Question 1025100
A)
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*[illustration pp4.JPG].
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B) {{{C[max]=2.5}}}{{{mg/L}}}
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C) Goes to zero as t goes to {{{infinity}}}
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D){{{(5t)/(t^2+1)=0.3}}}
{{{50t=3(t^2+1)}}}
{{{3t^2-50t+3=0}}}
{{{3(t^2-(50/3)t+(50/6)^2)+3=3(50/6)^2}}}
{{{3(t-25/3)^2=625/3-9/3}}}
{{{3(t-25/3)^2=616/3}}}
{{{(t-25/3)^2=616/9}}}
{{{t-25/3=0 +- sqrt(616)/3}}}
{{{t=25/3 +- (2/3)sqrt(154)}}}
{{{t=25/3+(2/3)sqrt(154)}}}
So, {{{C<0.3}}}{{{mg/L}}} when {{{t>25/3+(2/3)sqrt(154)}}}