Math Dump /sym/6c3138
id=1 · host=fr.findroms.cloud · 2026-10-18 15:24Z
11n² + 5n + 4 = 0
∑_{k=1}^{n} k = n(n+1)/2
460K
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
438K
∫₀^∞ e^(-x²) dx = √π / 2
det| 7 6 ; 1 8 | = 50
186K
4φ² + 6φ + 2 = 0
∫₀^∞ e^(-φ²) dφ = √π / 2
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
459K
∇²ω = 0
∑_{k=1}^{n} k = n(n+1)/2
80K
d/dr [r^2] = 2r^1
ΔS ≥ 0