Question

Uma bactéria cresce num tempo de 5 dias exponencialmente durante o inverno, cuja função é:
b(t)=4,25+(1/6)^-t

Answer 1

b(t)=4,25+(1/6)$b(t)=4,25+ ( \frac{1}{6}) ^{-t} \\T=5\\ b(5)=4,25+(\frac{1}{6})^{-5}\\ b(-5)=4,25+\frac{{1}^{-5}}{{6}^{-5}}\\ b(-5)=4,25+\frac{\frac{1}{5}}{\frac{6}{5}}\\b(-5)=4,25+\frac{1}{5} \times \frac{6}{5}\\b(-5)=4,25+\frac{6}{25}\\b(-5)=\frac{106,25+6}{25}\\b(-5)=\frac{112,25}{25}\\b(-5)=4,49$