R.I.P. 2016

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(C)Copyright 2016, C. Burke.

I got nothing.

I've said before that I didn't want to be an "obituary comic". I have noted the passing of a number of people over the past decade, but I've had to pick and choose. Generally, what matters are how much the person affected me or were a part of my life, my artistic ability and limitations and time -- as it both "Is it possible to get something meaningful done in a short time?" and "Do I have the time?" There's been less and less of the latter.

I might also add that when I kept to my 3x/week schedule, those occasional comics were a smaller percentage of the overall output. Keeping an irregular schedule, it's harder to justify "filler" comics. But I don't think this one needs justification. People have been complaining about 2016 for at least a month now, waiting for it to end already. Sadly, next year won't prove to be any different in this regard. People die. That won't change.




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Merry Christmas 2016: Christmas Reflections

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(C)Copyright 2016, C. Burke.

With a tail as big as a kite. With a tail as big as a kite.

Which is a four-sided figure with two pairs of consecutive sides congruent.




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Christmath 2016: Find the Area (UPDATED)

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(C)Copyright 2016, C. Burke.

This comic brought to you courtesy of The Law of Sines. Have a Happy Holiday. It's the Law.

I drew this on the white board and took a picture of it with the iPad in the classroom. I posted the original photo (with my title and (c) notice typed over it) on the blog early in the day, but I had to wait until I got home to clean up the photo a little. Funny thing was that I could lighten it up, but I couldn't auto-adjust the contrast -- too much junk from Whiteboard Past became suddenly visible!

The original photo is posted below.

Have a Merry Christmas, a Happy Holiday and Seasons Greetings to everyone out there.

Update: This came in email from longtime reader, friend and Scotch aficionado (which has nothing to do with this, just sayin'), Bill Ricker who wrote:

My favorite photo editor Fotoxx (only for Linux, best with Ubuntu) can do it too. IIRC it only rotates colors ... so Red->Green and Blue->Red instead of the Blue<->Green swap i did earlier.

So now we have a green tree with red lights, which fits the season a little better. Thanks, Bill!

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Holly Jolly

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(C)Copyright 2016, C. Burke.

And then watch ''March of the Wooden Soldiers'' (aka ''Babes in Toyland'')

I was actually going to use a Green Arrow icon, but I didn't know if that would be as obvious, and GA has nothing to do with Christmas.




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Repeat the Sounding Joy!

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(C)Copyright 2016, C. Burke.

Couldn't decide on one, so I did both. Choose A or B at your own leisure/peril.

Sorry, I hadn't expected such a long delay between comics. And if I had, that last one wouldn't have been my first choice for top of my page for so long. I mean, I thought it was cute, but c'mon now.




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Limited Edition!

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(C)Copyright 2016, C. Burke.

Forget the obvious trademark violations. Can you imagine the problems with trying to market a ''Ken doll''?

On the other hand, it is FUN to play around with different styles! It's a mysterious affair, this art stuff.




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Ken-Do #12: K-8 And Beyond!

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(C)Copyright 2016, C. Burke.

There's a Doctor Who reference in there, too. Or just sci-fi, in general.

There is a trend in some smaller schools that are housed in bigger buildings to grow into the space so the kids don't have to leave if they don't wish to. If there are multiple schools in a building, one might outgrow their space and move to a new facility while the remaining one may expand. So a K-5 school may add a grade per year until it gets to K-8. Becoming a K-12 school is a bigger decision entirely, so that may take a few years to happen.

I'd given Ken-Do a rest, mostly because I still have to go back and re-do all the old ones. While it's fine to create images in this style, I shouldn't be using random character generators from the internet. I'll probably edit them over the Christmas break.




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SCIENCE! With Scott: Competition

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(C)Copyright 2016, C. Burke.

There's a learning curve for new webcomic writers, even if they're teachers, too.

I like doing the occasional Science comic, but I don't know if this feature in particular will be ongoing. What do you think? (What's your vote?) By the way, I decided on the teacher's name a while ago, even if I haven't used it. (He has a last name, too.)

Ken's comic should be making a comeback at some point. Problem is that I have to rework all of them, and put a little bit more of my style (or one of my styles, at any rate) into them.

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Volume Displacement

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(C)Copyright 2016, C. Burke.

One method: Put the turkey in first. Fill the fryer with oil. Remove turkey. Heat the oil.

This one should've been up the day after Thanksgiving, except a) wasn't ready yet because we were enjoying the holiday, and b) I was a little worried that the newspaper would be filled with tragedy from this happening, despite it being a sitcom staple.

I hope everyone's Thanksgiving was a happy and safe one. I'm having great memories of Leftover Sundays past.

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Thanksgiving Probability Problem

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(C)Copyright 2016, C. Burke.

It approaches 1 as more people approach the buffet.

