document.write( "Question 677820: In which direction does the hyperbola defined by the equation below open?
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\n" ); document.write( "\n" ); document.write( "(x-3)^2/4^2-(y+5)^2/2^2=1\r
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Algebra.Com's Answer #421073 by stanbon(75887)\"\" \"About 
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In which direction does the hyperbola defined by the equation below open?
\n" ); document.write( "Check all that apply.
\n" ); document.write( "(x-3)^2/4^2-(y+5)^2/2^2=1
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\n" ); document.write( "(x-3)^2/16 - (y+5)^2/4 = 1
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\n" ); document.write( "Comment:l
\n" ); document.write( "Notice that when y = -5, you can solve for \"x\"
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\n" ); document.write( "Notice that when x = 3, you cannot solve for \"x\"
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\n" ); document.write( "Notice that the center of the hyperbola is at (3,-5)
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\n" ); document.write( "Since the curve cannot cross the line x = 3, it must
\n" ); document.write( "open to the right and to the left.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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