Image:King's Valley II Coverart.png

From Wikipedia, the free encyclopedia

Image
File history
File links

No higher resolution available.
King's_Valley_II_Coverart.png‎ (256 × 337 pixels, file size: 177 KB, MIME type: image/png)

[edit] Summary

Non-free / fair use media rationale for King's Valley II
Description	King's Valley II cover art.
Source	From GameFAQs.
Article	King's Valley II
Portion used	Image shown in its entirety.
Low resolution?	It is a low resolution image.
Purpose of use	Illustrates the entire subject of the article.
Replaceable?	It's the official cover. No free alternative could possibly exist.
Other information	Copyright owned by Konami.

[edit] Licensing

This image is cover art for a video or computer game, and the copyright for it is most likely owned by either the game's publisher or developer. It is believed that the use of low-resolution images of game cover art

to illustrate the packaging of the game in question
on the English-language Wikipedia, hosted on servers in the United States by the non-profit Wikimedia Foundation,

constitutes fair use under United States copyright law. Other uses of this image, on Wikipedia or elsewhere, might be copyright infringement. See Wikipedia:Non-free content for more information.

To the uploader: please add a detailed fair use rationale for each use, as well as the source of the work and copyright information.

Usage:

This template accepts the platform(s) as parameter to better categorize the tagged image.

{{Non-free game cover|Macintosh|Linux|Windows}}

{{Non-free game cover|Macintosh}}

or just

{{Non-free game cover}}

if the platform is unknown.

File history

Click on a date/time to view the file as it appeared at that time.

	Date/Time	Dimensions	User	Comment
current	12:46, 6 January 2008	256×337 (177 KB)	Salavat (Talk \| contribs)	({{Non-free use rationale \| Article = King's Valley II \| Description = '''''King's Valley II''''' cover art. \| Source = From [http://www.gamefaqs.com/computer/msx/image/918128.html?box=90471 GameFAQs]. \| Portion = I)