## Sunday, August 31, 2014

### Real-World Math: Bushels of Fun

Author's Note: I started writing this yesterday (Saturday), but by the time I returned home, there wasn't time to finish it and clean it up. I worked on it this morning, but all references to "today" were really yesterday.

There are four pecks in a bushel. I know this because my mother loved me a bushel and a peck. She'd tell me this all the time in song. Okay, so I was never sure what a "peck" was. Something chickens do, right? And those little kisses on the cheek, too. So four pecks would be two kisses on each check. One extra would make a bushel and a peck. At least, it would to a six-year-old.

Seriously, I knew what a bushel was -- it was a large basket that you could fill with fruit or other farm produce. In particular, tomatoes, which I'll get back to.

Now, in this day of metric measure, you may ask how many liters of tomatoes are in a bushel? Or should it be kilograms?

The answer is: who cares? The more important question is "how many jars of sauce will it make?" The answer to which is, about 18, if you're lucky. Not overly precise and not very scientific but quite an accurate estimate. Then again, maybe it is scientific since this is empirical evidence: we got an average of 18 jars per bushel. That's correct: I said "average". It wasn't precisely the same for each bushel. How could it be? (And do we really know?)

Today was Tomato Jarring Day aka Sauce Day. We started with five bushels of tomatoes. When everything was cut, boiled, strained, and boiled again, we had put 90 jars of sauce to bed, wrapped in blankets. We could do the math on this and convert it, if we wanted to:

One bushel is 35.293 liters, although the decimal portion would be silly because there's no way to know that we had precisely 1.0 bushels. Let's call it 35.3 liters. That makes five bushels equal to 176.5 liters.

In the end, we filled 90 one-quart jars with sauce. Ninety quarts is about 85.2 liters. So 176.5 liters of tomatoes created 85.2 liters of tomato sauce? Wow! That's odd! Half of the tomatoes disappeared? Was there that much skin and garbage and rot that we tossed aside? Nah, couldn't be. It's just that we're comparing two different quantities ... with the same units.

That Metric System sure is complicated and confusing in Real Life, aint it! (And don't get me started that "metric" means a standard of measurement, so metric measure means . . . you get the picture).

Converting it to metric does nothing for me. It's very scientific, but not particularly practical for us non-scientists who just want to think about the next couple years' worth of pasta-based meals. (And chicken parm, of course.)

But moving on from this. I am kidding, of course, but not by much. There are those who will insist that there is inherent beauty in the metric system's sets of ten this and then that all connected in a unified theory of scientific terms that apply to no one's life outside of academia and rocket science. Yes, rocket science is important and I'm all for it! But I -- and just about everyone I know -- will never participate in it. Get me the Moon, and I might change my tune!

Speaking of tunes...

There was much Beauty in my mother's singing. And she wasn't alone and Bushel and a Peck was hardly a one-in-a-million tune. I knew the words to Five-Foot-Two, Eyes of Blue before I was three-feet tall! No one will ever sing "157 point 5 Centimeters, Eyes of Wavelength 475 Nanometers"

To quote something that I think I read in one of my grandmother's Readers Digests several decades ago: "Who would walk a million kilometers for one of your thermometers?" Not Al Jolson. (Or Fred Mertz, either!). Or the Proclaimers. Or Venessa Carlton. Or ... you get the picture. Or, rather, the song.

Feet, inches, yards, miles... we know what they are, and what they are are musical and memorable. They're all different? They're all beautiful, like the rolling hills across the countryside. Imagine every mountain was the same size. Would you still climb one because it was there, if it was like the one next to it? Would you fly halfway around the world to climb one?

I'll take the ten miles of smiles or even ten feet of sleet, over meters of liters or liters of meters. I'll even take bushels of pecks 'til I'm filled to the gills. All 297.89 of them. Now, excuse me, it's time for some pasta; we needed four pounds to feed everyone.

## Friday, August 29, 2014

### (x, why?) Is In TV-Tropes

As many of you are probably aware(*), there is a website called TV Tropes which details those cliches and shortcuts writers take when trying to add little plot twists or just get to the end of the story quicker. And they cover a lot more than just TV. In fact, they have pages detailing webcomics, and I've made no secret of the fact that I'd love to be listed in there or have my page appear as examples as any of the tropes pages. However, I didn't want to add my own material there.

And as of tonight, (x, why?) has its own page. And I didn't do it!!! I was alerted to it by this post on Twitter:

I encourage anyone with an account to add examples, good or bad. (I can take it! I think.) Or as Mr. Taylor (of Taylor's Polynomials fame) says, let him know, and he'll add them.

I can't wait to see what shows up. And I might sneak in a few examples of my own. It's okay! I'll have Greg do it.

(*) If you weren't aware of it, I am not responsible for the 3-6 hours (or more) you may lose, in a blink, from endless clicking on extra links.

## Thursday, August 28, 2014

### The "mini" comics? Good idea or just quick filler?

I didn't get any response to yesterday's mini-comic, but then, I rarely do. (And not to complain because I'm happy to get comments, but it's generally the same 2 or 3 people.)

When I did a little analysis of the previous comics and saw that a year was going by before I reached the next 100th comic, I realized something was wrong. It's taking more and more time that I don't seem to have. And unless I want it to turn into a soap opera among the teachers, the jokes are getting harder to write. Granted, school will be starting soon, and that always seems to provide material. It's not just things the students say, mind you, sometimes I'll blurt out things in the middle of an explanation and realize, "Hey, I can use that."

Part of the problem is working on the artwork. I'll still not an artist, and sometimes it seems like I'm playing with my old "Colorforms(tm)", even if my characters are a little more expressive. Using shapes was quicker -- the just have to look like shapes, not the shapes that were in the last 3 strips or anything. And leaving it in black and white (and shades of gray) puts some more tools in my hands without worrying as much about the pixelation between the colors. There are nights I spend most of the time working on the comic doing "clean up" work on something that should have been a quick cut-and-paste job.

So odds are that there will be more minis. Will they all be shapes? I don't know. If they develop personalities, they could become a fourth set of characters, after the humans, the numerals (I have a special name for those guys that I've never used because I've wanted to trademark it first -- but I don't know how) and the Trigonometry Jones characters. I could also make minis with typed characters, when I don't need (or want) to use my regular numeric characters.

Those are my thoughts on the matter.

What are yours?

## Wednesday, August 27, 2014

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

(Click on the comic if you can't see the full image.) Would becoming purely kinetic be realizing your potential?

This is a new thing that I'm trying, (x, why?) minis. "Mini" as in "minimum", so it has a mathematical context, along with "minimum artwork, background and coloring". I thought about calling them "quickies", but that lacked the math connection and, frankly, this wasn't as quick as I would have liked.

