You gotta dew what you gotta dew.
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You gotta dew what you gotta dew.
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Sometimes a good pun is just lost on them. And this math man would prefer to go a-whiskey.
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Trapezoid! I choose you!
Okay, let's get the math out of the way first.
Polygons: A closed shape, consisting of straight lines that don't overlap each other. They can be concave or convex. Regular polygons have all sides congruent, and all angles congruent. Thus, regular polygons must be convex, with interior angles less than 180 degrees.
You might want to avoid this Definition: Polygon. Or you might want to read the comments for your own amusement.
As mentioned in the comic above, if you draw all the diagonals from one point, you divide the polygon into triangles, each with 180 degrees, and the sum of those is the sum of the interior angles. There will always be two fewer triangles than there are sides, which is the source of the n less two line in the song. Note that a triangle has no diagonals and is composed of a single triangle of 180 degrees, the trivial case for this rule.
About the comic:
In case anyone reading this is new to this blog or my webcomic (did you come from a #PokemonGO or #TMC16 hashtag?), I did want to explain a little about the characters above. They are NOT supposed to be the characters from the show. They are semi-occurring characters made up to imitate the characters in the show ... a little ... sort of. Except for Melissa (aka Missy), she's new. I won't comment on Mr. Wayne's anime-styled hairdo.
The two in the bottom panel, as regular readers know, are math teachers who regularly appear in the strip, but it's summer break, so they're in the yard, having just been in the pool.
For the long-time readers who think they missed something: Judy is the other English teacher who is sometimes seen with Michele. Her boyfriend is Chuck, who appeared a few times in the early years. He works in an office with Frank, aka "Uncle Frank", who has actually appeared more often -- generally, any time I need someone old and tired to express something. Frank, a leap baby, was introduced as being 40, but he's apparently a bit older.
One of these days, I have to do a character map, or at least update a wiki page somewhere.
As for the Polygons in the picture: only two are supposed to be something. The beaked one on the bottom came from comic #957.
I wanted to include a Pentagon with an army theme, but I've done that before. (In actuality, it just wasn't going to fit. You understand, right?)
Finally, I was going to have one of the kids catch a Polygon in a Polyball, and show the figure (the above-mentioned Pentagon) inscribed inside the circle. But I couldn't fit "inscribe" or "circumscribe" in the song! There aren't a lot of good rhymes.
While I'm at it, there aren't a whole lot of useful rhymes for shape either!
Forgive the long missive. I'm not usually this wordy. If you're new here, click on the "Comic" tag to see more. Or just explore. Thanks for stopping by.
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Even ''What's the 411?'' is outdated now. Anyone still use it?
This is a leftover joke that I didn't work into the original sequence of strips that started with Six Quits, or The Special Calculator, if you want to know why he quit.
Back then, I didn't run strips consecutively, so there will be some extra strips in between the episodes in the series. Enjoy.
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33. Given: Quadrilateral ABCD is a parallelogram with diagonals AC and BD intersecting at E.
Prove: Triangle AED = Triangle CEB
You can write a two-column proof or a paragraph proof. I am going to use a paragraph proof to show that they are congruent using SSS. That means that I need to show three sets of sides are congruent. You can also show this using ASA or SAS, but remember to show all three pairs. Don't leave any out.
ABCD is given as a parallelogram. AD is congruent to BC because the opposite sides of a parallelogram are congruent. AC and BD are given as the diagonals of the parallelogram. AC and BD bisect each other because the diagonals of a parallelogram bisect each other. This make AE congruent to CE and BE congruent to DE because of the definition of bisect. Therefore, triangle AED is congruent to triangle CEB because of SSS.
If you wanted to use ASA, you could do so by pointing out that AC and BD are transversals across parallel lines. After that, use Alternate Interior Angles to show EAD = ECB and EDA = EBC.
For SAS, you need a combination of the steps above.
Also, you know that there are vertical angles at point E which are congruent.
Many approaches to a correct answer.
Maybe not so much as a conversation as a response to a mini-rant.
I made a note at that point to turn this into a comic. I recently just found that note. Thankfully, I was able to find the original tweets in question.
