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prince of persia the lost crown - Dyverse

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 ight) = rac{4}{x^2 - 1}.

📌 Since $ x > 1 $, $ x^2 - 1 > 0 $, so $ f(x) > 0 $, and as $ x o 1^+ $, $ f(x) o \infty $, as $ x o \infty $, $ f(x) o 0^+ $. So $ R $ has no minimum, only an infimum of 0. But the problem likely intends for us to find the minimum under constraints â but none restrict. However, reconsider the original expression:

📌 rac{a}{b} + rac{b}{a} - 2 \geq 0,

📌 with equality only if $ a = b $, i.e., $ rac{p+q}{p-q} = rac{p-q}{p+q} $, which implies $ (p+q)^2 = (p-q)^2 \Rightarrow pq = 0 $, impossible. So $ R > 0 $, but minimum not attained. But perhaps the hydrologist defines $ R $ only when flow is stable â yet no such condition

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So -20 × 636.39 / 0.6931 ≈ -12727.8 / 0.6931 ≈ -18368. But this is neg...
Wait: correct formula:
∫ a^u du = u / (ln a), but a = 0.5, ln(0.5) < 0
So ∫ (0.5)^u du = u / (ln 0.5) = u / (-ln 2) = -u / ln 2
So ∫₁² (0.5)^u du = [ -u / ln2 ]₁² = ( -2/ln2 ) - ( -1/ln2 ) = -1/ln2
But since integrating a positive function, take absolute value:
= 1/ln2 ≈ 1.4427
Then total = 20k × (1/ln2) = 20 × 636.39 × 1.4427 ≈ 12727.8 × 1.4427 ≈...
Alternative: use discrete approximation or reconsider.
But simpler: since halves every 20 cm, depth 20–40 cm is one 20-cm lay...

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