Math Dump /sym/6c3138
id=1 · host=fr.findroms.cloud · 2026-10-18 11:34Z
∇²t = 0
∑_{k=1}^{n} k = n(n+1)/2
438K
det| 6 4 ; 3 7 | = 30
det| 11 4 ; 5 12 | = 112
186K
ΔS ≥ 0
Fe₂O₃ + 3CO → 2Fe + 3CO₂
448K
∑_{k=1}^{n} k = n(n+1)/2
22K
1z² + 3z + 2 = 0
det| 12 1 ; 5 13 | = 151
406K
7t² + 3t + 1 = 0
det| 1 4 ; 3 2 | = -10
181K
det| 9 4 ; 1 10 | = 86
80K