Math Dump /sym/6c3138

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

11n² + 5n + 4 = 0

e^{i\pi} + 1 = 0

460K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

438K

∫₀^∞ e^(-x²) dx = √π / 2

11t² + 9t + 7 = 0

186K

det| 7 4 ; 4 8 | = 40

411K

4φ² + 6φ + 2 = 0

CaCO₃ → CaO + CO₂

448K

9ψ² + 7ψ + 3 = 0

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

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

459K

N₂ + 3H₂ ⇌ 2NH₃

406K

∇ × E = −∂B/∂t

181K

∫₀^∞ e^(-t²) dt = √π / 2

80K

F = G m₁m₂ / r²

359K