Math Dump /sym/922513
id=607807 · host=jp.commstar.world · 2026-09-01 21:14Z
E = mc²
det| 4 1 ; 6 5 | = 14
402K
2φ² + 8φ + 5 = 0
det| 5 3 ; 7 6 | = 9
398K
λ = h/p
∇ × E = −∂B/∂t
\binom{n}{k} = \frac{n!}{k!(n-k)!}
det| 2 4 ; 1 3 | = 2
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\binom{n}{k} = \frac{n!}{k!(n-k)!}
12γ² + 6γ + 5 = 0
det| 7 4 ; 7 8 | = 28
∑_{k=1}^{n} k = n(n+1)/2
AgNO₃ + NaCl → AgCl↓ + NaNO₃