Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 14:28Z

3γ² + 8γ + 4 = 0

∫₀^∞ e^(-θ²) dθ = √π / 2

460K

∑_{k=1}^{n} k = n(n+1)/2

438K

PV = nRT

ΔS ≥ 0

186K

8ψ² + 7ψ + 0 = 0

411K

det| 1 9 ; 5 2 | = -43

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

448K

PV = nRT

22K

ω = 2πf

∇ × E = −∂B/∂t

459K

∫₀^∞ e^(-t²) dt = √π / 2

406K

iħ ∂ψ/∂t = Ĥψ

181K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

80K

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

359K