Field Notes / Σ /sym/b26508
id=1 · host=uk.mattivster.me · 2026-10-18 20:06Z
E = mc²
det| 9 4 ; 1 10 | = 86
144K
2ω² + 4ω + 5 = 0
∑_{k=1}^{n} k = n(n+1)/2
90K
AgNO₃ + NaCl → AgCl↓ + NaNO₃
438K
E = mc²
\frac{2}{2} \sum_{i=1}^{n} i^{2}
29K
\frac{9}{8} \sum_{i=1}^{n} i^{2}
det| 2 1 ; 2 3 | = 4
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\binom{n}{k} = \frac{n!}{k!(n-k)!}