Tuesday, August 15, 2017

Happy Pythagorean Triple Day!

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(C)Copyright 2017, C. Burke.

This is the first one this century that didn't happen during the school year, unless you count 8/6/10.

I was teaching for 3/4/5, 6/8/10, 5/12/13, and 9/12/15 (and their permutations), although I'd have to check which of those were actually school days and not weekends or holidays.

The ones that we have left to go in this century are 12/16/20, 7/24/25 and 10/24/26. Sadly, I won't be retired by then.




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Friday, August 11, 2017

Science Rocks!

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(C)Copyright 2017, C. Burke.

He'll try again tomorrow with a clean slate.




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Wednesday, August 09, 2017

Nine-Point Convex

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(C)Copyright 2017, C. Burke.

RIP Glen Campbell, Country Boy with True Grit

I really wanted to write a couple of verses to this, but I also really wanted this up today. So I might come back to it later. I might even ask for help on this one.




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Monday, August 07, 2017

Mama's Preferences

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(C)Copyright 2017, C. Burke.

p: you like the way you look that much; q: you should go and love yourself.




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Saturday, August 05, 2017

(x, why?) Mini: Split Into Four

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(C)Copyright 2017, C. Burke.

Any conclusions drawn are left as an exercise to the reader.




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Tuesday, August 01, 2017

Slap on the Back

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(C)Copyright 2017, C. Burke.

Numeric practical jokes. You have to figure that ''Kick Me'' doesn't work well when you don't have feet(*). Or meters.

This is a very rare rear glimpse of the numeric characters. I've only done this, I believe, once before, and that was part of the joke.

And I haven't really used these guys in quite a while. I need to use them more often. And give more of them actual names.




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Wednesday, July 26, 2017

Bibi

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(C)Copyright 2017, C. Burke.

There's always that kid that has to jump in to get part of the credit.

I use "Jeopardy" rules -- if you didn't buzz in (raise your hand) and get called on, the person currently answering can steal the answer and get the credit.

And if you think I inserted a girl into the class last week just so I could do this, well, you'd be incorrect, but it certainly helped. I haven't been waiting my whole career for this one, but I did write it in a notepad quite a while ago!




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Monday, July 24, 2017

(x, why?) Mini: Shapes of Things To Come

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(C)Copyright 2017, C. Burke.

But can you predict which shape will be around the corner?




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Friday, July 21, 2017

Cosmic Microwaves

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(C)Copyright 2017, C. Burke.

What it says on the Tin

If the Microwave was the greatest invention since the invention of popcorn, the Popcorn button was the greatest invention since the microwave!




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Wednesday, July 19, 2017

With A Song in My Math

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(C)Copyright 2017, C. Burke.

I could go on ALL DAY!!!! ... but I won't

Alphabetically speaking, you're OK. No, you're Buddy Kaye!




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Monday, July 17, 2017

The Doctor is Still In

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(C)Copyright 2017, C. Burke.

When is Who on?

Joke's on Michele: she'll have to sit through a whole Christmas special with two old male Doctors before we get a quick introduction to the new Who.

Oh, and "Spoiler Alert!" if you somehow missed the news.




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Thursday, July 13, 2017

Just What the Teacher Inscribed!

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(C)Copyright 2017, C. Burke.

I would've put 'Mom and Dad', but I had space limitations. Space: the final frontier.

Plus, from real life, there's a good choice that it's a single parent carrying on the good traditions while the other is poisoning their minds with 'weird stuff'.




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Tuesday, July 11, 2017

Don't Think, So?

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(C)Copyright 2017, C. Burke.

I've heard variations enough to think we put Descartes in front of a dead horse, beaten with the same puns.

Of course, if we set the Wayback Machine to 2010, we've tackled Descartes before.




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Sunday, July 09, 2017

(x, why?) Mini: Think Proof

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(C)Copyright 2017, C. Burke.

Not to be confused with 'Thinkproof' or even 'Thoughtproof', which almost seems to describe some average adults I know.




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Thursday, July 06, 2017

Like, Totally Independent

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(C)Copyright 2017, C. Burke.

I really hope this reads like it did in my head.

Math joke: I can imagine an 'x' separated from the other letters saying he was 'totally independent' from the other variables.
Maybe that can be a future joke(?).




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Wednesday, July 05, 2017

June 2017: Common Core Geometry Regents, Part 1

The following are the questions and answers (and commentary) for part of the New York State Algebra Regents exam. If you have any questions or comments (or corrections), please add them in the Comments section.

My apologies for typos, particularly if they are in the questions, because then the answers are subject to change.

Answers to Part II can be found here.

Answers to Part III can be found here.

Answers to Part IV can be found here.

June 2017, Geometry (Common Core), Part I

1.In the diagram below, triangle ABC = triangle DEF. (image omitted)
Which sequence of transformations maps triangle ABC onto triangle DEF?

(2) a reflection over the y-axis followed by a translation. ABC is flipped over the y-axis, so it is facing the opposite direction. Then it sinks down below the x-axis through a translation.


2.On the set of axes below, the vertices of triangle PQR have coordinates P(-6,7), Q(2,l), and R(-1, -3). (image omitted)
What is the area of triangle PQR?

(3) 25. If you notice, PQ is perpendicular to QR. The slope of PQ is -3/4, and the slope of QR is 4/3. Pythagorean theorem -- as well as knowledge of Pythagorean Triples -- will tell you that PQ = 10 and QR = 5. The area of a triangle is (1/2)(b)(h) = (1/2)(5)(10) = 25.
(You might also remember for several of my comics that the area of a triangle could be written as "one-half a b".)


3.In right triangle ABC, m<C = 90°. If cos B = 5/13, which function also equals 5/13?

(3) sin A. If C is the right angle, then A and B are the acute angles, which are complimentary. The sine of one acute angle will always equal the cosine of the other acute angle, but what is opposite one angle is adjacent to the other.


4.In the diagram below, m<ABC = 268°. (image omitted)
What is the number of degrees in the measure of <ABC?

(4) 46°. The complete circle is 360°. If major arc is ABC = 268, then AC = 360 - 268 = 92. The inscribed angle is half the size of the central angle, so 92 / 2 = 46°.


5.Given triangle MRO shown below, with trapezoid PTRO, MR = 9, MP = 2, and PO = 4. (image omitted)
What is the length of TR?

(4) 6. PO / MO = TR / MR. MO = 6, so 4 / 6 = TR / 9. Cross-multiply and get TR = 6.


6.A line segment is dilated by a scale factor of 2 centered at a point not on the line segment. Which statement regarding the relationship between the given line segment and its image is true?

(3) The line segments are parallel, and the image is twice the length of the given line segment. The image will also be twice as far away from the center than the original line was.


7.Which figure always has exactly four lines of reflection that map the figure onto itself?

(1) Square. Lines through the midpoints of each pair of parallel sides and the two diagonals. An octagon has eight lines: through the four pairs of parallel lines and lines through the opposites vertices.


8.In the diagram below of circle 0, chord D F bisects chord BC at E. (image omitted)
If BC = 12 and FE is 5 more than DE, then FE is

(2) 9. BC is bisected, so BE = CE = 6. The rule for intersecting chords tells us that

(BE)(CE) = (DE)(FE).
(6)(6) = (x)(x + 5)
36 = x2 + 5x

Stop here. If you did this, you're doing too much work for a multiple choice question.

