If you want to check it out, the name of the blog is ylGarris.
Once again, this post is several hours late, as I had to wait until I got home and used my PC to finish it as the new iOS 7 system on my iPad is NOT compatible with Blogger's "new post" page.
First Rule: If you don't have a bathroom key, cut back on the coffee.
Second Rule: Immediately make friends with someone who has a bathroom key.
It's been my experience that some people who espouse crazy theories don't actually believe them, but are really just trying to annoy you for something else you might've said.
Ever notice when you're someplace unfamiliar, you'll gravitate toward the one person you know... even if you don't particularly get along with that person.
EDIT 10/07/13: I started at a new school this morning. One of the first questions the principal asked me was how I was with Chemistry.
I took it in high school. And a half semester in college. Haven't seen it since, really, except for reference material for, say, a "mole" joke.
The odds are 100% or else I wouldn't have had a comic.
C'mon, everyone in the ATR* Pool!
(* ATR: Absentee Teacher Reserve)
Earlier Rant has been pushed down a bit.
okay. This is now a Rant. I hate the new iOS on my iPad which will no longer allow me to upload from my camera roll. I just spent 20 minutes, and most of that was waiting for it to load images that are already online at blogger. And worse than that, there is no option for my iPad, just for my phone -- but this isn't a phone! And the iOS should know that it isn't. If I click on the phone option, it find one video and ignores literally 100s of images.
So you'll have to wait until I find a PC or a laptop to show you the gravesite marker of H.P. Lovecraft. Maybe there's some unspeakable horror preventing the image from uploading. Yeah, let's go with that!
until later.
Partial credit for mis-applied knowledge.
And I know today is Talk-Like-a-Pirate Day, but I didn't have another pi joke ready.
My mistake, but remember: to ARRRRR is human.
You can look at the exam at the Regents website.
1. Given: Triangle ABD, BC is the perpendicular bisector of AD. Which cannot be proven?
The length of BC has nothing to do with AD, AC or CD. THe answer is (2).
2. In the diagram of circle O shown below, chord CD is parallel to diameter AOB and the measure of arc CD = 110. What is the measure of arc DB?
I swear that this is a repeat -- not just a similar problem. Maybe I've used it in class before, I don't know.
DB is half of the difference of 180 and 110, which is 35. Chord AC measures the other 35, for the full 180 degrees.
3. paraphrased What is the negation [of "One is a prime number."] and the truth value of the negation?
Just add "Not". Do not change the words. "One is not a prime number." This (the negation) is true.
4. Co-ordindates are given for a triangle, and then it is rotated 180 degrees. What kind of triangle is it?
The fact that a triangle has been rotated doesn't change what kind of triangle it is. The length of its sides and the size of its angles remain unchanged. There is NO NEED to find new co-ordinates. Use the original ones.
Eliminate "Isosceles" as a choice because every isosceles triangle has to be either acute, right or obtuse, which are the other three choices.
Looking at the points, you'll see that A and B have the same y-value, so it's a horizontal line. B and C have the same x-value, making it a vertical line. Therefore, AB is perpendicular to BC and ABC is a right triangle.
Sidenote: if it hadn't been "Right", if you couldn't find that two sides were perpendicular -- that is, had inverse reciprocal slopes -- how did they expect you to find if the triangle was either acute or obtuse? Were you supposed to find the lengths of the three sides and then determine if a^2 + b^2 < c^2, where a, b, and c are the three sides in order from smallest to biggest? If so, that's a lot of work for a multiple-choice problem.
5. What is an equation of the circle with center (-5,4) and a radius of 7?
The first of many questions about the equation of the graph of a circle: (x - h)^2 + (y - h)^2 = r^2
Plug in (-5) for h and 4 for k and 7 for r, you get (x + 5)^2 + (y - 4)^2 = 49.
If you forgot to flip the signs, you lost 2 points here and most likely 6 more points elsewhere because of one formula.
6. In triangle ABC, ∠A is congruent to ∠B, and ∠C is an obtuse angle. Which statement is true?
If angle C is obtuse, it is the biggest angle and the side across from it (AB) is the longest side of the triangle.
