Checkpoint 1.1.1. Adding Single-Digit Integers.
Answer.
\(7\)
Solution.
\(6 + 1 = {7}\text{.}\)
Appendix A Hints, Answers, and Solutions
1 Instructive Examples
1.1 Arithmetic
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.
\(3\)
Checkpoint 1.1.7. Local PG.
Answer.
\(6\)
Checkpoint 1.1.8. Local PG File.
Answer.
\(4\)
1.2 The Quadratic Formula
Checkpoint 1.2.2. Solving Quadratic Equations.
1.2.2.a Identify Coefficients.
Answer 1.
\(2\)
Answer 2.
\(-5\)
Answer 3.
\(-12\)
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 \(a = {2}\text{,}\) \(b = {-5}\text{,}\) \(c = {-12}\text{.}\)
1.2.2.b Use the Quadratic Formula.
Answer.
\(\left\{4,\frac{-3}{2}\right\}\)
Solution.
Recall that the quadratic formula is given in Theorem 1.2.1.
You already identified \(a = {2}\text{,}\) \(b = {-5}\text{,}\) and \(c = {-12}\text{,}\) so the results are:
\begin{equation*}
x = {\frac{-\left(-5\right)+\sqrt{\left(-5\right)^{2}-4\cdot 2\cdot \left(-12\right)}}{2\cdot 2}} = {4}
\end{equation*}
or
\begin{equation*}
x = {\frac{-\left(-5\right)-\sqrt{\left(-5\right)^{2}-4\cdot 2\cdot \left(-12\right)}}{2\cdot 2}} = {-{\frac{3}{2}}}
\end{equation*}
Checkpoint 1.2.3. Nested tasks.
1.2.3.a Identify Coefficients.
1.2.3.a.i
Answer.
\(3\)
Solution.
Take the coefficient of \(x^2\) for the value of \(a\text{.}\) In this case, \(a = {3}\text{.}\)
1.2.3.a.ii
Answer.
\(-16\)
Solution.
Take the coefficient of \(x\) for the value of \(b\text{.}\) In this case, \(b = {-16}\text{.}\)
1.2.3.a.iii
Answer.
\(-12\)
Solution.
Take the constant term for the value of \(c\text{.}\) In this case, \(c = {-12}\text{.}\)
1.2.3.b Use the Quadratic Formula.
Answer.
\(\left\{6,\frac{-2}{3}\right\}\)
Solution.
Recall that the quadratic formula is given in Theorem 1.2.1.
You already identified \(a = {3}\text{,}\) \(b = {-16}\text{,}\) and \(c = {-12}\text{,}\) so the results are:
\begin{equation*}
x = {\frac{-\left(-16\right)+\sqrt{\left(-16\right)^{2}-4\cdot 3\cdot \left(-12\right)}}{2\cdot 3}} = {6}
\end{equation*}
or
\begin{equation*}
x = {\frac{-\left(-16\right)-\sqrt{\left(-16\right)^{2}-4\cdot 3\cdot \left(-12\right)}}{2\cdot 3}} = {-{\frac{2}{3}}}
\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.
-
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 \(\textrm{m}^3\text{.}\)
-
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.
\(28.3885180688169\)
Solution.
SOLUTION
\begin{equation*}
\begin{array}{rcl} \displaystyle \int_0^{2} (4 e^x+2 \sin x)\, dx \amp =\amp
\displaystyle 4 e^x-2 \cos x \Big]_0^{2}
\\ \amp =\amp (4 e^{2} - 2 \cos 2) - (4 e^0 - 2 \cos0 )
\\ \amp =\amp 4 e^{2} - 2 \cos 2 - 2
\end{array}
\end{equation*}
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, 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.
\(\text{Choice 1, Choice 3, Choice 5}\)
Solution.
The correct answer is Choice 1, Choice 3, Choice 5.
Checkpoint 1.6.4. Choose a Subset of Options with Automated Labeling.
Answer.
\(\text{A, B, D}\)
Solution.
The correct answer is A, B, D.
Checkpoint 1.6.5. Choose a Subset of Options with Explicit Labeling.
Answer.
\(\text{SUSHI, TACO, PIZZA}\)
Solution.
The correct answer is SUSHI, TACO, PIZZA.
1.7 Tables
Checkpoint 1.7.1. Complete this Table.
Answer 1.
\(48\)
Answer 2.
\(36\)
Answer 3.
\(72\)
Answer 4.
\(54\)
Solution.
| \(\times\) | \(8\) | \(6\) |
| \(6\) | \(48\) | \(36\) |
| \(9\) | \(72\) | \(54\) |
1.8 Graphics in Exercises
Checkpoint 1.8.1. A static <latex-image> graph.
Answer.
\(\mathop{\rm C}\nolimits\mathopen{}\left(n+1,2\right)\hbox{ or }\frac{\left(n+1\right)n}{2}\)
Checkpoint 1.8.2. A randomized <latex-image> graph.
Answer.
\(24\ {\rm cm^{2}}\)
Checkpoint 1.8.3. A <latex-image> graph affected by <latex-image-preamble>.
Answer.
\(-1, 1, 2\)
Checkpoint 1.8.5. Solve using a graph.
Answer.
\(\left\{2\right\}\)
Solution.
The graph reveals that the solution set to \(f(x)=1\) is \({\left\{2\right\}}\text{.}\)
Exercises
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.
\(-2\)
Answer 2.
\(\frac{7-2x^{7}}{x}\)
Solution.
Solution explanation goes here.
Checkpoint 2.2.2. Answer is a function with domain issues.
Answer 1.
\(\sqrt{x-4}\)
Answer 2.
\(\ln\mathopen{}\left(\left|\frac{x}{x-4}\right|\right)\)
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. PreTeXt 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.
\(2\)
