Skip to main content
Logo image

Intermediate Algebra Functions and Graphs

Katherine Yoshiwara

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

1.3.6.23.
Answer.
  1. \(~ t ~\) \(0\) \(5\) \(10\) \(15\) \(20\)
    \(~ h ~\) \(-400\) \(-300\) \(-200\) \(-100\) \(0\)
    šŸ”—
  2. \(\displaystyle h=-44+20t \)
    šŸ”—
    šŸ”—
  3. grid
    šŸ”—
  4. \((0,-400) \text{:}\) The diver starts at a depth of 400 feet. \((20,0) \text{:}\) The diver surfaces after 20 minutes.
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—
1.3.6.25.
Answer.
  1. $\(2.40x,\) $\(3.20y\)
    šŸ”—
    šŸ”—
  2. \(\displaystyle 2.40x + 3.20y = 19,200\)
    šŸ”—
    šŸ”—
  3. line
    šŸ”—
  4. The \(y\)-intercept, 6000 gallons, is the amount of premium that the gas station owner can buy if he buys no regular. The \(x\)-intercept, 8000 gallons, is the amount of regular he can buy if he buys no premium.
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—

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

1.5.6.19.
Answer.
  1. \(\displaystyle M = 7000 - 400w\)
    šŸ”—
    šŸ”—
  2. The slope tells us that Tammyā€™s bank account is diminishing at a rate of $400 per week, the vertical intercept that she had $7000 (when she lost all sources of income).
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—
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 \(h = 0\)).
šŸ”—
šŸ”—
1.5.6.27.
Answer.
  1. \(55\degree\)F
    šŸ”—
    šŸ”—
  2. 9840 ft
    šŸ”—
    šŸ”—
  3. temperature vs altitude
    šŸ”—
  4. \(m=\dfrac{-3}{820}\text{.}\) The temperature decreases 3 degrees for each increase in altitude of 820 feet.
    šŸ”—
    šŸ”—
  5. \((19,133 \frac{1}{3}, 0)\text{.}\) At an altitude of \(19,133 \frac{1}{3}\) feet, the temperature is \(0\degree\)F. \((0,70)\text{.}\) At an altitude of 0 feet, the temperature is \(70\degree\)F.
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—

1.6 Chapter Summary and Review
1.6.3 Chapter 1 Review Problems

1.6.3.2.

Answer.
  1. \(t\) \(~ 5 ~\) \(~10~\) \(~15~\) \(~20~\) \(~25~\)
    \(R\) \(1960\) \(1820\) \(1680\) \(1540\) \(1400\)
    šŸ”—
  2. \(\displaystyle R = 2100 - 28t\)
    šŸ”—
    šŸ”—
  3. \((75, 0)\text{,}\) \((0, 2100)\)
    šŸ”—
    grid
    šŸ”—
  4. \(t\)-intercept: The oil reserves will be gone in 2080; \(R\)-intercept: There were \(2100\) billion barrels of oil reserves in 2005.
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—

1.6.3.18.

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

1.6.3.19.

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

1.6.3.21.

Answer.
\(m=\dfrac{1}{2} \text{,}\) \(b= \dfrac{-5}{4} \)
šŸ”—
šŸ”—

1.6.3.22.

Answer.
\(m=\dfrac{3}{4} \text{,}\) \(b= \dfrac{5}{4} \)
šŸ”—
šŸ”—

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 \(20.8\degree\text{C}\text{.}\)
šŸ”—
šŸ”—
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.
šŸ”—
šŸ”—

Applications

2.1.5.19.
Answer.
  1. scatterplot and regression line
    šŸ”—
  2. The graph is above.
    šŸ”—
    šŸ”—
  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.
    šŸ”—
    šŸ”—
  4. 44.5 days
    šŸ”—
    šŸ”—
  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.
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—

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

2.4 Gaussian Reduction
2.4.6 Problem Set 2.4

Applications

2.4.6.19.
Answer.
\(\dfrac{1}{2} \) lb Colombian, \(\dfrac{1}{4} \) lb French, \(\dfrac{1}{4} \) Sumatran
šŸ”—
šŸ”—

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.
    šŸ”—
    šŸ”—
  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.
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—

2.6 Chapter Summary and Review
2.6.3 Chapter 2 Review Problems

2.6.3.7.

