user: ### NOW ANSWER:
Question: What 2009 has 39th as the 2011?
Columns: ['2013', '2012', '2011', '2010', '2009', '2008', '2007']
Types: ['text', 'text', 'text', 'text', 'text', 'text', 'text']
Sample row: ['TBA', 'TBA', '39th', '42nd', '=33rd', '27th', '28th']
SQL:

assistant: SELECT "2009" FROM "Table" WHERE "2011" = '39th';