Spectra Ledger /sym/2e6562
id=1 · host=commstar.world · 2026-10-19 20:09Z
∑_{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
det| 10 3 ; 4 11 | = 98
139K
det| 1 6 ; 5 2 | = -28
det| 9 4 ; 1 10 | = 86
91K
ω = 2πf
4r² + 4r + 6 = 0
9r² + 1r + 4 = 0