Happy Joe Burke Day!, the anniversary of my father's birth, which until recently (and somewhat still today) was celebrated by friends and family members congregating at Denny's -- not the national chain, but the family bar/restaurant in Brooklyn, where Dad had his own barstool, along with a place for the dog to rest. They even had a spare walker for him in the later years. We'll be going to toast his memory after paying him (and Mom) a visit. ## Tuesday, August 26, 2014

### N-RN.3 (Real Number System) - Irrational Behavior

A short entry for tonight.

Moving on to another Common Core Algebra standard brings me to N-RN.3, which reads

Explain why the sum or product of two rational numbers is rational; that the sum of a rational number and an irrational number is irrational; and that the product of a nonzero rational number and an irrational number is irrational.

The key word here is "explain". Before we do that, let's establish that the rest of the standard is true: if you add two rational numbers, you will get an irrational number; the sum or product of a rational and irrational number is irrational. (Leaving out the obvious case of multiplying by zero.)

If we remember that the word "rational" comes from "ratio" and that ratios can be written as fractions with numerators and denominators that are integers. Then the sum of any two rational numbers can be written as the sum of two fractions. All we need to add these two fractions is a Common Denominator, which can be obtained by multiplying the denominators. The set of integers is closed under multiplication; the product will be another integer. If we add the numerators, the set of integers is also closed under addition. The total will be another ratio of two integers, which must be a rational number.

To show that the sum of a rational number and an irrational number is irrational, we can use contradiction. Supposed that r is rational and x is irrational. (Let's not use i as it has a different meaning, which may cause confusion.) Assume there is a sum r + x that is rational. If we add -r, which is also rational, we get r + x - r, or just x, which must be rational because the sum of two rational expressions is rational. This is a contradiction because we started with x as irrational. So the sum of a rational number and an irrational number must be irrational.

The same contradiction can be used to show that rx cannot be rational if x is irrational, by dividing both sides by r.

Now that we have that out of the way, the fortunate thing for Algebra students is knowing when the product or sum is rational or irrational. Don't be quick to assume that a radical sign indicates irrational. They love using the square root of 64 or the cube root of 27, both of which are perfectly rational. However, it helps if you can explain why it is so other than to say, "Well, you know, it's obvious." (And, hopefully, it is.)

## Sunday, August 24, 2014

### What Makes the Golden Ratio so Golden?

Now that I’ve concluded my Golden Ratio-themed comic serial, I wanted to get into a little what the Golden Ratio actually is. Just saying that it’s some number that’s approximately 1.618, doesn’t quite do it justice. What’s so special about that number? And what makes that a ratio?

Second point, first. A ratio is a comparison of two numbers, so it can cause confusion if only a single number is written, even if “to 1” is implied.

What makes it so special is what is being compared. Look at these two images, for example. They are basically the same thing, only the notation for the lengths are different.

Suppose we wanted to find a ratio of the length of the square to the length of the rectangle, and we wanted that ratio to equal the ratio of the length of the smaller rectangle to that of the bigger one, what value of x would accomplish that?

It terms out that we could set up the problem in either of these two ways (and many more, besides!). On the left, the square has a length of 1, the smaller rectangle has a width of x, so the bigger rectangle has a total length of x + 1. One the right, the square still has a length of 1, but the bigger rectangle has a total length of x, so the smaller rectangle has a length of x – 1.

Setting up the proportions and cross-multiplying we find:

In either case, we get the same quadratic equation: x2 – x – 1 = 0.
That’s not something easily factorable, so we have to use Ye Olde Quadratic Formula, after which we discover that the roots are

If we take the positive value, we get x = 1.618033988749894848204586... But what about the negative value? If we subtract root 5 from 1 and divide by 2, we get x = -0.618033988749894848204586...

Look at the decimal portion. They are the same! Keep in mind, that we’re dealing with irrational numbers here, which aren’t supposed to conform to patterns, but this one is just brilliant. And, yes, there is a reason for it.

Take a look at that second proportion, above, on the right, with the removable 1 in the denominator.

This is saying that one less than the number is the reciprocal of that number! If you divide 1 by 1.618033988749894848204586..., you will get 0.618033988749894848204586..., the same as if you just subtracted one.

This works for the negative value as well: If you divide 1 by -0.618033988749894848204586..., you will get -1.618033988749894848204586..., the same as if you just subtracted one.

One last point, and a hat tip to William Ricker for mentioning it, how else can we write the reciprocal of x, 1/x? What exponent gives us the reciprocal of x? An exponent of -1. That means that this equation, this proportion, can be written as:

Absolutely brilliant. And quite Golden, if I say so myself.

## Saturday, August 23, 2014

### A Little Bit of Fibonacci ...

Today's column isn't going to happen the way I wanted it to, so I'll try it for tomorrow. In the meantime, I'll leave you with this:

I just finished posted a series of comics with crazy numbers in the titles after the word "part", instead of the normal "Part 1/10, Part 2/10, ..." etc.

This is because those numbers were intended to be ratios, and not just any ratios, but Fibonacci number ratios. As you might have guessed from the first episode in that sequence, I would be working with Fibonacci numbers. You might also have guessed that I would have been done by the time I got to 21 -- That's okay! I was expecting to be done a lot sooner, too!

As I hope to get into a little more depth tomorrow, when you divide two consecutive Fibonacci numbers, you get a ratio that is close to the Golden Ratio, which is also known as "phi", and is approximately 1.6180339887. The larger the terms, the smaller the difference between the two ratios.

Here is a table showing you the first 15 terms:

But that's not the only thing interesting about the Golden Ratio. (Well, of course not! Books could -- and have -- been written!) There should be more about it in tomorrow's entry.

## Friday, August 22, 2014

### Trigonometry Jones and the Golden Ratio, Part 1.618033988749894848204586...

(Click on the comic if you can't see the full image.) And if the horse were here, they could ride off into the sunset.

For anyone who joined in the past five years and missed out, or are wondering about "Part 144/89", check out the following:

For a quick read, check out the Comics-Only site, starting here, and then continuing here.

Or if you might want to comment, check out these blog entries:
1/1, 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/34, 89/55, 144/89, 233/144, 377/233, 610/377. ## Thursday, August 21, 2014

### Trigonometry Jones and the Golden Ratio, Part 610/377

(Click on the comic if you can't see the full image.) So the Golden Ratio was actually a rare gem the whole time? Who knew?!

The gem came as close to the golden ratio as I could manage in Paint. Actually, it was closer, but I hated the way it looked, so I altered it a little to make it more esthetic.

It does seem like they've been looking for years. And now it's almost done.

For anyone who joined in the past five years and missed out, or are wondering about "Part 144/89", check out the following:

For a quick read, check out the Comics-Only site, starting here, and then continuing here.

Or if you might want to comment, check out these blog entries:
1/1, 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/34, 89/55, 144/89, 233/144, 377/233. ## Wednesday, August 20, 2014

### Trigonometry Jones and the Golden Ratio, Part 377/233

(Click on the comic if you can't see the full image.) Trigonometry is referring to the last time he was in the Mountie's Hall.

