Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:17Z

11n² + 5n + 4 = 0

N₂ + 3H₂ ⇌ 2NH₃

460K

e^{i\pi} + 1 = 0

438K

∫₀^∞ e^(-x²) dx = √π / 2

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

186K

\binom{n}{k} = \frac{n!}{k!(n-k)!}

411K

4φ² + 6φ + 2 = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

448K

\binom{n}{k} = \frac{n!}{k!(n-k)!}

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

N₂ + 3H₂ ⇌ 2NH₃

459K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

406K

∇²ω = 0

∑_{k=1}^{n} k = n(n+1)/2

181K

N₂ + 3H₂ ⇌ 2NH₃

80K

d/dr [r^2] = 2r^1

det| 7 4 ; 4 8 | = 40

359K

∇²r = 0

455K

8β² + 5β + 0 = 0