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.
\(3\)
1.2 The Quadratic Formula
Checkpoint 1.2.2. Solving Quadratic Equations.
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 \(a = {5}\text{,}\) \(b = {-6}\text{,}\) \(c = {-8}\text{.}\)
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.
Answer.
\(6\)
Solution.
Take the coefficient of \(x^2\) for the value of \(a\text{.}\) In this case, \(a = {6}\text{.}\)
Answer.
\(-31\)
Solution.
Take the coefficient of \(x\) for the value of \(b\text{.}\) In this case, \(b = {-31}\text{.}\)
Answer.
\(-30\)
Solution.
Take the constant term for the value of \(c\text{.}\) In this case, \(c = {-30}\text{.}\)
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.
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 \(a = {2}\text{,}\) \(b = {-5}\text{,}\) \(c = {-25}\text{.}\)
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.
\(593.23432548299\)
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.
\(-\cos\mathopen{}\left(x\right)+C\)
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, 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.
1
en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic#Canonical_representation_of_a_positive_integerCheckpoint 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.
\(\mathop{\rm C}\nolimits\mathopen{}\left(n+1,2\right)\hbox{ or }\frac{\left(n+1\right)n}{2}\)
Checkpoint 1.8.2. A randomized <latex-image> graph.
Answer.
\(48\ {\rm cm^{2}}\)
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 \({\left\{1\right\}}\text{.}\)
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.
\({}\)
Answer 2.
\({}\)
Answer 3.
\({}\)
Answer 4.
\(\text{ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789}\)
Answer 5.
\({}\)
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.
\(\ln\mathopen{}\left(\left|\frac{x}{x-2}\right|\right)\)
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.
Solution.
Hint.
Solution.
2.5 Runestone Assignment Testing
Exercises
2.5.1.
Answer.
\(2\)