Unlike a previous Thanksgiving with the Family comic, I didn't try to squeeze everyone into this "extended" family gathering. However, Thanksgivings are big affairs. Now that the torch has passed from mother to daughter (who is married, so nee Keegan), there are more guests with nieces and nephews and assorted relatives who are alone for the holiday because their immediate families no longer live around here.

Yeah, there are a lot of complaints, and sometimes it gets a little heated -- the oven has been on all day, after all -- it's still family. They're the only ones you have. And when they're gone, they're gone.

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June 2016 Common Core Algebra II Regents, Part 2

I don't usually post the answers to the Algebra II exam, but I've received enough requests that I've decided to post some of the open-ended questions.

June 2016, Algebra II

25. Solve for x: (1 / x) - (1 / 3) = -(1 / 3x)

The answer is best given in picture form, shown below. Basically, you need to find a common denominator to combine the fractions. You can then solve by cross-multiplying.
Alternatively, you could just multiply all the terms by 3x because x, 3, and 3x are all factors of 3x. The multiplication would remove all the fractions from the problem.

x = 4.

26. Describe how a controlled experiment can be created to examine the effect of ingredient X in a toothpaste.

This was really is open-ended. You can write a lot of different things, but there are a few points which will be mandatory. Basically, you needed to have a number of people, (pick a number, say 10 or 20), and assign them randomly to two groups. One of these groups will try the new toothpaste, and the other is the control group which will not use the new toothpaste.

27. Determine if x - 5 is a factor of 2x³ - 4x² - 7x - 10. Explain your answer.

One way to figure out if it is a factor is to divide the polynomial by (x - 5). If there is no remainder, then it is a factor.
Another is to say, if x - 5 is a factor of the polynomial, then 2x³ - 4x² - 7x - 10 = 0 when x = 5. Substitute 5 for x and see if you get 0. (You won't.)

The image below shows the result of the division. It does not divide evenly. There is a remainder. So it is NOT a factor.

28. On the axes below, graph one cycle of a cosine function with amplitude 3, period pi/2, midline y = -1, and passing through the point (0,2).

You needed several things for this two points. You would lose 1 point for one mistake, and 2 points for 2 or more mistakes.
You needed to have a scale to the graph. It needed to pass through (0, 2) and have the proper midline (y = -1)

29. A suburban high school has a population of 1376 students. The number of students who participate in sports is 649. The number of students who participate in music is 433. If the probability that a student participates in either sports or music is 974/1376, what is the probability that a student participates in both sports and music

The total number of students who play either sports or music is equal to the number who play sports PLUS the number who play music MINUS the ones who play BOTH. If you just add the first two numbers, then you will double count the students who participate in both activities.

You are given the total that play one or the other. You need to calculate the number who play both. Add the two numbers (649 + 433) and subtract 974 from that sum of 1082, which gives you 108. However, since this is a probability question, you need to write it as a fraction of the student population, or 108/1376.

The image shows what to do:

30. The directrix of the parabola 12(y + 3) = (x - 4)² has the equation y = -6. Find the coordinates of the focus of the parabola.

The directrix of a parabola is a horizontal line running below the vertex. (Or above, if it opens down.) The focus is a point such that the distance from it to the parabola is equal to the vertical distance from the parabola to the directrix.

To find the coordinates of the focus, you need to find the coordinates of the vertex, which would be exactly in the middle on a straight line. The focus will have the same x-coordinate as the vertex. The focus will be as far above the vertex as the directrix is below.

Rewrite the equation 12(y + 3) = (x - 4)²
Divide both sides by 12: y + 3 = (1/12)(x - 4)²
Subtract 3 from both sides: y = (1/12)(x - 4)² - 3
The vertex (h, k) is at (4, -3)
The directrix is y = -6. (-6) - (-3) = 3.
Add 3 to the vertex y-coordinate: (-3) + 3 = 0, which must be the y-coordinate of the focus.
The focus is at (4, 0).

31. Algebraically prove that (x³ + 9)/(x³ + 8) = 1 + (1)/(x³ + 8), where x =/= -2.

Change 1 into (x³ + 8)/(x³ + 8). Do the addition. The left and right sides are equal.

32. A house purchased 5 years ago for $100,000 was just sold for $135,000. Assuming exponential growth, approximate the annual growth rate, to the nearest percent.

Set up the compound interest equation, A = P(1 + r/n)^(nt).
Substitute what you know, A = 135,000, P = 10000, n = 1 (annual), t = 5 years: 135000 = 100000(1 + r/1)^((1)(5))
Solve for r:

135000 = 100000(1 + r)^5
1.35 = (1 + r)^5
1.35^(1/5) = 1 + r
1.06185875879 = 1 + r
0.06185875879 = r
r = 6%

End of Part II

How did you do?
Comments, questions, corrections and concerns are all welcome.
Typos happen.

Irrational Fractions

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(C)Copyright 2016, C. Burke.

He's irrational, two.

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