Category Archives: Announcements

Khan Academy

All students are expected to join the Khan Academy Algebra II class. There will be assignments; students are also encouraged to use this resource for further study.

To join Mr. Hartzer’s class, login at and select “Coaches”. Then enter this under “Join a class”: VR3JMK94

Note that you will not automatically receive assignments that were given before you joined; I’ll have to assign them to you manually.

The first required assignment is on matrices. You must complete a total of four points worth of exercises for full credit; additional exercises earn extra credit! Note that for this assignment, you must complete the exercise, but you don’t have to get 100%.

This is due Friday, October 13.

1 point exercises
Matrix Dimensions
Matrix Elements
Add & Subtract Matrices

2 point exercises
Multiply Matrices by Scalars
Matrix Equations: Addition & Subtraction

4 point exercise
Multiply Matrices

Welcome Back!

Welcome back to Hamtramck High School. This year, I’ll be teaching Algebra II, so I already know quite a few of you from last year, in Geometry. For those of you in Honors Algebra II, the main difference is that we’ll be digging in deeper, while keeping to the same pace.

On this website, you’ll find the daily PowerPoints (after you’ve seen them in class), as well as daily topics and information on any classwork or homework.

The Volume of a Sphere

Here is a quick explanation for the formula of the volume of a sphere. This is based on the proof known to the Ancient Greeks.

For this, consider three objects:

  1. A cylinder;
  2. A cone with the point  on the bottom;
  3. A hemisphere with the flat side down.

Each object has a height of r and a radius of r.

Take a slice of each object at some height h. The exposed surface (cross-section) will be a circle.

For the cylinder, the radius of this exposed circle will be r, because the radii of all circular cross-sections is r. So the area of the circle for the cylinder at height h is πr2.

For the cone, the radius of this exposed circle will be h, so the area is πh2.

By definition, each point on the hemisphere is r units away from the center. Each point on the exposed circle is at a height of h. Using the Pythagorean Theorem, the radius of the exposed circle is the square root of (r2-h2), so the area is π(r2-h2).

Note that this is the difference between the cylinder and the cone: This is the key.

Since, for each cross-section of the three objects, the area for the cylinder is equal to that of the cone plus that of the hemisphere, it must be the case that the volume of a cylinder is equal to the volume of a cone and the volume of a hemisphere, when all objects have the same radius and height.

We know the formula for the volume of a cylinder: πr2h. If h = r, then πr3.

We know the volume of a cone is one-third of this, so the volume of a hemisphere is two-thirds of this, 2πr3/3. The volume of a sphere is twice that of a hemisphere, that is, 4πr3/3.

Khan Academy

I have updated the NWEA information for all students that have created a Khan Academy account. Remember: You get extra credit for work that you complete on Khan Academy.

You can still create an account. To do so:

  1. Sign up at (or log in with an existing accounts)
  2. Visit
  3. There, in the “Add a coach” field, enter the appropriate class code
    • First hour: 2626QT
    • Second hour: P26X8H
    • Third hour: WMJE9C
  4. Comment here or let me know in person that you have done so, and I will add your NWEA scores