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Appendix A Hints, Answers, and Solutions

1 Instructive Examples
1.1 Arithmetic

Checkpoint 1.1.1. Adding Single-Digit Integers.

Answer.
\(7\)
Solution.
\(6 + 1 = {7}\text{.}\)

Checkpoint 1.1.2. Declaring a Problem Seed.

Answer.
\(13\)
Solution.
\(5 + 8 = {13}\text{.}\)

Checkpoint 1.1.3. Controlling Randomness.

Answer.
\(3\)
Solution.
\(1 + 2 = {3}\text{.}\)

Checkpoint 1.1.4. Special Answer Checking.

Answer.
\(x^{8}\)
Solution.
We add the exponents as follows, while including a gratuitous reference to the quadratic formula:
\begin{equation*} \begin{aligned} {x^{5}x^{3}}\amp =x^{5+3}\amp \text{Theorem 1.2.1}\\ \amp =x^{8} \end{aligned} \end{equation*}

Checkpoint 1.1.5. Using Hints.

Hint.
Factor the number inside the radical.
Answer.
\(4\sqrt{3}\)
Solution.
\begin{equation*} {\sqrt{48}}=\sqrt{4^2\cdot 3}=4\sqrt{3} \end{equation*}

Checkpoint 1.1.6. No Randomization.

Answer.
\(3\)

1.2 The Quadratic Formula

Checkpoint 1.2.2. Solving Quadratic Equations.

1.2.2.a Identify Coefficients.
Answer 1.
\(5\)
Answer 2.
\(-6\)
Answer 3.
\(-8\)
Solution.
Take the coefficient of \(x^2\) for the value of \(a\text{,}\) the coefficient of \(x\) for \(b\text{,}\) and the constant for \(c\text{.}\) In this case, they are \(a = {5}\text{,}\) \(b = {-6}\text{,}\) \(c = {-8}\text{.}\)
1.2.2.b Use the Quadratic Formula.
Answer.
\(\left\{2,\frac{-4}{5}\right\}\)
Solution.
Recall that the quadratic formula is given in Theorem 1.2.1.
You already identified \(a = {5}\text{,}\) \(b = {-6}\text{,}\) and \(c = {-8}\text{,}\) so the results are:
\begin{equation*} x = {\frac{-\left(-6\right)+\sqrt{\left(-6\right)^{2}-4\cdot 5\cdot \left(-8\right)}}{2\cdot 5}} = {2} \end{equation*}
or
\begin{equation*} x = {\frac{-\left(-6\right)-\sqrt{\left(-6\right)^{2}-4\cdot 5\cdot \left(-8\right)}}{2\cdot 5}} = {-{\frac{4}{5}}} \end{equation*}

Checkpoint 1.2.3. Nested tasks.

1.2.3.a Identify Coefficients.
1.2.3.a.i
Answer.
\(6\)
Solution.
Take the coefficient of \(x^2\) for the value of \(a\text{.}\) In this case, \(a = {6}\text{.}\)
1.2.3.a.ii
Answer.
\(-31\)
Solution.
Take the coefficient of \(x\) for the value of \(b\text{.}\) In this case, \(b = {-31}\text{.}\)
1.2.3.a.iii
Answer.
\(-30\)
Solution.
Take the constant term for the value of \(c\text{.}\) In this case, \(c = {-30}\text{.}\)
1.2.3.b Use the Quadratic Formula.
Answer.
\(\left\{6,\frac{-5}{6}\right\}\)
Solution.
Recall that the quadratic formula is given in Theorem 1.2.1.
You already identified \(a = {6}\text{,}\) \(b = {-31}\text{,}\) and \(c = {-30}\text{,}\) so the results are:
\begin{equation*} x = {\frac{-\left(-31\right)+\sqrt{\left(-31\right)^{2}-4\cdot 6\cdot \left(-30\right)}}{2\cdot 6}} = {6} \end{equation*}
or
\begin{equation*} x = {\frac{-\left(-31\right)-\sqrt{\left(-31\right)^{2}-4\cdot 6\cdot \left(-30\right)}}{2\cdot 6}} = {-{\frac{5}{6}}} \end{equation*}

Checkpoint 1.2.4. Copy a Problem with Tasks.

