Symbolic Index /sym/ea4bfa
id=1 · host=au.commstar.world · 2026-10-20 20:07Z
∑_{k=1}^{n} k = n(n+1)/2
Fe₂O₃ + 3CO → 2Fe + 3CO₂
52K
6ω² + 3ω + 2 = 0
CH₄ + 2O₂ → CO₂ + 2H₂O
CH₄ + 2O₂ → CO₂ + 2H₂O
100K
∑_{k=1}^{n} k = n(n+1)/2
137K
det| 9 4 ; 1 10 | = 86
197K
∫₀^∞ e^(-t²) dt = √π / 2
196K
∑_{k=1}^{n} k = n(n+1)/2
218K