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Appendix C Answers to Selected Exercises

1 Linear Models
1.1 Creating a Linear Model
1.1.3 Problem Set 1.1

Applications

1.2 Graphs and Equations
1.2.6 Problem Set 1.2

Applications

1.3 Intercepts
1.3.6 Problem Set 1.3

Skills Practice

Applications

1.4 Slope
1.4.5 Problem Set 1.4

Applications

1.5 Equations of Lines
1.5.6 Problem Set 1.5

Skills Practice

Applications

1.5.6.21.
Answer.
\(m = -0.0018\) degree/foot, so the boiling point drops with altitude at a rate of 0.0018 degree per foot. \(b = 212\text{,}\) so the boiling point is \(212\degree\) at sea level (where the elevation \(h = 0\)).

1.6 Chapter Summary and Review
1.6.3 Chapter 1 Review Problems

1.6.3.18.

Answer.
\(d\) \(V\)
\(-5\) \(-4.8\)
\(-2\) \(-3 \)
\(1\) \(-1.2\)
\(6\) \(1.8 \)
\(10\) \(4.2\)

1.6.3.19.

Answer.
\(q\) \(S\)
\(-8\) \(-8\)
\(-4\) \(36 \)
\(3\) \(168\)
\(5\) \(200 \)
\(9\) \(264\)

2 Applications of Linear Models
2.1 Linear Regression
2.1.5 Problem Set 2.1

Skills Practice

2.1.5.5.
Answer.
The slope is \(-2.5\text{,}\) which indicates that the snack bar sells 2.5 fewer cups of cocoa for each \(1\degree\text{C}\) increase in temperature. The \(C\)-intercept of 52 indicates that 52 cups of cocoa would be sold at a temperature of \(0\degree\text{C}\text{.}\) The \(T\)-intercept of 20.8 indicates that no cocoa will be sold at a temperature of \(20.8\degree\text{C}\text{.}\)
2.1.5.11.
Answer.
2 min: \(21\degree\)C; 2 hr: \(729\degree\)C; The estimate at 2 minutes is reasonable; the estimate at 2 hours is not reasonable.

Applications

2.1.5.19.
Answer.
  1. scatterplot and regression line
  2. The graph is above.
  3. The slope tells us that the time it takes for a bird to attract a mate decreases by 0.85 days for every additional song it learns.
  4. 44.5 days
  5. The \(C\)-intercept tells us that a warbler with a repretoire of 53 songs would acquire a mate immediately. The \(B\)-intercept tells us that a warbler with no songs would take 62 days to find a mate. These values make sense in context.

2.2 Linear Systems
2.2.5 Problem Set 2.2

Applications

2.3 Algebraic Solution of Systems
2.3.5 Problem Set 2.3

Applications

2.3.5.15.
Answer.
  1. Rani’s speed in still water: \(x\)
    Speed of the current: \(y\)
    Rate Time Distance
    Downstream \(x+y\) \(45\) \(6000\)
    Upstream \(x-y\) \(45\) \(4800\)
  2. \(\displaystyle 45(x+y)=6000 \)
  3. \(\displaystyle 45(x-y)=4800 \)
  4. Rani’s speed in still water is 120 meters per minute, and the speed of the current is \(13\dfrac{1}{3} \) meters per minute.

2.4 Gaussian Reduction
2.4.6 Problem Set 2.4

2.5 Linear Inequalities in Two Variables
2.5.5 Problem Set 2.5

Warm Up

2.5.5.3.
Answer.
  1. The graph of the equation is a line, and the graph of the inequality is a half-plane. The line is the boundary of the half-plane but is not included in the solution to the inequality.
  2. The graph of \(x + y\ge 10,000\) includes both the line \(x + y=10,000\) and the half-plane of the corresponding strict inequality.

2.6 Chapter Summary and Review
2.6.3 Chapter 2 Review Problems

2.6.3.40.

