Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:01Z
∑_{k=1}^{n} k = n(n+1)/2
8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
129K
∑_{k=1}^{n} k = n(n+1)/2
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
473K
det| 7 4 ; 4 8 | = 40
188K
det| 3 6 ; 0 4 | = 12
416K