Monday, July 28, 2014

Retro Math TV

(Click on the comic if you can't see the full image.)
(C)Copyright 2014, C. Burke.

The 90s version, The Rad Squad, wasn't as successful.

Oscar and Fivex appeared together in another classic show.

You know how those numeric actors always like to work in pairs.

Interesting (to me) Trivia!: This is comic #888 and it's appearing on 7/28/14, a day of sevens! Not that this was planned, my Internet was actually out for a day or so last week, and I wasn't feeling all that great anyway....




Wednesday, July 23, 2014

Because There Aren't Enough Vampires on TV...

(Click on the comic if you can't see the full image.)
(C)Copyright 2014, C. Burke.

They call him The Count because he loves that old cloak thingy he wears.

You might remember the Count from one of his previous appearances.




Monday, July 21, 2014

Coming Soon: More New Math TV

(Click on the comic if you can't see the full image.)
(C)Copyright 2014, C. Burke.

And wait until you see the New Five! He's like the old Five, but new.




Saturday, July 19, 2014

Math/Educator Website Links

The following are a bunch of links from pages I ripped out of teacher magazines. I haven't visited any of them despite having these pages for a very long time. So I'm sharing them here so I can look at them sometime in the future and clean up some of the random clutter on my desk at home.

CC.BetterLesson.com/mtp: More than 3000 classroom-ready lessons that can be integrated into any curriculum.
Lessons created by more than 130 Master Teachers, with step-by-step instructions, videos, thoughts on how the strategies help to implement Common Core, sample work, etc.

Share My Lesson: "Numerous resource aligned to the Common Core State Standards". A reminder that you are not alone.

10 Free Things! A page from NEA.org, where you can find books, activities, posters, lessons, videos, and more (or so they say!).
Some of the 10 things listed

This list may be amended at any time if I should find more resources I like, if any of these links are dead, or if I just don't like them.

Thursday, July 17, 2014

Useless Junk

(Click on the cartoon to see the full image.)
(C)Copyright 2014, C. Burke.

I just have to make sure that I don't make it on her list.




Tuesday, July 15, 2014

Math Lunch

(Click on the cartoon to see the full image.)
(C)Copyright 2014, C. Burke.

I'm hungry. Can I get sum of that?




Thursday, July 10, 2014

Null Set

(Click on the cartoon to see the full image.)
(C)Copyright 2014, C. Burke.

Gives new meaning to ''set the table''.

Poor guy in the corner heard "set ... zero" and thought he was being called. And then realized he wasn't. Hate when that happens.

Assignment for math students: define or describe each of these four sets in words and in set-builder notation, or however your teacher taught you that you're supposed to remember for that test.




Monday, July 07, 2014

Tens

(Click on the cartoon to see the full image.)
(C)Copyright 2014, C. Burke.

Get it? Two tens! Now you know the score!




Sunday, July 06, 2014

Stating the Case: When is a Parallelogram Not Also a Rhombus?

When is a Parallelogram not also a Rhombus? That's not a riddle, although if I think of a good punchline for it, I could make a comic out of it.
No, this is a serious Geometry question. In fact, it was the proof on the June 2014 Geometry Regents exam.

Specifically, question 38 read as follows:

The vertices of quadrilateral JKLM have coordinates J(-3,1), K(1,–5), L(7,-2), and M(3,4).
Prove that JKLM is a parallelogram.
Prove that JKLM is not a rhombus.
EDIT: Co-ordinates fixed.
EDIT: Link added -- June 2014 Geometry Regents, question 38.
Additionally, a grid was provided, but its use was optional. What wasn't given was enough space to write the answer, unless you thought to flip to the following page. The good news was that you did NOT have to write a two-column proof, but you did have to show your work and give the details.

This is a question my students should be able to answer easily. They probably didn't, but they should have. There are reasons why I say this.

When I approach any question that contains the word "Prove", I try to get the students to think of any TV show courtroom drama they've ever scene. (Sit-coms are a totally different animal, here.) Few of them might know what it's like to be in a real courtroom, so I settle for the simplified version. Each side will give an opening statement, and they will state what they are setting out to prove. They were present their evidence and build a case out of all the evidence they bring forward. There isn't one magic witness or exhibit that will hand-wave the case away. (If there were, the case likely would never have been brought.) Then, in the end, there are summaries, which include the lawyers' conclusions based on the evidence that they've presented, and they implore the jury to reach the same conclusion.