Answer.
\(\left(\dfrac{1}{2}, \dfrac{7}{2} \right) \)
šŸ”—
šŸ”—

2.6.3.10.

Answer.
\(\left(\dfrac{1}{2}, \dfrac{3}{2} \right)\)
šŸ”—
šŸ”—

2.6.3.26.

Answer.
20 to Boston, 25 to Chicago, 10 to Los Angeles
šŸ”—
šŸ”—

2.6.3.39.

Answer.
\(20p+9g \le 120\text{,}\) \(10p+10g \le 120\text{,}\) \(p\ge 0\text{,}\) \(g\ge 0\)
šŸ”—
system of inequalities
šŸ”—

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
šŸ”—
system of inequalities
šŸ”—

3 Quadratic Models
3.1 Extraction of Roots
3.1.9 Problem Set 3.1

Warm Up

Skills Practice

Applications

3.1.9.25.
Answer.
  1. \(\displaystyle A=1600(1+r)^2\)
    šŸ”—
  2. \(~r~\) \(0.02\) \(0.04\) \(0.06\) \(0.08\)
    \(~A~\) \(1664.64\) \(1730.56\) \(1797.76\) \(1866.24\)
    šŸ”—
  3. 11.8%
    šŸ”—
šŸ”—
šŸ”—

3.2 Intercepts, Solutions, and Factors
3.2.6 Problem Set 3.2

Warm Up

Applications

3.2.6.29.
Answer.
c. 306.5 ft at 0.625 sec d. 1.25 sec e. 5 sec
šŸ”—
šŸ”—
3.2.6.34.
Answer.
a. \(l=6, ~w=1-2x, ~h=x, ~V=6x(1-2x)~~\) e. \(6x(1-2x)=\dfrac{3}{4};~~\dfrac{1}{4}~\) ft
šŸ”—
šŸ”—

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

3.4.4.17.
Answer.
  1. \(\displaystyle (\dfrac{1}{4} \pm \sqrt{\dfrac{13}{15}}, 0)\)
    šŸ”—
  2. \(\displaystyle (\dfrac{1}{4}, -3\dfrac{1}{4})\)
    šŸ”—
šŸ”—
šŸ”—

3.5 Chapter 3 Summary and Review
3.5.3 Chapter 3 Review Problems

3.5.3.35.

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

3.5.3.36.

Answer.
\(A_1=\pi (x+y)^2- \pi x^2 - \pi y^2 = 2 \pi xy;~~A_2=\pi y (2x) = 2 \pi xy\)
šŸ”—
šŸ”—

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
    šŸ”—
  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.
    šŸ”—
šŸ”—
šŸ”—

Applications

4.1.6.27.
Answer.
b. \(10h(2h-6)=2160~~~\)c. 12 ft by 18 ft by 10ft
šŸ”—
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/
šŸ”—
šŸ”—
šŸ”—

4.2 The Vertex
4.2.5 Problem Set 4.2

Warm Up

4.2.5.1.
Answer.
  1. \(t\) \(0\) \(0.25\) \(0.5\) \(0.75\) \(1\) \(0.25\) \(1.5\)
    \(h\) \(-12\) \(-5\) \(0\) \(3\) \(4\) \(3\) \(0\)
    šŸ”—
  2. grid
    šŸ”—
  3. \(\displaystyle (1,4)\)
    šŸ”—
  4. The wrench reaches its greatest height of 4 feet.
    šŸ”—
  5. The wrench reaches its greatest height 1 second after Francine throws it.
    šŸ”—
šŸ”—
šŸ”—

