Spectra Ledger /sym/2e6562
id=1 · host=commstar.world · 2026-10-19 20:11Z
∑_{k=1}^{n} k = n(n+1)/2
12ω² + 6ω + 7 = 0
d/dn [n^4] = 4n^3
∑_{k=1}^{n} k = n(n+1)/2
376K
det| 9 1 ; 1 10 | = 89
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
344K
det| 1 6 ; 5 2 | = -28
det| 9 4 ; 1 10 | = 86
91K
e^{i\pi} + 1 = 0
ω = 2πf