Showing posts with label #MTBoS. Show all posts
Showing posts with label #MTBoS. Show all posts

Monday, January 01, 2018

2017: The Return of the Itinerant Teacher ... To Being Itinerant

2017: A Teaching Year in Review

There's an adage that annoys teachers: Those who can, do; those who can't, teach. And yet we will also ponder the fate of those who "can't teach" -- often they become "consultants" after a year and a half in the classroom, becoming self-appointed experts at telling the rest of us how to do it.

In my case, it's not a matter of "can't teach" as much as "won't let me". Now, I'm not assigning blame, for a couple of reasons: first, I don't wish to burn bridges, nor kill my own career; second, I'm not entirely blameless in the situation. Stuff happens.

Looking back, 2017 started as a year of promises, and in five and a half months, many planted seeds seemed to be taking root and flowering. Bearing fruit, if you will. And then someone razed the garden.

In December 2017, after a few months in the ATR (Absentee Teacher Reserve) Pool, subbing for a few different schools, I received an inquiry about a math position in Park Slope. My previous school year had been dreadful, resulting in a "Developing" rating and the loss of nearly 20 pounds. I had been happy to be in the pool for those months, subbing for other teachers, occasionally pushing into and assisting in math classes, and generally covering whatever needed covering. As restful as this was, however, I couldn't see this being a career path for me, so I looked into the school that had contacted me.

There was several pluses that stuck out: it was in my old neighborhood, which was easy enough to get to; it was a small school inside a big building, shared with other small schools; and I had previously worked in the building, so I knew that there wasn't a history of "trouble" in the halls and stairwells.

The principal and AP were nice people, answered my questions as well as they could. Note that my first question, considering previous experiences, was "Is this position provisional or permanent?" Was I being hired to teach math at the school or to fill a void until the end of the year?

Honestly, I couldn't tell you at the time which answer I expected or even which I would've preferred, so long as I had an answer and knew what to expect. I was told that he had been told by "downtown" that whomever he hired that's who he hired. That was it.

So in January 2017, I was making progress getting to know my new Algebra and Geometry students, just as they were making progress. I found strengths in some low-performing students. A few others were suddenly "turned on" to math, which they hadn't been in middle school. And I made important connections with a couple of students who seemed troubled in their own ways. I found ways to reach them, to be able to talk to them (somewhat). They knew they could talk to me if anything was troubling them, disrupting their class time, distracting them from learning.

I made enough progress that I was once able to say to a reluctant talker, "I'm sorry but I only have a minute, so can we jump to the point where you start speaking to me?"

It was abrupt, but she opened up right away and asked what she had to, and I was able to answer her. Next time, I let her go back to her natural process before speaking, but I noted she spoke up a little sooner.

By spring, we could practically have a conversation without prompting. By June, she told me that she saw me as a kind of "mentor" figure. That same day I was called into the office for a meeting. I thought it was a post-observation conference or the end-of-year review (a little early). When I saw the ELL teacher in the room, I wondered if there had been an issue with one of my students.

No, the ELL teacher was also the union rep. She was there because I was being excessed. Let go. From a "permanent" position. I chose my words carefully because I didn't want to end my career. And I might have. I felt betrayed. And I felt I was betraying my students who had made connections with me and whom I thought I'd see again the following school year.

In the principal's mind, the position hadn't been filled yet. He was still searching, not that he'd ever mentioned this in the prior six months. Any protest or argument from me was cut off with a simple question: would I like to still be considered for the position?

As much as I wanted to tell him to shove it -- OF COURSE, I WANTED TO BE CONSIDERED FOR THE POSITION. I thought it was MY position. I was already making plans for next year, what to change in the curriculum, how to approach Geometry with my current Algebra students, how I wanted to redo the classroom. That ended it. Except I knew I wasn't going to be considered or we wouldn't be having the conversation. I'm too expensive, and new teachers are easy on the budget.

So I started making my good-byes to some of the students. I hope they do well. A couple of them were following me on social media and still contacted me early in the fall semester, but the contact has fallen off. I hope they found new mentors, ones that aren't going anywhere.

This would be a good point for a musical interlude, in place of my summer break, so I can stop rambling and recollect my thoughts.

The end of the summer brought another plot twist: the city and the union agreed to change the rules about placement. If I didn't find a position, if a school did not hire me, then there was a good chance that the city would select an open position somewhere in Brooklyn (possibly beyond) and place me there anyway, whether or not I wished to go there and without regard to the school's desires or ability to fit me in their budget. And that placement would be permanent unless -- here's the catch -- the teacher received a developing or unsatisfactory rating!

So what did this mean for me as a teacher: I could find myself in a horrible school, in a horrible neighborhood, in a horrible situation like the one that caused me to drop to a weight beginning with the number 1 (which I hadn't seen since the year started with the number 1). I might not be able to get out of it, if I couldn't find an open position. Moreover, if the school didn't want me there, then they could make my life miserable so I'd get an unsatisfactory rating!

Granted, it was no picnic for the schools, either, and it was no surprise that I got a few inquiries after this announcement, while at the same time, open positions started disappearing. I applied to quite a few places, but, sorry, I wasn't going back to middle school, and I wasn't traveling to the Bronx. (Note: it's a minimum two-hour travel, one-way, by subway to get to the Bronx from my house.)

