Question 1060080
The inverse variation would be:
{{{ y = k*( 1/x) }}}
Then they say:
{{{ x = 39 }}}
{{{ y = 13 }}}
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{{{ 13 = k*( 1/39 ) }}}
Multiply both sides by {{{ 39 }}}
{{{ k = 13*39 }}}
{{{ k = 507 }}}
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By the logic of an inverse variation, if I
double {{{ x }}} , then {{{ y }}} should be {{{ 1/2 }}}
of what it was before
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I'll say {{{ x = 78 }}} ( twice {{{ 39 }}} )
Then
{{{ y = k*( 1/78 ) }}}
{{{ y = 507*(1/78) }}}
{{{ y = 6.5 }}} ( which is half of what it was, {{{ 13 }}} )
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Hope this makes some kind of sense