Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-20 06:35Z

∑_{k=1}^{n} k = n(n+1)/2

∇ × E = −∂B/∂t

273K

iħ ∂ψ/∂t = Ĥψ

263K

e^{i\pi} + 1 = 0

∑_{k=1}^{n} k = n(n+1)/2

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

108K

2H₂ + O₂ → 2H₂O

∇²γ = 0

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det| 9 4 ; 1 10 | = 86

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det| 4 8 ; 0 5 | = 20

∑_{k=1}^{n} k = n(n+1)/2

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e^{i\pi} + 1 = 0

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det| 4 7 ; 7 5 | = -29

det| 2 3 ; 5 3 | = -9

10t² + 5t + 2 = 0

det| 1 6 ; 5 2 | = -28

F = ma