Question 1009912
It does check out with {{{ 4 }}}
{{{ f(x) = 4*(2/5)^x }}}
{{{ f(-2) = 4*(2/5)^(-2) }}}
{{{ f(-2) = 4*( 1 / (2/5)^2) }}}
{{{ f(-2) = 4*(5/2)^2 }}}
{{{ f(-2) = 4*(25/4) }}}
{{{ f(-2) = 25 }}}
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{{{ f(1) = 4*(2/5)^1 }}}
{{{ f(1) = 8/5 }}}
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As for how to get the {{{ 4 }}},
{{{ f(-2) = a*(2/5)^(-2) }}}
{{{ f(-2) = (25a)/4 }}}
{{{ 25 = (25a)/4 }}}
{{{ 1 = a/4 }}}
{{{ a = 4 }}}
and
{{{ f(1) =a*(2/5)^1 }}}
{{{ f(1) = (2a)/5 }}}
{{{ 8/5 = (2a)/5 }}}
{{{ 2a = 8 }}}
{{{ a = 4 }}}
Hope this helps