Answer.
\(x+y \le 32\text{,}\) \(2x+1.6y \ge 56\text{,}\) \(x\ge 0\text{,}\) \(y\ge 0\text{,}\) where \(x\) represents ounces of tofu, \(y\) the ounces of tempeh
system of inequalities

3 Quadratic Models
3.1 Extraction of Roots
3.1.9 Problem Set 3.1

Warm Up

Skills Practice

Applications

3.2 Intercepts, Solutions, and Factors
3.2.6 Problem Set 3.2

Warm Up

Applications

3.3 Graphing Parabolas
3.3.6 Problem Set 3.3

Applications

3.4 Completing the Square
3.4.4 Problem Set 3.4

Applications

3.5 Chapter 3 Summary and Review
3.5.3 Chapter 3 Review Problems

3.5.3.35.

Answer.
\(A_1=x^2-\left(\dfrac{1}{2}y^2+\dfrac{1}{2}y^2\right)=x^2-y^2;~~A_2=(x+y)(x-y)=x^2-y^2\)

4 Applications of Quadratic Models
4.1 Quadratic Formula
4.1.6 Problem Set 4.1

Skills Practice

4.1.6.13.
Answer.
  1. \((5,0)\) and \((1,0)\text{;}\) \((3,0)\text{;}\) no \(x\)-intercepts
  2. 1 and 5; 3; \(\dfrac{6 \pm i\sqrt{12}}{2}\text{.}\) The real-valued solutions are the \(x\)-intercepts of the graph. If the solutions are complex, the graph has no \(x\)-intercepts.

Applications

4.1.6.27.
Answer.
b. \(10h(2h-6)=2160~~~\)c. 12 ft by 18 ft by 10ft
Note 4.1.19.
According to the latest research and data, there has been an increase in the number of tigers, and now the total number of wild tigers worldwide is 5,574, according to the World Animal Foundation at https://worldanimalfoundation.org/advocate/animal-captivity-statistics/

4.2 The Vertex
4.2.5 Problem Set 4.2

Skills Practice

Applications

4.3 Curve Fitting
4.3.5 Problem Set 4.3

4.4 Quadratic Inequalities
4.4.6 Problem Set 4.4

Skills Practice

4.5 Chapter 4 Summary and Review
4.5.3 Chapter 4 Review Problems

4.5.3.26.

Answer.
  1. \(\displaystyle y=-0.05x^2-0.003x+234.2\)
  2. \((-0.03, 234.2)~\) The velocity of the debris at its maximum height of 234.2 feet. The velocity there is actually zero.

5 Functions and Their Graphs
5.1 Functions
5.1.7 Problem Set 5.1

Applications

5.2 Graphs of Functions
5.2.6 Problem Set 5.2

Applications

5.2.6.15.
Answer.
  1. \(f(1000) = 1495\text{:}\) The speed of sound at a depth of \(1000\) meters is approximately \(1495\) meters per second.
  2. \(d = 570\) or \(d = 1070\text{:}\) The speed of sound is \(1500\) meters per second at both a depth of \(570\) meters and a depth of \(1070\) meters.
  3. The slowest speed occurs at a depth of about \(810\) meters and the speed is about \(1487\) meters per second, so \(f(810) = 1487\text{.}\)
  4. \(f\) increases from about \(1533\) to \(1541\) in the first \(110\) meters of depth, then drops to about \(1487\) at \(810\) meters, then rises again, passing \(1553\) at a depth of about \(1600\) meters.

5.3 Some Basic Graphs
5.3.4 Problem Set 5.3

Warm Up

Applications

5.4 Direct Variation
5.4.6 Problem Set 5.4

Applications

5.5 Inverse Variation
5.5.4 Problem Set 5.5

Applications

5.6 Functions as Models
5.6.6 Problem Set 5.6

Warm Up

Applications

Absolute Value

5.7 Chapter 5 Summary and Review
5.7.3 Chapter 5 Review Problems

5.7.3.5.

Answer.
\(N(10) = 7000\text{:}\) Ten days after the new well is opened, the company has pumped a total of \(7000\) barrels of oil.

