Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:40Z
∑_{k=1}^{n} k = n(n+1)/2
8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
191K
\lim_{r\to 0} \frac{\sin r}{r} = 1
Fe₂O₃ + 3CO → 2Fe + 3CO₂
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∑_{k=1}^{n} k = n(n+1)/2
153K
det| 5 2 ; 3 6 | = 24
473K
∑_{k=1}^{n} k = n(n+1)/2
338K