Making my search more desperate, an email arrived stating that my first temporary assignment was at Cobble Hill High School for American Studies. Nice neighborhood, not a lot of teenagers in it. I spent one single week at that school and I rated it as probably the worst week of my teaching career. That week was an absolute disaster, and I didn't want to repeat it. I only name the school so I can give credit where it's due. I had maybe 1 or 2 difficult assignments during my eight weeks there in September and October. While the school still has its share of problems, I didn't encounter anything like the last time. And I'll give a shout-out to Stephanie (I hope I spelled it right), for taking good care of me and the other ATRs assigned there. On my last day there, I told her that I wish I could take her with me.

Okay, so what about that forced assignment? It might still be coming for all I know. We keep hearing that math is a shortage area. And there are openings, but schools are still playing games.

In mid-September, a former colleague, now an assistant principal reached out to me through the DOE email and through Facebook (we're not "friends" on Facebook) to let me know about an opening in Queens. I wasn't thrilled with the idea of Queens, but with the future uncertain, it paid to check it out. Basically, it was a temporary position for someone who might be coming back soon, might be filing an extension, or might be retiring (after exhausting extensions). I met with the principal, saw the classrooms, spoke with some students, and, honestly, couldn't think of a reason not to be there, except the location and the travel time wasn't that bad. I was "basically" hired right there and they were going to put the paperwork through. By the middle of the following week, I got an email from Datacation, saying that an account had been set up for me with the online grading system at the school. I checked online and saw the rosters for four of my five classes. I just waited for the call to report.

And I waited.

And I waited.

And I checked the online grading system. I was still in there, but I only had one class, and it didn't have any students in it.

I can only assume that the teacher returned to work. You might think that someone might've informed me of this, especially after they pursued me and framed it as doing them a big favor. I was more disappointed about the snub then about the loss of the position.

Speaking of snubs. When the summer was drawing to a close and prospects were dwindling, I reached out to my old AP at the school I taught at for a decade (most of that before the current AP was there). The school that had excessed me -- twice. Why would I do such a thing? Because I still go back to their end-of-year parties to say good-bye to retiring colleagues. Last June, that school lost two-thirds of (non-ISS) math teachers. That is, they lost two out of three. Now considering that the AP of Operations told the Summer School principal that I am the "go to" guy for math, you might think that they would give me a call about coming back there.

You would be wrong. The call never came. I am Facebook friends with my former AP (although I have reason to believe that she's "muted" me), so I knew that she and her husband were on a European vacation this summer. When she returned, I contacted her through DOE email (not through Facebook -- that would be tacky). No response. Not even, "we have somebody." And here's the thing, she was at the end-of-year party. We'd spoken. She heard about what had happened to me, so she knew that I was available. We didn't discuss it -- again, it was a party and that would be tacky.

One last one: One school I contacted in the summer didn't return my messages, and then it no longer listed an opening. Early October they send me an email telling me that they've scheduled me for an interview AND a 20-minute demo lesson with such-and-such parameters and expectations, and it would be the day after I read the email. EXCUSE ME? I emailed back, "Sorry, but tomorrow is my annual check-up, and I need to be in the doctor's office. Can we reschedule?" They never replied.

So that's the way my year was going. I'm currently at a nice school close to my home. The UFT representative even approached me about staying there on my first day. If he asks me on my last day, I might ask him to see what he can do.

Life can be easy for someone in my situation if you can roll with it, but I can't see keeping myself afloat like this for another ten to fifteen years.

So while I'm happy for a relaxing end to 2017, I'm hoping to a more satisfying 2018.

Thursday, February 09, 2017

January 2017 Common Core Algebra I 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.
Part II was posted here. Part III was posted here. Part IV was posted here.

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

This question was worth 6 credits

1.Which expression is equivalent to 16x2 - 36?

(2) 4(2x + 3)(2x - 3). This question was a giveaway. It the Difference of Squares which can be factored into Conjugates, that is, the same binomial except that one has a plus and the other a minus. Only one choice has that!
Additionally, if you forgot how to factor, you could just multiply the choices and see which gives you the original expression.
You could factor this by dividing by 4, getting 4(4x2 - 9) which becomes 4(2x + 3)(2x - 3).
Or you could have factored it into (4x + 6)(4x - 6), each of which could have a factor of 2 taken out of them.

2.What is the solution set of the equation (x - 2)(x - a) = 0?

(3) 2 and a. Flip the signs: 2 - 2 = 0 or a - a = 0. Don't let the a throw you off.

3.Analysis of data from a statistical study shows a linear relationship in the data with a correlation coefficient of -0.524. Which statement best summarizes this result?

(4) There is a moderate negative correlation between the variables. Hopefully, you immediately eliminated the "positive" choices. A coefficient of -0.5 (or +0.5) can best be described as "moderate", as opposed to "strong" or "weak".

4.Boyle's Law involves the pressure and volume of gas in a container. It can be represented by the formula
P1V1 = P2V2. When the formula is solved for P2, the result is

(3) P1V1 / V2. Divide both sides by V2 to isolate P2

5.A radio station did a survey to determine what kind of music to play by taking a sample of middle school, high school, and college students. They were asked which of three different types of music they prefer on the radio: hip-hop, alternative, or classic rock The results are summarized in the table below.

Hip-Hop Alternative Classic Rock
Middle School 28 18 4
High School 22 22 6
College 16 20 14
What percentage of college students prefer classic rock?

(2) 28%. The first two rows are irrelevant. Only the college row is important. There were a total of 50 college students surveyed and only 14 prefer classic rock. That's 28%.

6.Which function has zeros of -4 and 2?

(4) The graph shows the function crossing the x-axis at -4 and 2. Not a trick question.
The equation for the function would involve multiplying the factors (x + 4)(x - 2), which would have a middle term of 2x, not 7x nor -7x.
Likewise, if the zeros are -4 and 2, then the Axis of Symmetry must be -1, which is exactly in the middle. If you use the formula x = (-b/2a), it is obvious that b cannot be either 7 or -7.

7.Which expression is equivalent to 2(3g - 4) - (8g + 3)?

(4) -2g - 11.

2(3g - 4) - (8g + 3)
6g - 8 - 8g - 3
-2g - 11

8.In 2014, the cost to mail a letter was 49¢ for up to one ounce. Every additional ounce cost 21¢. Which recursive function could be used to determine the cost of a 3-ounce letter, in cents?

(1) a1 = 49; an = an-1 + 21. The first ounce is 49 cents, and each additional is 21.

9.A car leaves Albany, NY, and travels west toward Buffalo, NY. The equation D = 280 - 59t can be used to represent the distance, D, from Buffalo after t hours. In this equation, the 59 represents the

(2) speed of the car. Every hour, the car is 59 miles closer because it is traveling at 59 miles per hour.

10. Faith wants to use the formula C(f) = (5/9) (f - 32) to convert degrees Fahrenheit, f, to degrees Celsius, C(f). If Faith calculated C(68), what would her result be?

(1) 20o Celsius. First of all, you are converting from Fahrenheit to Celsius, so choices (2) and (4) are right out. Second, when it comes to regular temperatures (not extreme cold), Celsius are lower than Fahrenheit. (They are the same at 40 below zero!) So you didn't even have to do the math to answer this. If you've been to the Caribbean in the winter, it's been in the 20s there and it's beach weather!

If you didn't realize that, there's the mathematical answer:

C(68) = (5/9)(68 - 32) = (5/9)(36) = (5)(4) = 20, or
C(68) = (5/9)(68 - 32) = (5/9)(36) = (180)/(9) = 20

11.Which scenario represents exponential growth?

(3) A species of fly doubles its population every month during the summer. Doubling is exponential. The rest are linear.

12.What is the minimum value of the function y = |x+ 3| - 2?

(1) -2. The vertex of this absolute value graph is (-3, -2). The minimum value is -2.

13.What type of relationship exists between the number of pages printed on a printer and the amount of ink used by that printer?

(2) positive correlation, and causal. The more pages printed, the more ink is used. And one causes the other.
And non-causal positive relationship would be something like "ice cream sales go up and swimsuit sales go up". Both go up in the warmer weather.

14.A computer application generates a sequence of musical notes using the function f(n) = 6(16)n, where n is the number of the note in the sequence and f(n) is the note frequency in hertz. Which function will generate the same note sequence as f(n)?

(2) h(n) = 6(2)4n. Forget everything in this question about music. You don't need to know it.
What you need to know is that 6(16)n = 6(2)4n.
16 is the same as 24
So (16)n is the same as (24)n
Rules of exponents say that you multiply the 4 and the n, which gives (2)4n.

15. Which value of x is a solution to the equation 13 - 36x2 = - 12?

(4) -5/6
You can rewrite the equation as 36x2 - 25 = 0, which is a Difference of Squares.
That factors into conjugates: (6x + 5)(6x - 5). For either binomial to equal zero under the Zero Product Property, x would have to equal positive or negative 5/6.

16.Which point is a solution to the system below?

2y < -12x + 4
y < -6x + 4

(4) (-3,2)
Look for a point that fits both inequalities. You can plug the points into both equations, or you can graph the lines in your graphing calculator.
1/2 is NOT less than -6(1) + 4
6 is NOT less that -6(0) = 4
5 IS less than -6(-1/2) + 4, which is 7, BUT 2(5) = 10 is NOT less than -12(-1/2) + 4, which is also 10. (They're equal.)
Choice (4) is left. 2 IS less than (-6)(-3) + 4 which is 22 AND 2(2) = 4 IS less than (-12)(-3) + 4, which is 40.

17.When the function f(x) = x2 is multiplied by the value a, where a > 1, the graph of the new function, g(x) = ax2

(2) opens upward and is narrower. Because a > 1, it is positive, so the parabola opens upward. When a > 1, the parabola shoots up faster, and as a result, it will be narrower.

18.Andy has $310 in his account. Each week, w, he withdraws $30 for his expenses. Which expression could be used if he wanted to find out how much money he had left after 8 weeks?

(4) 280 - 30(w - 1). This is a silly answer, but it is mathematically correct.
The answer you would expect is 310 - 30w, because 30 is subtracted each week.
First notice that only two choices have w multiplied by 30, and one of them is adding money.
After 1 week, 310 becomes 280. The number of weeks after that is reduced by 1, so w - 1.

19.The daily cost of production in a factory is calculated using c(x) = 200 + 16x, where x is the number of complete products manufactured. Which set of numbers best defines the domain of c(x)?

(4) whole numbers. Complete products would have to be positive integers or zero, which is the set of whole numbers. Negative numbers and fractions (real and rational numbers) wouldn't make sense.

20. Noah conducted a survey on sports participation. He created the following two dot plots to represent the number of students participating, by age, in soccer and basketball .

Which statement about the given data sets is correct?

(4) The data for basketball players have a greater mean than the data for soccer players.
Without doing the calculations, you can see that the data for the soccer players skew left and while the data for the basketball players skew right. So the mean and the median are both going to be higher for the basketball players. (You can check the median by counting the dots quickly enough if you don't believe me.)

21.A graph of average resting heart rates is shown below. The average resting heart rate for adults is 72 beats per minute, but doctors consider resting rates from 60-100 beats per minute within normal range.

Which statement about average resting heart rates is not supported by the graph?

(1) A 10-year-old has the same average resting heart rate as a 20-year-old.

According to the graph, a 10-year-old's rate is higher (approximate 90 beats/min) than a 20-year-old (72 beats/min).

22.The method of completing the square was used to solve the equation 2x2 - 12x + 6 = 0. Which equation is a correct step when using this method?

(1) (x - 3)2 = 6

2x2 - 12x + 6 = 0
x2 - 6x + 3 = 0
x2 - 6x + 3 + 6 = 0 + 6
x2 - 6x + 9 = 6
(x - 3)2 = 6

To complete the square, first divide all the terms by 2 to get rid of the leading coefficient. Next, look at the middle term, which in this case is now 6. Half of 6 is 3, and 3 squared is 9. We need to have 9 as the constant to have a perfect square. To get 9, we have to add 6 to both sides of the equation. (I sometimes teach students to subtract 3 and then add 9 if they can't figure out the square that they need. It's an extra step, but they can get the correct solution, so it's worth it.)

Once we have x2 - 6x + 9, we can factor that into (x - 3)(x - 3) or simply (x - 3)2, which is the square we wanted to complete. Hint: Once we knew that half of 6 is 3, we also knew that (x - 3)2 would be in our answer.

23.Nancy works for a company that offers two types of savings plans. Plan A is represented on the graph below.

Plan B is represented by the function f(x) = 0.01 + 0.05x2, where x is the number of weeks. Nancy wants to have the highest savings possible after a year. Nancy picks Plan B.
Her decision is

correct, because Plan B is a quadratic function and will increase at a faster rate .
First, x2 is quadratic, not exponential. Second, check to make sure that choice (3), "Plan A will have a higher value after 1 year" is incorrect. Although the function will increase at a faster rate, it starts off with a much lower rate. Check that it will catch up.

f(52) = 0.01 + 0.05(52)2 = 135.21. On the graph, Plan A gives an amount between $100 and $110 at 52 weeks.

24. The 2014 winner of the Boston Marathon runs as many as 120 miles per week. During the last few weeks of his training for an event, his mileage can be modeled by M(w) = 120(.90)w-l, where w represents the number of weeks since training began. Which statement is true about the model M(w)?

(3) M (w) represents the total mileage run in a given week.
The number of miles is decreasing by 10% per week, or is 90% of the previous week. This may sound counterproductive in training if you don't know that a marathon is only about 26 miles. It is stated in the question that w is the number of weeks since training began and not the number of weeks left until the marathon.

End of Part I

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

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Wednesday, February 08, 2017

January 2017 Common Core Algebra I 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.
Part II was posted here. Part III was posted here.

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

This question was worth 6 credits

37.Ian is borrowing $1000 from his parents to buy a notebook computer. He plans to pay them back at the rate of $60 per month. Ken is borrowing $600 from his parents to purchase a snowboard. He plans to pay his parents back at the rate of $20 per month.
Write an equation that can be used to determine after how many months the boys will owe the same amount.

Determine algebraically and state in how many months the two boys will owe the same amount.
State the amount they will owe at this time.

Ian claims that he will have his loan paid off 6 months after he and Ken owe the same amount.
Determine and state if Ian is correct. Explain your reasoning.

First part, the two equations:
Ian: y = 1000 - 60x, where x is the number of months and y is the remaining balance owed.
Ken: y = 600 - 20x.

They will be equal at 1000 - 60x = 600 - 20x

Second part: Solve the equation. Note, if you're equation is incorrect, solve it anyway because you will still get some credit!

1000 - 60x = 600 - 20x
1000 = 600 + 40x
400 = 40x
x = 10 months
y = 600 - 20(10) = 600 - 200 = $400 still owed.

Third part: Solve Ian's equation for y = 0, when no money is owed.

1000 - 60x = 0
1000 = 60x
x = 16.66666.... or 17 months.

Ian will not be paid off 6 months after he and Ken owe the same amount. He will still owe money.

Alternatively, six months after 10 is 16 months,
and y = 1000 - 60(16) = 1000 - 960 = 40. He will still owe $40 at 16 months, so he is not paid off.

End of Part IV

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

Tuesday, January 31, 2017

January 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.
Part II was posted here.

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

33. Graph f(x) = | x | and g(x) = -x2 + 6 on the grid below.
Does f(-2) = g(-2)? Use your graph to explain why or why not.

Note that if you explain using substitution instead of graphing, you will not get credit.
The graph looks like this:

So f(-2) = g(-2) because f(-2) = 2 and g(-2) = 2. The two graphs intersect at the point (-2, 2).

34. Two friends went to a restaurant and ordered one plain pizza and two sodas. Their bill totaled $15.95. Later that day, five friends went to the same restaurant. They ordered three plain pizzas and each person had one soda. Their bill totaled $45.90.
Write and solve a system of equations to determine the price of one plain pizza.
[Only an algebraic solution can receive full credit.]

Let p = the price of one pizza and let s = the price of one soda. The system of equations would then be:

1p + 2s = 15.95
3p + 5s = 45.90

You can solve this system using elimination by multiplying the first equation by -3, or by substitution if you solve the first equation for p by subtracting 2s from both sides.

Elimination first:

-3p - 6s = -47.85
3p + 5s = 45.90
-s = -1.95
s = 1.95
Plug 1.95 into either equation:
p + 2(1.95) = 15.95
p + 3.90 = 15.95
p =12.05
The price of a pizza is $12.05.

Substitution:

p = -2s + 15.95
3(-2s + 15.95) + 5s = 45.90
-6s - 47.85 + 5s = 45.90
-s - 47.85 = 45.90
-s = -1.95
s = 1.95
Plug 1.95 into either equation:
p + 2(1.95) = 15.95
p + 3.90 = 15.95
p =12.05
The price of a pizza is $12.05.

Yes, that is an unusual amount for a pizza, but they didn't want to make it "obvious". $1.95 is a reasonable price for a soda.

35. Tanya is making homemade greeting cards. The data table below represents the amount she spends in dollars,f(x), in terms of the number of cards she makes, x.


Write a linear function, j(x), that represents the data.
Explain what the slope and y-intercept of j(x) mean in the given context.

You need to find the slope and the y-intercept, which means the rate of change and the initial value.

Pick any two points on the table. Let's say you hate decimals, so you pick (6, 9) and (10, 12). Rate of change is the difference of the y values divided by the difference of the x-values. That is, (12 - 9) / (10 - 6) = 3 / 4 or .75. (You might want to use the decimal for the calculator in the next step.)

Next, you need the y-intercept. Start with y = mx + b and plug in values for x, y and m:

12 = (.75)(10) + b
12 = 7.5 + b
4.5 = b

The function is f(x) = .75x + 4.5

You need to explain the answer in context of the problem. $4.50 is the initial value, the starting cost for making the cards, and $0.75 is the cost per card.

36. Alex launched a ball into the air. The height of the ball can be represented by the equation h = -8t2 + 40t + 5, where his the height, in units, and t is the time, in seconds, after the ball was launched. Graph the equation from t = 0 to t= 5 seconds.


State the coordinates of the vertex and explain its meaning in the context of the problem.

The graph will look like this:


Notice that the vertex occurs between 2 and 3, so it must be 2.5 (which is exactly in the middle).

To find the height at t=2.5, substitute into the equation, and enter it into your calculator:
h = -8(2.5)2 + 40(2.5) + 5 = 55
The vertex is (2.5, 55). At 2.5 seconds the ball reaches its highest point, which is 55 feet units in the air.

EDIT: the units are "units" not feet. You don't have to write "units", but if you write "feet" you will lose a point for misidentifying the units.
The problem is in neither feet nor meters. A gravity problem dealing using feet would usually start with -16t2. A problem with meters would usually start with a leading coefficient of -4.9 or -5, depending upon the desired precision.

End of Part III

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

Monday, January 30, 2017

January 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.

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

25. In attempting to solve the system of equations y = 3x - 2 and 6x - 2y = 4, John graphed the two equations on his graphing calculator. Because he saw only one line, John wrote that the answer to the system is the empty set. Is he correct? Explain your answer.

John is not correct. The answer is not the empty set. The answer is that the two lines are coincident (they overlap each other), so there are an infinite number of points that solve this system of equations.

Proof: 6x - 2y = 4
-2y = -6x + 4
y = 3x - 2, which is the other line.
The two lines are coincidental lines.

The empty set would mean that the lines never intersected; i.e., they are parallel and have no points in common.

26. A typical marathon is 26.2 miles. Allan averages 12 kilometers per hour when running in marathons. Determine how long it would take Allan to complete a marathon, to the nearest tenth of an hour. Justify your answer.

"Justify" means that you need to show your work or offer some kind of proof.
This problem is about unit conversion, and the conversion you need is in the back of the book: 1 mile = 1.60943 km
26.2 miles = 26.2(1.60943 km/mi) = 42.164708 km
42.164708 km / (12 km/hr) = 3.5137256666....
Which is 3.5 to the nearest tenth of an hour.

27. Solve the inequality below: 1.8 - 0.4y > 2.2 - 2y

Solve inequalities like you solve equations, but be mindful of dividing by negative numbers.


1.8 - 0.4y > 2.2 - 2y
-0.4y > 0.4 - 2y
-0.4y > 0.4 - 2y
1.6y > 0.4
y > 0.25

Unusual that they didn't multiply or divide by a negative, but watch out for it.

28. Jakob is working on his math homework. He decides that the sum of the expression 1/3 + 6*sqrt(5)/7 must be rational because it is a fraction. Is Jakob correct? Explain your reasoning.

Jakob is not correct. For a fraction to be rational, both the numerator and the denominator must be rational numbers. Six radical Five is an irrational number, and dividing it by 7 doesn't make it rational.

Finally, the sum of any rational number (e.g., 1/3) and an irrational number is always irrational.

29. Graph the inequality y > 2x - 5 on the set of axes below. State the coordinates of a point in its solution.

The graph will be a broken line with a slope of 2 and a y-intercept of -5 that is shaded ABOVE the line.
If you don't remember where to shade, here is the simple test: take the point (0, 0) and substitute it into the inequality. If the result a true statement? Is 0 > 2(0) - 5? Yes, 0 > -5, so that point is in the shaded region. Because that point is above the line, you will shade above the line. Like this:

Remember to state ANY point in the solution. (0, 0) would work. Or (-10, 0). Or (0, 10). Any point in the shaded section that is NOT on the line. The line is NOT part of the solution because it is broken.

30. Sandy programmed a website's checkout process with an equation to calculate the amount customers will be charged when they download songs. The website offers a discount. If one song is bought at the full price of $1.29, then each additional song is $.99. State an equation that represents the cost, C, when s songs are downloaded.
Sandy figured she would be charged $52.77 for 52 songs. Is this the correct amount? Justify your answer.

The linear equation that models this problem is (Note that the variables are up to you): C = 1.29 + .99(n - 1)
You pay $1.29 for the first song. If you buy n song, you will pay 99 cents for n - 1 of them.

To find the cost of 52 songs, substitute 52 for n: C = 1.29 + .99(52 - 1) = 51.78
Sandy is incorrect.

Not that they asked this, but it looks like her mistake is that she calculated C = 1.29 + .99(52), without subtracting 1 first.

31. A family is traveling from their home to a vacation resort hotel. The table below shows their distance from home as a function of time.

Time (hrs) 0 2 5 7
Distance (mi) 0 140 375 480

Determine the average rate of change between hour 2 and hour 7, including units.

Average rate of change between points (2, 140) and (7, 480) is the difference of the y values divided by the difference of the x values:
(480 - 140) / (7 - 2) = 340 / 5 = 68 mi / hr.

Note that the question said to include units, so you better include the units or you will lose a point!

32. Nora says that the graph of a circle is a function because she can trace the whole graph without picking up her pencil.
Mia says that a circle graph is not a function because multiple values of x map to the same y-value.
Determine if either one is correct, and justify your answer completely.

Neither one is correct. Nora is wrong because not lifting a pencil is not the definition of a function. Piecewise functions sometimes require you to lift your pencil. Mia is wrong because she has it backwards: there are multiple values of y for the same x, which is the vertical line test. What Mia described would be a horizontal line test, but there is no such thing for functions. You are allowed to have multiple values of x map to the same value of y. (For example: the absolute value function repeats y values for more than one x.)

End of Part II

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

Saturday, November 19, 2016

June 2016 Common Core Algebra II Regents, Part 2

I don't usually post the answers to the Algebra II exam, but I've received enough requests that I've decided to post some of the open-ended questions.

June 2016, Algebra II

25. Solve for x: (1 / x) - (1 / 3) = -(1 / 3x)

The answer is best given in picture form, shown below. Basically, you need to find a common denominator to combine the fractions. You can then solve by cross-multiplying.
Alternatively, you could just multiply all the terms by 3x because x, 3, and 3x are all factors of 3x. The multiplication would remove all the fractions from the problem.

x = 4.

26. Describe how a controlled experiment can be created to examine the effect of ingredient X in a toothpaste.

This was really is open-ended. You can write a lot of different things, but there are a few points which will be mandatory. Basically, you needed to have a number of people, (pick a number, say 10 or 20), and assign them randomly to two groups. One of these groups will try the new toothpaste, and the other is the control group which will not use the new toothpaste.

27. Determine if x - 5 is a factor of 2x3 - 4x2 - 7x - 10. Explain your answer.

One way to figure out if it is a factor is to divide the polynomial by (x - 5). If there is no remainder, then it is a factor.
Another is to say, if x - 5 is a factor of the polynomial, then 2x3 - 4x2 - 7x - 10 = 0 when x = 5. Substitute 5 for x and see if you get 0. (You won't.)

The image below shows the result of the division. It does not divide evenly. There is a remainder. So it is NOT a factor.

28. On the axes below, graph one cycle of a cosine function with amplitude 3, period pi/2, midline y = -1, and passing through the point (0,2).

You needed several things for this two points. You would lose 1 point for one mistake, and 2 points for 2 or more mistakes.
You needed to have a scale to the graph. It needed to pass through (0, 2) and have the proper midline (y = -1)

29. A suburban high school has a population of 1376 students. The number of students who participate in sports is 649. The number of students who participate in music is 433. If the probability that a student participates in either sports or music is 974/1376, what is the probability that a student participates in both sports and music

The total number of students who play either sports or music is equal to the number who play sports PLUS the number who play music MINUS the ones who play BOTH. If you just add the first two numbers, then you will double count the students who participate in both activities.

