Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-21 10:00Z

6y² + 3y + 3 = 0

∇ × E = −∂B/∂t

195K

det| 3 1 ; 7 4 | = 5

262K

ΔS ≥ 0

F = G m₁m₂ / r²

N₂ + 3H₂ ⇌ 2NH₃

244K

det| 4 5 ; 4 5 | = 0

∂n/∂θ = 3n

231K

det| 7 4 ; 4 8 | = 40

26K

iħ ∂ψ/∂t = Ĥψ

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

209K

∇²ω = 0

331K

∂t/∂α = 6t

\frac{8}{7} \sum_{i=1}^{n} i^{2}

163K

ΔS ≥ 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

∫₀^∞ e^(-φ²) dφ = √π / 2

9t² + 6t + 4 = 0