Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:31Z

11n² + 5n + 4 = 0

e^{i\pi} + 1 = 0

460K

∇ × E = −∂B/∂t

438K

∫₀^∞ e^(-x²) dx = √π / 2

\binom{n}{k} = \frac{n!}{k!(n-k)!}

186K

E = mc²

411K

4φ² + 6φ + 2 = 0

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

448K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

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

459K

8ψ² + 7ψ + 0 = 0

406K

5z² + 8z + 3 = 0

181K

N₂ + 3H₂ ⇌ 2NH₃

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