## Monday, September 30, 2019

### Hearsay

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Don't confuse this with the rem(n) function!

I did just see the band last week, so, yes, I was thinking about math and comics during a concert.

And the "rem()" function, whatever that is, will have to wait for another day. Not that I'm likely to see them in concert any time soon.

### Update: Bonus Comic

Hearsay being in the news right now, and my seeing REO Speedwagon in concert last week, several ideas came to mind, and I had to decide which one to use.
Here is another take on the song, more topical, but less mathematical.

Come back often for more funny math and geeky comics. ## Friday, September 27, 2019

### Dividing 9's By 9

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And you can use x + 1 for 11's.

Instead of FOIL, I'm trying the "Box Method" again, only this time calling it The Area Model, and stressing the area of box, and the fact that it's a multiplication table. (I abandoned it years ago, because students would fill up tables and say, "I'm done!" without completing the problem. "That's not the answer. That's the work that gives you the answer.")

Using this in reverse, which in some cases is the same as Factoring By Grouping, the Area Model does division in a much nicer way than the traditional Long Division method, at least when there isn't a remainder. Actually, it's not that bad then, either.

Oh, and I have Snapped my fingers to distract from tapping the board to make the screen advance to the next slide. Sleight of hand is usually appreciated, even when they're on to you, if only because the other teachers aren't doing that.

Come back often for more funny math and geeky comics. ## Thursday, September 26, 2019

### Hands

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That's what stands out the most, right?

If anyone is confused, they traced their hands, and cut them out.
The original activity would have included writing on the fingers, but that just too small to see.

New students here because I didn't want to use Ms. Graham's other students who we've seen in the past.

Come back often for more funny math and geeky comics. ## Tuesday, September 24, 2019

### (x, why?) Mini: Heptagon

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Do sides matter? If we're talking a side of fries or a side of kale, then, yes, they certainly matter!

Don't downvote me!

Amusing aside: many of the houses I use for background are actually heptagons.

Come back often for more funny math and geeky comics. ## Friday, September 20, 2019

### School Life #10: Girlfriend

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Students change. Group dynamics change. The cast changes, too, but often changes right back again.

School Life is a collection of strips about the students in school (or possibly outside of it) that don't fall into the usual scholastic-based humor. With enough backstory, I could write my own fanfic based on these guys. Granted, I'd have to give them all names first. Added one today. Up until now, I had the four new girls as Filipino (and loosely based on groups of students I knew), but the more I looked at Kyung, the more I thought that maybe she could be Korean (possibly North?) who is befriended by the other three.

So welcome to (temporarily non-canonical, until I say so) Kyung Mi Park. The other three girls have Spanish surnames that I haven't decided on (but I'm leaning toward some former coworkers to whom I owe a great deal), with the tall one possibly named "Katie". The other two may or may not have somewhat "saintly" names, or American variations thereof.

In the meantime, I'm still deciding on the stories to tell, how to tell them, and how much of them to tell, because it really isn't the main focus of the comic. And as much as I deal with these youngsters transforming into adults (I can't always call them "young adults" just yet), their stories aren't mine to tell. To adapt, maybe, in a way to bring them out to my readers, but not to tell outright.

There could be another ten of these before the year is out, or there might not even be ten more in the future. Time will tell.

But Daisy won't. She's good with secrets.

Come back often for more funny math and geeky comics. ## Wednesday, September 18, 2019

### Pause

(Click on the comic if you can't see the full image.) It's a *pregnant* pause!

So this may go on a while, and we probably won't see the baby before next summer break ... unless I get desperate for a story line.

The bonus is that not only wouldn't a student teacher return again, unless they were in a time lock, but in the Ten Years comic, we learned that she was one of the comic's original students ... All Growed Up!

Come back often for more funny math and geeky comics. ## Monday, September 16, 2019

### Bigons

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Spherical lunes, to be exact.

Not the Cube. He's just a loon.

Come back often for more funny math and geeky comics. ## Friday, September 13, 2019

### Friday Full Moon 13th

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You know he'll be vamping about, life of the party of the undead.

Complete on 9/15 for 9/13. So much for punctuality.

