Answer

$\newcommand{\arraycolsep}[3]{} \newcommand{\dd}[2][]{\frac{d #1}{d #2}} \newcommand{\pp}[2][]{\frac{\partial #1}{\partial #2}} \newcommand{\veci}{\vec{\hat{\bf\imath}}} \newcommand{\vecj}{\vec{\hat{\bf\jmath}}} \newcommand{\utan}{\vec{\hat{t}}} \newcommand{\unormal}{\vec{\hat{n}}} \newenvironment{amatrix}[1]{% \begin{bmatrix} }{% \end{bmatrix} }$

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\lipsumexp }[0]{} \newcommand {\ximera }[0]{XiMeRa} \newcommand {\d }[0]{\,d} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

An example of various uses of answer.

The command \answer must be in math mode.

Type $2$ . $\answer {2}$ , $\answer [given]{2}$ , $\answer [tolerance=.2]{\frac {1}{2}}$ , $\int _a^b f(x) \ dx$ , $-\frac {1}{\frac {12}{1}}\sum _{n=1}^{\answer {\infty }} n$

Is $\ln (x^2+4) = \ln (|x^2+4|)$ ? Well $\ln (x^2+4) = \answer { \ln (x^2+4) }$ and $\ln (|x^2+4|) = \answer { \ln (|x^2+4|) }$