Tuesday, February 22, 2022

Two Two Two Two Two

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(C)Copyright 2022, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

As the title suggests, this is a sequel to Two Two Two Two! I bet you knew that, too.

I hope you don't mind that I posted it on a TWOSDAY! (da-dum-dum ... CRASH!)

And now we'll have to wait 100 years to rerun it. Some of my youngest readers might be around for that. If science advances enough, they might even be around for 2/22/2222.

But if I'm still around for 1/11/11, I'll feel like I have won.

When I said you wouldn't have a repetitive Groundhog's Day comic, I never said you wouldn't get any kind of repetitive comic.



I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.





Come back often for more funny math and geeky comics.



Friday, February 18, 2022

School Life #26: Talking With Myself

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(C)Copyright 2022, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

There are times talking to oneself is the only intelligent choice, even in a crowded room of people. Of course, they may not think so.

So, yeah, there's back story here which might have been touched on more over that past two years if there hadn't been a pandemic going on. One of these days, I'll get to that. And more with Daisy, too. Or I may just have to write fanfic for my own comic at Archive of Our Own.

Okay, since I mentioned that -- I do have some stuff over on Archive of Our Own, mostly based on other people's properties, so I can't really do much with them. Enjoy The Twilight Zone and the Injustice League/Doofenschmirtz Evil, Inc. crossover. That lengthy Car Wars saga? Yikes, that was written a loooooooong time ago.



I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.





Come back often for more funny math and geeky comics.



Wednesday, February 16, 2022

Kissing Numbers

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(C)Copyright 2022, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

Who am I kidding? i would've tempted fate and brought it up had I the chance to.

Truth is (grumble, grumble) that I'm still settling into a new curriculum in a new semester in a new classroom ... with ancient tecnhology. The PC is probably older than my high school students. It's slow and the monitor is distorted. I can barely do my work let alone squeeze in a little extra. And then there's all the translating I have to do because there isn't anyone else there to do it. Did I mention "grumble, grumble"?

Okay, so "kissing numbers" was, in fact, the subject of the February 14 math calendar entry, and the writeup fell to me. (Okay, so I volunteered it.) Rather than a cartoon, I just did a straight explanation, so I didn't publish it as Monday's cartoon.

So six cirlces can kiss one circle at one time, and 12 spheres can kiss a sphere at one time.



I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.





Come back often for more funny math and geeky comics.



Friday, February 11, 2022

Number Substitution Puzzle

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(C)Copyright 2022, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

Are you primed for homework?

As a simple substitution problem, there are only 9 letters in the puzzle and 10 digits to choose from. We know that 0 must be one of the numbers in play.

As explained in the comic, we know that three of the letters must represent 9, 1, and 0. That leaves a number of possibilities for D and N which must add up to 9.

This is where the "meta-puzzle" comes in. If we ignore the actual words, there are 8 possibilities that can be found using 7 & 2, 2 & 7, 3 & 6, and 6 & 3. If you use 4 & 5, then you will have to repeat a digit.

If, according to the meta-puzzle, ODD must be odd, then D must be 7 or 3. (Again, 5 won't work, but feel free to try it.)

These now have four solutions because "O" and "V" will be interchangeable. That is, there is nothing to prevent you from switching those two place values.

Of these four solutions, there will actually be only two distinct totals (because switching O and V doesn't affect its sum). Of those two, one will be a prime number and one will not. There will be an easy divisibility test to see which one is composite.

A big Thank You has to go to Rebecca Rappaport and her math-a-day calendar, where this puzzle came from. In that puzzle, you were asked for the sum of P + R + E, which is 1 + 0 + 9 = 10. This puzzle appeared yesterday on the tenth. I wrote the "official" solution, which didn't require a full solution, but I wanted to do one anyway. I didn't take the ODDness or EVENess or PRIMEness into account when I mentioned 8 solutions.



I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.





Come back often for more funny math and geeky comics.



Thursday, February 10, 2022

Calendar Problems 2

As mentioned in other posts, I follow the "Math a Day" Calendar posts on Twitter, and I've even supplied the "official" answer a few times.

