Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 12:57Z
2ψ² + 5ψ + 4 = 0
∫₀^∞ e^(-φ²) dφ = √π / 2
479K
N₂ + 3H₂ ⇌ 2NH₃
det| 4 8 ; 3 5 | = -4
276K
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
129K
\lim_{φ\to 0} \frac{\sin φ}{φ} = 1
det| 9 4 ; 1 10 | = 86
191K
∇²z = 0
∑_{k=1}^{n} k = n(n+1)/2
1K
N₂ + 3H₂ ⇌ 2NH₃
det| 3 7 ; 2 4 | = -2
473K
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
8t² + 2t + 3 = 0
\lim_{θ\to 0} \frac{\sin θ}{θ} = 1
F = ma
∫₀^∞ e^(-φ²) dφ = √π / 2