Come back often for more funny math and geeky comics. ## Tuesday, September 10, 2019

### Palindrome Dates

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I shy away from a lot of obvious date jokes, but it's the last palindrome week of the century, so I don't think I'll be doing it again!

Come back often for more funny math and geeky comics. ## Friday, September 06, 2019

### Summer Expectations

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I wanted a beat panel, followed by uncontrollable laughter, and then sliding under, but it was too space. So I went with Ackbar.

This image was created on Saturday 9/7 for Friday's strip. It was late, and I've fallen behind a bit.
Hopefully, Monday's strip will appear on Monday.

Actually, this would have been a good, ahem, Labor Day strip, if only because three of them went back to work right afterward.

Come back often for more funny math and geeky comics. ## Wednesday, September 04, 2019

### Summer Travel

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I wanted to say it was a "Thai" for the best summer story, but I didn't have any others to run with.

This image was created on Friday 9/6 for Wednesday's strip. It was late, and I've fallen behind a bit.
Hopefully, Friday's strip will be created on Friday as well.

Come back often for more funny math and geeky comics. ## Tuesday, September 03, 2019

### August 2019 Geometry Regents Part 1 (multiple choice)

The following are some of the multiple questions from the recent June 2019 New York State Common Core Geometry Regents exam.

### June 2019 Geometry, Part I

Each correct answer is worth up to 2 credits. No partial credit. Work need not be shown.

1. On the set of axes below, AB is dilated by a scale factor of 5/2 centered at point P.

Which statement is always true?

Shape (and therefore slope) is preserved in a dilation, so the line segments are parallel.
Choice (4) is backward. If it had been 5/2(AB) = A'B', that would have been correct.

2. The coordinates of the vertices of parallelogram CDEH are C(-5,5), D(2,5), E(-1,-1), and H(-8,-1). What are the coordinates of P, the point of intersection of diagonals CE and DH?

In a parallelogram, the diagonals bisect each other, so the point of intersection must be the midpoint of both lines.
The midpoint of CE is at ( (-5+-1)/2, (5+-1)/2 ), which is (-3, 2).
Checking the other line, just to be sure:
The midpoint of DH is at ( (2+-8)/2, (5+-1)/2 ), which is (-3, 2).

3. The coordinates of the endpoints of QS are Q(-9,8) and S(9,-4). Point R is on QS such that QR:RS is in the ratio of 1:2. What are the coordinates of point R?

A ratio of 1:2 means one-third of the way along he line.
The difference in the x-coordinates is 9 - (-9) = 18. One-third of 18 is 6, and -9 + 6 = -3.
The eliminates all but one choice.
Check the y-coordinates to be sure:
The difference in the y-coordinates is -4 - 8 = -12. One-third of -12 is -4, and 8 + -4 = 4.

4. If the altitudes of a triangle meet at one of the triangle’s vertices, then the triangle is

There's no way for this to happen unless one of the angles is a right angle, which makes it a right triangle.

5. In the diagram below of triangle ACD, DB is a median to AC, and AB = DB
If m∠DAB = 32°, what is m∠BDC?

If m∠DAB = 32°, then m∠ADB = 32° because it is an isosceles triangle.
This makes m∠DBC = 64° by the Remote Angle Theorem.
Since B is a median, AB = BC, but since AB = DB, then BC = DB, making DBC an isosceles triangle.
This makes BDC = BCD, and the pair of base angles have a sum equal to 180 - 64 = 116.
Half of 116 is 58, so m∠BDC = 58°

6. What are the coordinates of the center and the length of the radius of the circle whose equation is x2 + y2 = 8x - 6y + 39?

To put is into standard form (x - h)2 + (y - k)2 = r2, you need to move the x and y terms to the left side and then Complete the Square, twice.
Remember, you need to take half of the coefficient of x (or y) and then square it to find out what to add.

x2 + y2 = 8x - 6y + 39
x2 - 8x + y2 + 6y = 39

7. The line 3x 4y 8 is transformed by a dilation centered at the origin. Which linear equation could represent its image?

Answer: (2) y = 3/4 x + 8
A line that has been dilated would keep its slope and be parallel to the original (unless it was the same line).
The slope of the original line can be found by isolating y.
-3x + 4y = 8
4y = 3x + 8
y = 3/4 x + 8

The slope must be 3/4, so the choice is (2).

