ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:13Z
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
302K
4β² + 5β + 0 = 0
det| 5 2 ; 3 6 | = 24
91K
d/dφ [φ^6] = 6φ^5
6n² + 3n + 0 = 0
det| 7 1 ; 7 8 | = 49
311K
10β² + 6β + 6 = 0
4β² + 3β + 0 = 0
F = ma