(C)Copyright 2019, C. Burke.
I'm happy to see that they retained the knowledge and applied it elsewhere (factorial)
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I'm happy to see that they retained the knowledge and applied it elsewhere (factorial)
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He needs to slink back to his con cave!
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It would serve him right, if he could serve at all.
Depends on whether these are clubs playing within the school or teams playing within the city under league rules.
Two more student names were to be revealed here -- and then one was cut for space. Off-screen, I've almost decided on two other names.
Formats are driving me crazy. I've pretty much given up on making the School Life strips black & white because it was actually more trouble than just using color. (Clicking the B&W option from the menu leaves everyone looking too dark and drab.) But the biggest problem here was writing and rewriting the text to say everything I wanted to say, but making it sparse enough to fit in the allotted space -- and then realizing that I was working with a differently-sized comic and not using the regular School Life frame.
O, the joys of comic creation! Some times I think it's growing pains, but after nearly 12 years, I should be done with those!
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That Terrible Toluene is a polycyclic offender.
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Claims is the previous ad may be conjecture and may not have been proven.
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I guess they need their base first. Whose base is first?
If you're a Daisy fan, you'll notice there were two things out of character in this strip.
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"Why does that say x-squared and not x-cubed?" "I meant to do that to see if you were paying attention. You are. Very good."
Of course, I had to have a mistake in a comic about making mistakes. Other than planning on spreading out the dialogue before realizing that I just didn't have the room.
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Floppy eyes.
They may be out of date, but the Save icon hasn't changed yet.
I thing that it funny is how I do these "minis" to save time when I don't have enough for a full comic -- and yet I spend so much time making these work.
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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.
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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.
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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.
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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.
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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.
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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!
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Spherical lunes, to be exact.
Not the Cube. He's just a loon.
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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.
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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!
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Jump to aqua-space on my mark.
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.
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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.
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The following are some of the multiple questions from the recent June 2019 New York State Common Core Geometry Regents exam.
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.
Answer: (2) AB | A'B'
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?
Answer: (3) ( 3,2)
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?
Answer: (3) (-3,4)
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
Answer: (1) a right triangle
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?
Answer: (3) 58°
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 x^{2} + y^{2} = 8x - 6y + 39?
Answer: (2) 128
To put is into standard form (x - h)^{2} + (y - k)^{2} = r^{2}, 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.
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
Answer: (2) cos 33°
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?
Answer: (1) 523.7
Volume of a sphere is V = (4/3) pi * r^{3}
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?
Answer: (1) 23.8
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.
Answer: (3) 16 and 7.5
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?
Answer: (2) 6
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.
Answer: (2) 30
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 8^{2} + ER^{2} = 17^{2}
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:
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?
Answer: (1) 19°
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?
Answer: (4) (4,0)
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 x^{2} + 8x + y^{2} - 12y = 144. What are the coordinates of the center and the length of the radius of the circle?
Answer: (4) center (-4,6) and radius 14
You need to Complete the Square. (Yes, that's back from Algebra).
Half of 8 is 4, and 4^{2} is 16
Half of -12 is -6, and (-6)^{2} is 36.
x^{2} + 8x + y^{2} - 12y = 144
x^{2} + 8x + 16 + y^{2} - 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.
Answer: (2) 80°
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?
Answer: (4) 56
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.
Answer: (3) 100
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.
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
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The following are some of the multiple questions from the August 2019 New York State Geometry Regents exam.
Each correct answer is worth up to 4 credits. Partial credit is available. Work must be shown. Correct answers without work receive only 1 point.
32. Triangle ABC is shown below. Using a compass and straightedge, construct the dilation of Triangle ABC centered at B with a scale factor of 2. [Leave all construction marks.]
Is the image of triangle ABC similar to the original triangle? Explain why.
Answer:
Starting at point B, open the compass to Point A. Then from point A, make an arc. Use the straightedge to extend BA to the new arc. This line will have twice the length of BA.
Repeat the process for BC.
Finally, using the straightedge, complete the triangle from the endpoints of the two new line segments you constructed.
The image is similar to the original triangle because a dilation preserves the shape of the object. Therefore, the angles are the same size and by the AA Theorem, the two triangles are similar.
