Symbolic Index /sym/ea4bfa
id=1 · host=au.commstar.world · 2026-10-20 20:51Z
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
244K
∫₀^∞ e^(-x²) dx = √π / 2
52K
6ω² + 3ω + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
90K
CH₄ + 2O₂ → CO₂ + 2H₂O
det| 8 3 ; 4 9 | = 60
det| 1 2 ; 7 2 | = -12
197K
det| 3 6 ; 0 4 | = 12
262K