Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 21:56Z

10y² + 2y + 6 = 0

∫₀^∞ e^(-θ²) dθ = √π / 2

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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

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∫₀^∞ e^(-t²) dt = √π / 2

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

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iħ ∂ψ/∂t = Ĥψ

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e^{i\pi} + 1 = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

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det| 12 1 ; 5 13 | = 151

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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

e^{i\pi} + 1 = 0

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N₂ + 3H₂ ⇌ 2NH₃

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N₂ + 3H₂ ⇌ 2NH₃

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PV = nRT

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8ψ² + 7ψ + 0 = 0

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