Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-19 07:56Z
ω = 2πf
∑_{k=1}^{n} k = n(n+1)/2
326K
det| 7 4 ; 4 8 | = 40
198K
7r² + 8r + 6 = 0
det| 1 4 ; 3 2 | = -10
155K
∑_{k=1}^{n} k = n(n+1)/2
det| 10 9 ; 3 11 | = 83
det| 7 9 ; 1 8 | = 47
319K
2ψ² + 6ψ + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
359K
ω = 2πf