Field Notes / Σ /sym/b26508
id=1 · host=uk.mattivster.me · 2026-10-18 20:28Z
E = mc²
\lim_{t\to 0} \frac{\sin t}{t} = 1
144K
2ω² + 4ω + 5 = 0
∑_{k=1}^{n} k = n(n+1)/2
90K
E = mc²
det| 7 7 ; 4 8 | = 28
29K
\frac{9}{8} \sum_{i=1}^{n} i^{2}
det| 2 7 ; 7 3 | = -43
94K
det| 2 1 ; 2 3 | = 4
\frac{7}{1} \sum_{i=1}^{n} i^{2}
390K
det| 9 4 ; 1 10 | = 86
69K
\binom{n}{k} = \frac{n!}{k!(n-k)!}
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
144K
\lim_{γ\to 0} \frac{\sin γ}{γ} = 1
352K