\def\wsn{1}
\input worksheet.tex

\item{1.}  Below is a list of some ``simple'' algebra problems.  Some
of the solutions are correct and some of them are {\bf wrong}!  For
each problem:

\smallskip
\itemitem{A.}  determine if the answer is correct;

\smallskip
\itemitem{B.}  determine if there are any mistakes made in solving the
problem and list them ({\bf note} that just because the answer is
correct does not mean there are no mistakes);

\smallskip
\itemitem{C.}  if there are mistakes, redo the problem correctly; if
there are no mistakes, devise {\bf ANOTHER} correct method to solve
the problem.

$$
\eqalign{
\pt a &{x^2-1\over x+1}={x^2+(-1)\over x+1}={x^2\over x}+{-1\over1}=x-1
\qquad\qquad\quad\>\,
\pt b (x+y)^2-(x-y)^2=x^2+y^2-x^2-y^2=0\cr\cr
\pt c &{9(x-4)^2\over3x-12}={3^2(x-4)2\over3x-12}={(3x-12)^2\over3x-12}
  =3x-12
\qquad
\pt d {x^2y^5\over2x^{-3}}=x^2y^5\cdot2x^3=2x^6y^5\cr\cr
\pt e &{(2x^3+7x^2+6)-(2x^3-3x^2-17x+3)\over(x+8)+(x-8)}={(2x)^2-17x+9\over2x}
  =2x-17x+9=-15x+9=-6x\cr\cr
\pt f &{x^{-1}+y^{-1}\over x^{-1}-y^{-1}}={(x+y)^{-1}\over(x-y)^{-1}}
  =\left({x+y\over x-y}\right)^{-1}=-{x+y\over x-y}={x+y\over y-x}\cr}
$$

\bigskip
\item{2.}  Find the equations of {\it all} lines through the point
$(a,a^2)$ which intersect the curve $y=x^2$ exactly once.   

\bigskip
\item{3.}  The following graph represents speed vs. time for two cars,
A and B, traveling the same direction along a road.  Assume the cars are at
the same position at time $t=0$ hours.

\medskip
\centerline{\epsfxsize=4truein\epsfysize=2.5truein
\epsfbox{/home/oehrtman/m210/cargraph.eps}}

\medskip
\itemitem{a)}  State the relationship between the position of car A
and car B at $t=1$ hour.

\medskip
\itemitem{b)}  State the relationship between the speed of car A and
car B at $t=1$ hour.

\medskip
\itemitem{c)}  State the relationship between the acceleration of car
A and car B at $t=1$ hour.  Explain.

\medskip
\itemitem{d)}  What is happening to the relative position of the two
cars during the time interval between $t=0.75$ hours and $t=1$ hour?
(e.g. is one car pulling away from the other?)  Explain.

\eject
\itemitem{4.\hskip12pt a)}  Use absolute values to define a formula
that gives the distance between two numbers $x_1$ and $x_2$ on the real
number line.

\medskip
\itemitem{b)}  What is the distance formula for two points $(x_1,y_1)$
and $(x_2,y_2)$ in two dimensions?

\medskip
\itemitem{c)}  What is the distance formula for two points $(x_1,y_1,z_1)$
and $(x_2,y_2,z_2)$ in three dimensions?

\medskip
\itemitem{d)}  Rewrite the distance formula for one dimension using
the same pattern as your answers for parts b) and c).  Is this the
same formula you wrote down in part a)?  Why or why not?

\bigskip
\itemitem{5.\hskip12pt a)}  Write an inequality which expresses the
following: ``The distance from $x$ to 3 is no more than 5.''

\medskip
\itemitem{b)}  Rewrite the inequality
$$
0<|x-2|<1
$$
as a sentence giving information about the distance between two points
on the number line.

\medskip
\itemitem{c)}  Using your rewritten version of the inquality
$0<|x-2|<1$, shade on a number line the possible values for $x$. 

\medskip
\itemitem{d)}  Solve the inequality $0<|x-2|<1$ {\bf algebraically}
for $x$. 

\medskip
\itemitem{e)}  Graph on the same axes the function $f(x)=|x-2|$ and
the lines $y=0$ and $y=1$.  Use this diagram to {\bf graphically}
solve the inequality $0<|x-2|<1$.

\bigskip
\item{6.}  Solve for $x$
$$
\pt a (x-4)(x+1)<0\hskip.6in
\pt b x>{1\over x}\hskip.6in
\pt c x^3<1\hskip.6in
\pt d x^2<0
$$

\medskip
\item{7.}  A working light bulb is in a closed room with no windows.
Outside the room, is a panel of three switches, one of which controls
the light inside (up is on, down is off.)  You may do anything you
like to the three switches and then enter the room to inspect the
light.  After this, without any further experimentation, you must
indicate which switch controls the light.  What do you do?




\bye