$~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$105000\\ P=\textit{original amount deposited}\dotfill & \$69000\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &6 \end{cases}$

$105000=69000e^{\frac{r}{100}\cdot 6}\implies \cfrac{105000}{69000}=e^{\frac{3r}{50}}\implies \cfrac{7}{46}=e^{\frac{3r}{50}} \\\\\\ \log_e\left( \cfrac{35}{23} \right)=\log_e\left( e^{\frac{3r}{50}} \right)\implies \ln\left( \cfrac{35}{23} \right)=\cfrac{3r}{50}\implies 50\cdot \ln\left( \cfrac{35}{23} \right)=3r \\\\\\ \cfrac{50\cdot \ln\left( \frac{35}{23} \right)}{3}=r\implies 6.9976\approx r\implies \stackrel{\%}{7.00}\approx r$

8 0

3 years ago

Determine if the following relations are functions. If they are functions, also state the domain and range.

lapo4ka [179]

Answer:

desculpa mas eu não sei a resposta mais eu vou ver se posso te ajudar taaa

8 0

3 years ago

Exponential Function Question. Help please :/

Furkat [3]

Answer:

Step-by-step explanation:

60.760944

8 0

3 years ago