svn commit: r43237 - head/en_US.ISO8859-1/books/faq

[Glen Barber]

Author: gjb
Date: Sun Nov 24 23:25:14 2013
New Revision: 43237
URL: http://svnweb.freebsd.org/changeset/doc/43237

Log:
  Fix build.
  
  Sponsored by:	The FreeBSD Foundation

Modified:
  head/en_US.ISO8859-1/books/faq/book.xml

Modified: head/en_US.ISO8859-1/books/faq/book.xml
==============================================================================
--- head/en_US.ISO8859-1/books/faq/book.xml	Sun Nov 24 22:44:32 2013	(r43236)
+++ head/en_US.ISO8859-1/books/faq/book.xml	Sun Nov 24 23:25:14 2013	(r43237)
@@ -1339,22 +1339,22 @@
 
 	<answer>
 	  <para>For FFS file systems, the largest file system is practically
-	    limited by the amount of memory required to &man.fsck.8 the file
-	    system.  &man.fsck.8 requires one bit per fragment, which with
-	    the default fragment size of 4&nbps;KB equates to 32&nbps;MB
+	    limited by the amount of memory required to &man.fsck.8; the file
+	    system.  &man.fsck.8; requires one bit per fragment, which with
+	    the default fragment size of 4 KB equates to 32 MB
 	    of memory per TB of disk.  This does mean that on architectures
-	    which limit userland processes to 2&nbps;GB (e.g., &i386;),
-	    the maximum &man.fsck.8'able filesystem is ~60&nbps;TB.</para>
+	    which limit userland processes to 2 GB (e.g., &i386;),
+	    the maximum &man.fsck.8;'able filesystem is ~60 TB.</para>
 
-	  <para>If there was not a &man.fsck.8 memory limit the maximum
-	    filesystem size would be 2&nbps;^&nbps;64 (blocks) * 32&nbps;KB
-	    => 16 Exa * 32&nbps;KB => 512 ZettaBytes.</para>
+	  <para>If there was not a &man.fsck.8; memory limit the maximum
+	    filesystem size would be 2 ^ 64 (blocks) * 32 KB
+	    => 16 Exa * 32 KB => 512 ZettaBytes.</para>
 
 	  <para>The maximum size of a single FFS file is approximately
 	    2 PB with the default block size of 32 KB.  Each
 	    32 KB block can point to 4096 blocks.  With triple
-	    indirect blocks, the calculation is 32&nbps;KB * 12 +
-	    32&nbps;KB * 4096 + 32&nbps;KB * 4096^2 + 32&nbps;KB *
+	    indirect blocks, the calculation is 32 KB * 12 +
+	    32 KB * 4096 + 32 KB * 4096^2 + 32 KB *
 	    4096^3.  Increasing the block size to 64 KB will increase
 	    the max file size by a factor of 16.</para>
 	</answer>