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Exercises 4.5 Chapter Test

1.

The population of a city is growing by 2,000 people per year. If the city has a population of 42,000 in 2018,

  1. Write a recursive formula for the population n years after 2018.

  2. Write an explicit formula for the population n years after 2018.

  3. Use one of the your formulas to predict the population in 2025.

Answer.

  1. \(\displaystyle P_n = P_{n-1} + 2000; P_0 = 42,000\)

  2. \(\displaystyle P_n = 42,000 + 2000n\)

  3. \(\displaystyle P_7=56,000\)

2.

The population of a city is growing by 15% per year. If the city has a population of 85,000 in 2015,

  1. Write a recursive formula for the population n years after 2015.

  2. Write an explicit formula for the population n years after 2015.

  3. Use one of the your formulas to predict the population in 2030.

Answer.

  1. \(\displaystyle P_n = (1.15)P_{n-1}; P_0 = 85,000\)

  2. \(\displaystyle P_n = 85,000(1.15)^n\)

  3. \(\displaystyle P_{25}=691,650\)

3.

The average temperature at the surface of the earth is \(20^{\circ}C\text{.}\) Five kilometers below the surface of the earth the temperature is \(70^{\circ}C\text{.}\)

  1. Write an explicit linear equation that models how temperature depends on depth (\(n\)) below the surface of the earth.

  2. Use your equation to find the temperature 25 km below the surface of the earth.

  3. Use your equation to predict the dpeth that will correspond with a temperature of \(1,000^{\circ}C\text{.}\)

Answer.

  1. \(\displaystyle P_n = 20+10n\)

  2. 270 degrees C

  3. 98 km

Solution.

  1. When \(n = 0\text{,}\) the temperature is 20 degrees, so \(P_0 = 20\text{.}\)

    To find the common difference, we need to find the change in temperature per km. Since the temperature increases by 50 degrees in 5 km, \(d = 50 \div 5 = 10\text{.}\)

    We can model linear growth with the formula, \(P_n = P_0 + dn\text{,}\) so the equation is \(P_n = 20 + 10n\text{.}\)

  2. \(n = 25\) at 25 km below the surface. \(P_{25}=20+10\cdot 25 = 270\text{.}\)

  3. We need to find the n-value that corresponds with \(P_n=10,000\)

    \begin{equation*} 10,000 = 20 + 10n \end{equation*}
    \begin{equation*} 9,980 = 10n \end{equation*}
    \begin{equation*} n = 9,980 \div 10 = 998 \end{equation*}
4.

In 2016, a gallon of milk was $2.15. Four years later, the price of a gallon of milk was $2.48. If inflation is a type of exponential growth, find the inflation rate.

Answer.
2.5%
Solution.

We can use the exponential model, \(P_n = P_0(1+r)^n\) to solve for \(r\text{.}\)

\begin{equation*} 2.35 = 2.15(1+r)^4 \end{equation*}
\begin{equation*} \frac{2.35}{2.15} = (1+r)^4 \end{equation*}
\begin{equation*} \sqrt[4]{\frac{2.35}{2.15}} = 1 + r \end{equation*}
\begin{equation*} 1.025 = 1 + r \end{equation*}
\begin{equation*} r = 0.025 = 2.5\% \end{equation*}
5.

A car is purchased for $18,500. Five year later, the value of the car is $9,900.

  1. Find a linear equation that models the the value of the car with time.

  2. Use your linear equation to predict the value of the car in 10 years.

  3. Find an exponential equation that models the value of the car with time.

  4. Use your exponential equation to predict the value of the car in 10 years.

Answer.

  1. \(\displaystyle P_n = 18500 -1720n\)

  2. $1300

  3. \(\displaystyle P_n = 18,500(0.882)^n\)

  4. $5,270.57

Solution.

  1. The value of the car goes down by \(18,500 - 9,900 = 8,600\) in five years. \(d = 8,600 \div 5 = 1720\text{.}\) Since the value of the car is decreasing, we need a negative common difference. That makes the equation, \(P_n = 18500 - 1720n\)

  2. \(\displaystyle P_{10} = 18500 - 1720 \cdot 10 = 1300\)

  3. Use the equation model, \(P_n = P_0(1+r)^n\) to solve for \(r\text{.}\)

    \begin{equation*} 9,900 = 18,500(1+r)^5 \end{equation*}
    \begin{equation*} \frac{9900}{18500} = (1+r)^5 \end{equation*}
    \begin{equation*} \sqrt[5]{\frac{9900}{18500}}= 1 + r \end{equation*}
    \begin{equation*} 0.0882 = 1 + r \end{equation*}
    \begin{equation*} P_n = 18500(0.882)^n \end{equation*}

  4. \(\displaystyle P_{10}=18500(0.822)^{10} = 5270.57\)

6.

Assum \(y\) varies directly as the square of \(x\text{.}\) If \(y = 10\) when \(x = 2\text{,}\) what is \(y\) when \(x = 4\text{?}\)

Answer.
40
Solution.

y varies directly as the square of x: \(y = k x^2\text{.}\)

Plug in y = 10 and x = 2 to solve for k:

\begin{equation*} 10 = k(2)^2 \end{equation*}
\begin{equation*} 10 = k\cdot 4 \end{equation*}
\begin{equation*} k = \frac{10}{4}=\frac{5}{2} \end{equation*}

Plug in x = 4 to answer the question:

\begin{equation*} y = \frac{5}{2}\cdot 4^2 \end{equation*}
\begin{equation*} y = \frac{5}{2} \cdot \frac{16}{1} \end{equation*}
\begin{equation*} y = \frac{80}{2}=40 \end{equation*}
7.

The time it takes to drive to Grand Junction is inversely proportional to your average speed. If Justin averages 60mph, it will take him 4 hours to make the drive.

  1. Write an equation that models this situation.

  2. If Justin hits traffic and averages 50 mph, how long will it take him to the make the drive?

Answer.

  1. \(\displaystyle T = \frac{240}{S}\)

  2. 4.8 hours

Solution.

  1. Time varies inversely as speed is the relationship, \(T = \frac{k}{S}\text{.}\) Plug in \(S = 60\) and \(T = 4\) and solve for k:

    \begin{equation*} 4 = \frac{k}{60} \end{equation*}
    \begin{equation*} 4 \cdot 60 = \frac{k}{60} \cdot 60 \end{equation*}
    \begin{equation*} k = 240 \end{equation*}
    \begin{equation*} T = \frac{240}{S} \end{equation*}

  2. Plug in \(S = 50\) and solve for \(T\text{.}\)

    \begin{equation*} T = \frac{240}{50}=4.8 \end{equation*}