Operator Journal /sym/f43c60
id=1 · host=nl.mattivster.me · 2026-10-19 07:48Z
ω = 2πf
∑_{k=1}^{n} k = n(n+1)/2
129K
10ω² + 6ω + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
366K
ω = 2πf
det| 5 9 ; 0 6 | = 30
224K
det| 7 4 ; 4 8 | = 40
452K
det| 7 1 ; 7 8 | = 49
33K
det| 6 6 ; 3 7 | = 24
96K
CH₄ + 2O₂ → CO₂ + 2H₂O
273K