That happened in the final trailer for Trigonometry Jones and the Lost Compass.

One other note: Even though I storyboarded this, it's still not turning out exactly as I planned. But it is coming to a conclusion!
And that will be the last time I get this involved in the details!

For anyone who joined in the past five years and missed out, or are wondering about "Part 144/89", check out the following:

For a quick read, check out the Comics-Only site, starting here, and then continuing here.

Or if you might want to comment, check out these blog entries:
1/1, 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/34, 89/55, 144/89, 233/144. ## Tuesday, August 19, 2014

### Trigonometry Jones and the Golden Ratio, Part 233/144

(Click on the comic if you can't see the full image.) Dum da dum dum. Dum da dum. Dum da dum dum. Dum da dum dum dum.

For anyone who joined in the past five years and missed out, or are wondering about "Part 144/89", check out the following:

For a quick read, check out the Comics-Only site, starting here, and then continuing here.

Or if you might want to comment, check out these blog entries:
1/1, 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/34, 89/55, 144/89. ## Monday, August 18, 2014

### Trigonometry Jones and the Golden Ratio, Part 144/89

(Click on the comic if you can't see the full image.) Will Trigonometry do the Right Thing or the Reich Thing?

Yes, the gang is back! That chilling cliffhanger to which you've waited nearly FIVE years for its resolution!
... which will hopefully be happening, even though I've been dreading this.

For anyone who joined in the past five years and missed out, or are wondering about "Part 144/89", check out the following:

For a quick read, check out the Comics-Only site, starting here,

Or if you might want to comment, check out these blog entries:
1/1, 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/34, 89/55. ## Sunday, August 17, 2014

### Sunday Night, Thinking About the Numbers

Be forewarned: this is the obligatory filler post to keep my every-day-in-August streak going. Any ideas bouncing around my head all day would take too long to plot out, type up and semi-illustrate. So I'm punting. This may be boring, or it could be fun. Let's see.

Some numbers have been on my mind this past week or so. Those were the numbers 2000 and 900. Odd numbers to be sure -- except that they're both even. Actually, they're both sort of "anniversary" numbers, with one being a centenary and the other a millennial.

I noticed this past Monday, that I had reached more than 1900 tweets on my twitter account. I was wondering if I might do something silly for 2000. Or maybe 2001 would be better. Or maybe 2001 could say, "Hey, that was my 2000th tweet!" (First-world problems, right?) Hey, it's all about "branding", or so I'm led to believe. Funny thing is that this was on my mind during the last Weird Ed twitter chat. I didn't want to randomly comment on so many things that I'd fly over the top. I also didn't retweet a few funny or interesting links, and I hesitated on the reply button.

I checked the counter before tweeting every link to my posts and comics.

And then I had some fun, and the geek within created these:   I haven't gotten to Tweet 2010: the Tweet I Make Contact yet. Or maybe I have -- I wasn't paying attention earlier today.

The second number: 900.

Since 2007, I have posted comics on a sometimes regular basis. It's the reason why some of you are here. Math puns, teacher jokes, general geekiness. That's me.

I'm closing in on comic #900. Every 100th comic has been a little special, generally (but not always) having something to do with that number specifically.

Sadly, I noticed that it took over a year to get from 600 to 700, and from 700 to 800. For a while I was posting four times per week, over and above the promised three times per week. In the past couple years, there were weeks with only one update. Then again, this has been a year from Hell.

Finally, there were a couple more Number-Specific comics that come to mind that I want to add:

• 99: Luftballoon, read it anyway if you got the joke
• 666: Devil's Staircase, which, co-incidentally, fell on St. Valentines Day.

So what's coming up for Number 900? I guess I'll know in five more comics!

• ## Saturday, August 16, 2014

### Lines in Planes

(Click on the comic if you can't see the full image.) You've been skewered.

Skew lines are neither parallel nor intersecting, occuring in three-dimensional space, not two. Imagine skew lines as an overpass and a roadway passing beneath. They don't intersect, but they're not parallel, and they aren't traveling in the same direction.

Or think of the front-right edge of a shoebox and the top-left edge. They each join two planes, but there isn't a common plane containing both of them. ## Friday, August 15, 2014

### August 2014 Integrated Algebra Regents Exam, Multiple Choice

The problem with these exams is that they all happen at the same time and leave a lot of time to work on blog posts (and have a life, too!). Then again, whoever said that teachers have lives?

Here are the multiple-choice questions, the quick-and-dirty edition, which may or may not get updated in the near future.

1. What is the product of 3a2b and -2ab3?

(4) -6a32b4. Multiply the co-efficients, add the exponents (no exponent = 1).

2. The value of the expression |-20| - |6| is

(2) 14. (20 - 6 = 14)

3. When 9x2 - 100 is factored, it is equivalent to (3x - b)(3x + b). What is the value of b?

(2) 10. Difference of Squares Rule.

4. Which equation represents the line that passes through the points (1, 1) and (-2, 7)?

(4) y = -1/2 x + 6. Find the slope of the line first, (7-1)/(-2-1) = 6/(-3) = -1/2. Plug in x and y and check the two equations with that slope.

5. The graph below represents the parabolic path of a ball kicked by a young child. What are the vertex and the axis of symmetry for the parabola?

(1) vertex (3, 8); axis x = 3.

6. Which relationship can best be described as causal?

(3) The snow is falling and the stores run out of snow shovels. The others aren't related at all.

7. In a class, which data can be classified as qualitative?

(4) Hair color of students. Qualitative are descriptive, quantitative are amounts. I don't like this question because shoe size is more descriptive than an actual measurement. It is a measurement, but you'll never see it added, subtracted or averaged -- what would the point of knowing the "average" shoe size by, unless by "average" you meant "mode". But that's for another day.

8. Given the following:
..........................................................
What is the intersection of sets A, B, and C?

(4) Jade and Kyle are in all three sets, so they are the intersection.

9. The sun of (3x - 4)/(x + 3) and (2x - 5)/(x + 6) is

(1) (5x - 9)/(x + 3). Add the numerators, leave the denominators alone.

10. If Rosa's age is represented by R, which inequality represents the statement "Rosa is at most 29 years old"?

(3) R < 29

11. What is the slope of a line passing through points (-7, 5) and (5, -3)?

(2) Another slope question! Rise over run gives you (-2/3)

12. A positive correlation always exists on a scatter plot when

(4) y increases as x increases. Also, y decreases as x decreases. It trends up and to the right.

13. A sandwich consists of one type of meat, one type of condiment and one type of cheese. The possible choices are listed below:
.......................................................
In the sample space of all the possible different sandwiches consisting one type of meat, one type of condiment and one type of cheese, how many sandwiches do NOT include provolone cheese?

