# Give your students feedback,meaningfully and efficiently.

## Students Show Step-by-Step Work

Students can start with a blank Free Math document, copying down and working through problems just as they would in paper notebooks.

Students save their work as a file and submit it through an LMS in response to an assignment.

## Embrace Visual Learning

Students can include images in their solutions.

Including quickly snapping a picture of written work with their webcam.

## Simultaneously Review All Assignments

Complete solutions are shown, grouped by similar final answer.

You can award partial credit and give feedback to students that need help.

You don't need to type in an answer key, Free Math just provides an organized view of all student work.

## Analytics Show Where Students Struggled

Give feedback on the most impactful problems first,
everything else gets completion points. Assignments and grading sessions save directly from the browser to files in your downloads folder. From there you can store the files in any cloud system like Google Drive, Dropbox, OneDrive, etc. ## Office Hours

Have questions about how to get started with Free Math?
Want to talk with the development team about a feature suggestion?
Interested in meeting other teachers improving their classrooms with Free Math?

Come to office hours on Google Meet, held Monday, Wednesday and Friday at 8:30-9:30am CST

Help us bring simple freeform digital math assignments to the world's classrooms.

## Great for Many Areas of Math

### Algebra

$\frac{1}{x-4}+\frac{2}{x^2-16}=\frac{3}{x+4}$
$\frac{1}{x-4}+\frac{2}{\left(x-4\right)\left(x+4\right)}=\frac{3}{x+4}$
$\frac{1}{x-4}\cdot\left(\frac{x+4}{x+4}\right)+\frac{2}{\left(x-4\right)\left(x+4\right)}=\frac{3}{x+4}\cdot\left(\frac{x-4}{x-4}\right)$
$\frac{1\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{2}{\left(x-4\right)\left(x+4\right)}=\frac{3\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}$
$1\left(x+4\right)+2=3\left(x-4\right)$
$x+6=3x-12$
$x+18=3x$
$18=2x$
$9=x$

### Calculus

$\int x\ln xdx$
$u=\ln x$
$dv=xdx$
$du=\frac{1}{x}dx$
$v=\frac{x^2}{2}$
$\int x\ln sdx=\frac{x^2}{2}\ln x-\int\frac{x^2}{2}\cdot\frac{1}{x}dx$
$\frac{x^2}{2}\ln x-\frac{1}{2}\int xdx$
$\frac{x^2}{2}\ln x-\frac{1}{2}\left(\frac{x^2}{2}\right)+c$
$\frac{x^2}{2}\ln x-\frac{1}{4}x^2+c$

### Physics

$\text{A ball is thrown from 1 m above the ground.}$
$\text{It is given an initial velocity of 20 m/s}$
$\text{At an angle of 40 degrees above the horizontal}$
$\text{Find the maximum height reached}$
$\text{And velocity at that point}$
$x\left(t\right)=v\cos\left(\theta\right)t=20\cos\left(40\right)t=15.3t$
$y\left(t\right)=y_0+v\sin\left(\theta\right)t-\frac{9.8t^2}{2}$
$y\left(t\right)=1+20\sin\left(40\right)t-4.9t^2$
$y\left(t\right)=1+12.9t-4.9t^2$
$v_y\left(t\right)=v\sin\left(\theta\right)-9.8t$
$v_y\left(t\right)=12.9-9.8t$
$\max\ height\ at\ v_y\left(t\right)=0$
$12.9-9.8t=0$
$-9.8t=-12.9$
$t=\frac{-12.9}{-9.8}=1.3$
$y\left(1.3\right)=1+12.9\left(1.3\right)-4.9\left(1.3\right)^2$
$y\left(1.3\right)=9.5\ m$
$y\ component\ of\ velocity\ is\ 0\ at\ highest\ pt$
$total\ velocity\ =v_x=15.3\ \frac{m}{s}$
Free Math is free software: you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version. Free Math is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details. You should have received a copy of the GNU General Public License along with Free Math. If not, see <http://www.gnu.org/licenses/>.