What are the factors of 36?
1, 36; 2, 18; 3, 12; 4, 9; 6, 6.
Which two have a difference of 5? 4 and 9, both of which are choices.
FE is the longer one (it's "5 more"), so the answer is 9.


9.Kelly is completing a proof based on the figure below. (image omitted)
She was given that <A = <EDF, and has already proven AB = DE. Which pair of corresponding parts and triangle congruency method would not prove triangle ABC = triangle DEF?

(2) BC = EF and SAS. SAS requires that the congruent angle be included between the two pairs of congruent sides. BC and EF would give you SSA instead, and that is not allowed (except in the special case of right triangles, where it is HL).


10. In the diagram below, DE divides AB and AC proportionally, m<C = 26°, m<A = 82°, and DF bisects LBDE. (image omitted)
The measure of angle DFB is

(2) 54°. DE divides the two sides proportionally, so DE is parallel to side BC. Angle AED = 26 because it is a corresponding angle. Using the Exterior Angle theorem, we know that <EDB = 82 + 26 = 108. (If you forgot that theorem, you could have found <ADE because a triangle has 180 degrees and then found <EDB because it is supplementary.)

DF bisects <BDE, so <BDF = <EDF = 54. Angle DFB is an alternate interior angle to EDF, so it equals 54°.

You also could have solved it by finding the angles of triangle DBF. Angle B = 180 - 82 - 26 = 72. So <DFB = 180 - 54 - 72 = 54


11.Which set of statements would describe a parallelogram that can always be classified as a rhombus?
I. Diagonals are perpendicular bisectors of each other.
II. Diagonals bisect the angles from which they are drawn.
III. Diagonals form four congruent isosceles right triangles.

Update:
(4) I, II and III. The diagonals of a rhombus will divide it into four congruent isosceles triangles. This can only happen if they are perpendicular to each other, bisect each other and bisect the angles.
Note: There is some important information in this question that could have helped you with the proof in Part IV!

(2) I and III. The diagonals of a rhombus are NOT angle bisectors, unless it's a square.
Note: There is some important information in this question that could have helped you with the proof in Part IV!

I don't know where my head was when I typed the above. Obviously, it's a rectangle where the diagonals only bisect the angles when it is also a square. Just shows you that you should rush yourself, and to always double check. If the angles weren't bisected, how could III by true? Looks like I would not have scored 100 on this test.


12.The equation of a circle is x2 + y2 - 12y + 20 = 0. What are the coordinates of the center and the length of the radius of the circle?

(2) A = 1000(1 + 0.013)2. Interest increases your value, so choices (1) and (3) are right out. The percent 1.3% must be converted to a decimal, which is 0.013. The correct answer is (2). You need to complete the square. Half of -12 is -6, so we need to get to (y - 6)2:

x2 + y2 - 12y + 20 = 0
x2 + y2 - 12y + 36 + 20 = 36
x2 + y2 - 12y + 36 = 16
x2 + (y - 6)2 = 42
This make the center (0, 6) and the radius 4.


13. In the diagram of triangle RST below, m<T = 90°, RS = 65, and ST = 60. (image omitted) What is the measure of <S, to the nearest degree?

(1) 23°. First of all, I hope you realized that RT is the shortest side of the triangle (not because of looks, but because of Pythagorean Triples), so choices (3) and (4) are too big.
We have the adjacent side and the hypotenuse so we need to use cosine to find the angle.
cos S = 60 / 65
S = cos-1 (60 / 65) = 22.62..., which rounds to 23.


14. Triangle A'B'C' is the image of triangle ABC after a dilation followed by a translation. Which statement(s) would always be true with respect to this sequence of transformations?
I. Triangle ABC = triangle A'B'C'
II. Triangle ABC ~ triangle A'B'C'
III. AB || A'B'
IV. AA'= BB'

As a result of discrepancies in the wording, Questions 14 does not have one clear and correct answer.
Either (1) II, only or (3) II and III were accepted.
I don't know what the discrepancy might be that would make someone not think that AB was not parallel to A'B', unless there were printing errors in some books that left out one of the accent marks.

A dilation will preserve the shape of the original object, so the original and the image must be similar.
The center point is not given, but regardless, the orientation will not change, so the slopes of the sides will not change, so the sides will be parallel.
Update: (3) is an acceptable answer because it is possible, since exact values were not given, that after the translation, the dilation could cause AB and A'B' to become part of the same line, and not parallel. (Thank you to those who commented and emailed me.)


15. Line segment RW has endpoints R(-4,5) and W(6,20). Point P is on RW such that RP:PW is 2:3. What are the coordinates of point P?

(2) (0,11). Add 2 + 3 = 5, so RP is 2/5 the length of RW. Find difference of the x-values and y-values of RW and multiply by 2/5.
2/5(6 - (-4)) = 2/5(10) = 4
2/5(20 - 5) = 2/5(15) = 6
Add +4,+6 to point R(-4, 6). P(-4 + 4, 5 + 6) = P(0, 11).


16. The pyramid shown below has a square base, a height of 7, and a volume of 84. (image omitted)
What is the length of the side of the base?

(1) 6. Volume = (1/3)(Area of Base)(height)
84 = (1/3)(B)(7)
252 = 7B
36 = Area of the Base.
The base is a square, then the length of one side is the square root of 36, which is 6


17.In the diagram below of triangle MN 0, LM and LO are bisected by MS and OR, respectively. Segments MS and OR intersect at T, and m<N = 40°. (image omitted)
If m<TMR = 28°, the measure of angle OTS is

(4) 70°. Again, we can use the Exterior Angle Theorem to figure this out.
TMR = 28, but since MS bisects <M, <TMO is also 28, and <OMN = 56°. Since <N = 40°, <MON = 180 - 40 - 56 = 84°. Since OR bisects <MON, then <MOT and <SOT are each 42°.
Look at triangle MOT. If <TMO = 28 and <MOT = 42, then <OTS = 28 + 42 = 70°.


18.In the diagram below, right triangle ABC has legs whose lengths are 4 and 6. (image omitted)
What is the volume of the three-dimensional object formed by continuously rotating the right triangle around AB?

(1) 32 pi. First, rotating the triangle creates a cone. The Volume of a cone is (1/3)(Area of the Base)(height). The area of the base is (pi)(r)2, where the radius is 4. The height is 6 because AB is the axis the triangle is rotated about.
V = (1/3)(pi)(4)2(6) = (1/3)(16)(pi)(6) = 32 pi.

If you look at the incorrect answers: (2) assumes you used 6 as the radius (rotating about AC), (3) assumes you forgot the 1/3, and (4) assumes that you made both mistakes.


19.What is an equation of a line that is perpendicular to the line whose equation is 2y = 3x - 10 and passes through (-6,1)?

(2) y = -2/3 x - 3. The slope of the line in the question is 3/2. The slope of a perpendicular line must be -2/3 because it is the inverse reciprocal, so choices (3) and (4) are out.
Substitute -6 for x in choices (1) and (2).
(1) y = -2/3 (-6) - 5 = 4 - 5 = -1. No good. We don't want (-6, -1)
(2) y = -2/3 (-6) - 3 = 4 - 3 = 1. Check.


