of the card after the second year is 72. Now, to find the total percent decrease over the two years, we compare the final value to the original value: Original value: 100 Final value: 72 Decrease in value: 100 - 72 = 28 Total percent decrease = (Decrease in value / Original value) * 100 Total percent decrease = (28 / 100) * 100 Total percent decrease = 0.28 * 100 Total percent decrease = 28% So, the total percent decrease of the card's value over the two years is boxed{28%} .

answer:1. **Define the Problem:** - Let D be a point on side BC of triangle ABC. - The circumcircle of triangle ABD intersects line AC at point E. - The circumcircle of triangle ADC intersects line AB at point F. - Let O be the circumcenter of triangle AEF. - We need to prove that OD perp BC. 2. **Reflection Over D:** - Reflect point F over D to get point F'. This means that DF = DF' and FF' parallel BC. 3. **Angle Chasing:** - Since E and F lie on the circumcircles of triangle ABD and triangle ADC respectively, we have: [ angle CDE = angle BAC = angle FDB ] - This implies BC bisects angle EDF externally. 4. **Point F' on Line DE:** - Given that DF = DF' and FF' parallel BC, the construction of F' makes F' lie on line DE. 5. **Angle Relationships:** - Following the reflection and angle properties: [ angle FFD = angle CDFF' = angle CDE = angle BAC = angle FAE ] - Thus, point F' lies on the circumcircle of triangle AEF. 6. **Reflection Property in Circumcircle:** - Since O is the circumcenter of triangle AEF, it must also be the circumcenter of triangle AEFF'. Therefore, OF = OF'. 7. **Orthogonality Conclusion:** - Combining all these properties gives us: [ DF' = DF quad text{and} quad OF = OF' ] - Hence, OD is perpendicular to line FF'. - Since FF' parallel BC, we thus have: [ OD perp BC ] # Conclusion: (boxed{OD perp BC})

answer:This is a combination problem where the order of selection does not matter. We need to form a 5-person committee from the remaining 27 team owners (30 total - 3 who do not wish to serve). The number of ways to choose 5 people from the 27 eligible owners is given by the combination formula binom{n}{k} = frac{n!}{k!(n-k)!}. For our case, n = 27 and k = 5: [ binom{27}{5} = frac{27!}{5!(27-5)!} = frac{27 times 26 times 25 times 24 times 23}{5 times 4 times 3 times 2 times 1} ] Calculating the above: - (5! = 120) - (27 times 26 times 25 times 24 times 23 = 7,893,600) [ frac{7,893,600}{120} = 65,780 ] Thus, there are boxed{65,780} ways to choose the 5-person rules committee from the eligible team owners.

answer:First, we draw rectangle PQRS with PQ = 150. Point T is the midpoint of overline{PS}, and lines PT and QT are perpendicular. Since PT and QT are perpendicular, triangle PTQ is a right triangle at T. Let's denote the length of PS as x. Since T is the midpoint of overline{PS}, PT = frac{x}{2}. Now, because PQRS is a rectangle, PQ = QR = 150 and PS = QR = x. Triangle PTQ being right means that by Pythagoras's theorem: [ PT^2 + QT^2 = PQ^2 ] Substituting the known sides: [ left(frac{x}{2}right)^2 + left(frac{x}{2}right)^2 = 150^2 ] [ frac{x^2}{4} + frac{x^2}{4} = 22500 ] [ frac{x^2}{2} = 22500 ] [ x^2 = 45000 ] [ x = sqrt{45000} ] [ x = 150sqrt{2} ] The greatest integer less than PS is the floor of 150sqrt{2}, which is approximately: [ 150sqrt{2} approx 212.13 ] Therefore, the greatest integer less than PS is boxed{212}.