answer:1. **Understanding the sequence definition and inequality**: The sequence {a_n} is defined such that: [ a_{n+1} = frac{1}{2 - a_n} ] We are given that for any positive integer n, a_{n+1} > a_{n}. We must find the range of the initial term a_1. 2. **Start by expressing the inequality**: [ a_{n+1} - a_n > 0 ] Substitute a_{n+1}: [ frac{1}{2 - a_n} - a_n > 0 ] 3. **Combine the terms under a common denominator**: [ frac{1 - a_n(2 - a_n)}{2 - a_n} > 0 ] Simplify the numerator: [ 1 - 2a_n + a_n^2 ] Rewrite: [ frac{a_n^2 - 2a_n + 1}{2 - a_n} > 0 ] Factor the numerator as: [ frac{(a_n - 1)^2}{2 - a_n} > 0 ] 4. **Analyze the fraction**: The numerator (a_n - 1)^2 geq 0 because it is a square. For the fraction to be positive: [ 2 - a_n > 0 quad Leftrightarrow quad a_n < 2 ] And since (a_n - 1)^2 neq 0, a_n neq 1: [ a_n neq 1 ] 5. **Case analysis on a_1**: - **Case 1: a_1 < 1**: [ a_{n} < 1 quad Rightarrow quad 2 - a_{n} > 1 quad Rightarrow quad a_{n+1} = frac{1}{2 - a_n} < 1 ] Hence, if a_1 < 1, then forall n, a_n < 1 and a_{n+1} > a_n is satisfied. - **Case 2: a_1 > 1**: [ text{Let } a_1 > frac{m + 1}{m} text{ for some integer } m ] If a_n > frac{p + 1}{p}, then: [ 2 - a_n < 2 - frac{p + 1}{p} = frac{p - 1}{p} ] Then: [ a_{n+1} = frac{1}{2 - a_n} > frac{p}{p - 1} ] By repeating this, we can find m such that: [ a_{m} > 2 quad Rightarrow quad a_{m+1} < a_{m} ] So, if a_1 > 1, eventually the sequence will break the condition a_{n+1} > a_{n}. Conclusion: Given the conditions, the value range for a_1 such that a_{n+1} > a_n for all positive integers n is: [ boxed{a_1 < 1} ]

answer:# Problem: 设点 D, E, F 分别在 triangle ABC 的三边 BC, CA, AB 上，且 triangle AEF, triangle BFD, triangle CDE 的内切圆有相等的半径 r，又以 r_{0} 和 R 分别表示 triangle DEF 和 triangle ABC 的内切圆半径。求证: r + r_{0} = R。 : 1. 设 O_{1}, O_{2}, O_{3} 分别是 triangle AEF, triangle BFD, triangle CDE 的内心，O 是 triangle O_{1}O_{2}O_{3} 的内心，L 和 l_{0} 分别是 triangle ABC 和 triangle DEF 的周长。用 [ABC] 表示 triangle ABC 的面积。 2. 我们知道，面积关系为： [ [ABC] = [AEF] + [BFD] + [CDE] + [DEF] ] 3. 同时根据内切圆半径公式，面积还可以表示为： [ [ABC] = R cdot L ] [ [AEF] = r cdot (text{周长 }AEF) ] [ [BFD] = r cdot (text{周长 }BFD) ] [ [CDE] = r cdot (text{周长 }CDE) ] [ [DEF] = r_{0} cdot l_{0} ] 4. 将这些面积加起来： [ [AEF] + [BFD] + [CDE] + [DEF] = r cdot (text{周长 }AEF) + r cdot (text{周长 }BFD) + r cdot (text{周长 }CDE) + r_{0} cdot l_{0} ] 5. 由于 triangle AEF, triangle BFD, triangle CDE 的内切圆半径均为 r，我们有（将周长的和表示为 L + l_{0}）： [ [ABC] = r cdot (L + l_{0}) + r_{0} cdot l_{0} ] 6. 根据面积关系公式： [ R cdot L = r cdot (L + l_{0}) + r_{0} cdot l_{0} ] 7. 整理方程： [ R cdot L - r cdot (L + l_{0}) = r_{0} cdot l_{0} ] 8. 提取 L 和 l_{0}： [ (R - r) cdot L = (r + r_{0}) cdot l_{0} ] 9. 我们只需证明： [ (R - r) cdot L = R cdot l_{0} ] 10. 带入结论： [ (R - r) / R = l_{0} / L ] 11. 知道 triangle OBC sim triangle OO_2O_3 等比关系，所以按比例我们可以推导出 (R - r) / R 的比率。结合上述步骤，我们可以推出 r + r_{0} = R。 [ boxed{r + r_0 = R} ]

answer:To find out how many stickers Mika has, we need to add up all the stickers she received and had initially. She started with 20.0 stickers. She bought 26.0 stickers. She got 20.0 stickers for her birthday. Her sister gave her 6.0 stickers. Her mother gave her 58.0 stickers. So, the total number of stickers Mika has is: 20.0 (initial) + 26.0 (bought) + 20.0 (birthday) + 6.0 (sister) + 58.0 (mother) = 130.0 stickers Mika has boxed{130.0} stickers in total.

answer:Let C represent the set of adults who drink coffee and T represent the set of adults who drink tea. We are given: - P(C) = 90% - P(T) = 85% - P(neg (C cup T)) = 15%, where neg (C cup T) represents the adults drinking neither coffee nor tea. Since (P(neg (C cup T)) = 15%), therefore (P(C cup T) = 100% - 15% = 85%). Using the principle of inclusion-exclusion on C and T: [ P(C cup T) = P(C) + P(T) - P(C cap T) ] Substituting the given values yields: [ 85% = 90% + 85% - P(C cap T) ] This simplifies to: [ P(C cap T) = 90% + 85% - 85% ] [ P(C cap T) = 90% ] This calculation shows that at least 90% of adults must drink both coffee and tea, considering that everyone drinking either coffee or tea or both encompasses 85% of the population. Here, no subtraction occurs due to the coverage of non-drinkers. Conclusion with boxed answer: [boxed{90%}]