2025-04-10

a) $A=406.2236$
b) $P=\frac{1125}{32}$
c) $$
\begin{align}
& 500 = N_{0}1.25^{1.25} \
& 500 = N(1.321714079) \
& \frac{500}{1.321714079} = 375.2966436
\end{align}

d) $$ \begin{align} & 625 = P(0.71)^9 \\ & \frac{625}{0.71^9} = P = 13631.9 \end{align}

Function	Growth or decay	Initial	Rate
$20(1.02{)}^t$	growth	20	1.02
$0.8^n$	decay	1	0.8
$0.5(3{)}^x$	growth	0.5	3
$600{\left(\frac{5}{8}\right)}^w$	decay	600	5/8

3. the growth in population of a small town since 1996 is given by the function $P(n)=1250(1.03{)}^n$
   a) What is the initial population? Explain how you know.
   The initial population is 1250 as it’s the value when n = 0
   b) What is the growth rate? Explain how you know.
   The growth rate is 1.03 or 3%. we can see that as it’s the b value in the growth function
   c) Determine the population in the year 2007.

\begin{align}
& P = 1250(1.03)^{(2007-1996)} \
& P = 1730
\end{align}

$$d)Inwhichyeardoesthepopulationreach2000people?t$$

\begin{align}
& 2000 = 1250(1.03)^n \
& \frac{2000}{1250} = 1.03^n \
& \frac{8}{5} = 1.03^n \
& \ln\left( \frac{8}{5} \right) = \left( \ln(1.03^n) = n * \ln(1.03) \right) \
& \frac{\ln\left( \frac{8}{5} \right)}{\ln(1.03)} = n \
& \boxed{n = 15.9}
\end{align}

4. A computer loses its value each month after it is purchased. Its value as a function of time, in months, is modelled by .$V(m) = 1500(0.95)^m$ a) What is the initial value of the computer? Explain how you know. 1500. You can tell by setting m=0 b) What is the rate of depreciation? Explain how you know. b = 0.95 c) Determine the value of the computer after 2 years. $$ \begin{align} & 1500(0.95)^{(2*12)} \\ & = 437.98 \end{align}

d) In which month after it is purchased does the computer’s worth fall
below $900?

$$\begin{array}{rl}& \text{ chat it’s 10 months}\end{array}$$

5. . In 1990, a sum of $1000 is invested at a rate of 6% per year for 15 years.
   a) What is the growth rate?
   6%
   b) What is the initial amount?
   1000
   c) How many growth periods are there?
   15
   d) Write an equation that models the growth of the investment, and use it
   to determine the value of the investment after 15 years

$$\begin{array}{rl}& 1000(1.06{)}^{15}\\ & =2396.6\end{array}$$

6. species of bacteria has a population of 500 at noon. It doubles every 10 h.
   The function that models the growth of the population, P, at any hour, t, is $$
   P(t) = 500\left( \frac{t}{2^{10}} \right)

a) Why is the exponent $2^{10}$ b) Why is the base 2? c) Why is the multiplier 500? That is the base population d) Determine the population at midnight. e) Determine the population at noon the next day.

\begin{align}
& 500\left( \frac{24}{2^{10}} \right) \
& \frac{375}{32} = 11
\end{align}

$$f)Determinethetimeatwhichthepopulationfirstexceeds2000.$$