Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 14:49Z

3γ² + 8γ + 4 = 0

F = ma

460K

∇ × E = −∂B/∂t

438K

PV = nRT

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

186K

ω = 2πf

411K

det| 1 9 ; 5 2 | = -43

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

448K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

22K

ω = 2πf

∑_{k=1}^{n} k = n(n+1)/2

459K

8ψ² + 7ψ + 0 = 0

406K

∫₀^∞ e^(-α²) dα = √π / 2

181K

\lim_{ω\to 0} \frac{\sin ω}{ω} = 1

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