Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-19 01:19Z

CH₄ + 2O₂ → CO₂ + 2H₂O

4t² + 3t + 3 = 0

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∑_{k=1}^{n} k = n(n+1)/2

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6α² + 6α + 5 = 0

det| 7 4 ; 4 8 | = 40

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∑_{k=1}^{n} k = n(n+1)/2

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PV = nRT

9α² + 7α + 5 = 0

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

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CH₄ + 2O₂ → CO₂ + 2H₂O

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∇ × E = −∂B/∂t

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∂n/∂θ = 3n

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∫₀^∞ e^(-t²) dt = √π / 2

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F = G m₁m₂ / r²

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iħ ∂ψ/∂t = Ĥψ

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