\documentstyle[12pt,fleqn]{article}
\pagestyle{empty}
\setlength{\topmargin}{-1in}
\setlength{\oddsidemargin}{-.3in}
\setlength{\textheight}{9in}
\setlength{\textwidth}{7.2in}

\begin{document}

Key for: TEST 1 (Math 101 - 2)

\vspace{.1in}

\begin{enumerate}

\item  F ($\frac{1}{2}\%=.005$ but $50\%=.5$

T (they have the same slope (3))

F (they need to have negative reciprocal slopes)

T (we talked about this, look at the graph)

\item  5 miles.  Draw a picture.  One runner goes 4 miles and the other 3.
These are the legs of a right triangle.  
Use pythagoean theorem now: distance $=\sqrt{4^2+3^2}$

\item $g(x+1)-g(1)=(x+1)^2-(1)^1=(x+1)^2-1=x^2+2x$

\item  Multiply everything by 2: $-6 < 2x-3 \leq 6$,
add 3: $-3 < 2x \leq 9$, now divide by 2:
$-\frac{3}{2} < x \leq \frac{9}{2}$.  
The graph is everything between $-\frac{3}{2}$ and $\frac{9}{2}$, 
including $\frac{9}{2}$.

\item  When you have absolute value, think 2 answers.
So there are two equations to solve:
$x+2=3x-1$ or $x+2=-(3x-1)$.  
Solving these gives 2 solutions: $x=\frac{3}{2}$ or $x=-\frac{1}{4}$

\item  Draw a picture, and use similar triangles to get the equation:
$\frac{x}{6}=\frac{42}{9}$.  Solve this to get $x=28$.

\item  Use the distance formula:
$D=\sqrt{(-2-1)^2+(7-3)^2}=5$

\item  $m=\frac{-2-6}{5-3}=-4$

\item  First solve for $y$: $y=\frac{1}{2}x-1$.  
This shows right away that the $y$-intercept is -1.
Put $y=0$ and solve for $x$ to get the $x$-intercept (2).
Now plot these points and connect with a line.

\item  Yes, vertical line test.


\end{enumerate}

\end{document}