20. In quadrilateral BLUE shown below, BE ~ UL. (image omitted)
Which information would be sufficient to prove quadrilateral BLUE is a parallelogram?

(2) LU || BE. If a pair of line segments in a quadrilateral are both parallel and congruent, then the quadrilateral in a parallelogram. This information could be used in the proof in Part IV.
Note that choices (3) and (4) only show that three sides are congruent, not four. That could be a trapezoid.


21. A ladder 20 feet long leans against a building, forming an angle of 71° with the level ground. To the nearest foot, how high up the wall of the building does the ladder touch the building?

(4) 19 You know the length of the ladder, which is the hypotenuse of the right triangle formed by the ladder, the ground and the wall. The wall is opposite the 71° angle. So you need to use sine to find the answer.

So sin 71° = x / 20
x = 20 sin 71° = 18.91 = 19 feet.


22.In the two distinct acute triangles ABC and DEF, <B = <E. Triangles ABC and DEF are congruent when there is a sequence of rigid motions that maps

As a result of discrepancies in the wording, Questions 22 does not have one clear and correct answer. As a result, all choices were accepted.




23.A fabricator is hired to make a 27-foot-long solid metal railing for the stairs at the local library. The railing is modeled by the diagram below. The railing is 2.5 inches high and 2.5 inches wide and is comprised of a rectangular prism and a half-cylinder. (image omitted)
How much metal, to the nearest cubic inch, will the railing contain?

(1) 151. A reminder: there is no simple one-step formula that will give you this answer!
First you have to differentiate how much of the 2.5 inch height is part of the rectangular prism and what is part of the half-cylinder. The diameter of the half-cylinder is 2.5 in, which means that the radius is 1.25 in. Therefore the height of the prism is 1.25.

So the Volume of the railing = the Volume of the prism + the Volume of the half-cylinder
V = L * W * H + (1/2) (pi) (r)2 h
V = (2.5)(1.25)(27) + (1/2)(pi)(1.25)2(27)
V = 84.375 + 66.268..
V = 150.643... = 151




24.In the diagram below, AC = 7.2 and CE = 2.4. (image omitted)
Which statement is not sufficient to prove triangle ABC ~ triangle EDC?

(2) DE = 2.7 and AB = 8.1. In Choice (1), the parallel lines create alternate interior angles that are congruent, so the triangles are similar by AA (or AAA). The other three choices are show ratios that are 3:1, so the numbers themselves are not important. What is important is which sides are given. Keep in mind that there are vertical angles, so we do have one pair of congruent angles. So the other choices need to give us SAS or SSS. Choice (2) gives us SSA, which isn't allowed.

Update:
As noted in the comments below, this question, as written, has NO answer. It has a "best" answer, which could be argued only means that it's the "least wrong", which is still wrong. The Regents Board has stated that they are not rescoring the exams. Actually, I don't blame them -- the test is bad enough without making it worse. Allow me to explain my point of view:

Two other questions on this test had problems. One was invalidated because there was no positive correct answer, so everyone received credit. In another case, they accepted two answers for which credit was given. The passing grade for this exam is already artificially low. I would prefer a less confusing exam (I wouldn't say "less rigorous", but maybe that, too) with a higher threshold for passing, and a curve which doesn't punish the high-achievers.

I won't try to suggest that the Regents exam models real life any more than the SATs do. But test taking itself is a skill, and sometimes you do have to find the "best" solution, or even the "least worst", and not the "right" or the "one true" answer.

Now, I could go either way, depending upon the question. If the wording is so ambiguous as to leave a student guessing between two different choices depending on how you read something, that's a no-brainer: toss it. And if something is left out so that there is no positive answer, that should be immediately thrown out as well. Question 24, on the other hand, is a little different. First of all, you aren't looking for a positive answer at all; you want a negative answer. Which one is NOT sufficient, which means that you should be able to show that at least three of them are. Keeping in mind that this is a multiple-choice question, students are not expected to show any work at all, though they are encouraged to do so in the margins of the book. They are NOT expected to write two-page proofs in the margins, and most would not.

Now, let me stop and say this: I applaud any student that has the determination to write a two-page proof in the margins.

If you were to answer this Geometry question about similarity, the first step would be use the Laws of Similarity and Congruence. These are quick to use and quickly eliminate three of the four choices. At this point, one might select the fourth option (Choice 2) as being the correct response. At the very least, it is the one the Regents were looking for. (Yes, I hate when teachers make you guess the solution that they are looking for, whether or not it is the one that makes the most sense -- However, in this case, it is the one that makes the most sense.)

I was troubled by this question when I first saw it, because it gave exact numbers. I originally thought it would be a calculation problem, but the numbers proved little use in the methods shown above. That doesn't mean that they couldn't be useful for different methods.

In particular, you could use the Law of Sines to compare the ratios and find the angle between the sides. Given that there is a pair of vertical angles in the diagram, that is enough to prove similarity by AA. Instead of a logic-based argument, you could do a calculation-based argument.

But here's the thing: asking a student to do that in an Open-Ended problem would be difficult enough that I wouldn't even expect to find it in Part II. That would be a Part III question, at least, worth a minimum of 4 points. No one would expect any student, particularly one who "reviewed for the test" -- sadly, a necessary skill these days -- to do any of that on a multiple-choice question. Again, to those of you who go these extra lengths: More Power to You! But, sadly, using the Law of Sines in this situation probably wouldn't occur to most students even if, a) they were taught the concept properly, and b) they could execute it to its conclusion.

Finally, this isn't a case of confusion, but one of omission. There actually is a quick fix for this problem. Had the question stated "through the rules for Similarity and Congruency" (or something similar), their answer would be valid, but I suspect that such an addition would be deemed as giving too much away.

Oddly, it also could have been corrected had they switched the numbers around so that AB and DE were not the longest sides, which is an important part of Benjamin Catalfo's proof, but I doubt that they expected anyone to go this far on a multiple-choice question. Hopefully, they are now thus informed and won't make the same mistake twice. Likewise, I hope that Mr. Catalfo, upon finding no correct answer, was wise enough to choose the one that they were looking for. Sadly, sometimes this is how math -- and education, in general -- models the "real world".

End of Part I

How did you do?
Comments, questions, corrections and concerns are all welcome.
Typos happen.

Tuesday, July 04, 2017

Happy Fourth of July 2017!

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(C)Copyright 2017, C. Burke.

Can't be more like it than actually being it.

Mike's wife is thinking, "No one better be taking pictures from behind me."
I know this from experience.

I probably should add that I took that photo during a family vacation some years ago, so I'm the copyright holder. It is not in public domain.




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Wednesday, June 28, 2017

Pi vs. Tau

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(C)Copyright 2017, C. Burke.

Would you rather have
Pi in the Sky or Tau in the Scow?




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Monday, June 26, 2017

June 2017: Common Core Algebra Regents, Part 1

The following are the questions and answers (and commentary) for part of the New York State Algebra Regents exam. If you have any questions or comments (or corrections), please add them in the Comments section.

My apologies for typos, particularly if they are in the questions, because then the answers are subject to change.

Answers to Part II can be found here.

Answers to Part III can be found here.

Answers to Part IV can be found here.