6 Powers and Roots
6.1 Integer Exponents
6.1.5 Problem Set 6.1

Skills Practice

Applications

6.2 Roots and Radicals
6.2.9 Problem Set 6.2

Skills Practice

Applications

6.2.9.36.
Answer.
  1. \(6.5\times 10^{-13}\) cm; \(1.17\times 10^{-36} \text{ cm}^3\)
  2. \(\displaystyle 1.8\times 10^{14} g/\text{cm}^3\)
  3. Element Carbon Potassium Cobalt Technetium Radium
    Mass
    number, \(A\)
    \(14\) \(40\) \(60\) \(99\) \(226\)
    Radius, \(r\)
    (\(10^{-13}\) cm)
    \(3.1\) \(4.4\) \(5.1\) \(6\) \(7.9\)
  4. cube root

6.3 Rational Exponents
6.3.7 Problem Set 6.3

Skills Practice

6.3.7.31.
Answer.
\(x\) \(0\) \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)
\(f(x)\) \(0\) \(1\) \(2.5\) \(4.3\) \(6.4\) \(8.5\) \(10.9\)
\(g(x)\) \(0\) \(1\) \(2.8\) \(5.2\) \(8\) \(11.2\) \(14.7\)
power functions

Applications

6.4 Working with Radicals
6.4.6 Problem Set 6.4

Skills Practice

6.5 Radical Equations
6.5.6 Problem Set 6.5

6.6 Chapter 6 Summary and Review
6.6.3 Chapter 6 Review Problems

6.6.3.10.

Answer.
  1. Planet Density
    Mercury 5426
    Venus 5244
    Earth 5497
    Mars 3909
    Jupiter 1241
    Saturn 620
    Uranus 1238
    Neptune 1615
    Pluto 2355
  2. Mercury, Venus, Earth, and Mars

7 Exponential Functions
7.1 Exponential Growth and Decay
7.1.7 Problem Set 7.1

Skills Practice

7.1.7.19.
Answer.
The growth factor is \(1.2\text{.}\)
\(x\) \(0\) \(1\) \(2\) \(3\) \(4\)
\(Q\) \(20\) \(24\) \(28.8\) \(34.56\) \(41.47\)
7.1.7.20.
Answer.
The decay factor is \(0.8\text{.}\)
\(w\) \(0\) \(1\) \(2\) \(3\) \(4\)
\(N\) \(120\) \(96\) \(76.8\) \(61.44\) \(49.15\)
7.1.7.21.
Answer.
The decay factor is \(0.8\text{.}\)
\(t\) \(0\) \(1\) \(2\) \(3\) \(4\)
\(C\) \(10\) \(8\) \(6.4\) \(5.12\) \(4.10\)
7.1.7.22.
Answer.
The growth factor is \(1.1\text{.}\)
\(n\) \(0\) \(1\) \(2\) \(3\) \(4\)
\(B\) \(200\) \(220\) \(242\) \(266.2\) \(292.82\)

Applications

7.2 Exponential Functions
7.2.6 Problem Set 7.2

Skills Practice

7.2.6.19.
Answer.
\(x\) \(-3\) \(-2\) \(-1\) \(0\) \(1\) \(2\) \(3\)
\(f(x)=3^x \) \(\frac{1}{27} \) \(\frac{1}{9} \) \(\frac{1}{3} \) \(1\) \(3\) \(9\) \(27\)
\(g(x)=\left(\frac{1}{3} \right)^x \) \(27\) \(9\) \(3\) \(1\) \(\frac{1}{3} \) \(\frac{1}{9} \) \(\frac{1}{27} \)
two exponentials
7.2.6.21.
Answer.
Because they are defined by equivalent expressions, (b), (c), and (d) have identical graphs.

