Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:50Z

11n² + 5n + 4 = 0

∫₀^∞ e^(-θ²) dθ = √π / 2

460K

∇ × E = −∂B/∂t

438K

∫₀^∞ e^(-x²) dx = √π / 2

PV = nRT

186K

∇²t = 0

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4φ² + 6φ + 2 = 0

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

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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

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∫₀^∞ e^(-φ²) dφ = √π / 2

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

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8ψ² + 7ψ + 0 = 0

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iħ ∂ψ/∂t = Ĥψ

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∑_{k=1}^{n} k = n(n+1)/2

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