Skills Practice

Applications

4.2.5.25.
Answer.
  1. graph
    šŸ”—
  2. \(\displaystyle (-4,0),~(0,2)\)
    šŸ”—
  3. \(~~~~\) \(x \lt -4\) \(-4 \lt x \lt 2\) \(x \gt 2\)
    \(Y_1\) \(-\) \(+\) \(+\)
    \(Y_2\) \(-\) \(-\) \(+\)
    \(Y_3\) \(+\) \(-\) \(+\)
    šŸ”—
šŸ”—
šŸ”—

4.3 Curve Fitting
4.3.5 Problem Set 4.3

4.4 Quadratic Inequalities
4.4.6 Problem Set 4.4

Skills Practice

4.4.6.19.
Answer.
\((-\infty, \dfrac{-1}{2}) \cup (4, \infty)\)
šŸ”—
šŸ”—

4.5 Chapter 4 Summary and Review
4.5.3 Chapter 4 Review Problems

4.5.3.26.

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

4.5.3.32.

Answer.
\((-\infty,-\sqrt{3}) \cup (\sqrt{3},\infty)\)
šŸ”—
šŸ”—

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

5.2.6.13.
Answer.
  1. \(\displaystyle 0, ~\dfrac{1}{2}, ~0\)
    šŸ”—
  2. \(\displaystyle \dfrac{5}{6}\)
    šŸ”—
  3. \(\displaystyle \dfrac{-5}{6},~\dfrac{-1}{6},~\dfrac{7}{6},~\dfrac{11}{6}\)
    šŸ”—
  4. \(\displaystyle 1,~-1\)
    šŸ”—
  5. Max at \(x=-1.5,~0.5\text{,}\) min at \(x=-0.5,~1.5\)
    šŸ”—
šŸ”—
šŸ”—
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.
    šŸ”—
    šŸ”—
  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.
    šŸ”—
    šŸ”—
  3. The slowest speed occurs at a depth of about \(810\) meters and the speed is about \(1487\) meters per second, so \(f(810) = 1487\text{.}\)
    šŸ”—
    šŸ”—
  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.
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—

5.3 Some Basic Graphs
5.3.4 Problem Set 5.3

Warm Up

Applications

5.3.4.25.
Answer.
(b): reflected about \(x\)-axis, (c): reflected about \(y\)-axis
šŸ”—
šŸ”—

5.4 Direct Variation
5.4.6 Problem Set 5.4

Applications

5.4.6.9.
Answer.
  1. \(\text{Price of item} \) \(18\) \(28\) \(12\)
    \(\text{Tax} \) \(1.17\) \(1.82\) \(0.78\)
    \(\text{Tax}/\text{Price} \) \(\alert{0.065}\) \(\alert{0.065}\) \(\alert{0.065}\)
    Yes; \(6.5\%\)
    šŸ”—
    šŸ”—
  2. \(\displaystyle T = 0.065p\)
    šŸ”—
    šŸ”—
  3. direct variation
    šŸ”—
šŸ”—
šŸ”—

5.5 Inverse Variation
5.5.4 Problem Set 5.5

Warm Up

5.5.4.1.
Answer.
\(R=\dfrac{1}{3}I\text{.}\) Not inverse variation.
šŸ”—
šŸ”—

Applications

5.5.4.13.
Answer.
  1. \(\text{Width (feet)} \) \(2\) \(2.5\) \(3\)
    \(\text{Length (feet)} \) \(12\) \(9.6\) \(8\)
    \(\text{Length}\times \text{width} \) \(\alert{24}\) \(\alert{24}\) \(\alert{24}\)
    \(24\) square feet
    šŸ”—
    šŸ”—
  2. \(\displaystyle L = \dfrac{24}{w} \)
    šŸ”—
    šŸ”—
  3. inverse variation
    šŸ”—
šŸ”—
šŸ”—

5.6 Functions as Models
5.6.6 Problem Set 5.6

Warm Up

Applications

5.6.6.21.
Answer.
\(y = x^3\) stretched or compressed vertically
šŸ”—
cubic
šŸ”—
5.6.6.23.
Answer.
\(y =\dfrac{1}{x} \) stretched or compressed vertically
šŸ”—
reciprocal
šŸ”—

Absolute Value

5.6.6.19.
Answer.
\(w=\dfrac{13}{2} \) or \(w=\dfrac{15}{2} \)
šŸ”—
šŸ”—

5.7 Chapter 5 Summary and Review
5.7.3 Chapter 5 Review Problems

5.7.3.1.

