WWSC Hints, Answers, and Solutions

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 $a = {5}\text{,}$ $b = {-6}\text{,}$ $c = {-8}\text{.}$

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1.2.2.b Use the Quadratic Formula.

Answer.

$\left\{2,\frac{-4}{5}\right\}$

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Solution.

Recall that the quadratic formula is given in Theorem 1.2.1.

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You already identified $a = {5}\text{,}$ $b = {-6}\text{,}$ and $c = {-8}\text{,}$ so the results are:

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\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*}

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\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*}

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Checkpoint 1.2.3. Nested tasks.

1.2.3.a Identify Coefficients.

1.2.3.a.i

Answer.

$6$

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Solution.

Take the coefficient of $x^2$ for the value of $a\text{.}$ In this case, $a = {6}\text{.}$

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1.2.3.a.ii

Answer.

$-31$

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Solution.

Take the coefficient of $x$ for the value of $b\text{.}$ In this case, $b = {-31}\text{.}$

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1.2.3.a.iii

Answer.

$-30$

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Solution.

Take the constant term for the value of $c\text{.}$ In this case, $c = {-30}\text{.}$

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1.2.3.b Use the Quadratic Formula.

Answer.

$\left\{6,\frac{-5}{6}\right\}$

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Solution.

Recall that the quadratic formula is given in Theorem 1.2.1.

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You already identified $a = {6}\text{,}$ $b = {-31}\text{,}$ and $c = {-30}\text{,}$ so the results are:

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\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*}

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\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*}

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Checkpoint 1.2.4. Copy a Problem with Tasks.

1.2.4.a Identify Coefficients.

Answer 1.

$2$

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Answer 2.

$-5$

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Answer 3.

$-25$

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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 $a = {2}\text{,}$ $b = {-5}\text{,}$ $c = {-25}\text{.}$

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1.2.4.b Use the Quadratic Formula.

Answer.

$\left\{5,\frac{-5}{2}\right\}$

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Solution.

Recall that the quadratic formula is given in Theorem 1.2.1.

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You already identified $a = {2}\text{,}$ $b = {-5}\text{,}$ and $c = {-25}\text{,}$ so the results are:

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\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*}

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\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*}

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1.3 Open Problem Library

Checkpoint 1.3.1. Cylinder Volume.

Answer 1.

$360\pi\ {\rm m^{3}}$

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Answer 2.

$1130.97\ {\rm m^{3}}$

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Solution.

We use $r$ to represent the base’s radius, and $h$ to represent the cylinder’s height.

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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 $A = \pi r^{2}\text{.}$

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Putting together these two formulas, we have a cylinder’s volume formula:

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$\displaystyle{ V= \pi r^{2} h }$

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Throughout these computations, all quantities have units attached, and we only show them in the final step.

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Using the volume formula, we have:
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$\displaystyle{\begin{aligned} V \amp = \pi r^{2} h \\ \amp = \pi \cdot 6^{2} \cdot 10 \\ \amp = \pi \cdot 360 \\ \amp = 360 \pi \textrm{ m}^3 \end{aligned}}$
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Don’t forget the volume unit $\textrm{m}^3\text{.}$
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To find the decimal version, we replace $\pi$ with its decimal value, and we have:
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$\displaystyle{\begin{aligned}[t] V\amp = 360 \pi \\ \amp \approx 360 \cdot 3.14\ldots \\ \amp \approx {1130.97\ {\rm m^{3}}} \end{aligned}}$
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Don’t forget the volume unit $\textrm{m}^3\text{.}$
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1.4 Antidifferentiation
1.4.2 WeBWorK Exercises

1.4.2.1. Antiderivatives.

Answer.

$593.23432548299$

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Solution.

SOLUTION

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\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*}

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1.4.2.4.

Answer.

$-\cos\!\left(x\right)+C$

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1.4.2.5.

Answer.

$e^{x}+C$

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1.4.2.6. Show Your Work.

Answer.

$2x$

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1.6 Multiple Choice

Checkpoint 1.6.1. Drop-down/Popup.

Answer.

$\text{is not}$

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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

en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic#Canonical_representation_of_a_positive_integer

, 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.

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