Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 21:54Z

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

det| 3 7 ; 2 4 | = -2

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

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e^{i\pi} + 1 = 0

∑_{k=1}^{n} k = n(n+1)/2

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10t² + 9t + 5 = 0

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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

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

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iħ ∂ψ/∂t = Ĥψ

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∇ × E = −∂B/∂t

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

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8ψ² + 7ψ + 0 = 0

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