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%TCIDATA{Created=Mon Sep 27 12:17:07 1999}
%TCIDATA{LastRevised=Mon Sep 27 12:27:02 1999}
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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}

\begin{document}
\vspace{18pt}


\bigskip 

\begin{center}
RESPUESTAS
\end{center}

\begin{quotation}

\begin{itemize}

\item[1.] $28$ envases

\item[2.] $2$

\item[3.] $a=-\frac{7}{4}$

\item[4.] $x=3$

\item[5.] $c=\frac{5}{7}$

\item[6.] $r=\frac{st}{s+t}$

\item[7.] $x=5$ o $x=1$

\item[8.] $x+y-40=xy$, d\'{o}nde $x$ y $y$ son los n\'{u}meros

\item[9.] $3,\;-\frac{10}{7},\;\sqrt[5]{-32}$

\item[10.] $f\left( 2+h\right) =\frac{1}{3+2h}$

\item[11.] No existen n\'{u}meros reales que verifiquen las dos condiciones.

\item[12.] Los reales del intervalo $(-1,4)$.

\item[13.] $\frac{16}{9}$

\item[14.] perdi\'{o} $\$10.000$

\item[15.] $0.03\times 10^{-1}<33\times 10^{-4}<3\times 10^{-2}<\frac{1}{3}$

\item[16.] $\frac{9}{2}$ y $9$ centímetros.

\item[17.] Si $x$ representa la longitud del lado de la base el costo es, en pesos,
de $4.000x^{2}+\frac{576.000}{x}$

\item[18.] $\log _{a}\frac{3}{2}=0.176$

\item[19.] $x=-5\pm \sqrt{3}$

\item[21.] $\frac{1}{a}\geq \frac{2}{5}$

\item[22.] $50^{\circ }$

\item[23.] $AD$ mide $\frac{12\sqrt{5}}{5}$ unidades

\item[24.] $CD$ mide $\frac{36}{5}$ unidades

\item[25.] Identificados por las longitudes de los lados: $1\times 24,\;2\times
12,\;3\times 8$ y $4\times 6$. El de mayor perímetro es el primero: de $1\times 24$

\item[26.] El volumen de la esfera $A$ es igual a $8$ veces el volumen de la esfera 
$B$ 

\item[27.] Si es posible

\item[28.] sen $\alpha =\frac{4\sqrt{41}}{41},\;\cos \left( \pi -\alpha \right) =-%
\frac{5\sqrt{41}}{41},\;\tan \left( \pi +\alpha \right) =\frac{4}{5}$

\item[29.] sen $\gamma =\frac{3}{5},\;\cos \gamma =\frac{-4}{5}$

\item[30.] Las soluciones son $0,\;\frac{2\pi }{3},\;\frac{4\pi }{3},\;2\pi $
radianes.

\item[31.] Las diagonales miden $2\sqrt{37}$ y $2\sqrt{93}$ unidades.

\item[32.] $y=mx+\left( 3-2m\right) ,\;$y $y=-\frac{1}{m}x+\left( 3+\frac{2}{m}%
\right) $ para algun $m\neq 0$ o $x=2,\;y=3$

\item[33.] $\left( x-1\right) ^{2}+\left( y-\frac{1}{2}\right) ^{2}=\frac{25}{4}$

\item[34.] V\'{e}rtices $(0,3)$, $(0,-3)$, focos $(0,\sqrt{5}),\;(0,-\sqrt{5})$

\item[35.] Corresponde a la ecuaci\'{o}n $b$.

\end{quotation}

\end{itemize}

\bigskip

{\color{gray}{\hrule height 1pt}}
\docLink{index.html}{\buttonbox{\includegraphics{../back_but.gif}} Retorno

\end{document}