(1). Use the Counting Principle and multiple 3 * 3 * (4-1) = 27. There were four cheese, remove 1 for provolone.

14. The graph of the equation y = x2 is shown below.
....................................................
Which statement best describes the change in this graph when the coefficient of x2 is multiplied by 4?

(2) The parabola becomes narrower because it would increase higher, faster.

15. A parking lot is 100 yards long. What is the length of 3/4 of the parking lot, in feet. 1 yard = 3 feet.

(2) (3/4) * (100) * 3 = 75 feet

16. What is the solution of the equation (12/7x) + (3/2x) = 15/14

(3) Multiply the entire equation by 14x and you get 24 + 21 = 15x, so 45=15x and x = 3.

17. The expression (2x2 + 10x - 28)/(4x + 28) is undefined when x is

(2) -7, only. The fraction is undefined only if the denominator = 0. If the numerator is 0 (and not the denominator), the fraction is 0, not undefined.

18. In right triangle JKL in the diagram below, KL = 7, JK = 24, JL = 25, and angle K = 90 degrees.
...............................................
Which statement is not true?

(2) cos L = 24/25. That is false because cos L = 7 / 25.

19. A teacher asked the class to solve the equation 3(x + 2) = 21. Robert wrote 3x + 6 = 21 as his first step. Which property did he use?

(3) Distributive property.

20. If the roots of a quadratic equation are -4 and 2, the equation is equivalent to

(1) (x + 4)(x - 2) = 20. Flip the signs.

21. Kelsey scored the following points in her first six basketball games: 22, 14, 19, 22, 8 and 17. What is the relationship between the measures of central tendency?

(1) mode > median > mean

22. Sheba opened a retirement account with \$36,500. Her account grew at a rate of 7% per year compounded annually. She made no deposits or withdrawals on the account. At the end of 20 years, what was the account worth, to the nearest dollar?

(3) \$141,243. Exponential growth/compounded interest: Balance = 36500 (1.07)20

23. Which equation represents a vertical line?

(4) x = 12. Set x = to any constant, and you have a vertical line. In this case the line contains the points {... (12, -2), (12, -1), (12, 0), (12, 1), (12, 2), ... }

24. Byron has 72 coins in his piggy bank. The piggy bank contains only dimes and quarters. If he has \$14.70 in his piggy bank, which equation can be used to determine q, the number of quarters he has?

(3) 0.10(72 - q) + 0.25q = 14.70

25. Which graph represents the equation y = |x - 2|?

(4) The V shape in on the x-axis at (2,0). The -2 shifts the graph to the right.

26. If ax + 3 = 7 - bx, what is x expressed in terms of a and b?

(3) Subtract 3 from both sides. Add bx to both sides: ax + bx = 4.
Factor: (a + b)x = 4.
Divide: x = 4/(a + b)

27. Which equation represents a line that is parallel to the line whose equation is y = -3x?

(3) 6x + 2y = 4. Another slope question, but this time you had to work out which of the choices has a slope of -3 so it can be parallel to the given line.

28. What is the result when 6x2 - 13x + 12 is subtracted from -3x2 + 6x + 7?

(4) When you subtract 6 from -3 you get -9, and only choice (4) has that.

29. What is the solution set of the equation x / 3 = 8 / (x + 2).

(2). Cross multiply to get x2 + 2x = 24, subtract 24 from each side, factor and solve. Or plug in the choices and see what works. {-6, 4} both work.

30. Which set of integers is included in (-1, 3]?

(1) -1 is NOT included, but 3 IS included because of the square bracket. (I just went over this when I subbed last week. I hope they remembered!)

That's it for part one.
I will try to clean this up a bit over the weekend.

## Thursday, August 14, 2014

### August 2014 Integrated Algebra Regents Exam, Open-Ended Questions

This morning was the Integrated Algebra Regents exam, and it was either the last one or next to last one. (There may be one final one in January.)

I've decided to start with the Open-Ended questions because they're quicker to type and I can scan in some of the answers I worked out.
There were three two-point questions, three three-point questions, and three fout-point questions. Partial credit was available for all of them, and work was required.

### Part 2

31. Using his data on annual deer population in a forest, Noj found the following information:

25th percentile: 12
50th percentile: 15
75th percentile: 22
Minimum population: 8
Maximum population: 27

Using the number line below, construct a box and whisker plot to display the data.

They gave you the Five-Number Summary. All you had to do was graph it. Start by using the number line, which had 30 hash marks, so you could've gone from 1 to 30, if you wanted to. You MUST have a scale designated, don't put one number here and one number there just because. Don't skip one here and skip two there.

32. The diagram below consists of a square with a side of 4 cm, a semicircle on the top, and an equilateral triangle on the bottom. Find the perimeter of the figure to the nearest tenth of a centimeter. [Diagram omitted.]

First of all, realize that if you didn't round correctly, you lost one point on a two-point problem. Don't throw away points!

The square has two sides showing, each side is 4. The equilateral triangle has two sides showing, and the third side is the side of the square, so those two sides are also 4. So far we have 4 + 4 + 4 + 4 or 16. Finally, the length of a semicircle (half the circumference) is 1/2 times pi times the diameter, which is the side of the square, which gives us 1/2 * 3.14 * 4, which is 6.28, rounded to 6.3, as per the question. Add 16 + 6.3 to get 22.3, which is the final answer.

NOTE: In this question, you could get away with using 3.14 for pi because the numbers were small and you had to round. That won't be the case in the next problem.

33. A thermos in the shape of a cylinder is filled to 1 inch from the top of the cylinder with coffee. The height of the cylinder is 12 inches, and its radius is 2.5 inches. State, to the nearest hundredth of a cubic inch, the volume in the thermos.

Note: If you used either 3.14 or 3.1416 for pi, you got the wrong answer. I am serious. If you didn't use the Pi key on the calculator, or at least use 3.141592, you answer will not round correctly. Will you lose points for this? You're supposed to, but it might slip by, depending on the scorer.

The formula for Volume is pi * r2 * h, where h is the height of the coffee, not the thermos, so h = 11, not 12.
The correct equation is V = (3.141592...)(2.5)2(11) = 215.98, to the nearest hundredth.

### PART 3.

34. The top of a lighthouse, T, is 215 feet above sea level, L, as shown in the diagram below. The angle of depression from the top of the lighthouse to a boat, B, at sea is 26o. Determine, to the nearest foot, the horizontal distance, x, from the boat to the base of the lighthouse.

The angle of depression from the top of the lighthouse is equal to the angle of elevation from the boat. We have a right triangle, and we know the side opposite angle B and we want the side adjacent to angle B. This means that we need to use tangent to solve the problem.

Tan 26 = opp/adj = 215/x. So x = 215/tan 26 = 440.815..., or 441, to the nearest foot.