Applications

7.3 Logarithms
7.3.7 Problem Set 7.3

Skills Practice

7.4 Properties of Logarithms
7.4.5 Problem Set 7.4

Warm Up

Skills Practice

Applications

7.5 Exponential Models
7.5.5 Problem Set 7.5

Applications

7.6 Chapter 7 Summary and Review
7.6.3 Chapter 7 Review Problems

8 Polynomial and Rational Functions
8.1 Polynomial Functions
8.1.10 Problem Set 8.1

Skills Practice

Applications

8.2 Algebraic Fractions
8.2.5 Problem Set 8.2

Warm Up

Skills Practice

Applications

8.3 Operations on Algebraic Fractions
8.3.9 Problem Set 8.3

Skills Practice

Applications

8.4 More Operations on Fractions
8.4.5 Problem Set 8.4

Skills Practice

Applications

8.4.5.25.
Answer.
\(\overline{PQ}\) and \(overline{RS}: ~\dfrac{b}{a}\text{;}\) \(\overline{QR}\) and \(overline{SP}: ~\dfrac{b}{a}\)

8.5 Equations with Fractions
8.5.8 Problem Set 8.5

Applications

8.5.8.39.
Answer.
Because \(x=1\text{,}\) dividing by \(x-1\) in the fourth step is dividing by \(0\text{.}\)

8.6 Chapter 8 Summary and Review
8.6.3 Chapter 8 Review Problems

9 Equations and Graphs
9.1 Properties of Lines
9.1.4 Problem Set 9.1

Warm Up

9.1.4.3.
Answer.
Number Negative
reciprocal
Their
product
\(\dfrac{2}{3} \) \(\dfrac{-3}{2} \) \(-1 \)
\(\dfrac{-5}{2} \) \(\dfrac{2}{5}\) \(-1\)
\(6 \) \(\dfrac{-1}{6} \) \(-1\)
\(-4\) \(\dfrac{1}{4} \) \(-1\)
\(-1 \) \(1\) \(-1\)

Applications

9.2 The Distance and Midpoint Formulas
9.2.7 Problem Set 9.2

Applications

9.3 Conic Sections: Ellipses
9.3.5 Problem Set 9.3

Skills Practice

9.4 Conic Sections: Hyperbolas
9.4.6 Problem Set 9.4

Skills Practice

9.4.6.9.
Answer.
  1. \(\displaystyle \dfrac{x^2}{9} - \dfrac{y^2}{3} =1\)
  2. \(x\) \(0\) \(\dfrac{\pm3\sqrt{5}}{2} \) \(5\) \(\dfrac{\pm 15}{4} \)
    \(y\) undefined \(\pm 2\) \(\dfrac{\pm 16}{3} \) \(-3\)
9.4.6.25.
Answer.
Hyperbola; center \(\left(\dfrac{-1}{8},-1 \right) \text{,}\) transverse axis vertical, \(a^2=\dfrac{15}{65} \text{,}\) \(b^2=\dfrac{15}{16} \)

9.5 Nonlinear Systems
9.5.4 Problem Set 9.5

Skills Practice

9.6 Chapter 9 Summary and Review
9.6.3 Chapter 9 Review Problems

9.6.3.41.

Answer.
\((1,-2) \text{,}\) \((-1,2) \text{,}\) \(\left(2\sqrt{3}, \dfrac{-1}{\sqrt{3}} \right) \text{,}\) \(\left(-2\sqrt{3}, \dfrac{1}{\sqrt{3}} \right) \)

9.6.3.42.

Answer.
\(\left(\dfrac{\sqrt{34}}{2}, \dfrac{\sqrt{34}}{2} \right) \text{,}\) \(\left(\dfrac{-\sqrt{34}}{2}, \dfrac{-\sqrt{34}}{2} \right) \)

10 Logarithmic Functions
10.1 Logarithmic Functions
10.1.7 Problem Set 10.1

Skills Practice

Applications

10.1.7.29.
Answer.
  1. data points and log curve
  2. The graph resembles a logarithmic function. The function is close to the points but appears too steep at first and not steep enough after \(n = 15\text{.}\) Overall, it is a good fit.
  3. \(f\) grows (more and more slowly) without bound. \(f\) will eventually exceed \(100\) per cent, but no one can forget more than 100% of what is learned.

