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Home » 10 Winter-Backward Summer Bucket List Secrets You Must Save for This Year! - Abbey Badges

10 Winter-Backward Summer Bucket List Secrets You Must Save for This Year! - Abbey Badges

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Solution: Start with $ b_1 = 2 $. Compute $ b_2 = Q(2) = 2^2 - rac{2^4}{4} = 4 - rac{16}{4} = 4 - 4 = 0 $. Then compute $ b_3 = Q(0) = 0^2 - rac{0^4}{4} = 0 - 0 = 0 $. Thus, $ b_3 = oxed{0} $.

📌 Question: A high-performance computing algorithm processes data in nested loops, where the number of operations after $ k $ iterations is modeled by $ R(k) = 3^k - rac{3^{2k}}{2} $. Find the smallest positive integer $ k $ for which $ R(k) < -1 $.

📌 Solution: Solve $ 3^k - rac{3^{2k}}{2} < -1 $. Let $ x = 3^k $ (note $ x > 0 $). The inequality becomes $ x - rac{x^2}{2} < -1 $, or $ -rac{x^2}{2} + x + 1 < 0 $. Multiply by $-2$ (reversing inequality): $ x^2 - 2x - 2 > 0 $. Solve $ x^2 - 2x - 2 = 0 $: roots are $ x = 1 \pm \sqrt{3} $. Since $ x > 0 $, critical point is $ x = 1 + \sqrt{3} pprox 2.732 $. The quadratic is positive when $ x > 1 + \sqrt{3} $. Since $ x = 3^k $, find smallest $ k $ such that $ 3^k > 1 + \sqrt{3} $. Test $ k = 1 $: $ 3^1 = 3 > 2.732 $. Check $ R(1) = 3 - rac{9}{2} = -1.5 < -1 $. Thus, $ k = oxed{1} $.

📌 Question: A plant biologist models the growth efficiency of a drought-resistant crop with $ G(t) = rac{t^2 - 4}{t - 2} $. Simplify $ G(t) $ and determine its domain.

📌 Solution: Factor numerator: $ t^2 - 4 = (t - 2)(t + 2) $. Thus, $ G(t) = rac{(t - 2)(t + 2)}{t - 2} $. For $ t

📌 eq 2 $, cancel $ t - 2 $: $ G(t) = t + 2 $. The domain excludes $ t = 2 $, so $ t \in (-\infty, 2) \cup (2, \infty) $. Simplified form: $ G(t) = t + 2 $, domain: $ oxed{(-\infty, 2) \cup (2, \infty)} $.Question: If a linguist randomly selects three distinct letters from the English alphabet, what is the probability that they form a placeholder typographic symbol, given that there are 10 such symbols and 26 letters?

📌 Solution: The total number of ways to choose 3 distinct letters is $inom{26}{3}$. The favorable cases involve selecting 3 letters that correspond to exactly one placeholder symbol. Assuming each symbol is defined by a unique combination of 3 letters (with no overlap in symbol definitions), there are 10 favorable combinations. Thus, the probability is $rac{10}{inom{26}{3}} = rac{10}{2600} = rac{1}{260}$. $oxed{\dfrac{1}{260}}$

📌 Question: If a historian randomly arranges 5 historical documents, 2 of which are letters and 3 are manuscripts, what is the probability that the two letters are adjacent?

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