I like them so much that I starting using them as warmups in my own classes. The problem was that sometimes the question was at a level beyond my students' current abilities or they were just trivial in nature. Moreover, if an AP walked in, I'd think that they'd prefer to see a "Do Now" that related to what was being covered that day. To that extent, I started writing some of my own.

What follows are a handful of the calendar problems that I created. If any of Rebecca Rappaport's snuck through, allow me to apologize in advance for the error. The answers to each should be a whole number between 1 and 31. The ones below are not necessarily sequential.

If this is popular, I can make it a regular thing. Maybe every other Thursday or so, assuming I have enough to post (and I remember to schedule the post).

As always, math teachers are welcome to use these in their classrooms. If you put them online, please credit this blog. And a link would be nice. Maybe a comment saying "Howdy"? Please refrain from adding them from things that you intend to sell.

Enjoy!

Wednesday, February 09, 2022

Lewis & Clark

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(C)Copyright 2022, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

K? Shouldn't it be a big red S? Or maybe it's K for Kryptonite.

I haven't used Mr. Ibsen in a while, so I thought I'd check in with our History class. This isn't the first time Lewis & Clark made it into this comic either. That theme was explored in four comics, starting with one about their Guide. You have to skip over a couple comics because the parts weren't sequential. Also, Google stored my photos in odd ways back then, so they're shrunken. (It's on my list of things to fix.) But you can click on the images to enlarge them.

And there's also Manifest Destiny which appeared on the 800th anniversary of the signing of the Magna Carta.



I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.





Come back often for more funny math and geeky comics.



Tuesday, February 08, 2022

Algebra Problems of the Day (Integrated Algebra Regents, January 2011)



Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones. The Integrated Algebra Regents covered most of the same material as the current Algebra Regents, with a few differences.

More Regents problems.

Integrated Algebra Regents, January 2011

Part I: Each correct answer will receive 2 credits.


11. A student correctly graphed the parabola shown below to solve a given quadratic equation.


What are the roots of the quadratic equation associated with this graph?

1) -6 and 3
2) -6 and 0
3) -3 and 2
4) -2 and 3

Answer: 4) -2 and 3


The roots are the zeroes of the function. For what values of x is the function (the y value) equal to 0?

The x-intercepts occur at -2 and 3. This is Choice (4).

The y-intercept is -6, which in this graph is NOT the minimum, so it isn't really an important number on the graph, compared to the roots and the vertex.(in my opinion)





12. Which value of x is the solution of the equation 2/3 x + 1/2 = 5/6

1) 1/2
2) 2
3) 2/3
4) 3/2

Answer: 1) 1/2


Plugging in choices may be messy if you don't like dealing with fractions. It may be easier to solve the equation.

(2/3 x + 1/2 = 5/6 ) * 6

(6)(2/3)x + (6)(1/2) = (6)(5/6)

4x + 3 = 5

4x = 2

x = 1/2

The correct answer is Choice (1).





13. What is the range of the data represented in the box-and-whisker plot shown below?



1) 40
2) 45
3) 60
4) 100

Answer: 3) 60


The range is the highest number minus the lowest number. The highest number is 75 and the lowest is 15. And 75 - 15 = 60, which is Choice (3).

Choice (1), 40, is the median, the middle of the data. If is also the length of the box portion of the graph, also known as the Interquartile Range.

Choice (2), 45, is nothing of significance. It's the middle of the number line, but for purposes of statistics, that is irrelevant.

Choic (4), 100, is the range shown in the graph, but the data doesn't cover the entire number line. The line could have stopped at 90 or 80 and nothing on the graph would have changed.





14. Which equation illustrates the associative property?

1) x + y + z = x + y + z
2) x(y + z) = xy + xz
3) x + y + z = z + y + x
4) (x + y) + z = x + (y + z)

Answer: 4) (x + y) + z = x + (y + z)


The Associative Property tells us that we can group the terms differently, like in Choice (4).

Choice (1) shows the Identity Property. The left side is the same as the right side of the equal sign.

Choice (2) shows the Distributive Property of Multiplication Over Addition.

Choice (3) shows the Commutative Property, because the order of the operations has been switched around.





15. Josh and Mae work at a concession stand. They each earn $8 per hour. Josh worked three hours more than Mae. If Josh and Mae earned a total of $120, how many hours did Josh work?