June 2017, Algebra I (Common Core), Part I

1.To keep track of his profits, the owner of a carnival booth decided to model his ticket sales on a graph. He found that his profits only declined when he sol between 10 and 40 tickets. Which graph could represent his profits?

(3) (image omitted). If his profits declined, then the y values will decrease in that interval -- in other words, the graph will go down, but only between x = 10 and x = 40. In choice (1), his profits are flat; that is, the remain the same. In choice (4), there is a decline, but it is in the wrong place.


2.The formula for the surface area of a right rectanguar prism is A = 2lw + 2hw + 2lh, where l, w, and h represent the length, width, and height, respectively. Which term of this formula is not dependent on the height?

(3) 2lw. There is no height, h, in the term.


3.Which graph represents y = SQRT(x - 2)?

(4) image omitted. Hopefully, this was an easy one. Besides the fact that you had a graphing calculator available to you, only one of the four graphs was a square root graph. The "- 2" under the radical shifts the graph two units to the right away from the origin.

You should make a note of this. Shifting functions is a common theme and may show up again


4.A student plotted the data from a sleep study as shown in the graph below. (image omitted)
The student used the equation of the line y = -0.09x + 9.24 to model the data. What does the rate of change represent in terms of these data?

(2) The average number of hours of sleep per day increases 0.09 hour per year of age.
Negative correlation. Sleep goes down as age goes up.


5. Lynn, Jude, and Anne were given the function f(x) = -2x2 + 32, and they were asked to find f(3). Lynn's answer was 14, Jude's answer was 4, and Anne's answer was +4. Who is correct?

(1) Lynn, only. Calculate (-2)(3)2 + 32 = (-2)(9) + 32 = -18 + 32 = 14.
The other two tried to solve for the zeroes of the function instead substitution 3 into the function.


6.Which expression is equivalent to 16x4 - 64?

(3) (4x2 + 8)(4x2 - 8). Difference of Squares rule.
If you multiply the choices, choice (1) will have a middle term of -64x2. Choice (2) doesn't even produce the terms in the original expression, plus it will have an x2 term. Choice (4) also yields incorrect coefficients, but no middle term.


7.Vinny collects population data, P(h), about a specific strain of bacteria over time in hours, h, as shown in the graph below. (image omitted)
Which equation represents the graph of P(h)?

(1) P(h) = 4(2)h. If you look at the graph, you'll notice that the y-values are doubling. It's exponential with a base of 2.
If you look at the choices, there is only one exponential function. The others are linear, quadratic and cubic (3rd power).
If you weren't sure if the graph was a parabola, then you could have looked at the y-intercept. In Choice (3), if you substitute 0 for h, then P(h) = 3(0)2 + 0.2(0) + 4.2 = 4.2. However, the graph shows a point at (0, 4), so choice (3) is incorrect.


8.What is the solution to the system of equations below?

y = 2x + 8
3(-2x + y) = 12

(1) no solution. Since "no solutions" and "infinite solutions" are options, it might make sense to rewrite the second equation in slope-intercept form first.

3(-2x + y) = 12
-2x + y = 4
y = 2x + 4

The first equation has a slope of 2 and y-intercept of 8. The second has the same slope but a different y-intercept.
That makes them parallel, and there will be no solutions.


9.A mapping is shown in the diagram below. (image omitted)
This mapping is

(3) not a function, because Feb has two outputs, 28 and 29. Each element in the domain (input) can map to only one element of the range (output). It is okay for Jan and Mar to both map to 31.


10. Which polynomial function has zeroes at -3, 0 and 4?

(3) f(x) = x(x + 3)(x - 4). Flip the signs. Which of the choices will give you a result of 0 when you enter each of those values?
Choices (1) and (2) don't work for 0 (you can stop checking). In choice (4), -3 - 3 =/= 0, nor is 4 + 4.


11.Jordan works for a landscape company during his summer vacation. He is paid $12 per hour for mowing lawns and $14 per hour for planting gardens. He can work a maximum of 40 hours per week, and would like to earn at least $250 this week. if m represents the number of hours mowing lawns and g represents the number of hours planting gardens, which system of inequalities could be used to represent the given conditions?

(1) m + g < 40, 12m + 14g > 250.
The hours must be less then or equal to 40. The amount he makes is the sum of the number of hours times the pay for those hours for each of these jobs, and that should be greater than or equal to 250.


12.Anne invested $1000 in an account with a 1.3% annual interest rate. She made no deposits or withdrawals on the account for 2 years. If interest was compounded annually, which equation represents the balance in the account after the 2 years?

(2) A = 1000(1 + 0.013)2. Interest increases your value, so choices (1) and (3) are right out. The percent 1.3% must be converted to a decimal, which is 0.013. The correct answer is (2).


13. Which value would be a solution for x in the inequality 47 - 4x < 7?

(4) 11. You could solve the inequality, or you can plug in the choices to see what works. Note that 47 - 4(10) = 7, it is not less than 7. Don't let them catch you on that.

47 - 4x < 7
-4x < -40
x > 10
11 is greater than 10.



14.Bella recorded data and used her graphing calculator to find the equation for the line of best fit. She then used the correlation coefficient to determine the strength of the linear fit.
Which correlation coefficient represents the strongest linear relationship?

(1) 0.9. The closer to 1 or -1, the stronger the correlation, the closer to a straight line you get.


15.The height, in inches, of 12 students are listed below
61, 67, 72, 62, 65, 59, 60, 79, 60, 61, 64, 63
Which statement best describes the spread of these data?

(4) 79 is an outlier, which would affect the standard deviation of these data.


16.The graph of a quadratic function is shown below. (image omitted)
An equation that represents the function could be

(4) q(x) = -1/2 (x - 15)2 + 25. The four choices are written in vertex form. The vertex is (15, 25).
Vertex form is y = a(x - h)2 + k. h = 15 and k = 25. (Note that the minus is part of the form.)


17.Which statement is true about the quadratic function g(x), shown in the table below, and f(x) = (x - 3)2 + 2?

(3) They have the same axis of symmetry. The vertex of the equation is (3, 2). The vertex of the table is (3, -5). These points are different but they both lie on the axis of symmetry x = 3.


18.Given the function f(n) defined by the following:

f(1) = 2
f(n) = -5f(n - 1) + 2
Which set could represent the range of this function?

(2) {2, -8, 42, -208, ...}. Frankly, the other choices make no sense. The first number has to be 2, so choices (3) and (4) are right out. Choice (1) is all positive, but the negative multiplier. Substitute 2, 3, and 4, and you'll get the rest of the numbers in the list.


19.An equation is given below.

4(x - 7) = 0.3(x + 2) + 2.11
The solution to the equation is

(1) 8.3

4(x - 7) = 0.3(x + 2) + 2.11
4x - 28 = 0.3x + 0.6 + 2.11
3.7x = 2.71 + 28
3.7x = 30.71
x = 8.3



20.A construction worker needs to move 120 ft3 of dirt by using a wheelbarrow. One wheelbarrow load holds 8 ft3 of dirt and each load takes him 10 minutes to complete. One correct way to figure out the number of hours he would need to complete his job is

(4) (image omitted)
Choice (4) is the only one where all the units cancel out, except for hours.


21.One characteristic of all linear functions is that they change by

(3) equal differences over equal intervals. In other words, a constant rate of change.


