ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 21:26Z
\binom{n}{k} = \frac{n!}{k!(n-k)!}
∫₀^∞ e^(-r²) dr = √π / 2
∑_{k=1}^{n} k = n(n+1)/2
125K
∫₀^∞ e^(-n²) dn = √π / 2
91K
12γ² + 2γ + 2 = 0
det| 5 2 ; 3 6 | = 24
17K
det| 7 4 ; 4 8 | = 40
310K
E = mc²
det| 7 1 ; 7 8 | = 49
264K
λ = h/p
\frac{3}{8} \sum_{i=1}^{n} i^{2}
\binom{n}{k} = \frac{n!}{k!(n-k)!}
∇ × E = −∂B/∂t
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O