Students are different. Students, who a year earlier in Algebra would have me solve a five-step problem in 17 steps, and repeat the solution and check twice because they still weren't "getting it", will look at a Geometry "proof" and say, "It's true because, you know, Math."

They're not sure what the "Math" is, but the Math is there, so it must be true.
The point is that they need to be sure.

This led to some interesting discussions online about the test question. How do you know, for example, if you gave enough information for only 2 points and not for 4 points?

Let's get to the basics

Parallelograms and Rhombuses

Prove that JKLM is a parallelogram. Prove that JKLM is not a rhombus.

How do you prove that a quadrilateral is also a parallelogram?

You have several choices: if the opposite sides are parallel, or if the opposite sides are congruent, or if the diagonals bisect each other, then the figure is a parallelogram.
This statement, whichever one or ones you use, has to be your conclusion, but you have to back those up with evidence:

  • To prove opposite sides are parallel, you need to find the slopes of the four sides
  • To prove that the opposite sides are congruent, you need to find the lengths of the four sides
  • To prove that the diagonals bisect each other, you have to find the midpoints of both diagonals.

Any of those are easy to do, although I'd say that slopes and midpoints are quicker to find than the lengths. Which one you do is up to you, but you can save time if you keep the second part of the question in mind. Using the slopes for the parallelogram is fine, but it won't help you with the rhombus.

That being said, I'd probably find the slopes first just because it's pretty much second-nature to me to do that first.

How do you prove that a quadrilateral is rhombus?

You have a couple of choices: if all four sides are congruent, or if the diagonals perpendicularly bisect each other, then the figure is a rhombus.
This statement, whichever one or ones you use, has to be your conclusion, but you have to back those up with evidence:

  • To prove that the opposite sides are congruent, you need to find the lengths of, at least, two consecutive sides. If you know it's a parallelogram, then you know that the opposite sides are congruent. It would've helped if you did the distance formula before now, instead of slopes, but you aren't penalized for doing extra work.
  • To prove that the diagonals perpendicularly bisect each other, you have to find the slopes of the diagonals. If you know it's a parallelogram, then you know that the diagonals bisect each other. You don't need to prove this, even if you didn't show it earlier.

So how much work is actually required for this problem?

Believe it or not, you could have solved this problem simply by finding the lengths of the four sides. Is that worth 6 points? No, that was worth 2 points. The rest of the points came from you conclusions and your reasons. Just because you found the lengths, you haven't (and this isn't meant to be a plotting reference) you haven't connected the dots yet. You haven't given a conclusion, nor stated under what rules this proves your case.

JKLM is a parallelogram because the opposite sides are congruent. (Work is shown for this.) JKLM is NOT a rhombus because all four sides are NOT congruent.

or

JKLM is a parallelogram because the diagonals bisect each other. (Work is shown for this.) JKLM is NOT a rhombus because the diagonals are not perpendicular. (Work is shown for this.)

Either of these would be complete answers good for full credit.

By contrast "They're parallel." ... well, just isn't.

As you can see, the slopes of the sides aren't needed for the problem, but it isn't incorrect to find them. Other work left for the reader: find the lengths of the sides, and the midpoints and slopes of the diagonals.

Friday, July 04, 2014

Happy Fourth of July 2014

(Click on the cartoon to see the full image.)
(C)Copyright 2014, C. Burke.

As far as I know, there aren't rules for displaying a flag at 45 degrees. It probably shouldn't. You probably will never have a radical-two length flag, either.




Thursday, July 03, 2014

Celsius, Fahrenheit and Kelvin

(Click on the cartoon to see the full image.)
(C)Copyright 2014, C. Burke.

Which one is the favorite? Check out the independent rankine!




Tuesday, July 01, 2014

Zero And One

(Click on the cartoon to see the full image.)
(C)Copyright 2014, C. Burke.

In math and in life, there's always some constant.

And in computer science, there's !=, which means "not equal to", in case you didn't know.