Math Dump /sym/6c3138

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

11n² + 5n + 4 = 0

5φ² + 1φ + 1 = 0

460K

∇ × E = −∂B/∂t

438K

∫₀^∞ e^(-x²) dx = √π / 2

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

186K

N₂ + 3H₂ ⇌ 2NH₃

411K

4φ² + 6φ + 2 = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

448K

det| 10 7 ; 3 11 | = 89

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

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

459K

∂y/∂γ = 8y

406K

d/dt [t^3] = 3t^2

181K

det| 7 2 ; 7 8 | = 42

80K

\lim_{ω\to 0} \frac{\sin ω}{ω} = 1

359K

8ψ² + 7ψ + 0 = 0

455K