Symbolic Index /sym/ea4bfa
id=1 · host=au.commstar.world · 2026-10-20 20:13Z
∑_{k=1}^{n} k = n(n+1)/2
CH₄ + 2O₂ → CO₂ + 2H₂O
244K
det| 9 4 ; 1 10 | = 86
52K
6ω² + 3ω + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
91K
CH₄ + 2O₂ → CO₂ + 2H₂O
∫₀^∞ e^(-n²) dn = √π / 2
50K
det| 8 3 ; 4 9 | = 60
∫₀^∞ e^(-φ²) dφ = √π / 2
197K
9x² + 3x + 6 = 0
∑_{k=1}^{n} k = n(n+1)/2
262K