Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-19 20:19Z

∑_{k=1}^{n} k = n(n+1)/2

6t² + 6t + 4 = 0

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

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det| 8 9 ; 3 9 | = 45

8ψ² + 7ψ + 0 = 0

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∂z/∂φ = 7z

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C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

∇ × E = −∂B/∂t

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F = ma

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det| 2 3 ; 3 3 | = -3

det| 4 2 ; 6 5 | = 8

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det| 1 4 ; 3 2 | = -10

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

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

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e^{i\pi} + 1 = 0

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