Answer.
A function: Each \(x\) has exactly one associated \(y\)-value.
šŸ”—
šŸ”—

5.7.3.3.

Answer.
Not a function: The IQ of \(98\) has two possible SAT scores.
šŸ”—
šŸ”—

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.
šŸ”—
šŸ”—

5.7.3.6.

Answer.
\(H(16) = 3\text{:}\) At 16 mph, the trip takes 3 hours.
šŸ”—
šŸ”—

5.7.3.48.

Answer.
  1. \(x\) \(0.25\) \(0.50\) \(1.00\) \(1.50\) \(2.00\) \(4.00\)
    \(y\) \(4.00\) \(2.00\) \(1.00\) \(0.67\) \(0.50\) \(0.25\)
    šŸ”—
  2. \(\displaystyle y = \dfrac{1}{x}\)
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—

6 Powers and Roots
6.1 Integer Exponents
6.1.5 Problem Set 6.1

Skills Practice

Applications

6.1.5.31.
Answer.
  1. \(x\) \(1\) \(2\) \(4.5\) \(6.2\) \(9.3\)
    \(g(x)\) \(1\) \(0.125\) \(0.011\) \(0.0042\) \(0.0012 \)
    šŸ”—
  2. they decrease
    šŸ”—
    šŸ”—
  3. \(x\) \(1.5\) \(0.6\) \(0.1\) \(0.03\) \(0.002\)
    \(f(x)\) \(0.30\) \(4.63\) \(1000\) \(37,037\) \(125 \times 10^6\)
    šŸ”—
  4. they increase
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—

6.2 Roots and Radicals
6.2.9 Problem Set 6.2

Skills Practice

6.2.9.11.
Answer.
\(\dfrac{1}{4}x^{1/2}-2x^{-1/2}+\dfrac{1}{\sqrt{2}}x\)
šŸ”—
šŸ”—

Applications

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

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
šŸ”—

Applications

6.4 Working with Radicals
6.4.6 Problem Set 6.4

Warm Up

6.4.6.5.
Answer.
\begin{equation*} ~\sqrt{ab} =\sqrt{a}\sqrt{b}~ \end{equation*}
\begin{equation*} ~\sqrt{\dfrac{a}{b}} =\dfrac{\sqrt{a}}{\sqrt{b}}~ \end{equation*}
šŸ”—
šŸ”—

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
    šŸ”—
  2. Mercury, Venus, Earth, and Mars
    šŸ”—
šŸ”—
šŸ”—

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

Applications

7.1.7.24.
Answer.
  1. Years after 1983 \(0\) \(5\) \(10\) \(15\) \(20\)
    Value of house \(20,000\) \(25,526\) \(32,578\) \(41,579\) \(53,066\)
    šŸ”—
  2. \(\displaystyle V(t)=200,000(1.05)^t\)
    šŸ”—
  3. house value
    šŸ”—
  4. $359,171.27; $458,403.66
    šŸ”—
šŸ”—
šŸ”—

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
šŸ”—
7.2.6.21.
Answer.
Because they are defined by equivalent expressions, (b), (c), and (d) have identical graphs.
šŸ”—
šŸ”—

