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If ( y ) varies inversely as ( x ), then we can express this relationship as ( y = \frac{k}{x} ) for some constant ( k ). Given that ( y = 5 ) when ( x = 3 ), we can find ( k ) by substituting these values: ( 5 = \frac{k}{3} ), which gives ( k = 15 ). Now, to find ( x ) when ( y = 15 ), we substitute into the inverse variation equation: ( 15 = \frac{15}{x} ). Solving for ( x ) gives ( x = 1 ).

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1w ago

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