Symbolic Index /sym/ea4bfa
id=1 · host=au.commstar.world · 2026-10-20 20:26Z
∑_{k=1}^{n} k = n(n+1)/2
CH₄ + 2O₂ → CO₂ + 2H₂O
244K
∑_{k=1}^{n} k = n(n+1)/2
52K
6ω² + 3ω + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
90K
CH₄ + 2O₂ → CO₂ + 2H₂O
∑_{k=1}^{n} k = n(n+1)/2
50K
det| 7 4 ; 4 8 | = 40
137K
det| 5 6 ; 4 6 | = 6
262K