ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:05Z
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
312K
4β² + 5β + 0 = 0
d/dz [z^11] = 11z^10
125K
d/dφ [φ^6] = 6φ^5
6n² + 3n + 0 = 0
det| 3 7 ; 2 4 | = -2
311K
10β² + 6β + 6 = 0
F = ma
d/dα [α^1] = 1α^0