10.2 Logarithmic Scales
10.2.7 Problem Set 10.2

Applications

10.2.7.19.
Answer.
\(1\text{,}\) \(80\text{,}\) \(330\text{,}\) \(1600\text{,}\) \(7000\text{,}\) \(4\times 10^7\)
10.2.7.21.
Answer.
Proxima Centauri: \(15.5\text{;}\) Barnard: \(13.2\text{;}\) Sirius: \(1.4\text{;}\) Vega: \(0.6\text{;}\) Arcturus: \(-0.4\text{;}\) Antares: \(-4.7\text{;}\) Betelgeuse: \(-7.2\)
10.2.7.27.
Answer.
A: \(a\approx 45\text{,}\) \(p \approx 7.4\%\text{;}\) B: \(a \approx 400\text{,}\) \(p \approx 15\%\text{;}\) C: \(a\approx 6000\text{,}\) \(p\approx 50\%\text{;}\) D: \(a \approx 13000\text{,}\) \(p \approx 45\%\)

10.3 The Natural Base
10.3.7 Homework 10.3

Skills Practice

10.3.7.1.
Answer.
\(x\) \(-10\) \(-5\) \(0\) \(5\) \(10\) \(15\) \(20\)
\(f(x)\) \(0.135\) \(0.368\) \(1\) \(2.718\) \(7.389\) \(20.086\) \(54.598\)
growth
10.3.7.3.
Answer.
\(x\) \(-10\) \(-5\) \(0\) \(5\) \(10\) \(15\) \(20\)
\(f(x)\) \(20.086\) \(4.482\) \(1\) \(0.223\) \(0.05\) \(0.011\) \(0.00248\)
decay
10.3.7.11.
Answer.
\(P (t) = 20\left(e^{0.4} \right)^t \approx 20\cdot 1.492^t\text{;}\) increasing; initial value \(20\)
10.3.7.13.
Answer.
\(P (t) = 6500\left(e^{-2.5} \right)^t \approx 6500\cdot 0.082^t\text{;}\) decreasing; initial value \(6500\)
10.3.7.15.
Answer.
  1. \(x\) \(0\) \(0.5\) \(1\) \(1.5\) \(2\) \(2.5\)
    \(e^x\) \(1 \) \(1.6487\) \(2.7183\) \(4.4817\) \(7.3891\) \(12.1825\)
  2. Each ratio is \(e^{0.5} \approx 1.6487\text{:}\) Increasing \(x\)-values by a constant \(\Delta x = 0.5\) corresponds to multiplying the \(y\)-values of the exponential function by a constant factor of \(e^{\Delta x}\text{.}\)
10.3.7.17.
Answer.
  1. \(x\) \(0\) \(0.6931\) \(1.3863\) \(2.0794\) \(2.7726\) \(3.4657\) \(4.1589\)
    \(e^x\) \(1 \) \(2\) \(4\) \(8\) \(16\) \(32\) \(64\)
  2. Each difference in \(x\)-values is approximately \(\ln 2\approx 0.6931\text{:}\) Increasing \(x\)-values by a constant \(\Delta x = \ln 2\) corresponds to multiplying the \(y\)-values of the exponential function by a constant factor of \(e^{\Delta x} = e^{\ln 2} = 2\text{.}\) That is, each function value is approximately equal to double the previous one.
10.3.7.33.
Answer.
  1. \(n\) \(0.39\) \(3.9\) \(39\) \(390\)
    \(\ln {(n)}\) \(-0.942 \) \(1.361 \) \(3.664 \) \(5.966 \)
  2. Each difference in function values is approximately \(\ln 10\approx 2.303\text{:}\) Multiplying \(x\)-values by a constant factor of 10 corresponds to adding a constant value of \(\ln 10\) to the \(y\)-values of the natural log function.
10.3.7.35.
Answer.
  1. \(n\) \(2\) \(4\) \(8\) \(16\)
    \(\ln n\) \(0.693 \) \(1.386 \) \(2.079 \) \(2.773 \)
  2. Each quotient equals \(k\text{,}\) where \(n = 2^k\text{.}\) Because \(\ln {(n)} = \ln {(2^k)} = k\cdot \ln {(2)}\text{,}\) \(k = \dfrac{\ln {(n)}}{\ln {(2)}}\text{.}\)

Applications

10.4 Chapter 10 Summary and Review
10.4.3 Chapter 10 Review Problems