Math Dump /sym/6c3138

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

11n² + 5n + 4 = 0

F = ma

460K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

438K

∫₀^∞ e^(-x²) dx = √π / 2

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

186K

det| 7 4 ; 4 8 | = 40

411K

4φ² + 6φ + 2 = 0

Fe₂O₃ + 3CO → 2Fe + 3CO₂

448K

∑_{k=1}^{n} k = n(n+1)/2

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

6t² + 6t + 4 = 0

459K

iħ ∂ψ/∂t = Ĥψ

406K

∇ × E = −∂B/∂t

181K

8x² + 2x + 7 = 0

80K

F = G m₁m₂ / r²

359K