1) 6
2) 9
3) 12
4) 15

Answer: 2) 9


Use the equation 8(x + x + 3) = 120 to find x, the number of hours that MAE worked, and then x + 3, which is how many JOSH worked. (You could also use 8(x + x - 3) = 120 to find Josh immediately, if you realize that you can use subtraction here.)

What is inside the parentheses are the total number of hours worked. The 8 is their pay since they are both paid the same amount. You don't need to distribute it -- you can just divide the equation by 8!

8(x + x + 3) = 120

2x + 3 = 15

2x = 12

x = 6

MAE worked 6 hours, so JOSH work 6 + 3 = 9 hours.

If you didn't and the 3, you got Choice (1). If you just divided by 8, you got Choice (4). If you divided by 8 and then subtracted 3, you got Choice (3). Those would all be bad.




More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.



Monday, February 07, 2022

Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2011)



Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Algebra 2/Trigonometry Regents, January 2011

Part I: Each correct answer will receive 2 credits.


6. What is the solution set of the equation |4a + 6| − 4a = −10?

1) ∅
2) {0}
3) {1/2}
4) {0, 1/2}

Answer: 1) ∅


You can plug in the last 3 choices, or you can work backward.

If should be obvious that 0 cannot be a choice because |6| =/= -10.

So you can check 1/2 or solve.

|4 (1/2) + 6| - 4(1/2) = |2 + 6| - 2 = 8 - 2 = 6, not -10.

While that isn't enough to prove that the answer is the Null Set, ∅, it does eliminate all of the choices we were given.

Working backward:

|4a + 6| − 4a = −10

|4a + 6| = -10 + 4a

4a + 6 = -10 + 4a OR 4a + 6 = -(-10 + 4a)

6 = -10 (impossible) or 4a + 6 = 10 -4a

Impossible or 8a = 4

Impossible or a = 1/2

But we have already check a = 1/2, and it is an extraneous solution that was created when we removed the absolute value symbols. If you substitute a = 1/2 into the second line above, you would have an absolute value equaling a negative number, which isn't possible.

So the answer is Null.





7. If sin A = 2/3 where 0° < A < 90°, what is the value of sin 2A?

1) 2√(5) / 3
2) 2√(5) / 9
3) 4√(5) / 9
4) -4√(5) / 9

Answer: 3) 4√(5) / 9


If 0° < A < 90° then 0° < 2A < 180°, so sin 2A is positive. Eliminate Choice (4).

If sin A = 2/3, then imagine a right triangle with one leg = 2 and the hypotenuse = 3. The other leg would be √(9 - 4) = √(5). So the cos A = √(5) / 3.

Sin 2A = 2 sin A cos A = 2 (2/3) (√(5)/3) = 4√(5) / 9, which is Choice (3).

You could have also found the approximate value for sin-2 2/3, which is about 41.81 degrees. Multiply that by 2 to get 83.62. The sine of that value would be close to 1. It's 0.993.

If you evaluate Choices (1) and (3), you'll see that (3) is the correct answer.





8. A dartboard is shown in the diagram below. The two lines intersect at the center of the circle, and the central angle in sector 2 measures 2Ï€/3.




1) 1/6
2) 1/3
3) 1/2
4) 2/3

Answer: 2) 1/3


The entire circle is 2Ï€. Sector 2 is 2Ï€/3, so Sector 4 is 2Ï€/3. That means that the total of Sectors 1 and 4 must be 2Ï€ - (2Ï€/3 + 2Ï€/3) = 2Ï€/3. Each of the two sectors would be half of that, or 2Ï€/6 each.

So the probability that a dart hitting the circle randomly will fall in 1 or 3 is 1/3 because 1 and 3 together have the same area as 2 and 4 have individually. This is Choice (2).

x3 + x2 − 2x = 0

x(x2 + x − 2 = 0

x(x + 2)(x - 1) = 0

x = 0 or x = -2 or x = 1





9. If f(x) = x2 − 5 and g(x) = 6x, then g(f(x)) is equal to

1) 6x3 − 30x
2) 6x2 − 30
3) 36x2 − 5
4) x2 + 6x − 5

Answer: 2) 6x2 − 30


Since g(x) = 6x, if the input the g is f(x), then multiply f(x) by 6.

g(f(x)) = g(x2 − 5) = 6(x2 − 5) = 6x2 − 30.