22.What are the solutions to the equation x2 - 8x = 10

(2) 4 + SQRT(26)

x2 - 8x = 10
x2 - 8x + 16 = 10 + 16
(x - 4)2 = 26
x - 4 = SQRT(26)
x = 4 + SQRT(26)



23.The formula for blood flow rate is given by F = (P1 - P2) / r, where F is the flow rate, P1 the initial pressure, P2 the final pressure, and r the resistance created by blood vessel size. Which formula can not be derived from the the given formula?

(3) r = F(p2 - p1). To solve for r, you would have to multiply the original equation by r, but then you would have to divide by F, which would put it in the denominator of a fraction.




24.Morgan throws a ball up into the air. The height of the ball above the ground, in feet, is modeled by the function h(t) = -16t2 + 24t, where t represent the time, in seconds, since the ball was thrown. What is the appropriate domain for this situation?

(1) 0 < t < 1.5. The domain is the t axis, so choices (3) and (4) are no good.
If you substitute t = 1.5, then h(1.5) = 0, which is the ground. Everything after that would yield a negative number meaning that the ball went below the ground.

End of Part I

How did you do?
Comments, questions, corrections and concerns are all welcome.
Typos happen.

Sunday, June 25, 2017

June 2017: Common Core Geometry Regents, Part 4

The following are the questions and answers (and commentary) for part of the New York State Algebra Regents exam. If you have any questions or comments (or corrections), please add them in the Comments section.

My apologies for typos, particularly if they are in the questions, because then the answers are subject to change.

The answers to Part II can be found here.

The answers to Part III can be found here.

June 2017, Geometry (Common Core), Part IV

35.Quadrilateral PQRS has vertices P(-2, 3), Q(3, 8), R(4, 1) and S(-1, -4).
Prove that PQRS is a rhombus.
[The use of the set of axes is optional.]

This is NOT a two-column proof. You could write one, but it will not be worth any points if you don’t use the coordinates and the formulas to show the work. You need to back up everything you say with numbers. (And those numbers have to be correct.)

There are a number of ways to show that a quadrilateral is a rhombus. The easiest is that the four sides have equal lengths. You must have a concluding statement that says this and show the work. (Radical 50 or 5 radical 2)

You can show that the diagonals are perpendicular bisectors of each other. So you need to find the slopes of the lines and the midpoint of each line. The slopes must be inverse reciprocals and therefore are perpendicular (that last part is important!) and the midpoints must be the same point.

If you only prove that the slopes are perpendicular, you MUST also show that the figure is a parallelogram. Otherwise, your proof is incomplete. A kite, for example, has perpendicular diagonals.

To show that it is NOT a square, you can show that two sides don’t have right angles: find the slopes and show that they are not perpendicular. You can show that the diagonals are not congruent: find the lengths of the two diagonals. You can even use Pythagorean Theorem to show that two sides and one diagonal do not form a right triangle.

Note that you did not have to simplify your radicals when you use the distance formula, but you did have to have accurate numbers.

Seem the image below.

Points: Basically, you got 4 points for proving and stating the figure was a rhombus and 2 points for proving that it was not a square, for a total of 6 points.




36. Freda, who is training to use a radar system, detects an airplane flying at a constant speed and heading in a straight line to pass directly over her location. She sees the airplane at an angle of elevation of 15o and notes that it is maintaining a constant altitude of 6250 feet. One minute later, she sees the airplane at an angle of elevation of 52o. how far has the airplane traveled, to the nearest foot?

Determine and state the speed of the airplane, to the nearest mile per hour.

You have to find the horizontal distance to the plane from the first sighting, and then the horizontal distance of the second sighting. And then subtract the two values to find how far the plane traveled.
Finally, you have to convert the number of feet it moved per second instead miles per hour.

This could also be done with the Law of Sines but I'll save that for another post, where I can go more in-depth into the problem.

Sketch a figure, or two, to show the plane's location. In each right triangle, you have the height of 6250, which is opposite the angle (either 15 or 52). And you need the ground distance, which will be adjacent to the angle. You don't need the hypotenuse in either triangle. That means using tangent.

So Tan 15o = 6250 / x and Tan 52o = 6250 / y. And the distance traveled will be x - y.


x = 6350 / tan 15 and y = 6250 / tan 52
x = 23325/32 and y = 4883.03
x - y = 18442 feet

No matter what number you got in the top half of the problem, you can get point for converting it into milers per hour correctly.
To change feet per second into miles per hour, multiply by 60 and divide by 5280.

18442.28... x 60 / 5280 = 1106536.94... / 5280 = 297.57... = 210 mph.

You might have run into problems if your calculator was in Radians mode. Hopefully, the negative numbers alerted you to a problem.

If you used sine or cosine (other than using the Law of Sines), you would have lost two points from the top portion of the problem.


End of Part IV

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Comments, questions, corrections and concerns are all welcome.
Typos happen.

Saturday, June 24, 2017

June 2017: Common Core Geometry Regents, Part 3

The following are the questions and answers (and commentary) for part of the New York State Algebra Regents exam. If you have any questions or comments (or corrections), please add them in the Comments section.

My apologies for typos, particularly if they are in the questions, because then the answers are subject to change.

The answers to Part II can be found here.

June 2017, Geometry (Common Core), Part III

32. Triangle ABC has vertices A(-5, 2), B(-4, 7), and C(-2, 7), and triangle DEF has vertices at D(3, 2), E(2, 7), and F(0, 7). Graph and label triangle ABC and triangle DEF on the set of axes below.

Determine and state the single transformation where triangle DEF is the image of triangle ABC

Use your transformation to explain why triangle ABC = triangle DEF.

(image will be uploaded soon)
If you graph the two triangles, you will see that one is the reflection of the other. (Mirror images.) However, they are NOT reflected over the y-axis. To reflect over the y-axis, you would have to translate it first. But they want a single move.

You have to find the reflection line, which will be the vertical line halfway between points C and F, which occurs at x = -1.
So the transformation is rx = -1.

Because a reflection is a rigid motion (which preserves distance and shape), DEF is congruent to ABC.


33. Given: RS and TV bisect each other at point X
TR and SV are drawn (image omitted)
Prove: TR || SV

Here’s the approach you need to take: If the lines are parallel, then the alternate interior angles along the transversals will be congruent. You can show that they are congruent by proving that the two triangles are congruent. SAS looks like the easiest approach.

StatementReason
1. RS and TV bisect each other at point X Given
2. RX = XS and TX = XV Definition of bisect
3. <TXR = <VXS Vertical Angles are congruent
4. Triangle TXR = Triangle VXS SAS
5. <T = <V CPCTC
(Corresponding Parts of Congruent Triangles are Congruent)
6. TR || SV If two lines are crossed by a transversal and the alternate interior angles are congruent, then the lines are parallel.



34. A gas stations has a cylindrical fueling tank that hold the gasoline for its pumps, as modeled below. The tank holds a maximum of 20,000 gallons of gasoline and has a height of 34.5 feet. (image omitted)
A metal pole is used to measure how much gas is in the tank. To the nearest tenth of a foot, how long does the pole need to be in order to reach the bottom of the tank and still extend one foot outside the tank? Justify your answer. [I ft3 = 7.48 gallons]

Before you start, what are you looking for? The height of the stick, with is the diameter plus 1 foot. When you use the Volume formula, you will get the radius. So you need to find the radius, then double it and add 1, and then round it to the nearest tenth of a foot. Do not round to the nearest tenth in the middle of the problem. You don’t need to carry as many decimal places as I’m showing. However, I left the numbers in the calculator, so I’m showing exactly what I did. There may be minor differences in your numbers if you round that won’t affect the final answer.

