Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-19 23:41Z
11t² + 7t + 4 = 0
∑_{k=1}^{n} k = n(n+1)/2
198K
E = mc²
∑_{k=1}^{n} k = n(n+1)/2
63K
\binom{n}{k} = \frac{n!}{k!(n-k)!}
∫₀^∞ e^(-φ²) dφ = √π / 2
366K
det| 7 4 ; 4 8 | = 40
201K
CH₄ + 2O₂ → CO₂ + 2H₂O
AgNO₃ + NaCl → AgCl↓ + NaNO₃
ΔS ≥ 0
det| 10 1 ; 4 11 | = 106
e^{i\pi} + 1 = 0
∑_{k=1}^{n} k = n(n+1)/2
iħ ∂ψ/∂t = Ĥψ