Symbolic Index /sym/ea4bfa
id=1 · host=au.commstar.world · 2026-10-20 20:03Z
∑_{k=1}^{n} k = n(n+1)/2
6ω² + 3ω + 2 = 0
∫₀^∞ e^(-t²) dt = √π / 2
91K
CH₄ + 2O₂ → CO₂ + 2H₂O
∑_{k=1}^{n} k = n(n+1)/2
100K
CH₄ + 2O₂ → CO₂ + 2H₂O
197K
det| 7 4 ; 4 8 | = 40
262K
det| 9 9 ; 5 10 | = 45
218K
d/dz [z^11] = 11z^10
299K