7. There is an illustration showing a triangle with two medians meeting at point F.
The length of AF is twice the size of FB.
8. In circle O, diameter AB intersects chord CD at E. If CE is congruent to ED, then ∠CEA is which type of angle?
The chord must be perpendicular to the diameter to be bisected by it.
9. paraphrase: three congruent triangles If ABC = JKL = RST, then BC must be congruent to ____?
A notation question. BC must be congruent to KL and ST. ST is listed as a choice.
10. A triangle with one side extended. The exterior angle and the two remote angles are labeled with algebraic expressions.
11. The bases of a right triangular prism are triangles ABC and DEF. Angles A and D are right angles, AB = 6, AC = 8, and AD = 12. What is the length of edge BE?
Visualize the prism. This is NOT a Pythagorean Theorem problem. They are not looking for BC or EF. BE is the same height as AD, which is 12.
12. A second circle equation question. Center is (-4, 1), radius is 3. Use the formula from question 5.
13. The illustration shows corresponding angles are congruent.
14. The lateral area of a right circular cone is equal to 120(pi) cm^2. If the base of the cone has a diameter of 24 cm, what is the length of the slant height, in centimeters?
The reference table lists Lateral Area (L) for a Right Circular Cone as L = (pi)r l where l is the slant height
Half of 24 is 12, so 120(pi) = (12)r(pi), and r = (120*pi)/(12*pi) = 10.
15. Is the system of equations parallel, perpendicular, the same line or something else? Find the slope of each of the lines. You can rewrite the equations in slope-intercept form (y=mx+b) or in Standard form (Ax + By = C) and use m=-A/B.
3y + 6 = 2x has a slope of 2/3.
2y - 3x = 6 has a slope of 3/2.
Those are reciprocals, but they are not inverse (or negative) reciprocals. So they intersect, but not at right angles.
16. In a coordinate plane, the locus of points 5 units from the x-axis is the
First off, I have a problem with the use of the word "is", which indicates a singular answer, when two of the choices are plural, including the correct choice.
Five units above or below the x-axis (y=0) are the lines y = 5 or y = -5, respectively.
17. The sides of a triangle are 8, 12, and 15. The longest side of a
similar triangle is 18. What is the ratio of the perimeter of the smaller triangle to the perimeter of the larger triangle?
The ratio between corresponding sides of the smaller to the larger is 15/18 or 5/6, or 5:6.
The perimeter is the sum of the sides and will have the same ratio.
18. What is the converse of the statement “If lines m and n are parallel, then lines m and n do not intersect”?
q -> p: If lines m and n do not intersect, then lines m and n are parallel.
19. When the system of equations y + 2 = (x - 4)^2 and 2x + y - 6 = 0
is solved graphically, the solution is?
You could work backward from the choices, or you could expand (x-4)^2 and solve the system of equations to get the answer. Neither (6, 6) nor (-2, 2) work in the linear equation, so they could be eliminated immediately. (4, -2) works in the linear equation and the parabola.
20. shortened: If VW = 7x - 3 and AB = 3x + 1, what is the length of VC?
AB is the midsegment of a triangle, VW is the parallel side, VC is half of VW and is congruent to AB.
Solve for x, using: 7x - 3 = 2(3x + 1), which yields x = 5, which again is NOT the final answer.
Substitute 5 for x in 3x + 1, and the answer is 16.
21. Two prisms have equal heights and equal volumes. The base of one
is a pentagon and the base of the other is a square. If the area of the pentagonal base is 36 square inches, how many inches are in the length of each side of the square base?
Seriously?
Volume = Area of Base X height. The volumes are the same, and the heights are the same, so the areas of the bases are the same; i.e., 36.
Since it's a square, the length of a side is the square root of 36, or 6. (Do NOT divide by 4 and get 9, which is choice (2)!)
22. What is the difference between the sum of the measures of the interior angles of a regular pentagon and the sum of the measures of the exterior angles of a regular pentagon?
The interior adds up to 540 degrees, and the exterior to 360. That's a difference of 180.
