Field Notes / Σ /sym/b26508
id=1 · host=uk.mattivster.me · 2026-10-18 20:43Z
E = mc²
2ω² + 4ω + 5 = 0
E = mc²
\frac{9}{8} \sum_{i=1}^{n} i^{2}
AgNO₃ + NaCl → AgCl↓ + NaNO₃
211K
det| 2 1 ; 2 3 | = 4
det| 7 4 ; 4 8 | = 40
390K
\binom{n}{k} = \frac{n!}{k!(n-k)!}
AgNO₃ + NaCl → AgCl↓ + NaNO₃
144K
\binom{n}{k} = \frac{n!}{k!(n-k)!}
352K