Many teachers of younger grades are frustrated with long division. It's not easy to teach, and it seems to be an outdated skill since we have calculators. (That said, I've used that skill in Algebra 2 classes with dividing polynomials, which has become more difficult because many of them haven't done long division by hand in five or more years!)
This particular rant (meant to be humorous, from the start) came from Doug Robertson, author of He's the Weird Teacher and The Teaching Text (You're Welcome)
The response was purely me.
For the doubters, here are the tweets in question.
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Shut up and Math with me.
This was the alternative to a similar comic I did last summer. I did the other because it was different. To be honest, I wasn't entirely happy with that final comic, which is probably why this one stayed in my head ever since then.
The only reasons that I didn't do it sooner were because I needed to make new art (Yes, it's similar to This Is Crazy) that fit the song, and had a couple of blank lines. I've reworked these lyrics many times over the past year and yet ... I never wrote the words down!
If I did, it's in a notebook, but it's not on my hard drive or flash drive. Honestly, I thought I'd typed up at least one version so I wouldn't forget it. Also, having it electronically makes it easier to create the text on the comic.
In any case, it's finally done and posted. Hopefully, someone enjoys it enough to comment here or somewhere.
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There are more elements to T, but the flag is only so big.
The Cardinality of R is at least 3, according to the text "among these are".
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What follows is a portion of the Common Core Geometry exam. Illustrations will be added at a later time when they become available.
Part II is posted here.
Part III and IV are posted here.
1. A student has a rectangular postcard that he folds in half lengthwise. Next, he rotates it continuously about the folded edge. Which three-dimensional object below is generated by this rotation?
(3). The cylinder. A rectangular postcard folded over is still a rectangle. Roll it along the edge and it becomes a cylinder. (The label of a soup can is a rectangle.)
2. A three-inch line segment is dilated by a scale factor of 6 and centered at its midpoint. What is the length of its image?
(4) 18 inches. If it is centered in the middle, call the left endpoint -1.5 and the right endpoint +1.5. Multiply those by 6 and you get -9 and +9. The distance between those two points is 18.
3. Kevin's work for deriving the equation of a circle is shown below.
(2) Step 4. +4 was added to the left, but -4 was added to the right.
4. Which transformation of OA would result in an image parallel to OA?
(1) a translation of two units down. Translations do not affect orientation. The lines will be parallel.
5. Using the information given below, which set of triangles can not be proven similar?
(3). One triangle has sides 2 and 8, the other has 16 and 32. They are not proportional.
Choices (1) and (4) are similar because of SAS. Choice (2) has similarity due to AA.
6. A company is creating an object from a wooden cube with an edge length of 8.5 cm. A right circular cone with a diameter of 8 cm and an altitude of 8 cm will be cut out of the cube. Which expression represents the volume of the remaining wood?
(4) (8.5)^{3} - (1/3) pi(4)^{2}(8). Volume of a cube is s^{3}. Volume of a cone is (1/3)pi(r)^{2}(h). The diameter is 8, so the radius is 4.
7. Two right triangles must be congruent if
(3) the corresponding legs are congruent. If the legs are congruent with a right angle between them (all right angles are congruent), they are congruent due to SAS. Likewise, if the legs are congruent, then the hypotenuses are congruent, so the triangles are congruent by SSS.
8. Which sequence of transformations will map Triangle ABC onto Triangle A'B'C'?
(4) dilation and rotation. It got bigger, so there's a dilation. Its orientation has changed because of a rotation.
9. In parallelogram ABCD, diagonals AC and BD intersect at E. Which statement does not prove parallelogram ABCD is a rhombus?
(1) AC = DB. Congruent diagonals prove that a parallelogram is a rectangle, not that it is a rhombus.
10. In the diagram below of circle 0, OB and OC are radii, and chords AB, BC, and AC are drawn.
(2) m<BAC = (1/2)m<BOC. The size of the inscribed angle is half of the central angle. There is nothing that says that AB and AC are congruent, so BAC does not have to be isosceles, even if it looks like it in the image.
11. A 20-foot support post leans against a wall, making a 70° angle with the ground. To the nearest tenth of a foot, how far up the wall will the support post reach?
