Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-19 07:59Z
ω = 2πf
∑_{k=1}^{n} k = n(n+1)/2
198K
7r² + 8r + 6 = 0
∫₀^∞ e^(-n²) dn = √π / 2
63K
∑_{k=1}^{n} k = n(n+1)/2
det| 3 6 ; 0 4 | = 12
366K
det| 10 9 ; 3 11 | = 83
2ψ² + 6ψ + 0 = 0
det| 9 4 ; 1 10 | = 86
263K
ω = 2πf
∂ψ/∂t = 3ψ
6θ² + 3θ + 6 = 0
det| 8 4 ; 5 9 | = 52