That is Choice (2).





10. Which arithmetic sequence has a common difference of 4?

1) {0, 4n, 8n, 12n, . . .}
2) {n, 4n, 16n, 64n, . . .}
3) {n + 1, n + 5, n + 9, n + 13, . . .}
4) {n + 4, n + 16, n + 64, n + 256, . . .}

Answer: 3) {n + 1, n + 5, n + 9, n + 13, . . .}


If there is a common difference of 4, then each term should be 4 more than the term before it (except for the first term, of course).

Choice (1) looks like it has a common ratio of 4, not a common difference. However, it's not a geometric series because of the 0 first term. You can't multiply 0 by any number and get anything other than 0. Eliminate it.

Choice (2) has a common ratio of 4, not a common difference. Eliminate it.

Choice (3) has a common difference of 4. The expression n + 5 is 4 more than n + 1, or n + 5 - 4 = n + 1, and n + 9 - 4 = n + 5, etc. Choice (3) is the answer.

Choice (4) has neitehr a common difference nor a common ratio. The constant has a ratio of 4, but the variable has a ratio of 1. This isn't allowed. Eliminate it.




More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.



Sunday, February 06, 2022

Geometry Problems of the Day (Geometry Regents, January 2011)



Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Geometry Regents, January 2011

Part I: Each correct answer will receive 2 credits.


6. A line segment has endpoints A(7,−1) and B(−3,3). What are the coordinates of the midpoint of AB?

1) (1,2)
2) (2,1)
3) (-5,2)
4) (5,-2)

Answer: 2) (2,1)


The coordinates of the midpoint will be halfway between 7 and -3, and halfway between -1 and 3.

( (7 + -3)/2, (-1 + 3)/2) ) = (2, 1)





7. What is the image of the point (−5,2) under the translation T3,−4?

1) (-9,5)
2) (-8,6)
3) (-2,-2)
4) (-15,-8)

Answer: 3) (-2,-2)


T3,−4 is a translation of the point 3 units to the right and 4 units down.

(-5 + 3, 2 - 4) --> (-2,-2)

Which is Choice (3).





8. When writing a geometric proof, which angle relationship could be used alone to justify that two angles are congruent?

1) supplementary angles
2) linear pair of angles
3) adjacent angles
4) vertical angles

Answer: 4) vertical angles


Of the four types of angles listed, only vertical angles MUST BE congruent. The others could be congruent, but aren't likely to be.





9. Plane R is perpendicular to line k and plane D is perpendicular to line k. Which statement is correct?

1) Plane R is perpendicular to plane D.
2) Plane R is parallel to plane D.
3) Plane R intersects plane D.
4) Plane R bisects plane D.

Answer: 2) Plane R is parallel to plane D.


Consider an elevator shaft running through all the floors of a building. It is perpendicular to all the floors. Now consider planes P and R as two of the floors. They will be parallel to each other and both perpendicular to the line k.

If R were perpendicular to P, the k could not be perpendicular to both. Also Choice (1) makes no sense because if it were true, Choice (3) would also be true, which isn't allowed on a test.

Choice (4) makes no sense because of Choice (3) and because you can't bisect a plane, not even with another plane. It extends forever.





10. The vertices of the triangle in the diagram below are A(7,9), B(3,3), and C(11,3).


What are the coordinates of the centroid of △ABC?

1) (5,6)
2) (7,3)
3) (7,5)
4) (9,6)

Answer: 3) (7,5)


The centroid is where the three medians meet. The medians are lines drawn from one vertex to the midpoint of the opposite side.

The midpoint of BC is (7,3). Call that D. This means that the median AD is a vertical line of length 6.

The centroid cuts a median into two pieces with a ratio of 2:1. That means that the centroid is 4 units down from (7,9), which is (7,5).

After finding D, you pretty much had the answer, because it couldn't be (7,3).




More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.



Algebra Problems of the Day (Integrated Algebra Regents, January 2011)



Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones. The Integrated Algebra Regents covered most of the same material as the current Algebra Regents, with a few differences.

More Regents problems.

Integrated Algebra Regents, January 2011

Part I: Each correct answer will receive 2 credits.


