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Created August 5, 2010 16:47
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Latex Equation Challenge
\documentclass[12pt]{article}
\usepackage{fullpage}
\usepackage[fleqn]{amsmath}
\begin{document}
\begin{abstract}
\begin{center}
Equation Challenge
\end{center}
\end{abstract}
\begin{enumerate}
\item Evaluate $ \left(a + 3b\right)^2 \div \left(-ab\right) $ when $a = 3$ and $b = -2$.
\item Simplify: $ \displaystyle{ \frac{x^2 + 2x - 8}{x^2 + 6x + 8} \div \frac{x^2 - 5x + 6}{5x + 10}} $
\item Simplify: $ \cfrac{\cfrac{5}{3x-2} - 10}{\cfrac{4}{3x-2} + 4} $
\item Solve $V = \pi r (r + h)$ for $h$.
\item Graph the solution for $x + 1 < 5$.
\item Solve:
\begin{align*}
2x - y + 3z & = 9 \\
-x + 2y + 2z & = 9 \\
x + y + z & = 6
\end{align*}
\item Simplify: $\sqrt[5]{32 x^8 y^{10}}$
\item Simplify: $\sqrt[2]{81 x^3} - 3x \sqrt{16x}$
\item Simplify: $\left(2^{-3} x^2 y^{-1}\right) \left(2^{-1} x y^2\right)^{-2}$
\item Solve for $x$: $ \log_6 x + \log_6 (x + 1) = 1 $
\item Write $ \sin^2 x - \cos^2 x $ using only cosines.
\item Given $\displaystyle{f(x) = \frac{2x^2}{x - 2}}$, evaluate $f(-2)$.
\item Let $f(x) = (2x - 1)^3$, find $f'(x)$.
\item Integrate: $\displaystyle{ \int_{\pi/4}^{\pi}{\frac{\sin x \, dx}{\cos x}}}$
\item Find $\displaystyle{\sum_{n=1}^5{\frac{2^n}{n!}}}$
\end{enumerate}
\end{document}
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