Also, when written in Standard Form Ax + By = C, the slope of the line is -A/B, which in this case is -3/-4 = 3/4.

8. In the diagram below, AC and BD intersect at E.
Which information is always sufficient to prove triangle ABE = triangle CDE?

Answer: (4) BD and AC bisect each other.
Bisecting each other means that the two triangles will be congruent because of SAS, usually the vertical angles between the sides.
Choice (1) is enough to prove that they are similar by AAA because of Alternate Interior Angles, but not that they are congruent.
Choice (2) gives you SSA, which isn't a postulate or theorem. (And don't read it backward!)
Choice (3) isn't enough information. You'll have one side and one angle.

9. The expression sin 57° is equal to

The sine of an angle (n) is equal to the cosine of the complementary angle (90 - n).

10. What is the volume of a hemisphere that has a diameter of 12.6 cm, to the nearest tenth of a cubic centimeter?

Volume of a sphere is V = (4/3) pi * r3
A hemisphere is half of that. The radius is half of the 12.6 diameter, or 6.3
So V = (1/2)(4/3)(3.141592)(6.3)3 = 523.6971...

A quick guess of the wrong answers would be (2) forget the 1/2, (3) took 1/2 the volume but used the diameter, (4) made both mistakes.

11. In the diagram below of triangle ABC, D is a point on BA, E is a point on BC, and DE is drawn.
If BD = 5, DA = 12, and BE = 7, what is the length of BC so that AC || DE?

For the lines to be parallel, the following proportion needs to be true. Notice that it says EC and NOT BC.
BD / DA = BE / EC
5 / 12 = 7 / EC
5 EC = 84
EC = 16.8
BC = BE + EC = 7 + 16.8 = 23.8

You could have done BD/BA = BE/BC if you preferred.

12. A quadrilateral must be a parallelogram if

Answer: (3) one pair of sides is both parallel and congruent
If the lines are parallel and congruent, then the quadrilateral must be a parallelogram.
Choices (1), (2) and (4) could be an isosceles trapezoid.

13. In the diagram below of circle O, chords JT and ER intersect at M.

If EM = 8 and RM = 15, the lengths of JM and TM could be

If two chords of a circle intersect, then the products of their segments will be equal.
So (JM)(MT) = (RM)(ME) = (15)(8) = 120
Only Choice (3) has two factors with a product of 120.

14. Triangles JOE and SAM are drawn such that ∠E = ∠M and EJ = MS. Which mapping would not always lead to triangle JOE = triangle SAM?

Answer: (4) JO maps onto SA

Choice (1) gives you ASA. Choice (2) gives you AAS. Choice (3) gives you SAS. Choice (4) gives you SSA, which is not a theorem nor postulate (as stated in an earlier question).

15. 5 In triangle ABC shown below, ∠ACB is a right angle, E is a point on AC, and ED is drawn perpendicular to hypotenuse AB.
If AB = 9, BC = 6, and DE = 4, what is the length of AE?

The triangles are similar because they each have a right angle and they share angle A. That means that the corresponding sides are proportional. However, be careful not to mix up the sides because of the way it is drawn. AE is the hypotenuse of the smaller triangle.

AB / BC = AE / DE
9 / 6 = AE / 4
6 AE = (9)(4) = 36
AE = 6

16. Which equation represents a line parallel to the line whose equation is -2x + 3y = -4 and passes through the point (1,3)?

Answer: y - 3 = (2/3)(x - 1)

The choices are all in Point-slope form, which means y - 3 = m(x - 1), where m is the same slope as the original equation.
Immediately, cross off choices (3) and (4).
Spoiler alert: the slope is going to be positive, so (1) is wrong, too, but let's continue.

-2x + 3y = -4
3y = 2x - 4
y = 2/3x - 4, slope = 2/3.

As stated in an earlier question, when the equation is in Standard Form Ax + By = C, the slope is -A/B.
So the slope was -(-2)/3 = 2/3.

17. In rhombus TIGE, diagonals TG and IE intersect at R. The perimeter of TIGE is 68, and TG = 16.

What is the length of diagonal IE?

