Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:02Z
∑_{k=1}^{n} k = n(n+1)/2
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
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∫₀^∞ e^(-z²) dz = √π / 2
479K
8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
191K
\lim_{r\to 0} \frac{\sin r}{r} = 1