6. What is the value of x in the equation 2(x − 4) = 4(2x + 1)?

1) -2
2) 2
3) -1/2
4) 1/2

Answer: 1) -2


You xan try each of the choices or you can solve the problem.

2(x − 4) = 4(2x + 1)

x - 4 = 2(2x + 1)

x - 4 = 4x + 2

-6 = 3x

-2 = x

That is Choice (1).





7. WThe rectangle shown below has a diagonal of 18.4 cm and a width of 7 cm.
To the nearest centimeter, what is the length, x, of the rectangle?



1) 11
2) 17
3) 20
4) 25

Answer: 2) 17


READ THE CHOICES!!! NOW READ THEM AGAIN!!!!!

Only one choice makes ANY sense, so it must be the answer.

We know that x CANNOT be 11 because 7 - 11 - 18.4 does NOT make a triangle. The two legs only add up to 18, which is less than 18.4. Eliminate Choice (1).

We also know that x CANNOT be greater than or equal to the hypotenuse of the right triangle, so it cannot be 20 or 25. Eliminate (3) and (4).

That leaves Choice (2) as the answer.

If you didn't see that, you had to do Pythagorean Theorem:

x2 + 72 = 18.42

x2 = 289.56

x2 = 289.56

x = 17.016...





8. When a3 − 4a is factored completely, the result is

1) (a - 2)(a + 2)
2) a(a - 2)(a + 2)
3) a2(a - 4)
4) a(a - 2)2

Answer: 2) a(a - 2)(a + 2)


Eliminate Choice (1) because it's only quadratic and not cubic. You can eitehr factor the expression or multiply the choices.

a(a2 - 4)

a(a - 2)(a + 2)

This is Choice (2).





9. Which ratio represents sin x in the right triangle shown below.



1) 28/53
2) 28/45
3) 45/53
4) 53/28

Answer: 1) 28/53


The sine ratio is the length of the side opposite the angle, which is 28, over the length of the hypotenuse, which is 53.

So the sin x = 28/53.

Choice (2) shows tan x, which is the opposite over the adjacent side.

Choice (3) shows cos x, which is the adjacent over the hypotenuse.

Choice (4) shows the cosecant ratio, which is the reciprocal of the sine. It is useful to know sometimes, but not here.





10. What is the value of the expression (a3 + b0)2 when a = −2 and b = 4?

1) 64
2) 49
3) -49
4) -64

Answer: 2) 49


The last thing that happens when you do Order of Operations, after substituting, is SQUARING whatever is in the parentheses. That means that the answer cannot be negative. Eliminate Choice (3) and (4).

Also notice that b has an exponent of 0, so it doesn't matter what b is, b0 = 1.

So (-2)3 = -8, and -8 + 1 = -7. (-7)2 = 49. Which is Choice (2).




More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.



Friday, February 04, 2022

Not Teaching

(Click on the comic if you can't see the full image.)
(C)Copyright 2022, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

Warning: don't try this at school.

Don't try it with your spouse, either.

Take your kid and let them do it to your spouse. Hilarious. (Or so I'm told.)

For longtime readers, yes, this appears to be the sister of a student Mike had a few years ago.



I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.





Come back often for more funny math and geeky comics.



Wednesday, February 02, 2022

Two Two Two Two

(Click on the comic if you can't see the full image.)
(C)Copyright 2022, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

Two two two two!!

Does anything else need to be said?

I know I generally avoid "date-based" comics, but some you just can't pass up.

Hey, it beats a repetitive Groundhog's Day comic, right?



I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.





Come back often for more funny math and geeky comics.



Tuesday, February 01, 2022

Quack!

(Click on the comic if you can't see the full image.)
(C)Copyright 2022, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

We all need a little down time, right?

Speaking of which, I can't point to any specific reason why I haven't posted a comic in the past week, other than the usual "the idea I had would take too long so I never started it." Just the time of year, I guess. And a blizzard.

The image above started with a proof regarding a circle and two congruent right triangles (but you had to show they were right and congruent).

Happy Year of the Tiger! Mathematical as that is with its 2022 - 12N, we've seen how well I draw ragged tigers.



I also write Fiction!


You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Preorder the softcover or ebook at Amazon.

Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.





Come back often for more funny math and geeky comics.