Symbolic Index /sym/ea4bfa
id=1 · host=au.commstar.world · 2026-10-20 20:17Z
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
244K
∫₀^∞ e^(-t²) dt = √π / 2
52K
6ω² + 3ω + 2 = 0
CH₄ + 2O₂ → CO₂ + 2H₂O
det| 4 6 ; 4 5 | = -4
100K
det| 8 3 ; 4 9 | = 60
det| 9 4 ; 1 10 | = 86
137K
9x² + 3x + 6 = 0
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
262K
CH₄ + 2O₂ → CO₂ + 2H₂O
196K
∑_{k=1}^{n} k = n(n+1)/2
218K