(4) 18.8. The hypotenuse is 20. The opposite is x. The angle is 70°. Opposite and hypotenuse means use sine.
sin(70) = x / 20, so x = 20 * sin(70) = 18.79... = 18.8
12. Line segment NY has endpoints N(-11,5) and Y(5, -7). What is the equation of the perpendicular bisector of NY?
(1) y + 1 = (4/3)(x + 3). The slope of the line NV is (-7 - 5) / (5 + 11) = -3/4. That means that a line perpendicular to this one has a slope of 4/3. This eliminates choices (2) and (4).
The midpoint is ( (-11+5)/2, (5-7)/2 ) = (-3, -1). Flip the signs to positive, and you get choice (1).
13. In Triangle RST shown below, altitude SU is drawn to RT at U.
If SU = h, UT = 12, and RT = 42, which value of h will make Triangle RST a right triangle with LRST as a right angle?
(2) 6*SQRT(10). (6 times radical 10).
Notice that they give RT and not RU. RU = 42 - 12 = 30.
h^{2 = (30)(12) = 360
h = Sqrt(360) = Sqrt(36)*Sqrt(10) = 6*Sqrt(10).
}
14. In the diagram below, Triangle ABC has vertices A(4,5), B(2,l), and C(7,3).
What is the slope of the altitude drawn from A to BC?(4) -5/2. The altitude from A to BC has a slope that is the inverse reciprocal of the slope of BC, because the altitude is perpendicular to BC. The slope of BC is 2/5 (up 2 boxes and right 5). The inverse reciprocal is -5/2.
Note that (-5/2)(2/5) = -1, which is true for inverse reciprocals.
15. In the diagram below, Triangle ERM ~ Triangle JTM.
Which statement is always true?(4) tan E = TM/JM. Angle E corresponds to angle J. Tan J = opposite / adjacent = TM / JM.
16. On the set of axes below, rectangle ABCD can be proven congruent to rectangle KLMN using which transformation?
(3) reflection over the x-axis.
17. In the diagram below, DB and AF intersect at point C, and AD and FBE are drawn. If AC= 6, DC= 4, FC = 15, m<D = 65°, and m<CBE = ll5°, what is the length of CB?
(1) 10. If CDE = 115, then CDF = 65°, so the triangles are similar. (The vertical angles are congruent also.) This makes the corresponding sides proportional.
4 / 6 = x / 15
6x = (4)(15) = 60
x = 10
18. Seawater contains approximately 1.2 ounces of salt per liter on average. How many gallons of seawater, to the nearest tenth of a gallon, would contain 1 pound of salt?
(2) 3.5. Conversion: 1 pound = 16 ounces. 1 liter = 0.264 gallon. (This comes from the reference table in the back of the book.) Compare ounces to gallons with ounces to gallons.
1.2 / .264 = 16 / x
1.2x = (.264)(16) = 4.224
x = 3.52 = 3.5
19. Line segment EA is the perpendicular bisector of ZT, and ZE and TE are drawn.
Which conclusion can not be proven?(2) Triangle EZT is equilateral. EZT is isosceles, but not necessarily equilateral.
20. A hemispherical water tank has an inside diameter of 10 feet. If water has a density of 62.4 pounds per cubic foot, what is the weight of the water in a full tank, to the nearest pound?
(1) 16,336/ Density equals mass divided by Volume, in this case pounds per cubic foot. So mass(weight) = Density * Volume.
Volume = (1/2) * (4/3) * pi * r^{3} = (4 / 6) * pi * 5^{3} = 261.799...
Mass = 62.4 * 261.799 = 16336.2576 = 16,336 pounds.
21. In the diagram of Triangle ABC, points D and E are on AB and CB, respectively, such that AC || DE.
If AD = 24, DB = 12, and DE = 4, what is the length of AC ?(2) 12. The triangles are similar, so DB/DE = AB/AC. AB = AD + DB = 24 + 12 = 36.
36 / x = 12 / 4
12x = 144
x = 12
22. Triangle RST is graphed on the set of axes below.
(3) 45. The area of a triangles is 1/2*b*h. The base and height are perpendicular to each other. It is easy to see that angle S is a right angle by comparing the slopes. SR has a slope of 2 (you can see this by counting boxes). ST has a slope of -1/2. Multiply 2*(-1/2) = -1, so the lines are perpendicular, and RST is a right triangle.
