Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-19 18:37Z
∇²α = 0
∫₀^∞ e^(-t²) dt = √π / 2
326K
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
198K
6φ² + 8φ + 4 = 0
∫₀^∞ e^(-x²) dx = √π / 2
155K
∂φ/∂φ = 1φ
∑_{k=1}^{n} k = n(n+1)/2
201K
det| 4 8 ; 1 5 | = 12
\lim_{t\to 0} \frac{\sin t}{t} = 1
176K
\lim_{ω\to 0} \frac{\sin ω}{ω} = 1
det| 7 4 ; 4 8 | = 40
359K
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
∇²r = 0