35. There are six apples, five oranges and one pear in John's basket. His friend takes three pieces of fruit at random without replacement. Determine the probability that all three fruits taken are apples.

There are twelve pieces of fruit to begin with, six of which are apples. Each time he takes one, there is one fewer piece of fruit and one fewer apple.
The equation you needed to write was P = 6/12 * 5/11 * 4/10, which after cancelling factors (or multiplying and simplifying) leaves you with 1/11.

36. Express [THE FOLLOWING in simplest radical form.

I'll leave this for pictures as there are too many radical signs.

The y is there for no apparent reason, and the two radicals cannot be combined because they aren't alike. This was a bit of an unusual question, with three twists: the variable, the distribution of the negative sign and the different radicals in the answer.

### PART 4.

37. On the set of axes below, solve the following system of inequalities graphically:

y + 3 < 2x
-2y < 6x - 10
State the coordinates for a point in the solution set.

Again, I'll let the scan speak for me. Your answer should look similar if you rewrote the inequalities in slope-intercept form correctly. The "Ex", means "Example" and that was there for me as a scorer, as a reminder that there wasn't a single answer for that point.

Remember to label at least one of the inequalities. You can choose any point in the solution set, but don't pick a point on the broken line because that isn't part of the solution. It's a boundary.

38. The actual side of a square tile is 4 inches. The manufacturers allow a relative error of 0.025 in the area of a tile. Two machines are used to cut the tiles. Machine A produces a square tile with a length of 3.97 inches. Machine B produces a square tile with a length of 4.12 inches. Determine which machine produces a tile whose area falls within the allowed relative error.

This was another unusual question. It was easy in that they didn't have you calculate any surface areas or volumes. On the other hand, you had to calculate the relative error for both machines. This was also, arguably, a dumb question because the question implies strongly that the answer is Machine A. For this reason, if you said Machine A without any calculations, you will not get any points.

The work is shown in the scan below:

39. Solve the following system of equations algebraically:

y = x2 - 6x + 9
y = -9x + 19

Another scan. You didn't need to check your answers, but when I saw (-5, 64), I thought that that couldn't possibly be right, so I checked -- and it was. The other answer was a more reasonable (2, 1). You needed two find both x values and the y values associated with them.

And those are the open-ended questions.

Thanks for reading. I do this for you.

### August 2014 Geometry Regents, Part 2 and 3

Continuing with the answers to yesterday's exam. If you missed Part 1, it's here.
There were six two-point questions, three four-point questions, and one six-point question. Partial credit was available for all of them, and work was required.

### Part 2

29. Triangle ABC has coordinates A(-2, 1), B(3, 1) and C(0, -3). On the set of axes below [OMITTED], graph and label Triangle A'B'C', the image of triangle ABC after a dilation of 2.

You only needed the image, but it helped to graph the original first. A dilation of 2, centered on the origin (because nothing else was stated), means that you double all the x- and y-coordinates. You had to label them A', B', and C', but you didn't need to write the actual co-ordinates.
If you didn't do the graph, but wrote and labeled the coordinates of the image, you scored one point. If you Dilated by something other than 2, or did a different Transformation, you may get one point if you carried it through correctly to the end of the problem. (I saw kind a few T-2,-2.)

30. In the diagram below of triangle ABC, DE and DF are midsegments.

If DE = 9 and BC = 17, determine and state the perimeter of quadrilateral FDEC.

If DE is a midsegment then E is the midpoint of BC. That makes EC half of 17, which is 8.5. (Don't round!) Because DF is a midsegment, it is half the length of BC, so it is also 8.5. AC is twice as big as DE, so 2 X 9 = 18. CF is half of 18, which is 9 again. Quadrilateral FDEC is a parallelogram (hopefully, you knew that!), and the perimeter is the sum of its four sides: 9 + 9 + 8.5 + 8.5 = 35.

I saw a couple of 34 and 36 from people who apparently didn't like fractions or decimals.

31. The image of triangle ABC under a translation is triangle A'B'C'. Under this translation, B(3, -2) maps onto B'(1, -1). Using this translation, the coordinates of image A' are (-2, 2). Determine and state the coordinates of point A.

Note: They gave you A', and you need to work backward. First, calculate the translation: to go from 3 to 1, you subtract 2. To go from -2 to -1, you add 1. The transformation is T-2, 1. You have to do the opposite to A' to get A. (-2 + 2, 2 - 1) gets you A(0, 1). The most common incorrect answer I saw was (-4, 3).

32. As shown in the diagram below, quadrilateral DEFG is inscribed in a circle and m<D = 86. [Diagram coming soon.]
Determine and state mGFE (arc)
Determine and state m<F.

Angle D is an inscribed angle. It measure 86 degrees, so the arc GFE is double that amount or 172 degrees.

Angle F is also an inscribe angle. If arc GFE is 172, then arc EDG is 360-172 = 188 degrees. The angle is half as much, which is 94 degrees.

Note that the shape inscribed in the circle is neither a parallelogram nor a trapezoid. It's just an oddly-shaped quadrilateral. If you made a mistake on part one, but were consistent throughout the problem, you got half credit.

33. In the diagram below, QM is a median of triangle PQR and point C is the centroid of triangle PQR.

If QC = 5x and CM = x + 12, determine and state the length of QM.

The centroid is two-thirds of the way from the angle to the midpoint of the opposite side. The equation you would write is 5x = 2(x + 12). Don't forget to double the expression on the right because it's only half as long.

5x = 2x + 24
3x = 24
x = 8

So QM = 5(8) + (8) + 12 = 40 + 8 + 12 = 60

34. The sum of the interior angles of a regular polygon is 540 degrees. Determine and state the number of degrees in one interior angle of the polygon.

You should have known that the polygon is a pentagon because the sum of the interior angles of a pentagon is 540. You could have calculated any way you wanted to if you didn't know, making a chart if you needed to. (I don't think you needed to. You could have just stated it as a fact because you can't learn about interior angles and not know a pentagon has 540 degrees.)

Divide 540 by 5 and you get 108 degrees, which you also probably knew before you did the problem, but you needed to show the work. Did you have to state or show that the polygon had five sides? I don't know. I would say "No". There was nothing in the scoring chart that says so (other than a note that if you calculated that it had 5 sides, but did no further work, you got one point.)

### Part 3

35. Given: MT and HA intersect at B, MA || HT, and MT bisects BA.
Prove: MA = HT.

You had two choices here, one using AAS and another using ASA. Make sure you provided the correct reasons for the one you chose.
In paragraph form:
MA is parallel HT, which is given. This means that angle M is congruent to angle T AND angle A is congruent to angle H because they are pairs of alternate interior angles. MB = TB because MT is bisected by HA. So by AAS, triangle AMB is congruent to triangle HMT. Then by the definition of congruent polygons (or CPCTC, corresponding parts of congruent triangles are congruent), MA must be congruent to HT.