The length of RS = SQRT(3^{2} + 6^{2) and the length of ST is SQRT(122 + 62). Multiply 1/2 * SQRT(9 + 36) * Sqrt(144 + 36) = 45. (If you got 90, you forgot the 1/2.) }
23. The graph below shows AB, which is a chord of circle 0. The coordinates of the endpoints of AB are A(3,3) and B(3,-7). The distance from the midpoint of AB to the center of circle 0 is 2 units.
What could be a correct equation for circle O?(1) (x - 1)^{2} + (y + 2)^{2} = 29. The center of the circle must be at (1, -2) or (5, -2). The signs must be flipped in the equation of the circle, so any choice with +1, +5 or -2 can be eliminated. Choices (2) and (3) are out. The diameter has to be longer than the chord, so the radius has to be longer than 5.
24. What is the area of a sector of a circle with a radius of 8 inches and formed by a central angle that measures 60°?
(3) 32 * pi / 3. A = (60 / 360) * pi * (8)^{2}2 = 64 * pi / 6 = 32 * pi / 3
End of Part I
What follows is a portion of the Common Core Geometry exam. Other parts will be posted on other days. Illustrations will be added at a later time when they become available.
Part II is posted here.
32. A barrel of fuel oil is a right circular cylinder where the inside measurements of the barrel are a diameter of 22.5 inches and a height of 33.5 inches. There are 231 cubic inches in a liquid gallon. Determine and state, to the nearest tenth, the gallons of fuel that are in a barrel of fuel oil.
The Volume of a cylinder is pi*r^{2}*h. So V = pi(11.25)^{2}(33.5) = 13319.9
To convert cubic inches into gallons, divide by 231: 13319.9 / 231 = 57.66...
Answer: 57.7 gallons.
33. Given: Parallelogram ABCD, EFG, and diagonal DFB
You could write a two-column proof or a paragraph proof. For this blog, paragraph is a little easier.
Angle DFE is congruent to Angle BFG because they are vertical angles. AD is parallel to BC because opposite sides of a parallelogram are parallel. BD is a transversal. Angle ADB is congruent to CBD because they are alternate interior angles. Therefore Triangle DEF ~ triangle BGF because of AA (Angle-Angle Theorem).
34. In the diagram below, Triangle A'B'C' is the image of Triangle ABC after a transformation.
The transformation was a Dilation of scale factor 2.5 centered on (0, 0). Point A(-2, 4) -> A'(-5, 10). Point B(-2, -4) -> B'(-5, -10). Point C(4, -4) -> C'(-10,-10).
-5/-2 = 2.5. 10/4 = 2.5. -10/-4 = 2.5. 10/4 = 2.5
Dilations preserve shape so the angles are the same size. Therefor the triangles are similar.
35. Given: Quadrilateral ABCD with diagonals AC and BD that bisect each other, and < = <2
Make sure you restate the Given information. Make sure that you restate what you want to prove as your final statement. (In this case, you are proving two things, so one of them will be in the middle, right before you start proving the second half.) Do NOT use what you are trying to prove and a reason why something must be true.
Statement | Reason |
1. AC and BD bisect each other. | Given |
2. <1 = <2 | Given |
3. AB || CD | Opposite sides of a parallelogram are parallel. |
4. Angle DCA = Angle 1 | Alternate interior angles |
5. Angle DCA = Angle 2 | Transitive Property of Congruence |
6. Triangle ACD is isosceles | If the base angles of a triangle are congruent, then the triangle is Isosceles |
7. AD = CD | The sides opposite congruent base angles of an isosceles triangle are congruent |
8. ABCD is a rhombus | A parallelogram with consecutive sides congruent is a rhombus |
9. Angle AEB is a right angle. | Diagonals of a rhombus are perpendicular. |
10. Triangle AEB is a right triangle. | A triangle with a right angles is a right triangle. |
36. A water glass can be modeled by a truncated right cone (a cone which is cut parallel to its base) as shown below.
To find the height, you need to have similar triangles. To have similar triangles, the bases must be parallel so that the base angles are congruent.