Applications

7.2.6.32.
Answer.
\(x\) \(f(x)=x^2\) \(g(x)=2^x \)
\(-2\) \(4\) \(\frac{1}{4} \)
\(-1\) \(1\) \(\frac{1}{2} \)
\(0\) \(0\) 1
\(1\) \(1\) \(2\)
\(2\) \(4\) \(4\)
\(3\) \(9\) \(8\)
\(4\) \(16\) \(16\)
\(5\) \(25\) \(32\)
exponential growth and qudratic
  1. \(\displaystyle Three\)
    šŸ”—
    šŸ”—
  2. \(\displaystyle x = -0.77,~2,~4\)
    šŸ”—
    šŸ”—
  3. \(\displaystyle (-0.77, 2) \cup (4,\infty)\)
    šŸ”—
    šŸ”—
  4. \(\displaystyle g(x)\)
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—

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

7.4.5.43.
Answer.
\(k=\dfrac{\log{\left(\dfrac{N}{N_0}\right)}}{t \log {(a)}} \)
šŸ”—
šŸ”—
7.4.5.45.
Answer.
\(t=\dfrac{1}{k}\log{\left(\dfrac{A}{A_0}+1\right)} \)
šŸ”—
šŸ”—

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

7.6.3.27.

Answer.
\(\dfrac{\log {(5.1)}}{1.3}\approx 0.5433 \)
šŸ”—
šŸ”—

7.6.3.29.

Answer.
\(\dfrac{\log {(2.9/3)}}{-0.7}\approx 0.21 \)
šŸ”—
šŸ”—

7.6.3.35.

Answer.
\(\log_b {(x)} + \dfrac{1}{3} \log_b {(y)} - 2 \log_b {(z)}\)
šŸ”—
šŸ”—

7.6.3.37.

Answer.
\(\dfrac{4}{3} \log {(x)} -\dfrac{1}{3} \log {(y)}\)
šŸ”—
šŸ”—

7.6.3.39.

Answer.
\(\log{\right(\sqrt[3] {\dfrac{x}{y^{2}}}\left)} \)
šŸ”—
šŸ”—

7.6.3.43.

Answer.
\(\dfrac{\log {(63)}}{\log {(3)}}\approx 3.77 \)
šŸ”—
šŸ”—

7.6.3.45.

Answer.
\(\dfrac{\log {(50)}}{-0.3\log {(6)}}\approx -7.278 \)
šŸ”—
šŸ”—

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

Skills Practice

8.1.10.5.
Answer.
(b) and (c) are not polynomials; they have variables in a denominator.
šŸ”—
šŸ”—

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

8.3.9.13.
Answer.
\(\dfrac{6x(x-2)(x-1)^2}{(x^2-8)(x^2-2x+4)}\)
šŸ”—
šŸ”—

Applications

8.3.9.51.
Answer.
\(\dfrac{3q}{8\pi R}-\dfrac{a^2q}{8 \pi R^3}\)
šŸ”—
šŸ”—

8.4 More Operations on Fractions
8.4.5 Problem Set 8.4

Skills Practice

8.4.5.17.
Answer.
\(\dfrac{1}{2}x^2\dfrac{1}{2}-\dfrac{1}{3x^2}\)
šŸ”—
šŸ”—

Applications

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

8.5 Equations with Fractions
8.5.8 Problem Set 8.5

Applications

8.5.8.29.
Answer.
\(168=\dfrac{72p}{100-p}\text{;}\) \(p=70\%\)
šŸ”—
šŸ”—
8.5.8.39.
Answer.
Because \(x=1\text{,}\) dividing by \(x-1\) in the fourth step is dividing by \(0\text{.}\)
šŸ”—
šŸ”—

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
\(\dfrac{2}{3} \) \(\dfrac{-3}{2} \) \(-1 \)
\(\dfrac{-5}{2} \) \(\dfrac{2}{5}\) \(-1\)
\(6 \) \(\dfrac{-1}{6} \) \(-1\)
\(-4\) \(\dfrac{1}{4} \) \(-1\)
\(-1 \) \(1\) \(-1\)
šŸ”—

