✅ Após resolver os cálculos, concluímos que o novo preço do aluguel após os dois referidos aumentos sucessivos é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf N_{P} = R\$\:1188,00\:\:\:}}\end{gathered}$}[/tex]
Sejam os dados:
[tex]\Large\begin{cases} V_{A} = R\$\:1000,00\\t'= 10\% = 0,1\\t'' = 8\% = 0,08\end{cases}[/tex]
O novo preço "Np" do aluguel após o aumento é:
[tex]\Large\displaystyle\text{$\begin{gathered} \bf(I)\end{gathered}$}[/tex] [tex]\Large\displaystyle\text{$\begin{gathered} N_{P} = V_{A} + V_{A}\cdot A_{E}\end{gathered}$}[/tex]
Onde:
[tex]\Large\begin{cases} N_{P} = Novo\:prec_{\!\!,}o\\V_{A} = Valor\:aluguel\\A_{E} = Aumento\:equivalente\end{cases}[/tex]
Se:
[tex]\Large\displaystyle\text{$\begin{gathered} A_{E} = \left[(1 + t')\cdot(1 + t'') - 1\right]\end{gathered}$}[/tex]
Inserindo o valor do aumento equivalente na equação "I", temos:
[tex]\Large\displaystyle\text{$\begin{gathered} \bf(II)\end{gathered}$}[/tex] [tex]\Large\displaystyle\text{$\begin{gathered} N_{P} = V_{A} + V_{A}\cdot\left[(1 + t')\cdot(1 + t'') - 1\right]\end{gathered}$}[/tex]
Substituindo os valores na equação "II", temos:
[tex]\Large\displaystyle\text{$\begin{gathered} N_{P} = 1000 + 1000\cdot\left[(1 + 0,1)\cdot(1 + 0,08) - 1\right]\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 1000 + 1000\cdot\left[1,1\cdot 1,08 - 1\right]\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 1000 + 1000\cdot\left[1,188 - 1\right]\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 1000 + 1000\cdot 0,188\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 1000 + 188\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 1188\end{gathered}$}[/tex]
✅ Portanto, o valor do aluguel após os aumentos é:
[tex]\Large\displaystyle\text{$\begin{gathered} N_{P} = R\$\:1188,00\end{gathered}$}[/tex]
Saiba mais:
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✅ Após resolver os cálculos, concluímos que o novo preço do aluguel após os dois referidos aumentos sucessivos é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf N_{P} = R\$\:1188,00\:\:\:}}\end{gathered}$}[/tex]
Sejam os dados:
[tex]\Large\begin{cases} V_{A} = R\$\:1000,00\\t'= 10\% = 0,1\\t'' = 8\% = 0,08\end{cases}[/tex]
O novo preço "Np" do aluguel após o aumento é:
[tex]\Large\displaystyle\text{$\begin{gathered} \bf(I)\end{gathered}$}[/tex] [tex]\Large\displaystyle\text{$\begin{gathered} N_{P} = V_{A} + V_{A}\cdot A_{E}\end{gathered}$}[/tex]
Onde:
[tex]\Large\begin{cases} N_{P} = Novo\:prec_{\!\!,}o\\V_{A} = Valor\:aluguel\\A_{E} = Aumento\:equivalente\end{cases}[/tex]
Se:
[tex]\Large\displaystyle\text{$\begin{gathered} A_{E} = \left[(1 + t')\cdot(1 + t'') - 1\right]\end{gathered}$}[/tex]
Inserindo o valor do aumento equivalente na equação "I", temos:
[tex]\Large\displaystyle\text{$\begin{gathered} \bf(II)\end{gathered}$}[/tex] [tex]\Large\displaystyle\text{$\begin{gathered} N_{P} = V_{A} + V_{A}\cdot\left[(1 + t')\cdot(1 + t'') - 1\right]\end{gathered}$}[/tex]
Substituindo os valores na equação "II", temos:
[tex]\Large\displaystyle\text{$\begin{gathered} N_{P} = 1000 + 1000\cdot\left[(1 + 0,1)\cdot(1 + 0,08) - 1\right]\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 1000 + 1000\cdot\left[1,1\cdot 1,08 - 1\right]\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 1000 + 1000\cdot\left[1,188 - 1\right]\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 1000 + 1000\cdot 0,188\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 1000 + 188\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 1188\end{gathered}$}[/tex]
✅ Portanto, o valor do aluguel após os aumentos é:
[tex]\Large\displaystyle\text{$\begin{gathered} N_{P} = R\$\:1188,00\end{gathered}$}[/tex]
Saiba mais: