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igor_vitrenko [27]

3 years ago

14

t is -40°F on a cold night in the North Pole. When the sun is out the next day, it is -5°F. What is the difference in temperatur

es?

1 answer:

a_sh-v [17]3 years ago

4 0

Answer:

because at 40 its still warm but really cold but in -5 its cold wherever you go even in the warmest blanket

Step-by-step explanation:

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2 years ago

Read 2 more answers

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Harlamova29_29 [7]

<h2> Question # 3</h2>

Answer:

$x = 13.094$

Step-by-Step Explanation

Considering the exponential equation

$e^{x-9}-6=54$

Solving the exponentiation equation

$e^{x-9}-6=54$

$\mathrm{Add\:}6\mathrm{\:to\:both\:sides}$

$e^{x-9}-6+6=54+6$

$e^{x-9}=60$

$\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$

$\ln \left(e^{x-9}\right)=\ln \left(60\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\ln \left(e^{x-9}\right)=\left(x-9\right)\ln \left(e\right)$

$\left(x-9\right)\ln \left(e\right)=\ln \left(60\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(a\right)=1$

$\ln \left(e\right)=1$

$x-9=\ln \left(60\right)$

$x=\ln \left(60\right)+9$

$x=13.094$ ∵ $\ln \left(60\right)=4.094$

Therefore, x = 13.094

<h2> Question # 4</h2>

Answer:

$x=0.386$

Step-by-Step Explanation

Considering the exponential equation

$3e^{9x}-6=91$

Solving the exponentiation equation

$3e^{9x}-6=91$

$\mathrm{Add\:}6\mathrm{\:to\:both\:sides}$

$3e^{9x}=97$

$\mathrm{Divide\:both\:sides\:by\:}3$

$\frac{3e^{9x}}{3}=\frac{97}{3}$

$e^{9x}=\frac{97}{3}$

$\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$

$\ln \left(e^{9x}\right)=9x\ln \left(e\right)$

$9x\ln \left(e\right)=\ln \left(\frac{97}{3}\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(a\right)=1$

$\ln \left(e\right)=1$

$9x=\ln \left(\frac{97}{3}\right)$

$\mathrm{Divide\:both\:sides\:by\:}9$

$\frac{9x}{9}=\frac{\ln \left(\frac{97}{3}\right)}{9}$

$x=\frac{\ln \left(\frac{97}{3}\right)}{9}$

$x=\frac{3.476}{9}$ ∵ $ln\left(\frac{97}{3}\right)=3.476$

$x=0.386$

Therefore, x = 0.386

5 0

3 years ago

A population of bacteria is growing according to the exponential model P = Ce^kt, where P is the number of colonies and t is mea

Elza [17]

we know that initially there were 120 colonies or namely that C = 120, well, what time was that? that was the hour 0, or namely t = 0, so then

$P=Ce^{rt}\qquad \begin{cases} P=\textit{accumulated amount}\\ C=\textit{initial amount}\dotfill &120\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &0\\ \end{cases} \\\\\\ P=120e^{r\cdot 0}\implies P=120e^0\implies P=120\cdot 1\implies P=120$

we also know that when t = 4, C = 550

$\textit{Amount of Population Growth} \\\\ P=Ce^{rt}\qquad \begin{cases} P=\textit{accumulated amount}\dotfill&120\\ C=\textit{initial amount}\dotfill &550\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &4\\ \end{cases} \\\\\\ 120=550e^{4r}\implies \cfrac{120}{550}=e^{4r}\implies \cfrac{12}{55}=e^{4r}~\hfill \stackrel{\textit{using this log cancellation rule}}{\log_a(a^x)=x}$

$\log_e\left( \frac{12}{55}\right)=\log_e(e^{4r})\implies \log_e\left( \frac{12}{55}\right)=4r\implies \ln\left( \frac{12}{55}\right)=4r \\\\\\ \cfrac{\ln\left( \frac{12}{55}\right)}{4}=r\implies -0.381\approx r = k$

8 0

2 years ago