Symbolic Index /sym/ea4bfa
id=1 · host=au.commstar.world · 2026-10-19 00:35Z
11y² + 4y + 6 = 0
∑_{k=1}^{n} k = n(n+1)/2
479K
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
det| 9 4 ; 1 10 | = 86
102K
det| 12 1 ; 0 13 | = 156
det| 8 5 ; 3 9 | = 57
276K
CaCO₃ → CaO + CO₂
E = mc²
∑_{k=1}^{n} k = n(n+1)/2
CH₄ + 2O₂ → CO₂ + 2H₂O
2H₂ + O₂ → 2H₂O
\binom{n}{k} = \frac{n!}{k!(n-k)!}
11θ² + 5θ + 3 = 0