Applications

9.1.4.15.
Answer.
b. \(m_{PQ}=\dfrac{-5}{2} \text{;}\) \(m_{PR}=\dfrac{2}{5} \)
šŸ”—
šŸ”—

9.2 The Distance and Midpoint Formulas
9.2.7 Problem Set 9.2

Warm Up

9.2.7.3.
Answer.
\(x\) \(-4\) \(-3\) \(-2\) \(-1\) \(0\) \(1\) \(2\) \(3\) \(4\)
\(y\) \(0\) \(\pm\sqrt{7}\) \(\pm\sqrt{12}\) \(\pm\sqrt{15}\) \(\pm 4\) \(\pm\sqrt{15}\) \(\pm\sqrt{12}\) \(\pm\sqrt{7}\) \(0\)
circle of radius 4 centered at origin
šŸ”—

Skills Practice

9.2.7.7.
Answer.
distance: \(\sqrt{20} \text{;}\) midpoint: \(\left(0,-2 \right) \)
šŸ”—
šŸ”—
9.2.7.9.
Answer.
distance: \(8\text{;}\) midpoint: \(\left(-2,-1 \right) \)
šŸ”—
šŸ”—
9.2.7.13.
Answer.
center: \((-3,0) \text{;}\) radius: \(\sqrt{10} \)
šŸ”—
šŸ”—

Applications

9.2.7.29.
Answer.
\((x+4)^2+y^2=20 \text{;}\) center: \((-4,0) \text{;}\) radius: \(\sqrt{20} \)
šŸ”—
šŸ”—
9.2.7.31.
Answer.
\(x^2+(y-5)^2=27 \text{;}\) center: \((0,5) \text{;}\) radius: \(\sqrt{27} \)
šŸ”—
šŸ”—
9.2.7.33.
Answer.
\(\left(x-\dfrac{3}{2}\right)^2+(y+4)^2=\dfrac{29}{4}\)
šŸ”—
šŸ”—

9.3 Conic Sections: Ellipses
9.3.5 Problem Set 9.3

Skills Practice

9.3.5.13.
Answer.
  1. \(\displaystyle \dfrac{x^2}{9}+\dfrac{y^2}{4}=1\)
    šŸ”—
    šŸ”—
  2. \(x\) \(0\) \(\pm 3\) \(-2\) \(\dfrac{\pm 3\sqrt{3} }{2} \)
    \(y\) \(\pm 2\) \(0\) \(\dfrac{\pm 2\sqrt{5} }{2} \) \(1\)
    šŸ”—
šŸ”—
šŸ”—
9.3.5.27.
Answer.
\(\dfrac{(x-3)^2}{25}+\dfrac{(y-3)^2}{16}=1\)
šŸ”—
šŸ”—

9.4 Conic Sections: Hyperbolas
9.4.6 Problem Set 9.4

Skills Practice

9.4.6.9.
Answer.
  1. \(\displaystyle \dfrac{x^2}{9} - \dfrac{y^2}{3} =1\)
    šŸ”—
    šŸ”—
  2. \(x\) \(0\) \(\dfrac{\pm3\sqrt{5}}{2} \) \(5\) \(\dfrac{\pm 15}{4} \)
    \(y\) undefined \(\pm 2\) \(\dfrac{\pm 16}{3} \) \(-3\)
    šŸ”—
šŸ”—
šŸ”—
9.4.6.23.
Answer.
Parabola; vertex \((0,2)\text{,}\) opens downward, \(a=\dfrac{-1}{2} \)
šŸ”—
šŸ”—
9.4.6.25.
Answer.
Hyperbola; center \(\left(\dfrac{-1}{8},-1 \right) \text{,}\) transverse axis vertical, \(a^2=\dfrac{15}{65} \text{,}\) \(b^2=\dfrac{15}{16} \)
šŸ”—
šŸ”—
9.4.6.27.
Answer.
Parabola; vertex \((-4,2)\text{,}\) opens upward, \(a=\dfrac{1}{4} \)
šŸ”—
šŸ”—

9.5 Nonlinear Systems
9.5.4 Problem Set 9.5

Skills Practice

9.5.4.13.
Answer.
\((2,-1) \text{,}\) \((-2,1) \text{,}\) \((1,-2) \text{,}\) \((-1,2) \)
šŸ”—
šŸ”—

9.6 Chapter 9 Summary and Review
9.6.3 Chapter 9 Review Problems

9.6.3.35.

