Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:34Z

11n² + 5n + 4 = 0

∑_{k=1}^{n} k = n(n+1)/2

460K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

438K

∫₀^∞ e^(-x²) dx = √π / 2

det| 4 7 ; 2 5 | = 6

186K

det| 7 4 ; 4 8 | = 40

411K

4φ² + 6φ + 2 = 0

CH₄ + 2O₂ → CO₂ + 2H₂O

448K

∇²ω = 0

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

459K

\lim_{β\to 0} \frac{\sin β}{β} = 1

406K

∇ × E = −∂B/∂t

181K

N₂ + 3H₂ ⇌ 2NH₃

80K

F = G m₁m₂ / r²

359K