They chose the numbers carefully. There are four congruent right triangles making up that rhombus.
The perimeter is 68, meaning that each side is 17, which is the hypotenuse of the triangles. TG = 16, but since TG and IE bisect each other, TR and RG are each 8. This is one leg of the right triangles.
So 82 + ER2 = 172
If you've followed my advice before this, you've learned from Pythagorean Triples and know that these are 8-15-17 triangles.
If you didn't ... sigh, okay, let's do it:

82 + ER2 = 172
64 + ER2 = 289
ER2 = 225
ER = 15

ER = 15 = IR, so IE = 30.

18. In circle O two secants, ABP and CDP, are drawn to external point P. If mAC = 72°, and mBD = 34°, what is the measure of ∠P?

Angle P will be one-half of the difference between the two arcs.
So (72° - 34°) / 2 = 19°.

19. What are the coordinates of point C on the directed segment from A(-8,4) to B(10,-2) that partitions the segment such that AC:CB is 2:1?

To get a 2:1 ratio, C has to be 2/3 of the way along the line from A to B.
To get from -8 to 10, you have to move 18 units. Multiply 2/3 * 18 = 12. Add 12 to -8 to get 4.
To get from 4 to -2, you have to move 6 units. Multiply 2/3 * 6 = 4. Subtract 4 from 4 to get 0.
Subtract in the second case because you are moving down.

20. The equation of a circle is x2 + 8x + y2 - 12y = 144. What are the coordinates of the center and the length of the radius of the circle?

You need to Complete the Square. (Yes, that's back from Algebra).
Half of 8 is 4, and 42 is 16
Half of -12 is -6, and (-6)2 is 36.
x2 + 8x + y2 - 12y = 144
x2 + 8x + 16 + y2 - 12y + 36 = 144 + 16 + 36
(x + 4)2 + (y - 6)2 = 196
The center is (-4, 6) and the radius is SQRT(196) = 14.

21. In parallelogram PQRS, QP is extended to point T and ST is drawn.

If ST = SP and m∠R = 130°, what is m∠PST?

There is a parallelogram and an isosceles triangle.
Angle QPS = angle R = 130 degrees. Angle SPT is supplementary to QPS, so it is 50 degrees. Since PST is an isosceles triangle, the base angles are equal, so angle T is also 50 degrees. The vertex angle PST is 180 - 50 - 50 = 80 degrees.

22. A 12-foot ladder leans against a building and reaches a window 10 feet above ground. What is the measure of the angle, to the nearest degree, that the ladder forms with the ground?

Before starting, realize that the angle with the ground must be bigger than the angle with the building, which means it has to be over 45 degrees. So you can eliminate choices 1 and 2.
The wall is opposite the angle, and the ladder is the hypotenuse, so you need to use Sine to find the angle.
Sin x = opp/hyp
Sin x = (10/12)
x = sin-1(10/12) = 56.44...

If you got a ridiculously low decimal, then your calculator is in Radians mode.

23. In the diagram of equilateral triangle ABC shown below, E and F are the midpoints of AC and BC, respectively.

If EF = 2x + 8 and AB = 7x - 2, what is the perimeter of trapezoid ABFE?

Since EF is a midsegment, then it is half the length of AB.
Since ABC is an equilateral triangle, EA and FB are half the length of AB.

So first we need to find x and then EF.

2(2x + 8) = 7x - 2
4x + 16 = 7x - 2
16 = 3x - 2
18 = 3x
x = 6

Since x = 6, then EF = 2(6) + 8 = 12 + 8 = 20.
The perimeter is equal to FIVE TIMES the length of EF, so 20 * 5 = 100.

Where did I get FIVE TIMES from? EF = EA = FB = 1/2(AB), so AB = 2 EF.

24. Which information is not sufficient to prove that a parallelogram is a square?

Answer: (3) The diagonals are perpendicular and one pair of adjacent sides are congruent.
Choice (3) describes a rhombus which does not have to be a square. Each of the other choices have one condition that makes it a rhombus and another that makes it a rectangle. If it is both a rhombus and a rectangle, it must be a square.

End of Part I

How did you do?