33. In the diagram below, triangle ABE = triangle CBD.
Prove: triangle AFD = triangle CFE
Answer:
You need to prove that those two tiny triangles are similar.
You can see that Angles A and C are equal from the original triangles. And angles AFD and CFE are vertical angles. You need to have one pair of congruent sides.
You don't know that AD = CE directly, but you do know that AB = CB and DB = EB because of the congruency of the larger triangles. Therefore if you subtract the smaller segment from the larger one, you get the sides that we need to be congruent.
Your proof would go something like this:
Statement | Reason |
1. Triangle ABE = Triangle CBD | 1. Given |
2. Angle A = Angle C | 2. CPCTC |
3. Angle AFD = Angle CFE | 3. Vertical angles are congruent. |
4. AB = CB; DB = EB | 4. CPCTC |
5. AD = CE | 5. Subtraction Postulate |
6. triangle AFD = triangle CFE/td> | 6. AAS |
34. A cargo trailer, pictured below, can be modeled by a rectangular prism and a triangular prism. Inside the trailer, the rectangular prism measures 6 feet wide and 10 feet long. The walls that form the triangular prism each measure 4 feet wide inside the trailer. The diagram below is of the floor, showing the inside measurements of the trailer.
Answer:
Volume = Area of the Base times the Height.
The base is shaped like a pentagon, which, in this case, can be broken into a rectangle and a triangle. We have the base of the triangle, 6 feet; however, we need to solve for its height (lower case h). We can use the Pythagorean Theorem for this.
3^{2} + x^{2} = 4^{2}
9 + x^{2} = 16
x^{2} = 7
x = SQRT(7) = 2.65 (approximately)
Area of the base = (6)(10) + (1/2)(6)(2.65) = 67.95
Volume = Area * height = 67.95 * 6.5 = 441.675 = 442 cubic feet
A correct answer is worth up to 6 credits. Partial credit is available.
35. The coordinates of the vertices of Triangle ABC are A(1,2), B(-5,3), and C(-6,-3).
Prove that Triangle ABC is isosceles
[The use of the set of axes on the next page is optional.]
State the coordinates of point D such that quadrilateral ABCD is a square.
Prove that your quadrilateral ABCD is a square.
[The use of the set of axes below is optional.]
Answer:
Two separate questions.
First, to prove that ABC is isosceles, you will have to show that two of the sides have the same length.
It can be assumed from the second part of the question even before we start that it will be a right triangle, because that is the only way that you will be able to add one more point and have a square. Furthermore, AB and BC would have to be the congruent sides.
The second and third parts are really the same question. Once you have the square, you have to prove that it is a square. This is more rigorous than just an "explanation" or "justification". However, by this point you have already shown that two consecutive sides are congruent, so all you have to do is show that the sides are parallel, making it both a parallelogram and a rhombus, and that there is at least one right angle, which makes it a square.
First Part:
Length of AB = SQRT( (-5 - 1)^{2} + (3 - 2)^{2} ) = SQRT(37)
Length of BC = SQRT( (-6 - -5)^{2} + (-3 - 3)^{2} ) = SQRT(37)
AB = BC, so triangle ABC is isosceles.
(No reason to check the length of AC.
Second Part:
To get from point A to B, you have to go down 6 left and up 1. (This can be seen if you graph it, or if you worked out the math above.) CD would have to have the same slope, but to get from C to D, you would have to go 6 right and down 1. So D would have to be (-6 + 6, -3 - 1) or D(0, -4).
Third part:
The slope of AB = -1/6. The slope of CD = -1/6
The slope of BC = (-3 -3)/(-6 - -5) = -6/-1 = 6
The slope of DA = (-4 - 2)/(0 - 1) = -6 / -1 = 6
AB || CD, AD || BC because they have the same slopes.
AB is perpendicular to BC because their slopes are inverse reciprocals (have a product of -1).
So Angle B is a right angle.
A quadrilateral with opposite sides parallel, with consecutive sides equal and a right angle must be a square.
End of Exam
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Apparently, many (x, why?) songs are sung by shapes or symbols. And most of my Zero characters were *not* singers.
I thought it was going to be a few zeroes and tens when I started going through the old comics. And I never did, say, Boy Four-ge, even though a song was used for a truth table.
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