What follows is a portion of the Common Core Geometry exam. Other parts will be posted on other days. Illustrations will be added at a later time when they become available.
25. Describe a sequence of transformations that will map triangle ABC onto triangle DEF as shown below.
There are multiple possible answers. One possibility is a reflection over the x-axis, followed by a translation 4 units to the right. (The reverse order would work, too.)
26. Point P is on segment AB such that AP:PB is 4:5. If A has coordinates (4,2), and B has coordinates (22,2), determine and state the coordinates of P.
The distance between the two x-coordinates is 18 units. If AP:PB is 4:5, then the two segments are 4x and 5x in length, and 4x + 5x = 18. So 9x = 18, and x = 2.
4(2) = 8, 4 + 8 = 12. P is at (12, 2).
27. In triangle CED as shown below, points A and B are located on sides CE and ED, respectively. Line segment AB is drawn such that AE = 3.75, AC = 5, EB = 4.5, and BD = 6.
Explain why AB is parallel to CD.
As shown in the illustration below, if you can show that the sides of the smaller triangle and the larger triangle are proportional, then the triangles are similar. If they are similar, then the corresponding angles of the two triangles are congruent.
Angle EAB is congruent to angle ECD and they are corresponding angles on a transversal. Therefore, AB is parallel to CD.
28. Find the value of R that will make the equation sin 73° = cos R true when 0° < R < 90°.
Explain your answer.
The sine of an angle is equal to the cosine of the complementary angle. 90 - 73 = 27.
R = 27.
29. In the diagram below, Circle 1 has radius 4, while Circle 2 has radius 6.5. Angle A intercepts an arc of length pi, and angle B intercepts an arc of length 13*pi / 8.
The measure of angle A is arclength / radius, which is pi / 4 radians.
The measure of angle B is 13(pi)/8 / 6.5, which is (13 pi) / (8*6.5) = (13 pi) / 52 = pi / 4.
Angle A and B have the same measure.
Another approach, which is a little more old-school, is to use circumference to find arclength.
The circumference of a circle is pi*d or 2*pi*r.
The length of an arc of a circle is the circumference times the size of the angle/360.
In the first circle, (A / 360) (2) (4) (pi) = pi
So (A / 360) (2) (4) = 1
Use inverse operations to isolate A, and A = 45.
In the second circle, (B / 360) (2) (6.5) (pi) = 13(pi) / 8
So (B / 360) (2) (6.5) = 13 / 8
Use inverse operations to isolate B, and B = 45.
So A and B have the same measure and that is 45.
30. A ladder leans against a building. The top of the ladder touches the building 10 feet above the ground. The foot of the ladder is 4 feet from the building. Find, to the nearest degree, the angle that the ladder makes with the level ground.
The ladder makes a right triangle. The wall is the opposite side. The ground is the adjacent side. The ladder is the hypotenuse, but we don't know or need to find the length of the ladder. So we have opposite and adjacent, so we are using tan. More specifically, since we are looking for the size of the angle, we need tan^{-1}.
tan(x) = 10/4, so x = tan^{-1} (10/4) = 68 degrees.
31. In the diagram below, radius OA is drawn in circle 0. Using a compass and a straightedge, construct a line tangent to circle 0 at point A. [Leave all construction marks.]
Sorry, but I'm still not good with handling constructions electronically.
Use the straightedge to extend the radius. From the point A, measure off two points on other side. Then construct a perpendicular bisector. Let's call these first two marks B and C. Go to B and make an arc above and below the line. Do the same at C so that you make two X's. Draw the perpendicular bisector, it will be tangent to the circle.
End of Part II
Any questions?
What follows is a portion of the Common Core Integrate Algebra exam. Other parts will be posted on other days. Illustrations will be added at a later time when they become available.
Part I is posted here.
Part II is posted here.
33. The height, H, in feet, of an object dropped from the top of a building after t seconds is given by H(t) = -16t^{2} + 144.
How many feet did the object fall between one and two seconds after it was dropped?