1.2.4.a Identify Coefficients.
Answer 1.
\(2\)
Answer 2.
\(-5\)
Answer 3.
\(-25\)
Solution.
Take the coefficient of \(x^2\) for the value of \(a\text{,}\) the coefficient of \(x\) for \(b\text{,}\) and the constant for \(c\text{.}\) In this case, they are \(a = {2}\text{,}\) \(b = {-5}\text{,}\) \(c = {-25}\text{.}\)
1.2.4.b Use the Quadratic Formula.
Answer.
\(\left\{5,\frac{-5}{2}\right\}\)
Solution.
Recall that the quadratic formula is given in Theorem 1.2.1.
You already identified \(a = {2}\text{,}\) \(b = {-5}\text{,}\) and \(c = {-25}\text{,}\) so the results are:
\begin{equation*} x = {\frac{-\left(-5\right)+\sqrt{\left(-5\right)^{2}-4\cdot 2\cdot \left(-25\right)}}{2\cdot 2}} = {5} \end{equation*}
or
\begin{equation*} x = {\frac{-\left(-5\right)-\sqrt{\left(-5\right)^{2}-4\cdot 2\cdot \left(-25\right)}}{2\cdot 2}} = {-{\frac{5}{2}}} \end{equation*}

1.3 Open Problem Library

Checkpoint 1.3.1. Cylinder Volume.

Answer 1.
\(360\pi\ {\rm m^{3}}\)
Answer 2.
\(1130.97\ {\rm m^{3}}\)
Solution.
We use \(r\) to represent the base’s radius, and \(h\) to represent the cylinder’s height.
A cylinder’s volume formula is \(V= (\text{base area}) \cdot \text{height}\text{.}\) A cylinder’s base is a circle, with its area formula \(A = \pi r^{2}\text{.}\)
Putting together these two formulas, we have a cylinder’s volume formula:
\(\displaystyle{ V= \pi r^{2} h }\)
Throughout these computations, all quantities have units attached, and we only show them in the final step.
  1. Using the volume formula, we have:
    \(\displaystyle{\begin{aligned} V \amp = \pi r^{2} h \\ \amp = \pi \cdot 6^{2} \cdot 10 \\ \amp = \pi \cdot 360 \\ \amp = 360 \pi \textrm{ m}^3 \end{aligned}}\)
    Don’t forget the volume unit \(\textrm{m}^3\text{.}\)
  2. To find the decimal version, we replace \(\pi\) with its decimal value, and we have:
    \(\displaystyle{\begin{aligned}[t] V\amp = 360 \pi \\ \amp \approx 360 \cdot 3.14\ldots \\ \amp \approx {1130.97\ {\rm m^{3}}} \end{aligned}}\)
    Don’t forget the volume unit \(\textrm{m}^3\text{.}\)

1.4 Antidifferentiation
1.4.2 WeBWorK Exercises

1.4.2.1. Antiderivatives.

Answer.
\(593.23432548299\)
Solution.
SOLUTION
\begin{equation*} \begin{array}{rcl} \displaystyle \int_0^{5} (4 e^x+5 \sin x)\, dx \amp =\amp \displaystyle 4 e^x-5 \cos x \Big]_0^{5} \\ \amp =\amp (4 e^{5} - 5 \cos 5) - (4 e^0 - 5 \cos0 ) \\ \amp =\amp 4 e^{5} - 5 \cos 5 + 1 \end{array} \end{equation*}

1.4.2.4.

Answer.
\(-\cos\!\left(x\right)+C\)

1.4.2.5.

Answer.
\(e^{x}+C\)

1.4.2.6. Show Your Work.

Answer.
\(2x\)

1.6 Multiple Choice

Checkpoint 1.6.1. Drop-down/Popup.

Answer.
\(\text{is not}\)
Solution.
If \(\sqrt{2}\) were rational, then \(\sqrt{2}=\frac{p}{q}\text{,}\) with \(p\) and \(q\) coprime. But then \(2q^2=p^2\text{.}\) By the Fundamental Theorem of Arithmetic 1 , the power of \(2\) dividing the left side is odd, while the power of \(2\) dividing the right side is even. This is a contradiction, so \(\sqrt{2}\) is not rational.

Checkpoint 1.6.2. Choose one.