First, convert gallons to cubic feet: 20,000 / 7.48 = 2673.79679
V = (pi) (r2)(h)
2673.79679 = (pi) (r2)(34.5)
r2 = 2673.79679 / (34.5 * pi)
r2 = 24.66944789
r = 4.96683
d = 9.93366
The stick is 10.9 feet long.

End of Part III

How did you do?
Comments, questions, corrections and concerns are all welcome.
Typos happen.

Friday, June 23, 2017

June 2017: Common Core Geometry Regents, Part 2

The following are the questions and answers (and commentary) for part of the New York State Algebra Regents exam. If you have any questions or comments (or corrections), please add them in the Comments section.

My apologies for typos, particularly if they are in the questions, because then the answers are subject to change.

June 2017, Geometry (Common Core), Part II

25. Given: Trapezoid JKLM with JK || ML Using a compass and straightedge, construct the altitude from vertex J to ML. {Leave all construction marks.]

There are several methods that work. One of them is to start at point J. Make an arc that cuts across LM in two places. You are allowed to use the straightedge to extend LM, if necessary. (If your arc goes outside the trapezoid, you MUST extend the line.)

From each of the points where the arc intersects LM, make an arc. At the two points of intersection, use the straightedge to draw the altitude, which will intersect point J.
Note: you only need the arcs to intersect once because you already have J.




26.Determine and state, in terms of pi, the area of a sector that intercepts a 40o arc of a circle with a radius of 4.5.

The area of a sector of a circle is the area of the circle times the measure of the central angle divided by 360o. (In other words, you multiply the area by the fraction of the circle represented.)

A = (40/360) * pi * (4.5)2
A = 2.25 pi or A = 9/4 pi.

You could also have found the area by formula A = ½ r2 O, where O is the central angle, measured in radians.
40o = 40 (pi ) / (180) = 2 pi / 9
A = ½ (4.5)2 (2 pi / 9) = (1/2) (20.25) (2 / 9) pi = 2.25 pi, which is the same answer.


27. The diagram below shows two figures. Figure A is a right triangular prism and figure B is an oblique triangular prism. The base of figure A has a height of 5 and a length of 8 and the height of prism A is 14. The base of figure B has a height of 8 and a length of 5 and the height of prism B is 14.

Use Cavalieri’s Principle to explain why the volumes of these two triangular prisms are equal.
[Image omitted.]

First, yes, you needed to refer or appeal to the principle, even if you didn’t use the name.

Second, you had to be very specific about the language, particularly using words like base, length, and height.

Third, the triangle bases are NOT congruent, even if the area of the Bases is the same.

Fourth, the oblique prism has a height of 14. This is not the slant height, or the length of the sides. In other words, the surface areas are different, the prisms are not congruent, etc.

Those were some misunderstandings I came across speaking to students.

What was necessary to state was this: the areas of the base were equal and the heights of the prisms are the same, therefore the Volumes must be equal.

You could have mentioned that the area of the cross sections will be equal at every level, but it would not have been necessary.

You could have found the area of the triangles and the volumes of the prisms, but that was not necessary. And be careful if you calculate them incorrectly. (For instance, leaving out the ½, or using 1/3, which is for pyramids. A triangular prism is not a pyramid!)


28. When volleyballs are purchased, they are not fully inflated. A partially inflated volleyball can be modeled by a sphere whose volume is approximately 180 in3. After being fully inflated, its volume is approximately 294 in3. To the nearest tenth of an inch, how much does the radius increase when the volleyball is fully inflated?

You need to use the formula V = (4 / 3) (pi) (r3) for each volume, and solve for r.
Then subtract the two radii to find the increase. Do NOT find a ratio.
This was a lot of work for only 2 points, with plenty of places for an error to sneak in, but there was nothing “tricky” about it. It was just a lot of steps.
Look at the image below. The increase is 0.6 inches.




29. In right triangle ABC shown below (image omitted), altitude CD is drawn to hypotenuse AB.
Explain why triangle ABC ~ triangle ACD.

This is an explanation, not a proof. You need to have reasons, back up what you write, but you don’t need to be so formal.

Simplest answer: If you said it was true because of the Right Triangle Altitude Theorem (and stated what that says), that was sufficient. You didn’t have to prove it. You already know it’s a theorem. (You don’t have to prove Pythagorean Theorem every time you use it, right?)

Otherwise, you can prove it using AA, or AAA, but you didn’t need the third angle. If you do this, you need three things for two points: two pairs of congruent angles and a statement that the are similar because of AA.

Angle CDA is a right angle because of the altitude. Angle ACB is a right angle of the given triangle. Angle A is in both triangles. The two triangles have two pairs of angles that have the same measure so they are similar by AA.


30. Triangle ABC and triangle DEF are drawn below. (image omitted)
If AB = DE, AC = DF, and <A = <D, write a sequence of transformations that maps triangle ABC onto triangle DEF.

You can see that there needs to be a rotation and a translation. That answer isn’t good enough.
Consider this: if there was a coordinate plane, you would have been expected to give amounts, directions, etc. This is true here as well. You can do it using rigid motions.

Translate triangle ABC along vector CF, mapping C to point F. Then rotate ABC around point C until point A is mapped onto point D.


31. Line n is represented by the equation 3x + 4y = 20. Determine and state the equation of line p, the image of line n, after a dilation of scale factor 1/3, centered at the point (4, 2).
[The use of the set of axes below is optional.]

Explain your answer.

If you substitute (4, 2) into 3x + 4y = 20, you get 3(4) + 4(2) = 20, 12 + 8 = 20, 20 = 20.
Therefore, (4, 2) is a point on line n. (You also would have noticed this if you graphed the line.)

The dilation of a line centered at a point on the line will not affect the line at all. (One third of the infinite length is infinity. One third of its 0 width is zero. One third of its 0 distance from the center is still zero.)

So the equation for p is 3x + 4y = 20.
If you rewrote it in slope-intercept form for some reason, you would have y = -3/4 x + 5


End of Part II

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Typos happen.

End-of-Year Review 2017

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(C)Copyright 2017, C. Burke.

If those were your final words, begging a pardon might be an effective strategy. But likely, not.




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Thursday, June 22, 2017

June 2017: Common Core Algebra Regents, Part 4

The following are the questions and answers (and commentary) for part of the New York State Algebra Regents exam. If you have any questions or comments (or corrections), please add them in the Comments section.

My apologies for typos, particularly if they are in the questions, because then the answers are subject to change.

The answers to Part II can be found here.
The answers to Part III can be found here.

June 2017, Algebra I (Common Core), Part IV

36.Michael has $10 in his savings account. Option 1 will add $100 to his account each week. Option 2 will double the amount in his account at the end of each week.
Write a function in terms of x to model each option of saving.

Michael wants to have at least $700 in his account at the end of 7 weeks to buy a mountain bike . Determine which option(s) will enable him to reach his goal. Justify your answer.

