Does this graph show a constant or variable rate of change?

Mathematics

1 answer:

valentinak56 [21]2 years ago

5 0

Answer:

constant

Step-by-step explanation:

You might be interested in

a country's population in 1990 was 134 million. in 2000 it was a 139 million. estimate the population in 2014 using the exponent

yuradex [85]

F=ir^t

139=134r^10

139/134=r^10

r=(139/134)^(1/10) then:

f=134(139/134)^(t/10) so in 2014, t=24 so

f=134(139/134)^(2.4)

f≈146 million (to nearest million)

Some will say that you have to use the exponential function, but it really gives you the same answer...even for continuous compounding :)...

A=Pe^(kt)

139=134e^(10k)

139/134=e^(10k)

ln(139/134)=10k

k=ln(139/134)/10 so

A=134e^(t*ln(139/134)/10) when t=24

A=134e^(2.4*ln(139/134))

A≈146 million (to nearest million)

The only real reason or advantage to using A=Pe^(kt) is when you start getting into differential equations...

6 0

2 years ago

You deposit $7550 in an account that pays 7.25% interest, compounded continuously. How long to the

Afina-wow [57]

$~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$\stackrel{tripled}{(7550)3}\\ P=\textit{original amount deposited}\dotfill & \$7550\\ r=rate\to 7.25\%\to \frac{7.25}{100}\dotfill &0.0725\\ t=years \end{cases}$

$3(7550)=7550e^{0.0725\cdot t}\implies \cfrac{3(7550)}{7550}=e^{0.0725t}\implies 3=e^{0.0725t} \\\\\\ \log_e(3)=\log_e\left( e^{0.0725t} \right)\implies \ln(3)=0.0725t \\\\\\ \cfrac{\ln(3)}{0.0725}=t\implies 15.15327\approx t\implies 15\approx t$