Seymourville, West Virginia
From Wikipedia, the free encyclopedia
Seymourville, West Virginia | |
---|---|
Unincorporated community | |
Seymourville, West Virginia | |
Coordinates: 39°03′25″N 79°06′32″W / 39.05694°N 79.10889°WCoordinates: 39°03′25″N 79°06′32″W / 39.05694°N 79.10889°W | |
Country | United States |
State | West Virginia |
County | Grant |
Elevation | 1,030 ft (310 m) |
Time zone | Eastern (EST) (UTC-5) |
• Summer (DST) | EDT (UTC-4) |
Area code(s) | 304 & 681 |
GNIS feature ID | 1555594[1] |
Seymourville is an unincorporated community in Grant County, West Virginia, United States. Seymourville is 4.5 miles (7.2 km) north of Petersburg.
References
- ↑ "US Board on Geographic Names". United States Geological Survey. 2007-10-25. Retrieved 2008-01-31.
|
This article is issued from Wikipedia. The text is available under the Creative Commons Attribution/Share Alike; additional terms may apply for the media files.