How To Fill Leftover Space

Question

I am making a fairly space intensive LaTeX "cheat sheet." Right now, I have added a single table, as so:

However, there is a large gap to the right of this table that I would like to fill with a few equations if possible. How would I go about doing this? I've seen other cheat sheet LaTeX examples, but they all had columns, which will not work for me because the various tables and equations lists will be of varying lengths and widths.

This is my current code:

\documentclass{article}

\usepackage[margin=0.25in]{geometry}
\usepackage{tabu}
\usepackage{colortbl}
\usepackage[usenames,dvipsnames]{xcolor}
\usepackage{hhline}
\usepackage{boxedminipage}

\usepackage{graphicx}

\setlength\parindent{0pt}


\begin{document}

{\tabulinesep=3pt
\begin{tabu}{|c | c | c| c| c|}
\hline
\multicolumn{5}{|c|}{\bf \large Temperature Conversions} \\\hline

\bf  & \bf To K & \bf To C & \bf To F & \bf To R\\
\bf From K  &   --------    &   $C = K - 273.15$    &  $F = \frac{9}{5} K - 459.67$  &  $R = \frac{9}{5} K$ \\
\bf From C & $K = C + 273.15$ & -------- & \cellcolor{Goldenrod} $F = \frac{9}{5} C + 32$ & $R = \frac{9}{5} (C + 273.15)$\\
\bf From F & $K = \frac{5}{9} (F - 459.67)$ & \cellcolor{Goldenrod} $C = \frac{5}{9}(F - 32)$ & -------- & $R = F - 459.67$\\
\bf From R & $K = \frac{5}{9} R$ & $C = \frac{5}{9}(R - 491.67)$ & $F = R + 459.67$ & --------\\\hline
\end{tabu}
}   

\end{document}

Answer 1

Here is an example of what you can do. Adjusting \thistablewidth and the distance passed to \vspace may require some trial and error.

\documentclass{article}

\usepackage[margin=0.25in]{geometry}
\usepackage{tabu}
\usepackage{colortbl}
\usepackage[usenames,dvipsnames]{xcolor}
\usepackage{hhline}
\usepackage{boxedminipage}
\usepackage{graphicx}
\usepackage{calc}
\usepackage{amsmath}

\setlength\parindent{0pt}


\begin{document}
\newlength{\thistablewidth}
\setlength{\thistablewidth}{0.7\textwidth}
\begin{minipage}[t]{\thistablewidth}
\tabulinesep=3pt
\begin{tabu}{|c | c | c| c| c|}
\hline
\multicolumn{5}{|c|}{\bf \large Temperature Conversions} \\\hline

\bf  & \bf To K & \bf To C & \bf To F & \bf To R\\
\bf From K  &   --------    &   $C = K - 273.15$    &  $F = \frac{9}{5} K - 459.67$  &  $R = \frac{9}{5} K$ \\
\bf From C & $K = C + 273.15$ & -------- & \cellcolor{Goldenrod} $F = \frac{9}{5} C + 32$ & $R = \frac{9}{5} (C + 273.15)$\\
\bf From F & $K = \frac{5}{9} (F - 459.67)$ & \cellcolor{Goldenrod} $C = \frac{5}{9}(F - 32)$ & -------- & $R = F - 459.67$\\
\bf From R & $K = \frac{5}{9} R$ & $C = \frac{5}{9}(R - 491.67)$ & $F = R + 459.67$ & --------\\\hline
\end{tabu}   
\end{minipage}
\hfill
\begin{minipage}[t]{\textwidth-\thistablewidth}
\vspace{-5em}
\begin{gather*}
a=b\\
b=c\\
c=a\\
e = mc^2\\
\omega^2 = gk \tanh {kh}
\end{gather*}

\end{minipage}

\end{document}