Option 1: f(x) = 100x + 10
Option 2: g(x) = 10(2)x

f(7) = 100(7) + 10 = 710
g(7) = 10(2)7 = 10(128) = 1280
Both options will enable him to reach his goal.

Your answer to the second part is dependent upon the function you wrote in the first part. If you made a mistake in the beginning, you need to carry that through to the end.


37. Central High School had five members on their swim team in 2010. Over the next several years, the team increased by an average of 10 members per year. The same school had 35 members in their chorus in 2010. The chorus saw an increase of 5 members per year.

Write a system of equations to model this situation, where x represents the number of years since 2010.

Graph this system of equations on the set of axes below.

Explain in detail what each coordinate of the point of intersection of these equations means in the context of this problem.

Swim: y = 10x + 5
Chorus: y = 5x + 35

In the graph (below), the coordinates of the point of intersection are (6, 65). The six means six years after 2010, or 2016. The 65 means that there will be 65 members on the swim team and in chorus.




End of Part IV

How did you do?
Comments, questions, corrections and concerns are all welcome.
Typos happen.

Wednesday, June 21, 2017

June 2017: Common Core Algebra Regents, Part 3

The following are the questions and answers (and commentary) for part of the New York State Algebra Regents exam. If you have any questions or comments (or corrections), please add them in the Comments section.

My apologies for typos, particularly if they are in the questions, because then the answers are subject to change.

The answers to Part II can be found here.

June 2017, Algebra I (Common Core), Part III

33. The function r(x) is defined by the expression x2 + 3x - 18. Use factoring to determine the zeroes of r(x).
Explain what the zeroes represent on the graph of r(x).

x2 + 3x - 18 = 0
(x + 6)(x - 3) = 0
x + 6 = 0 or x - 3 = 0
x = -6 or x = 3 are the zeroes of the function

The zeroes of the function means that the graph will cross the x-axis at -6 and 3.


34. The graph below models Craig's trip to visit his friend in another state. In the course of his travels, he encountered both highway ad city driving.
Based on the graph, (image omitted) during which interval did Craig most likely drive in the city? Explain your reasoning.
Explain what might have happened in the interval between B and C.
Determine Craig's average speed, to the nearest tenth of a mile per hour, for his entire trip.

I would NOT want to grade this question. It assumes too much on the part of the students -- in particular, that you will travel more miles on the highway at a faster rate than in the city.

Second, you have to realize that the flat line between B and C means that the car is not moving at all (which is reasonable for the exam) but supposing why the car isn't moving. Did it stop on purpose? Is it stuck in traffic? Do cars get stuck in the city more than on the highway?

The answer that they are (probably) looking for is between points D and E, hours 5 and 7 when the rate of miles per hour has decreased, but the car is still moving. You wouldn't go as fast during city driving.

The 1.5 hours that the car was stopped was likely a stop in the trip and not driving at all.
Could be a rest stop. Could be a mall. Could be lunch. Could be a major traffic jam with a tree on the highway or a truck fire or a seven-car pile-up. I hope students get creative on this one!

Hint: to get the nearest tenth of a mile per hour, divide the total number of miles by the total number of hours: 230 miles / 7 hours = 32.8571428571, or 32.9 to the nearest tenth.


35. Given
g(x) = 2x2 + 3x + 10
k(x) = 2x + 16
Solve the equation g(x) = 2k(x) algebraically for x, to the nearest tenth.
Explain why you chose the method you used to solve this quadratic equation.

g(x) = 2 k(x)
2x2 + 3x + 10 = 2(2x + 16)
2x2 + 3x + 10 = 4x + 32
2x2 - x - 22 = 0

The Quadratic Formula is used in the image below:

Update: I cut the bottom from the image. Looks like I made a parenthesis error of some sort on the calculator and didn't catch it.

Calculate those fractions to the nearest tenth, and you get
x = 3.576... and x = -3.076, which round to x = 3.6 and x = -3.1

End of Part III

How did you do?
Comments, questions, corrections and concerns are all welcome.
Typos happen.

Tuesday, June 20, 2017

June 2017: Common Core Algebra Regents, Part 2

The following are the questions and answers (and commentary) for part of the New York State Algebra Regents exam. If you have any questions or comments (or corrections), please add them in the Comments section.

My apologies for typos, particularly if they are in the questions, because then the answers are subject to change.

Answers to Part III can be found here.

June 2017, Algebra I (Common Core), Part II

25. Express in simplest form: (3x2 + 4x - 8) - (-2x2 + 4x + 2)

5x2 - 10.
Show something on the paper to indicate where you got this: line them up vertically; distribute the "-1" and combine like terms. Something, so you'll be sure to get both points. (Frankly, this is the kind of question that you should be able to do without showing any work. It can all be done in your head.)

26. Graph the function f(x) = -x2 - 6x on the set of axes below.
State the coordinates of the vertex of the graph.

See the graph below. The vertex is at (-3, 9). You need to state the point and have a correct graph to get both points.

27. State whether 7 - SQRT(2) is rational or irrational. Explain your answer.

It is irrational because 7 is rational and SQRT(2) is irrational and the sum or difference of a rational and an irrational number is always irrational.

28. The value, v(t), of a car depreciates according to the function v(t) = P(.85)t, where P is the purchase price of the car and t is the time, in years, since the car was purchased. State the percent that value of the car decreases by each year. Justify your answer.

The car's value decreases by 15% each year because 1.00 - .85 = .15, which is 15%.

29. A survey of 100 students was taken. It was found that 60 students watched sports, and 34 of these students did not like pop music. Of the students who did not watch sports, 70% liked pop music. Complete the two-way frequency table.

Watch Sports Don't Watch Sports Total
Like Pop
Don't Like Pop
Total

Answer: see table below
Because 60 of 100 watched sports, then 40 did not, so the bottom row is 60, 40, 100.
Of 60, 34 did not like pop, so 26 did. First column is 26, 34, 60.
TWIST -- they used percentages in the next portion of the question.
Of the 40, 70% liked pop music. (.70)(40) = 28, and 40 - 28 = 12.
The second column is 28, 12, 40.
Add the totals for each row. Last column is 54, 46, 100.

Watch Sports Don't Watch Sports Total
Like Pop 26 28 54
Don't Like Pop 34 12 46
Total 60 40 100

30. Graph the inequality y + 4 < -2(x - 4) on the set of axes below.

See graph below.

If you recognize point-slope form then you know that the slope of the boundary (broken) line is -2 and (4, -4) is a point on that broken line.

If you didn't recognize that, you could subtract 4 from each side and put

y = -2(x - 4) - 4

into your graphing calculator, and get the table of values.

Or use the Distributive property and created your own table from the following:

y < -2(x - 4) - 4
y < -2x + 8 - 4
y < -2x + 4

31. If f(x) = x2 and g(x) = x, determine the value(s) of x that satisfy the equation f(x) = g(x).

Substitute x2 = x
Subtract x2 - x = 0
Factor x(x - 1) = 0
Find the zeroes: x = 0 or x - 1 = 0, so x = 0 or x = 1.

32. Describe the effect that each transformation below has on the function f(x) = |x|, where a > 0.
g(x) = |x - a|
h(x) = |x| - a

g(x) will shift f(x) a units to the right.
h(x) will shift f(x) a units down.
Both graphs will have the same shape.

