Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-19 23:37Z
11t² + 7t + 4 = 0
∑_{k=1}^{n} k = n(n+1)/2
198K
E = mc²
CH₄ + 2O₂ → CO₂ + 2H₂O
155K
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\frac{2}{2} \sum_{i=1}^{n} i^{2}
366K
CH₄ + 2O₂ → CO₂ + 2H₂O
AgNO₃ + NaCl → AgCl↓ + NaNO₃
det| 10 1 ; 4 11 | = 106
det| 5 3 ; 7 6 | = 9
412K
∑_{k=1}^{n} k = n(n+1)/2
iħ ∂ψ/∂t = Ĥψ