Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-20 06:39Z

∑_{k=1}^{n} k = n(n+1)/2

∇ × E = −∂B/∂t

273K

3z² + 5z + 1 = 0

263K

e^{i\pi} + 1 = 0

F = G m₁m₂ / r²

139K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

108K

2H₂ + O₂ → 2H₂O

∑_{k=1}^{n} k = n(n+1)/2

333K

det| 7 4 ; 4 8 | = 40

160K

det| 4 8 ; 0 5 | = 20

iħ ∂ψ/∂t = Ĥψ

439K

e^{i\pi} + 1 = 0

413K

det| 4 7 ; 7 5 | = -29

∇ × E = −∂B/∂t

297K

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

128K

10t² + 5t + 2 = 0

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

404K

det| 1 6 ; 5 2 | = -28