Answer.
\(\dfrac{(x+1)^2}{16}+\dfrac{(y-4)^2}{4}=1 \)
šŸ”—
šŸ”—

9.6.3.36.

Answer.
\(\dfrac{(x-3)^2}{4}+\dfrac{(y-1)^2}{25}=1 \)
šŸ”—
šŸ”—

9.6.3.37.

Answer.
\(\dfrac{(x-2)^2}{16}-\dfrac{(y+3)^2}{9}=1 \)
šŸ”—
šŸ”—

9.6.3.41.

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

9.6.3.42.

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

9.6.3.47.

Answer.
Morning train: 20 mph, evening train: 30 mph
šŸ”—
šŸ”—

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
    šŸ”—
  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.
    šŸ”—
    šŸ”—
  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.
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—

10.2 Logarithmic Scales
10.2.7 Problem Set 10.2

Applications

10.2.7.19.
Answer.
\(1\text{,}\) \(80\text{,}\) \(330\text{,}\) \(1600\text{,}\) \(7000\text{,}\) \(4\times 10^7\)
šŸ”—
šŸ”—
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: \(-7.2\)
šŸ”—
šŸ”—
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: \(a \approx 13000\text{,}\) \(p \approx 45\%\)
šŸ”—
šŸ”—

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
šŸ”—
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
šŸ”—
10.3.7.11.
Answer.
\(P (t) = 20\left(e^{0.4} \right)^t \approx 20\cdot 1.492^t\text{;}\) increasing; initial value \(20\)
šŸ”—
šŸ”—
10.3.7.13.
Answer.
\(P (t) = 6500\left(e^{-2.5} \right)^t \approx 6500\cdot 0.082^t\text{;}\) decreasing; initial value \(6500\)
šŸ”—
šŸ”—
10.3.7.15.
Answer.
  1. \(x\) \(0\) \(0.5\) \(1\) \(1.5\) \(2\) \(2.5\)
    \(e^x\) \(1 \) \(1.6487\) \(2.7183\) \(4.4817\) \(7.3891\) \(12.1825\)
    šŸ”—
  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 \(e^{\Delta x}\text{.}\)
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—
10.3.7.17.
Answer.
  1. \(x\) \(0\) \(0.6931\) \(1.3863\) \(2.0794\) \(2.7726\) \(3.4657\) \(4.1589\)
    \(e^x\) \(1 \) \(2\) \(4\) \(8\) \(16\) \(32\) \(64\)
    šŸ”—
  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.
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—
10.3.7.33.
Answer.
  1. \(n\) \(0.39\) \(3.9\) \(39\) \(390\)
    \(\ln {(n)}\) \(-0.942 \) \(1.361 \) \(3.664 \) \(5.966 \)
    šŸ”—
  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.
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—
10.3.7.35.
Answer.
  1. \(n\) \(2\) \(4\) \(8\) \(16\)
    \(\ln n\) \(0.693 \) \(1.386 \) \(2.079 \) \(2.773 \)
    šŸ”—
  2. Each quotient equals \(k\text{,}\) where \(n = 2^k\text{.}\) Because \(\ln {(n)} = \ln {(2^k)} = k\cdot \ln {(2)}\text{,}\) \(k = \dfrac{\ln {(n)}}{\ln {(2)}}\text{.}\)
    šŸ”—
    šŸ”—
šŸ”—
šŸ”—

Applications

10.4 Chapter 10 Summary and Review
10.4.3 Chapter 10 Review Problems

10.4.3.31.

Answer.
\(t=\dfrac{-1}{k}\ln{\left(\dfrac{y-6}{12} \right)} \)
šŸ”—
šŸ”—

10.4.3.41.

Answer.
Order 3: 17,000; Order 4: 5000; Order 8: 40; Order 9: 11
šŸ”—
šŸ”—
šŸ”—
PrevTopNext