If you used the fact that angle MBA is congruent to angle TBH because they are vertical angles, you could have used ASA, instead of AAS.

If you thought that MT and HA bisected each other and that MB was congruent to TB, you are incorrect. However, if you made only this mistake and then used SAS (NOT SSA) to prove the triangles congruent and finished with CPCTC, you should have gotten at least half credit for the problem. I don't think that thinking the lines bisected each other is a "conceptual" mistake or just a logical one, so I can't say if you would lose one or two points on that.

36. A right circular cone has an altitude of 10 ft and the diameter of the base is 6 ft as shown in the diagram below. [COMING SOON] Determine and state the lateral area of the cone, to the nearest tenth of a square foot.

The good news: the formula was in the back of the book. Lateral area of a cone is L = Pi * r * l, where l is the slant height. Note that r is NOT squared.
Now the bad news: if you used r = 6 or l = 10, you messed up. If you did both, you probably scored zero points. The radius is 3, which is half the diameter. The height of the cone is not the slant height, which is the length from the tip of the cone to the edge of the base. The slant height can be found by making a right triangle with the height and the radius, which is actually shown in the picture. Using Pythagorean Theorem, we find that the slant height is the square root of 109. If you thought you made a mistake because it was so terrible, you didn't. I hope you didn't erase it, thinking you got it wrong.

The equation you should have written was L = (3.141592)(3)(108)(1/2), which gives 98.4, to the nearest tenth.
Warning: if you used 3.14, instead of the pi key, you will be off by a tenth. Will you lose a point for that? Very likely, yes. Maybe not.

37. The construction question:

To divide the line into four equal segments using only a compass and straightedge (and NOT a ruler, measuring the line), you needed to construct a perpendicular bisector. Then construct a second between the left endpoint and the midpoint and finally construct a third one between the midpoint and the right endpoint. You have four equal pieces.

To be continued . . .

EDIT: continuing

38. I have to make another copy of the sketch of the locus problem. It was a little complicated, if only because they have NEVER asked a question like that in that way. I had told me review class the day before that questions like this one always have a vertical or a horizontal line, and to make sure they make the correct one. This one had a line with a slope of -1, and you have to write the equations.

I hope to have a sketch posted soon, but in the meantime:

The equation of the circle was (x - 3)2 + (x + 2)2 = 25
The two points given had a midpoint of (1, -5) and the locus was a line with a slope of -1. The equation of that line was y = -x - 4.
The line intersected the circle at the leftmost and bottommost points of the circle (-2, -2) and (3, -7), which you had to state, not simply mark with an X.

UPDATE: Here's a sketch. Obviously, your sketch doesn't have to have a perfect circle drawn. If you did, you might have noticed that the circle goes through (0, -6), but it is NOT one of the solutions.

The question stated On the set of axes, graph the locus of points 5 units from the point (3, -2). Write an equation that represents this locus. On the same set of axes, graph the locus of points equidistant from the points (0, -6) and (2, -4). Write an equation that represents this locus. State the coordinates of all points that satisfy both conditions.

If you sketched either incorrectly, you needed to state whatever points satisfied both conditions in your graph. If your graph didn't overlap, you had to say that No Points satisfied the conditions, and so on.

I don't know where all the points came from, but you basically got one point for each locus, one point for each equation, and one point from the two points satisfying the conditions. Where's the sixth point? I don't know, but if you had the other five, you had all the points.

## Wednesday, August 13, 2014

### August 2014 Geometry Regents, Part 1 (Multiple Choice)

I just spent a few hours grading the Geometry Regents exams at my old school. It was described as "doable". I think that the students who came to my tutoring classes the past few days will be happy ... assuming that they were all working out their own problems and not just copying my answers.

A few observations: the proof was pretty straightforward, with a choice of AAS or ASA, as long as you backed up the one you used; the locus question was very unusual in that you had to give an equation of the circle locus and the line was neither vertical nor horizontal; the construction question had you bisecting multiple times (and using your ruler and drawing the curves later didn't count). These issues with the Open-Ended problems with be discussed tomorrow (I hope).

Here are the multiple-choice questions from Part I. Keep in mind, that I have to type all of these, so the rest of the test may not show up on my blog as quickly as you may like. Questions are always welcome. Likewise, because I've been asked to hurry with this, there are no diagrams included. They may get added at a later time.

1. A rectangular prism is shown in the diagram below. [DIAGRAM IS BOX-SHAPED WITH DIAGONALS ON TOP AND BOTTOM DRAWN.]
Which pair of line segments would always be both congruent and parallel?

(4) DB and HF are parallel. Of the other three choices, two pair intersect and one pair is skew.

2. In parallelogram QRST, diagonal QS is drawn. Which statement must always be true?

(3) Triangle STQ is congruent to QRS. The diagonal of a parallel creates two congruent triangles by SSS. (Proof left as an exercise to the reader.)

3. In the diagram below of circle O, diameter AB and chord CD intersect at E. [DIAGRAM IS A CIRCLE AS DESCRIBED.]
If AB is perpendicular to CD, which statement is always true?

(4) Arc CB is congruent to arc BD. (Also arcs AC and AD are congruent, in case you were wondering.)

4. What is an equation of the line that passes through (-9, 12) and is perpendicular to the line whose equation is y = 1/3 x + 6?

(2) y = -3x - 15. If it is Perpendicular then the slopes are negative reciprocals. The slope of the new line must be -3. (Eliminate two choices.) Plugging in (-9, 12) into choice (2) shows that it is a point on that line. Likewise, substitution into equation (4) shows that it is not correct.

5. In the diagram below, under what transformation is triangles X'Y'Z' the image of triangle XYZ? [DIAGRAM SHOWS A CO-ORDINATE PLANE WITH A TRIANGLE IN QUADRANT I AND A CONGRUENT, BUT *ROTATED* TRIANGLE IN QUADRANT IV.]

(3) It's a rotation of 90 degrees clockwise.

6. What is the solution of the system of equations y - x = 5 and y = x2 + 5?

(1) You can check very quickly that (0, 5) is a solution. From the choices, the other solution has to be (1, 6) or (-1, 6). Check (1, 6); it works.

7. In the diagram below, parallelogram ABCD has vertices A(1, 3), B(5, 7), C(10, 7) and D(6, 3). Diagonals AC and BD intersect at e.

What are the coordinates of point E?

(3) (5.5, 5) The diagonals of a parallelogram bisect each other, so the coordinates of point E is the midpoint of either diagonal.

8. Right triangle ABC is shown in the graph below. [DIAGRAM SHOWS A CO-ORDINATE PLANE WITH RIGHT TRIANGLE ABC IN QUADRANT I.]
After a reflection over the y-axis, the image of triangle ABC is triangle A'B'C'. Which statement is not true?