Answer.
\(\text{The Fundamental ... of Calculus}\)
Solution.
The correct answer is The Fundamental ... of Calculus.

Checkpoint 1.6.3. Choose a Subset of Options.

Answer.
\(\text{Choice 2, Choice 4, Choice 5}\)
Solution.
The correct answer is Choice 2, Choice 4, Choice 5.

Checkpoint 1.6.4. Choose a Subset of Options with Automated Labeling.

Answer.
\(\text{B, C, D}\)
Solution.
The correct answer is B, C, D.

Checkpoint 1.6.5. Choose a Subset of Options with Explicit Labeling.

Answer.
\(\text{TACO, SUSHI, PIZZA}\)
Solution.
The correct answer is TACO, SUSHI, PIZZA.

1.7 Tables

Checkpoint 1.7.1. Complete this Table.

Answer 1.
\(24\)
Answer 2.
\(24\)
Answer 3.
\(48\)
Answer 4.
\(48\)
Solution.
\(\times\) \(6\) \(6\)
\(4\) \(24\) \(24\)
\(8\) \(48\) \(48\)

1.8 Graphics in Exercises

Checkpoint 1.8.1. A static <latex-image> graph.

Answer.
\(\mathop{\rm C}\nolimits\!\left(n+1,2\right)\hbox{ or }\frac{\left(n+1\right)n}{2}\)

Checkpoint 1.8.2. A randomized <latex-image> graph.

Answer.
\(48\ {\rm cm^{2}}\)

Checkpoint 1.8.3. A <latex-image> graph affected by <latex-image-preamble>.

Answer.
\(-3, 0, 3\)

Checkpoint 1.8.5. Solve using a graph.

Answer.
\(\left\{1\right\}\)
Solution.
The graph reveals that the solution set to \(f(x)=1\) is \({\left\{1\right\}}\text{.}\)
a plot of a curve on a cartesian set of axes; the x axis ranges from -1 to 4, and the y-axis ranges from -1 to 4; the curve enters from the left, below the x-axis, and curves upward and to the right until it reaches the point (0,0); from here it continues predominantly rightward for a bit, bending slightly upward more and more as it progresses; it passes through the points (1,1) and (1.25992,2) before leaving the graph moving more and more upward and to the right; a horizontal line segment moves rightward from y=1 on the y-axis until it reaches a point on the curve; a vertical line segment moves down from this point to x=1 on the x-axis.

Exercises

1.8.1.
Answer.
\(5\)
Solution.
1.8.2.
Answer.
\(13\)
Solution.

2 Technical Examples
2.1 PGML Formatting and Verbatim Calisthenics

Checkpoint 2.1.2.

Answer 1.
\({\verb!<>&'";!}\)
Answer 2.
\({\verb!#%&<>\^_`|~!}\)
Answer 3.
\({\text{\$\{\}}}\)
Answer 4.
\(\text{ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789}\)
Answer 5.
\({\text{!()*+,-./:=?@[]}}\)

2.2 Subject Area Templates

Checkpoint 2.2.1. Answer is a number or a function.

Answer 1.
\(-8\)
Answer 2.
\(\frac{7-8x^{7}}{x}\)
Solution.
Solution explanation goes here.

Checkpoint 2.2.2. Answer is a function with domain issues.

Answer 1.
\(\sqrt{x-2}\)
Answer 2.
\(\ln\!\left(\left|\frac{x}{x-2}\right|\right)\)
Solution.
Solution explanation goes here.

Checkpoint 2.2.3. Multiple Choice by Popup, Radio Buttons, or Checkboxes.

Answer 1.
\(\text{Blue}\)
Answer 2.
\(\text{Blue}\)
Answer 3.
\(\text{Blue}\)
Solution.
The correct answer is Blue.
The correct answer is Blue.
The correct answer is Blue.

Checkpoint 2.2.4.

Answer.
\(\text{Choice 3}\)
Solution.
The answer is Choice 3.

Checkpoint 2.2.5. Tables.

Answer.
\(5\)
Solution.
The missing number is 5.

2.3 Stress Tests

Checkpoint 2.3.1. PTX problem source with server-generated images.

2.3.1.a
Solution.
2.3.1.b
Hint.
Solution.

2.5 Runestone Assignment Testing

Exercises

2.5.1.
Answer.
\(2\)