Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:38Z

11n² + 5n + 4 = 0

∑_{k=1}^{n} k = n(n+1)/2

460K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

438K

∫₀^∞ e^(-x²) dx = √π / 2

λ = h/p

186K

det| 7 4 ; 4 8 | = 40

411K

4φ² + 6φ + 2 = 0

PV = nRT

448K

N₂ + 3H₂ ⇌ 2NH₃

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

\lim_{r\to 0} \frac{\sin r}{r} = 1

459K

∇ × E = −∂B/∂t

406K

∑_{k=1}^{n} k = n(n+1)/2

181K

\frac{2}{2} \sum_{i=1}^{n} i^{2}

80K