(4) AC will not be parallel to A'C' after the reflection. A'C' will have a positive slope.

9. What is an equation of circle O shown in the graph below? [DIAGRAM SHOWS A CO-ORDINATE PLANE WITH A CIRCLE CENTERED AT (-2,4) WITH RADIUS 4.]

(4) Flip the signs, square the radius.

10. In the diagram below of right triangle ABC, an altitude is drawn to the hypotenuse AB.

Which proportion would always represent a correct relationship of the segments?

(3) The Right-Triangle Altitude Theorem. Altitude z is used twice in the proportion.

11. Quadrilateral ABCD is graphed on the set of axes below. [DIAGRAM SHOWS A CO-ORDINATE PLANE WITH DIAMOND-SHAPED QUADRILATERAL IN QUADRANTS I AND iv.]

(3) It's a rhombus, but not a square.

12. Circle O is represented by the equation (x + 3)2 + (y - 5)2 = 48. The coordinates of the center and the length of the radius of circle O are...

(1) Second time the circle equation is used in this exam. Flip the signs and simplify the square root of 48. (Not that you have to, there's only one possibility in the choices.)

13. In the diagram below of circle O, chord AB is parallel to chord CE.

A correct justification for mAC = mBD in circle O is

(1) Parallel chords intercept congruent arcs.

14. What is theslope o fa line perpendicular to the line whose equation is 3x - 7y + 14 = 0?

(2) Perpendicular again. The slope of the given line is 3/7. The slope is the inverse reciprocal, which is -7/3.

15. Line segment AB has endpoint A located at the origin. Line segment AB is longest when the coordinates of B are ...

(2) The question essentially asks, which of the four choices is the farthest from the origin. You might also notice that if you add the absolute value of x and the absolute value of y, in each case, the sum is 10. However, that isn't how you find distance. The distance formula involves squaring numbers, so it shouldn't be a surprise that (2, -8) is the farthest point.

16. In triangle FGH, m<F = m<H, GF = x + 40, HF = 3x - 20 and GH = 2x + 20. The length of GH is ...

(3) Because angles F and H are congruent, it is an isosceles triangle with legs GF = GH. So x + 40 = 2x + 20. (Note: HF does not matter in this.) Subtract x from both sides, and subtract 20 from both sides, and you find that x = 20. THAT IS NOT THE ANSWER. Plug 20 in for x, and 2(20) + 20 = 60.

17. In the diagram of quadrilateral ABCD, diagonals AEC and BED are perpendicular at E. [DIAGRAM SHOWS A KITE-SHAPE QUADRILATERAL WITH TWO DIAGONALS.]
Which statement is always true based on the given information?

(4) If the lines are perpendicular, then four right angles are formed, which are all 90 degrees, and therefore all congruent to each other.

18. Which set of numbers could represent the lengths of the sides of a right triangle?

(4) {8, 15, 17} is the only Pythagorean Triple listed. You should know it just from seeing it. If you didn't know it, you had to check each using the Pythagorean Theorem, although you should have realized that {7, 7, 12} couldn't be correct because the hypotenuse of that isosceles triangle would have to be 7(root 2).

19. In quadrilateral ABCD, the diagonals bisect its angles. If the diagonals are not congruent, quadrilateral ABCD must be a

(3) It's a rhombus, but not a square. I think I wrote that once before.

20. Line m and point P are shown in the graph below. [DIAGRAM SHOWS A CO-ORDINATE PLANE WITH A STRAIGHT LINE WITH A POSITIVE SLOPE AND A P BELOW IT IN QUADRANT IV.]
Which equation represents the line passing through P and parallel to line m?

(2) The line has a slope of 2, so the parallel line also has a slope of 2. The equation was given in point-slope form: y - y0 = m(x - x0).

21. Which compound statement is true?

(1) A disjunction (OR) only needs one part to be true. A square has four sides.

22. In triangle CAT, m<C = 65, m<A = 40 and B is a point on side CA, such that TB is perpendicular to CA. Which line segment is shortest?

(2) In any triangle, the side opposite the smallest angle is the smallest side, but there's a complication here. The triangle is cut into two smaller triangles, and because CAT is NOT a right triangle (it's 65-40-75), we can't apply the Right Triangle Altitude Theorem, even if that would have been of some use. Because TB is an altitude, it creates two right angles, so we can find the size of the smaller angles that T has been cut into, 50 and 25. The side opposite the 25-degree angle is going to be the smallest of the four segments listed.

23. In the diagram of triangle ABC below, DE || BC, AD = 3, DB = 2, and DE = 6.

What is the length of BC?

(2) Because the lines are parallel, the triangles are similar and the sides are proportional. 3 : 6 :: (3 + 2) : x, or 3x = 30. So x = 10.

24. In triangle ABC, an exterior angle at C measures 50 degrees. If m<A > 30, which inequality must be true?

(1) It must be less than 20 because of the Exterior Angle Theorem.

25. Which graph represents the graph of the equation (x - 1)2 + y2 = 4?

(2) Third time they are asking about the equation of a circle.

26. The equations of lines k, p, and m are given below:

k: x + 2y = 6
p: 6x + 3y = 12
m: -x + 2y = 10

Which statement is true?

(1) Lines p and m have perpendicular slopes (-2 and 1/2).

27. Peach Street and Cherry Street are parallel. Apple Street intersects them, as shown in the diagram below:

If M<1 = 2x + 36 and M<2 = 7x - 9, what is M<1?

(4) The two angles are supplementary, so 2x + 36 + 7x - 9 = 180. (They are NOT congruent, equal to each other.)
Therefore 9x + 27 = 180, 9x = 153, and x = 17. Again, that is NOT the answer.
Substitute 17 for x: 2(17) + 36 = 34 + 36 = 70 degrees.

28. A regular pyramid has a height of 12 centimeters and a square base. If the volume of the pyramid is 256 cubic centimeters, how many centimeters are in the length of one side of its base?

(1) 256 * 3 = 768. 768 / 12 = 64. The square root of 64 is 8.

That's it for the multiple-choice. I hope you did well.

As always, if there are any typos in the above, please holler at me at your earliest convenience so I can adjust things.

### Math Can Be Painful

(Click on the comic if you can't see the full image.) That's my attempt at constructing the Gowanus Canal. It's a trendy area now, even though it still stinks. (Literally, the smell is nauseating.) However, the boats aren't really as big as they are on the map. Nor do they float on their sides, as seen in this aerial view. ## Tuesday, August 12, 2014

### Math Isn't Everything. It Isn't Even the Only Thing.

Math isn't everything. It isn't even the only thing.

That may sound shocking coming from me, but theoretical math (you know, a lot of that Algebra stuff) needs to be applied to real life using real life conditions, which sometimes (most times?) take you out of the scope of any classroom problem. This is why some people don't think they're using algebra (when, in fact, they are), or why they think it isn't really practical.

A quick example of what I mean: there's a joke floating around that goes something like this: only in a math class can you buy 36 oranges and 25 apples and not be thought crazy.

Here's a different example (not a joke). Each weekend in August, the Miller family barbecues six hot dogs. Buns come in packages of eight. Over four weekends, how many packages of buns should they buy?

The "correct" answer is four packages.

Now, wait a minute, you protest! The cook 24 franks, they need 24 buns, and you get 24 buns in three packages of eight.

That is certainly true. On the second weekend, you still have 2 buns leftover from the prior weekend. This brings the next question -- the real world question -- who gets the stale buns? Probably the shy, quiet one who doesn't speak up for himself. Or the youngest one who doesn't know any better. Most likely, Mom, who who sacrifice for her children, giving them the food from her mouth if need be, assuming she wanted two hot dogs to begin with. (Take better care of Mom, she's been good to you!)

In the real world, even if the bread hasn't reached it's expiration date, those leftovers still won't be as fresh as new rolls will be. Moreover, consider the fourth weekend. All of the bread is leftover, and no one gets a fresh roll. If you're not on a really tight budget, buy new bread each week.

What do you do with the extra bread? Feed the birds. Make breadcrumbs. Have a really funky looking sandwich on Monday.

What do I know? I'm a math teacher, not a cook.

## Sunday, August 10, 2014

### N-RN.2 (Real Number System) - Rationalizing the Denominator

This is the third and final column on Common Core Standard N-RN.2. If you missed the first two parts, Part 1 dealt with evaluating expressions with rational exponents, and Part 2 showed how to simplify using factor trees and how to add and subtract radicals. The last piece of this standard (and since I'm only dealing with part ".2", I could really call it a "substandard" if I wanted to, mockingly) is to "Simplify radical expressions by rationalizing the denominator (Algebra 1 - EE.2)".

Previously, we mentioned that you can multiply two radical numbers by multiplying their radicands. We also factored radical numbers in order to simplify them. Let's talk about division. When you divide, you multiply by the reciprocal; that is, you can create a fraction of the two numbers without relying on your early education "gazintas". (You remember, "2 gazinta 6 three times".) Likewise, when you take the square root of a fraction, you are actually dividing one radical number by another.

So if you wanted the square root of 1/4, you would take the square root of the numerator (radical 1 is 1) over the square root of the denominator (radical 4 is 2). The result would be 1/2. But suppose we wanted the square root of 1/2? Again, we can split it up into the square root of 1 (which is 1) over the square root of 2. Here's where we run into a problem because there are rules from fractions. One of them is that there cannot be any radicals in denominator. You have to get rid of them.

We haven't discussed this before, but there's really only one simply way to get rid of a square root sign: square the number. We need to multiply the denominator by radical 2. We are allowed to do this because it's a fraction and we won't change the value of the fraction at all as long as we multiply the numerator by the same amount as the denominator. The fractions (square root of 2 over square root of 2 looks scary to evaluate until you remember that any number, even an irrational, divided by itself is one, with the exception of zero. If you multiply a fraction by 1, it doesn't change its value, even if it looks different. The result is that the radical is gone from the denominator and has moved into the numerator, which is allowed. One more example. Try it yourself before scrolling down and looking at the image. What is the square root of 4/5?
Take the square root of each number. Rationalize the denominator. What's left in the numerator? What's left in the denominator? That's it for this standard. Time to move on to part 3, coming soon.

## Saturday, August 09, 2014

### N-RN.2 (Real Number System) - Dealing With Radicals

In a recent post, we explored evaluating expressions with rational exponents in them, but there's more to Common Core Standard N-RN.2. Don't worry, some of it's easier to deal with than what we already tackled.

These are the items listed below the standard, at least according to the IXL website, where I found the list:

• Simplify radical expressions (Algebra 1 - EE.1)
• Simplify radical expressions by rationalizing the denominator (Algebra 1 - EE.2)
• Multiply radical expressions (Algebra 1 - EE.3)
• Simplify radical expressions using the distributive property (Algebra 1 - EE.5)

Is there anything else to deal with? I don't know. Dealing with rational exponents isn't in this list, and yet I think that they might be encountered before Algebra 2. FYI, the notation "EE" stands for Expressions and Equations, which allows me to once again state that "Expressions don't have equal signs and are evaluated, and Equations do have equal signs and are solved."

Simplifying radical expressions is not a difficult task -- as long as you know that it does NOT mean pushing buttons on your calculator and coming up with an approximate decimal equivalent to 8 or 12 or 15 decimal places. Simplifying a radical is similar to reducing a fraction to its lowest terms. It makes it easier to deal with for computations (particularly adding and subtracting, when the radicals have to be "like terms") and comparisons. If the only thing you're planning to do with a radical number is square it, then, yes, simplifying it is a bigger waste of time than converting an improper fraction into a mixed number when it's only going to be used for slope.

There is a very straightforward method of simplifying square roots, but it seems to mystify some of my students who, apparently, never grasped the concept of what a square root (or a perfect square) was in the first place. They memorize steps, but uncertainty about the order causes them to mess up at the very end, removing radical signs from irrational numbers or leaving them in after taking a square root. (For example, they'll write that the square root of nine = the square root of three, instead of three.)

The simplest method involves finding the largest perfect square which is a factor of the radicand (i.e., the number under the radical sign). If it isn't the largest perfect square, then the radical hasn't been fully simplified. An example: One problem my students face is not understanding the concept of a perfect square, so instead of 25 and 2, then use 5 and 10. After that, they're stuck, or they just decide, for example, that the square root of 5 is the same as 5 without the radical sign.

Because of this, I tried a different approach, using factor trees. They remembered doing them in middle school, and actually liked using them again. (You see, your teacher was right! You are using them again!) The example looked something like this instead: After they have the prime factorization under the radical, I have them circle the pairs of numbers, cross them out and write one factor outside the radical. This has two downsides to it: first, if the number has a lot of factors, there will be a lot of extra work (but at least they will know, for certain, that they simplified their answer); second, if they don't complete the problem, they basically just drew a factor tree, which looks kinds childish and silly from a high school student.

Multiplying two radicals is as simple as multiplying two fractions. Just multiply the numbers under the radicand. For instance, radical 7 times radical 10 equals radical 70. If the number can be simplified, do it, according to the rules above. Obviously, if you square a radical, such as radical 6 times radical 6, the radical symbol goes away. In this case, you get radical 36, which is just 6.

As mentioned above, if you want to add or subtract radicals, they have to be alike. You can't add or subtract the following the way they are: They aren't alike. It's like two to add 52 + 42 and getting 92. (In other words, you don't.)

But if you simplify the radicals, how to combine them becomes much clearer: Finally, there is Division, but I'll save that for another column because of the standard, above, Simplify radical expressions by rationalizing the denominator.