Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:40Z
∑_{k=1}^{n} k = n(n+1)/2
det| 5 6 ; 7 6 | = -12
75K
det| 4 6 ; 4 5 | = -4
479K
8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
Fe₂O₃ + 3CO → 2Fe + 3CO₂
390K
\lim_{r\to 0} \frac{\sin r}{r} = 1
det| 9 4 ; 1 10 | = 86
1K
det| 12 1 ; 1 13 | = 155
∑_{k=1}^{n} k = n(n+1)/2
473K
det| 6 6 ; 3 7 | = 24
338K
N₂ + 3H₂ ⇌ 2NH₃