Fibonacci13 ,
[tex3]\text{X} = \text{Xo}.\text{e}^{\frac{k.t}{2}}[/tex3]
Fibonacci13 escreveu: ↑Qua 14 Abr, 2021 15:08
Se a quantidade inicial Xo se reduz à metade em 2 horas
[tex3]\frac{\text{Xo}}{2} = \text{Xo}.\text{e}^{\frac{k.2}{2}}[/tex3]
[tex3]\text{e}^k=\frac{1}{2}[/tex3]
Fibonacci13 escreveu: ↑Qua 14 Abr, 2021 15:08
em 5 horas existirá no sangue:
[tex3]\text{X} = \text{Xo}.\text{e}^{\frac{k.5}{2}}[/tex3]
[tex3]\text{X} = \text{Xo}.\text{e}^{k.\frac{5}{2}}[/tex3]
[tex3]\text{X} = \text{Xo}.(\text{e}^{k})^{\frac{5}{2}}[/tex3]
Porém, determinamos que [tex3]\text{e}^k=\frac{1}{2}[/tex3]
[tex3]\text{X} = \text{Xo}.\left(\frac{1}{2}\right)^{\frac{5}{2}}[/tex3]
[tex3]\text{X} = \text{Xo}.\left(\frac{1}{\sqrt{2^5}}\right)[/tex3]
[tex3]\text{X} = \text{Xo}.\left(\frac{1}{4.\sqrt{2}}\right)[/tex3]
[tex3]\text{X} = \text{Xo}.\left(\frac{1}{4.(1,41)}\right)[/tex3]
[tex3]\text{X} ≈ \text{Xo}.(0,177)[/tex3]
[tex3]{\color{red}\boxed{\text{X} ≈ 17,7\% \text{ de }\text{Xo}}}[/tex3]