Symbolic Index /sym/ea4bfa
id=1 · host=au.commstar.world · 2026-10-20 20:23Z
∑_{k=1}^{n} k = n(n+1)/2
CH₄ + 2O₂ → CO₂ + 2H₂O
52K
6ω² + 3ω + 2 = 0
det| 9 7 ; 3 10 | = 69
90K
CH₄ + 2O₂ → CO₂ + 2H₂O
∑_{k=1}^{n} k = n(n+1)/2
100K
det| 8 3 ; 4 9 | = 60
∫₀^∞ e^(-n²) dn = √π / 2
197K
9x² + 3x + 6 = 0
∑_{k=1}^{n} k = n(n+1)/2
262K
∑_{k=1}^{n} k = n(n+1)/2
196K