23. If line ℓ is perpendicular to distinct planes P and Q, then planes P and Q _________ ... are parallel.
24. Another equation of a circle problem, this time with graphs. Hint: the radius is only 2. Second hint: the center is (0, 2), not (0, -2).
25. There's a circle with center B and tangents AC and AD. AB, BC and BD are drawn. AB = 15 (read the problem, not the picture!) and AC = 12.
It's a right triangle, so the radius of the circle (BC and BD) is 9. (9-12-15 is a multiple of 3-4-5. You should KNOW that. You shouldn't need to calculate it again.)
26. Triangle ABC is a right triangle with altitude AD drawn to the hypotenuse BC. If BD = 2 and DC = 10, what is the length of AB?
AB is the hypotenuse of the smaller triangle, and 2 is the shorter leg of the smaller triangle. BC, which is 12, is the hypotenuse of the largest triangle, and AB is the base of it.
So BD:AB = AB:BC, or (AB)^2 = (BD)(BC)=(2)(12)=(24). So, (AB)=24^(.5), which is 2(6)^(.5).
Alternatively,
Solve this equation (AD)^2 = (BD)(DC) = (2)(10) = 20
You get the square root of 20.
Now (2)^2 + (20) = (AB)^2.
(AB)^2 = 24, again. Same answer.
27. Triangle ABC has vertices A(0,0), B(6,8), and C(8,4). Which equation
represents the perpendicular bisector of BC?
Find the slope of BC: (4 - 8) / (8 - 6) = (-4)/2 = -2.
Any line perpendicular to BC must have a slope of positive (1/2).
Only one choice has that, so the fact that the line is a bisector is meaningless in this context.
Had it been open-ended, you would have had find the midpoint of BC and plug those co-ordinates into the y = (1/2)x + b and solved for b, the y-intercept. Doesn't sound like a whole lot of fun.
28. Chords AB and CD intersect at point E in a circle with center at O.
If AE = 8, AB = 20, and DE = 16, what is the length of CE?
(AE)(EB) = (CE)(ED), so (8)(12) = (CE)(16)
CE = 6.
Notice that they gave you AB and not (EB).
End of Part I.
Most mistakes are likely typos that snuck in because I didn't proofread carefully and because I'm still not getting enough sleep (because I'm typing up things like this).
If you find mistakes that are typographical in nature, please point them out so I can adjust them.
If my math is wrong on a problem -- if I fell into one of those traps I warn my students about -- please, feel free to explain and give the correct solution.
Nothing is every as simple as it seems. Except Algebra. Algebra's easy. I could show you, but you'd need a vacancy first . . .
Filling vacancies is not a straightforward issue. It's obviously not just a matter of openings vs. teachers: they have to be the right teachers. That is, at the very least, in the correct subject area and with the background necessary for whatever the student population happens to be. And then, of course, are the internal politics that keep veterans out of a school because they just cost too darn much. (It wasn't that long ago in NYC that teachers were "revenue neutral" for the school's budget. Then the principals were forced to do what the mayor couldn't. But enough politics, Have a Nice Day!)
I did get to see the exam finally, and I read through it. I don't remember it off the top of my head, but I do remember thinking about it, and wanting to talk about it.
Is it too late for that now? Have you already reviewed these problems the first week of school?
Don't even get them started about ''un-real'' numbers. How can numbers not be real?
And, yes, this Venn is a lie -- I don't have nearly enough students who know what Klingons and Romulans are, let alone how irrational they can be. What do they teach in middle school?
EDIT (9/13/13): The point of the comic was the Comfort Zone that I'm constantly trying to push the kids out of, but it occurs to me that some people might be interested in what the actual Venn diagram should/could look like. Not surprisingly, it turns out that there are quite a few to be found doing a simple Google search of ''venn diagram real number system''.
A new Freedom Tower now rises higher than the Towers (1776 feet) like a Phoenix from the ashes. But we'll never forget those ashes and what they represent. I had a very long walk home that day, but, as my father might've said, it beat the alternative. Too many didn't get to take that long walk, and they will never be forgotten.
I thought about changing my name to Mr. VACA one year, just for fun.
