Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:59Z

11n² + 5n + 4 = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

460K

det| 3 1 ; 7 4 | = 5

438K

∫₀^∞ e^(-x²) dx = √π / 2

8ψ² + 7ψ + 0 = 0

186K

∇ × E = −∂B/∂t

411K

4φ² + 6φ + 2 = 0

\frac{8}{7} \sum_{i=1}^{n} i^{2}

448K

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

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

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

459K

∫₀^∞ e^(-α²) dα = √π / 2

406K

10t² + 9t + 5 = 0

181K