$\bf =ae^{kt}\qquad 
\begin{cases}
1994\impliedby \textit{year 0, starting point}\\
t=0\qquad P=182
\end{cases}\implies 182=ae^{k0}
\\\\\\
182=a\cdot e^0\implies 182=a\cdot 1\implies 182=a
\\\\\\
thus\qquad P=182e^{kt}\\\\
-------------------------------\\\\$

$\bf P=182e^{kt}\qquad 
\begin{cases}
2002\impliedby \textit{8 years later}\\
t=8\qquad P=186
\end{cases}\implies 186=182e^{k8}
\\\\\\
\cfrac{186}{182}=e^{8k}\implies ln\left( \frac{93}{91} \right)=ln(e^{8k})\implies ln\left( \frac{93}{91} \right)=8k
\\\\\\
\cfrac{ln\left( \frac{93}{91} \right)}{8}=k\implies 0.0027\approx k\implies \boxed{P=182e^{0.0027t}}$

what's the population in 2004? well, from 1994 to 2004 is 10 years later, so t = 10

plug that in, to get P for 2004

3 0

3 years ago

Find the sum of the interior angle measures of the polygon

pshichka [43]

<h3>Answer: 1260</h3>

Work Shown:

We have 9 sides of this polygon. See the diagram below. So n = 9

Plug that into the formula below to find the sum of the interior angles

S = 180(n-2)

S = 180(9-2)

S = 180*7

S = 1260

8 0

2 years ago