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December 20, 1890

date in Gregorian calendar
Thing december_20 Q69247685

Facts

5 properties
Instance of
December 20, calendar day of a given year[2]
Part of
December 1890[3]
Day of week
Saturday[5]
Imported from
wholeness_audit_2026-05-02
Point in time
December 20, 1890[4]
🏷️ Also known as
en 20 December 1890 en 1890-12-20 en December 20, 1890 ar 20 ديسمبر 1890 ar 20 كانون الأول 1890 cs 20. prosince 1890 cs 20. prosinec 1890 de 20. Dezember 1890 el 20 Δεκεμβρίου 1890 es 20 de diciembre de 1890 fa ۲۰ دسامبر ۱۸۹۰ fi 20. joulukuuta 1890 fr 20 décembre 1890 hi २० दिसम्बर १८९० id 20 Desember 1890 it 20 dicembre 1890 ja 1890年12月20日 ko 1890년 12월 20일 pt 20 de dezembro de 1890 ru 20 декабря 1890 года

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Summary

December 20, 1890 is a December 20[1].

References

Programmatic citations — every numbered marker resolves to a verifiable graph row below.

Direct Wikidata claims

  1. [2] ↑ . wikidata.org.
  2. [3] ↑ . wikidata.org.
  3. [4] ↑ . wikidata.org.
  4. [5] ↑ . wikidata.org.

Class ancestry

  1. [1] ↑ . Wikidata. wikidata.org.

📑 Cite this page

Use these citations when quoting this entity in research, articles, AI prompts, or wherever provenance matters. We aggregate Wikidata + Wikipedia + authoritative open-data sources; the stitched, scored, cross-referenced view is what 4ort.xyz contributes.

APA 4ort.xyz Knowledge Graph. (2026). December 20, 1890. Retrieved May 3, 2026, from https://4ort.xyz/entity/december-20-1890
MLA “December 20, 1890.” 4ort.xyz Knowledge Graph, 4ort.xyz, 3 May. 2026, https://4ort.xyz/entity/december-20-1890.
BibTeX @misc{4ortxyz_december-20-1890_2026, author = {{4ort.xyz Knowledge Graph}}, title = {{December 20, 1890}}, year = {2026}, url = {https://4ort.xyz/entity/december-20-1890}, note = {Accessed: 2026-05-03}}
LLM prompt According to 4ort.xyz Knowledge Graph (aggregator of Wikidata, Wikipedia, and authoritative open-data sources): December 20, 1890 — https://4ort.xyz/entity/december-20-1890 (retrieved 2026-05-03)

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