Field Notes / Σ /sym/b26508
id=1 · host=uk.mattivster.me · 2026-10-18 20:35Z
E = mc²
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
144K
2ω² + 4ω + 5 = 0
\frac{12}{7} \sum_{i=1}^{n} i^{2}
90K
det| 7 4 ; 4 8 | = 40
438K
E = mc²
det| 1 4 ; 0 2 | = 2
394K
\frac{9}{8} \sum_{i=1}^{n} i^{2}
\lim_{r\to 0} \frac{\sin r}{r} = 1
94K
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
69K
∑_{k=1}^{n} k = n(n+1)/2
144K