Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-20 04:51Z

∇²n = 0

10β² + 4β + 1 = 0

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Fe₂O₃ + 3CO → 2Fe + 3CO₂

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9r² + 4r + 1 = 0

E = mc²

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∑_{k=1}^{n} k = n(n+1)/2

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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

8ψ² + 7ψ + 0 = 0

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∇²γ = 0

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11γ² + 3γ + 6 = 0

∇ × E = −∂B/∂t

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∑_{k=1}^{n} k = n(n+1)/2

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∑_{k=1}^{n} k = n(n+1)/2

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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

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

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\lim_{ω\to 0} \frac{\sin ω}{ω} = 1

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