$~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$5150\\ P=\textit{original amount deposited}\dotfill & \$4500\\ r=rate\to 7.5\%\to \frac{7.5}{100}\dotfill &0.075\\ t=years \end{cases}$

$5150=4500e^{0.075\cdot t} \implies \cfrac{5150}{4500}=e^{0.075t}\implies \cfrac{103}{90}=e^{0.075t} \\\\\\ \log_e\left( \cfrac{103}{90} \right)=\log_e(e^{0.075t})\implies \log_e\left( \cfrac{103}{90} \right)=0.075t \\\\\\ \ln\left( \cfrac{103}{90} \right)=0.075t\implies \cfrac{\ln\left( \frac{103}{90} \right)}{0.075}=t\implies\stackrel{\textit{about 1 year and 291 days}}{ 1.8\approx t}$

4 0

1 year ago

What is the sum of the polynomials?<br> (6x+7+x^2)+(2x^2-3)

Oduvanchick [21]

Hai !
6x+7+x^2+2x^2-3
6x+7+x^2+2x^2+-3
combine like terms
6x+7+x^2+2x^2+-3
(x^2+2x^2)+(6x)+(7+-3)
=3x^2+6x+4
Cheers :3

4 0

3 years ago

What is 100 divided by 10?

siniylev [52]

100 divided by 10 is 10.

3 0

3 years ago

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