You are given the total that play one or the other. You need to calculate the number who play both. Add the two numbers (649 + 433) and subtract 974 from that sum of 1082, which gives you 108. However, since this is a probability question, you need to write it as a fraction of the student population, or 108/1376.

The image shows what to do:

30. The directrix of the parabola 12(y + 3) = (x - 4)2 has the equation y = -6. Find the coordinates of the focus of the parabola.

The directrix of a parabola is a horizontal line running below the vertex. (Or above, if it opens down.) The focus is a point such that the distance from it to the parabola is equal to the vertical distance from the parabola to the directrix.

To find the coordinates of the focus, you need to find the coordinates of the vertex, which would be exactly in the middle on a straight line. The focus will have the same x-coordinate as the vertex. The focus will be as far above the vertex as the directrix is below.

Rewrite the equation 12(y + 3) = (x - 4)2
Divide both sides by 12: y + 3 = (1/12)(x - 4)2
Subtract 3 from both sides: y = (1/12)(x - 4)2 - 3
The vertex (h, k) is at (4, -3)
The directrix is y = -6. (-6) - (-3) = 3.
Add 3 to the vertex y-coordinate: (-3) + 3 = 0, which must be the y-coordinate of the focus.
The focus is at (4, 0).

31. Algebraically prove that (x3 + 9)/(x3 + 8) = 1 + (1)/(x3 + 8), where x =/= -2.

Change 1 into (x3 + 8)/(x3 + 8). Do the addition. The left and right sides are equal.

32. A house purchased 5 years ago for $100,000 was just sold for $135,000. Assuming exponential growth, approximate the annual growth rate, to the nearest percent.

Set up the compound interest equation, A = P(1 + r/n)^(nt).
Substitute what you know, A = 135,000, P = 10000, n = 1 (annual), t = 5 years: 135000 = 100000(1 + r/1)^((1)(5))
Solve for r:


135000 = 100000(1 + r)^5
1.35 = (1 + r)^5
1.35^(1/5) = 1 + r
1.06185875879 = 1 + r
0.06185875879 = r

r = 6%

End of Part II

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

Saturday, October 22, 2016

August 2016 Common Core Algebra II Regents Part 4

I don't usually post the answers to the Algebra II exam, but I've received enough requests that I've decided to post some of the open-ended questions.
Some questions were posted earlier here, here, here and here

August 2016, Algebra II Part 4

37. Seth’s parents gave him $5000 to invest for his 16th birthday. He is considering two investment options. Option A will pay him 4.5% interest compounded annually. Option B will pay him 4.6% compounded quarterly.
Write a function of option A and option B that calculates the value of each account after n years.

Seth plans to use the money after he graduates from college in 6 years. Determine how much more money option B will earn than option A to the nearest cent.

Algebraically determine, to the nearest tenth of a year, how long it would take for option B to double Seth’s initial investment.

Not an overly difficult Part 4 question, but it is a little involved.

For the first part, be sure to write a function and not just an expression. The initial amount is $5000, the rates are 0.045 and 0.046, respectively. Because the second one is quarterly, you have to divide the rate by 4 and multiple the exponent, n, by 4. (You would not be incorrect if you did this in the first function, using 1.)
A = 5000(1 + 0.045)n
B = 5000(1 + 0.046/4)4n

Use the two functions you just wrote, with n = 6. Write down both amounts and then subtract.
A = 5000(1 + 0.045)6 = 6511.30062424 = 6511.30
B = 5000(1 + 0.046/4)(4)(6) = 6578.86985121 = 6578.87
6578.87 - 6511.30 = 67.57

Doubling the $5000 investment means the future value will be $10000. Put this in the second function, and solve for n.
10000 = 5000(1 + 0.046/4)4n -- divide both sides by 5000
2 = (1 + 0.046/4)4n -- change the rational expression into a decimal
2 = (1.0115)4n -- solve using logs
4n = log1.01152 -- put this into your calculator
4n = 60.6196... -- divided by 4
n = 15.1549
In 15.2 years, Seth's investment will double.


End of exam




Any questions?

Thoughts, comments, concerns?

Saturday, October 08, 2016

August 2016 Common Core Algebra II Regents Part 3

I don't usually post the answers to the Algebra II exam, but I've received enough requests that I've decided to post some of the open-ended questions.
Some questions were posted earlier here, here and here.

August 2016, Algebra II Part 3

33. Find algebraically the zeros for p(x) = x3 + x2 - 4x - 4.

On the set of axes below, graph y = p(x).

Group the first two terms and the last two terms of the expression. Note that this causes a sign change in the second group:

p(x) = (x3 + x2) - (4x + 4)

Factor each pair:
p(x) = x2(x + 1) - 4(x + 1)

Do you see that each pair has (x + 1) in it? That can be factored, and put up front:
p(x) = (x + 1)(x2 - 4)

Almost done. the x2 expression is the Difference of Squares and can be factored further:
p(x) = (x + 1)(x + 2)(x - 2)

Set p(x) = 0
0 = (x + 1)(x + 2)(x - 2)

(x + 1) = 0 or (x + 2) = 0 or (x - 2) = 0

x = -1 or x = -2 or x = 2

That's your answer.

