summaryrefslogtreecommitdiff
path: root/texmf-dist/doc/latex/eqexam/examples/hw02.tex
blob: 0f03398bef6ad1dcffda78aa6d1d7ead7bea819c (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage[pdf,forpaper,cfg=hw,nopoints,useforms,
% Try compiling this file under one of these options:
%     nosolutions,
%    answerkey,
    vspacewithsolns,
]{eqexam}

\subject[CA]{College Algebra}
\title[HW2]{HW \#2}
\author{Dr.\ D. P. Story}
\date{Spring 2011}
\duedate{03/05/11}
\keywords{Homework due \theduedate}

% Make the end of solution label blank
\def\exrtnlabelformat{}
\def\exrtnlabelformatwp{}
\solAtEndFormatting{\eqequesitemsep{3pt}}


\begin{document}

\maketitle

\begin{exam}{HW}

\begin{instructions}[]
In preparation for the quiz on Thursday, solve each of these short
problems in the space provided before looking at their solutions at the
end of the document.
\[
   \text{\url{http://faculty.nwfsc.edu/web/math/storyd}}
\]
All class assignments and other announcements will be posted on
this web site.
\end{instructions}

\begin{problem*}[4ea]
Let $P(-4,2)$ and $Q(2,-3)$ be two points in the plane.
\begin{parts}
\item Find the distance $d(P,Q)$ between $P$ and $Q$.
\begin{solution}[1in]
We use the distance formula
\[
    d(P,Q)=\sqrt{(2+4)^2+(-3-2)^2}=\sqrt{61}
\]
to obtained the required answer.
\end{solution}

\item Find the midpoint $M$ between $P$ and $Q$.
\begin{solution}[\sameVspace]
We use the midpoint formula
\[
    M=\left(\frac{-4+2}{2},\frac{2+(-3)}{2}\right)=\left(-1,-\dfrac{1}{2}\right)
\]
to obtained the required answer.
\end{solution}
\end{parts}
\end{problem*}

\begin{problem*}[3ea]
Complete each of the two sentences below with correct entries.
\begin{parts}
  \item The function $ g(x) = | x+2 | $ can be graphed from the library
  function $ f(x) = |x| $ by shifting it \fillin[u]{.25in}{2} units
  \fillin[u]{1.25in}{horizontally} (horizontally/vertically) \fillin[u]{.75in}{left} (left/right/up/down).
% Make the solution label blank.
\ifkeyalt
\begin{solution}[]
% We want this to be shown at the end of the file, but not if the author changes the option
% to answerkey
    The function $ g(x) = | x+2 | $ can be graphed from the library
    function $ f(x) = |x| $ by shifting it \fillin[u]{.25in}{2} units
  \fillin[u]{1.25in}{horizontally} (horizontally/vertically) \fillin[u]{.75in}{left}
  (left/\penalty0right/\penalty0up/\penalty0down). % Note: \penalty0 is inserted to help TeX break the line after the forward slash /
\end{solution}
\fi
  \item The function $ g(x) = 5 - x^2 $ can be graphed from the library
  function $ f(x) = x^2 $ by first reflecting it with respect to the
  \fillin[u]{.25in}{$x$} axis, then shifting it \fillin[u]{.25in}{5} units
  \fillin[u]{1.25in}{vertically} (horizontally/vertically) \fillin[u]{.75in}{upward} (left/right/up/down).
\ifkeyalt
\begin{solution}[]
% We want this to be shown at the end of the file, but not if the author changes the option
% to answerkey.
  The function $ g(x) = 5 - x^2 $ can be graphed from the library function
  $ f(x) = x^2 $ by first reflecting it with respect to the
  \fillin[u]{.25in}{$x$} axis, then shifting it \fillin[u]{.25in}{5} units
  \fillin[u]{1.25in}{vertically} (horizontally/vertically)
  \fillin[u]{.75in}{upward} (left/right/up/down).
\end{solution}
\fi
\end{parts}
\end{problem*}

\begin{problem}[5]
The circle $ x^2 + y^2 = 25 $ passes through
the point $P(3,4)$.  Let $\ell$ be the line passing though the origin and
the point $P$.  Find the equation of the line perpendicular to line $\ell$ and passing
through point $P$.
\begin{solution}[1in]
The slope of the line perpendicular to $\ell$ is $ m = -\tfrac{3}{4} $, the line
must pass through $(3,4)$; thus, the line is $ y-4 = -\tfrac{3}{4}(x-3) \implies
y = -\tfrac{3}{4} x + \tfrac{25}{4} $. Thus,
\[
    \text{Ans:}\quad\boxed{ y = -\tfrac{3}{4} x + \tfrac{25}{4} }
\]
This is the equation of the line tangent to the circle at $P(3,4)$.
\end{solution}
\end{problem}

\begin{problem}[3]
If the slope the a line is negative, then the line is
    \begin{answers}{4}
    \bChoices
        \Ans0 increasing\eAns
        \Ans1 decreasing\eAns
        \Ans0 constant\eAns
        \Ans0 none of these\eAns
    \eChoices
    \end{answers}

% The \texttt{answers} and \texttt{manswers} environments can now
% be copied and pasted into the solutions environment as well.
%
\ifkeyalt
\begin{solution}[.25in]
If the slope the a line is negative, then the line is
    \begin{answers}{4}
    \bChoices
        \Ans0 increasing\eAns
        \Ans1 decreasing\eAns
        \Ans0 constant\eAns
        \Ans0 none of these\eAns
    \eChoices
    \end{answers}
\end{solution}
\fi
\end{problem}
\end{exam}
\end{document}