End of Part II

How did you do?
Comments, questions, corrections and concerns are all welcome.
Typos happen.

a || b

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(C)Copyright 2017, C. Burke.

There are hand-writing issues, both with chalk and on electronic whiteboards.




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Thursday, June 15, 2017

Geometry Regents Questions? Post them here

As I type this, the New York State Geometry Regents is about 15 hours away

If you have a specific question about a topic that you would like to ask or have explained, post it as a comment to this thread. Then check in later for a response.

I'll check and update the comments periodically throughout the evening.

Update #1:
Anonymous said...
Could we go over completing the square for circles? Also, the equation of a circle?

Okay, "Anonymous". It's like this. The standard form of the equation of a circle is

(x - h)2 + (y - k)2 = r2
Where (h, k) are the coordinates of the center of the circle and r is the radius.
Note: you can use the Distance formula or Pythagorean Theorem to see if another point they give is on the circle if the distance from the center to the point is the same as the radius.

As you might remember from Algebra, (x - h)2 is the square of a binomial, and can be written as (x - h)(x - h). If you were to multiply that (think FOIL), you would get x2 - 2hx + h2. That expression would be a perfect square. If you have an incomplete expression, you need to complete it, so that you can factor it.

Say they give you x2 + y2 + 6y = 16
x2 is a perfect square. It's the same as (x - 0)2.
However, y2 + 6y is not a perfect square -- it doesn't have a constant term.
Do you see the 2hx in the expression above? The middle term of the complete square is double the number in the binomial, so you need to find half of it.
Half of 6 is 3. The final term of the completed square is h2, h is 3, so h2 is 9.
So you need to add 9 to both sides of the equation, which gives you

x2 + y2 + 6y + 9 = 16 + 9
This can be reduced to x2 + (y + 3)2 = 25
In this case, the center is (0, -3) -- because you flip the sign -- and the radius is the square root of 25, which is 5.
It is quite possible -- likely, even -- that they will give you a problem with a radius that has an irrational length.

Does this help/answer your question?

Update #2:
Anonymous said... Can you go over density problems

Density is mass divided by Volume. Imagine you have something that weighs 10 pounds. If it fits in the palm of your hand, it's pretty dense. If it's the size of your kitchen table, it's not very dense at all.
D = m / V, like those "dense" people you meet at the D.M.V. when you apply for a learner's permit.
They have to give you 2 of the three values, so that you can find the third one. However, they can make you figure out Volume.
Volume of a prism is the Area of its Base times its height.
A rectangle prism would be length X width X height
A cylinder would be pi * r2 * h, etc.

After that, it's likely to just be an Algebra problem.

I don't have a specific example I can give you or that.

Update #3:
Anonymous said...
Can you go over proofs

Not really. I could spend a week on proofs. If you want something specific, I would check my old Regents exam posts.
Here are some general guidelines.
Look at the image. What do you see? What do things look like? You CANNOT go based on looks, but it might give you a direction to go in.
Don't "assume" anything. Either it's given, or you derived it from what was given.
Make a plan. How are you going to get there?

Does it involve proving triangles are congruent? Then you'll need SSS, SAS, ASA, AAS or HL. (Don't make a backward SSA of yourself!)
If you use any of those, you need to specify three pairs of things that are congruent. In the case of HL, make sure you state that all right angles are congruent. Seriously -- it needs to be stated.
If you are proving that two sides of a triangle or two angles are congruent, then the last reason will probably be CPCTC (Corresponding parts of congruent triangles are congruent).

They won't give you anything that you can't figure out. Two of the biggies are vertical angles are congruent, and the reflexive property (for sides or angles).

If it's a circle, remember that you can add extra radii, and that all radii of a given circle are congruent. Tangents are perpendicular to the radius, so you might see a right triangle in the circle. Similar triangles (use AA) inside the circle are also possible.
Make sure you state all the given, and if there's an illustration, mark off everything you know. It might give you ideas, or it might remind you what you haven't explicitly stated yet.

Obviously, you need to know your theorems. And there are a lot of them.

Does this help/answer your question?

Update #3:
Heaven said... Could we go over finding a section of a circle? And those circle problems dealing with an external point?

I assume you mean a "sector" of a circle, like a slice of pie? Think of slice of pi, if that helps.
The area of the sector of a circle is a fraction of the area of the entire circle.
The fraction that you need to multiply by is the central angle over 360 degrees. (Times pi r square)

They can also ask the reverse. They can tell you the area of the sector and the radius and have you come up with the central angle by working backward.
Think Algebra: inverse operations.

I'm not sure what you mean by "those circle problems dealing with an external point".
Do you mean finding the size of an angle from the arcs the lines intersect?
Do you mean the relationship between the lengths of the secants or tangents from an external point?
Do you have an example?

Final Update ... It's Friday morning
Anonymous Anonymous said... Can you go over finding a point on a circle, and also ratios of line segments?

Suppose you are given an equation like (x - 3)2 + (y + 1)2 = 20.
If you wanted to know if a point is on the circle, say (5, 3), substitute those values into the equation.
(5 - 3)2 + (3 + 1)2 =?= 20
22 + 42 = 20 ?
4 + 16 = 20
20 = 20, check
Therefore, (5, 3) is a point on the circle. Had it equaled anything other than 20, it would not have been on the circle.

If a point in on a line somewhere that isn't the midpoint, you need to use ratios to find its position.
For example, if given A(-1, 2) and B(7, 8) and you and P was a point such that the ratio of the lengths AP:PB was 3:1, where would P be?
First, 3 + 1 = 4, so P is 3/4 of the way from A to B.
Find the difference of the x values 7 - (-1) = 8, multiply it by 3/4, and you get +6.
Find the difference of the y values 8 - 2 = 6, multiply it by 3/4, and you get +4.5
Add those values to the coordinates of A to get P. P(-1+6, 2+4.5) gives you P(5, 6.5).
Yes, you can get a decimal.

Monday, June 12, 2017

RIP Adam "Batman" West

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(C)Copyright 2017, C. Burke.

Riddle me this: Was that a 'bat-joke' or a 'dad-joke' hidden in there?

I had afternoon reruns of Batman and Superman growing up, but Batman's shows were newer, in color, and had super villains in them. Superman had bad guys, but not Lex Luthor or ... whoever.

Bat-this, Bat-that. It was a part of growing up. I still tell my students when the bell rings, even though they don't get the reference, that I'll see them tomorrow.
"Same bat-time, same bat-channel."




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Friday, June 09, 2017

Failure

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(C)Copyright 2017, C. Burke.

Witty/Inspirational quote about how good Failure is goes here. Like "Yeah! Summer School! No sand in my toes!" Or something.




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Wednesday, June 07, 2017

The 0'Factor, Episode 17

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(C)Copyright 2017, C. Burke.

And things might get a little improper before that happens!

I knew I hadn't done one of these in a while. Didn't realize that it had been more than TWO YEARS since the last one.
Especially when you consider the "promo" version, without other characters, are easy to create. Coming up with a throwaway line, on the other hand, takes a little more time.

It also might be time for a new "set" because that desk was one of the earliest things I ever drew for this strip (nearly 10 years ago) that I still use, and for the life of me, I have no idea what fonts I use -- or even if I still have them available!




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