Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:00Z
∑_{k=1}^{n} k = n(n+1)/2
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
75K
8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
\lim_{r\to 0} \frac{\sin r}{r} = 1
det| 12 1 ; 1 13 | = 155
det| 9 4 ; 1 10 | = 86
473K
N₂ + 3H₂ ⇌ 2NH₃
det| 5 5 ; 2 6 | = 20
e^{i\pi} + 1 = 0