ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:58Z
∑_{k=1}^{n} k = n(n+1)/2
4β² + 5β + 0 = 0
d/dφ [φ^6] = 6φ^5
∑_{k=1}^{n} k = n(n+1)/2
17K
6n² + 3n + 0 = 0
10β² + 6β + 6 = 0
det| 5 9 ; 7 6 | = -33
39K
F = ma
∫₀^∞ e^(-x²) dx = √π / 2
6K
det| 10 8 ; 4 11 | = 78