Determine, algebraically, how many seconds it will take for the object to reach the ground.
h(1) = -16(1)^{2} + 144 = 128
h(2) = -16(2)^{2} + 144 = 80
h(2) - h(1) = 128 - 80 = 48 feet between the 1st and 2nd second.
Solve for h(t) = 0.
34. The sum of two numbers, x and y, is more than 8. When you double x and add it to y, the sum is
less than 14.
Graph the inequalities that represent this scenario on the set of axes below.
Kai says that the point (6,2) is a solution to this system. Determine if he is correct and explain your reasoning.
The first inequality you need to graph is x + y > 8. The second one is 2x + y < 16.
The graph looks like this:
(graph will be uploaded later)
Looking at the graph, Kai is incorrect because (6, 2) is on a broken line which is not part of the solution set.
Note that if you drew the graph with solid lines, you lost a point for that. However, Kai would have been correct according to that mistaken graph. You have to be consistent.
35. An airplane leaves New York City and heads toward Los Angeles. As it climbs, the plane gradually increases its speed until it reaches cruising altitude, at which time it maintains a constant speed for several hours as long as it stays at cruising altitude. After flying for 32 minutes, the plane reaches cruising altitude and has flown 192 miles. After flying for a total of 92 minutes, the plane has flown a total of 762 miles.
Determine the speed of the plane, at cruising altitude, in miles per minute.
Write an equation to represent the number of miles the plane has flown, y, during x minutes at
cruising altitude, only.
Assuming that the plane maintains its speed at cruising altitude, determine the total number of
miles the plane has flown 2 hours into the flight.
There was much discussion over this question, but in the end, there was no arguing with the rubric for scoring this question. If you feel it's an unfair question, appeal to the state. In the meantime...
To find the speed at cruising altitude, use the two points given (32 minutes, 192 miles) and (92 minutes, 762 miles).
Speed in Miles per minute = changes in distance (miles) / change in time (minutes)
(762 - 192) / (92 - 32) = 9.5
If you showed your work, you have 1 point already.
The second part was where many students got caught up.
The word
only was meant to apply to both the x and the y values, not just the x. In other words, you did not need to account for the distance traveled before reaching cruising altitude.
Because of this, the correct equation was y = 9.5x.
If you included "+ 192", you didn't get credit.
For the last part, you not only need to remember the initial 192 miles, but also the first 32 minutes of the flight. The question states that it is 2 hours into the flight, not 2 hours at cruising altitude. Also remember that you are dealing with miles per minute, so you need to convert.
2 hours = 120 minutes
120 - 32 = 88 minutes at cruising altitude
y = 9.5(88) + 192 = 836 + 192 = 1028 miles
36. On the set of axes below, graph:
(Graph will be posted later)
It is important that you had a break in the line at x = -1. The linear portion ends, the quadratic portion needed to have an open circle.
Most of the mistakes I saw fit into these categories:
There was one solution because f(x) and g(x) only intersect one time.
You did not have to give the coordinates of the solution, and the solution was NOT a proper explanation. Seriously. You had to reference the fact that the lines only cross/intersect one time so there is one solution.
Also, if you had a graphing error, your final answer had to match the graph you drew. If, for example, you graph g(x) = -1/2x + 1, that would be one graphing error, but that line would intersect f(x) two times. Your answer had to match your graph.
37. Franco and Caryl went to a bakery to buy desserts. Franco bought 3 packages of cupcakes and 2 packages of brownies for $19. Caryl bought 2 packages of cupcakes and 4 packages of brownies for $24. Let x equal the price of one package of cupcakes and y equal the price of one package of brownies.
Write a system of equations that describes the given situation.
On the set of axes below, graph the system of equations.
Determine the exact cost of one package of cupcakes and the exact cost of one package of brownies in dollars and cents. Justify your solution.
The two equations were
(The graph will be loaded later.)
You could solve the system of equations using elimination. You could also solve them using the functions on the graphing calculator, but you needed to explain how you got your answer. A correct pair of answers without an explanation or procedure was only 1 point instead of 2.
Multiply the first equation by 2 and subtract
2(3.50) + 4y = 24
7.00 + 4y = 24
4y = 17
y = 4.25