Put the equation in your calculator to get a table of values. From part 1, you know you have these points:
the roots (-1, 0), (-2, 0) and (2, 0) as well as (0, -4), the y-intercept.
You need x = 1. p(1) = (1)3 + (1)2 - 4(1) - 4 = 1 + 1 - 4 - 4 = -6. So (1, -6) is a point on the curve.
Important: You need to see the curve, so use the calculator!





34. One of the medical uses of Iodine–131 (I–131), a radioactive isotope of iodine, is to enhance x-ray images. The half-life of I–131 is approximately 8.02 days. A patient is injected with 20 milligrams of I–131. Determine, to the nearest day, the amount of time needed before the amount of I–131 in the patient’s body is approximately 7 milligrams.

Given the fractions, exponents and subscripts in this problem, I'm going to use an image file to show my work.

Basically, you need to use the formula for exponential decay: y = a(b)x
In this case, y = 7, your target amount, a = 20, the initial amount, b = .5, which is half, and x = t/8.02 days because you are finding the number of 8.02 day increments.
Finding the exponent means using log.5 7/20 = 1.5145731728...
Multiply by 8.02 and round down. The answer is 12 days.






Any questions?

Thoughts, comments, concerns?

More to come.

Tuesday, October 04, 2016

August 2016 Common Core Algebra II Regents Part 2, continued

I don't usually post the answers to the Algebra II exam, but I've received enough requests that I've decided to post some of the open-ended questions.
Some questions were posted earlier here and here.

August 2016, Algebra II Part 2 (continued)

31. The distance needed to stop a car after applying the brakes varies directly with the square of the car’s speed. The table below shows stopping distances for various speeds.

Determine the average rate of change in braking distance, in ft/mph, between one car traveling at 50 mph and one traveling at 70 mph.
Explain what this rate of change means as it relates to braking distance.




32. Given events A and B, such that P(A) = 0.6, P(B) = 0.5, and P(A ∪ B) = 0.8, determine whether A and B are independent or dependent.

P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
0.8 = 0.6 + 0.5 - P(A ∩ B)
0.8 = 1.1 - P(A ∩ B)
-0.3 = - P(A ∩ B)
0.3 = P(A ∩ B)

If A and B are independent, then P(A ∩ B) = P(A) * P(B)
0.3 = (0.6)(0.5)
0.3 = 0.3
Therefore, A and B are independent.

End of Part II.

to be continued.

Any questions?

Comments, corrections welcome.

Monday, September 26, 2016

August 2016 Common Core Algebra II Regents Part 2, continued

I don't usually post the answers to the Algebra II exam, but I've received enough requests that I've decided to post some of the open-ended questions.
Some questions were posted earlier here.

August 2016, Algebra II Part 2 (continued)

28.Using the identity sin2 O + cos2O = 1, find the value of tan O, to the nearest hundredth, if cos O is –0.7 and O is in Quadrant II.

Copy the identity. Substitute the value for cosine and then square it. Solve for sine. Remember, because it is in Quadrant II where the y values are positive, we need a positive answer for sine. Once we know sine and cosine, we can divide to find tangent.
sin2 O + cos2O = 1
sin2 O + (-0.7)2 = 1
sin2 O + .49 = 1
sin2 O = 0.51
sinO = SQRT(0.51) = .7141...
tanO = sinO / cosO = 0.714 / -0.7 = -1.02

You could also have gotten this answer by converting the identity, by dividing all the terms by cos20, which would then give you an equation with tan 2O in it. Again, remember that because it is Quadrant II where x is negative and y is positive, that the tangent must be a negative value.



29. Elizabeth waited for 6 minutes at the drive thru at her favorite fast-food restaurant the last time she visited. She was upset about having to wait that long and notified the manager. The manager assured her that her experience was very unusual and that it would not happen again.

A study of customers commissioned by this restaurant found an approximately normal distribution of results. The mean wait time was 226 seconds and the standard deviation was 38 seconds. Given these data, and using a 95% level of confidence, was Elizabeth’s wait time unusual? Justify your answer.

A 95% level of confidence means that the value should be within 2 standard deviations from the mean.
The mean is 226. Plus 1 deviation = 226 + 38 = 264. Plus another deviation 264 + 38 = 302 seconds.
Six minutes is 360 seconds, which would be unusual because it is more than two standard deviations away from the mean.



30. The x-value of which function’s x-intercept is larger, f or h? Justify your answer.

f(x) = log(x - 4)
x
h(x)
-1
6
0
4
1
2
2
0
3
-2

The x-intercept of h(x) = 2, as shown in the table.
To find the x-intercept of f(x), set f(x) = 0.
log(x - 4) = 0
100 = x - 4
1 = x - 4
5 = x

f(x) has the higher x-intercept.

to be continued.

Any questions?

Comments, corrections welcome.

Saturday, September 24, 2016

August 2016 Common Core Algebra II Regents Part 2

I don't usually post the answers to the Algebra II exam, but I've received enough requests that I've decided to post some of the open-ended questions.

August 2016, Algebra II Part 2

25.The volume of air in a person’s lungs, as the person breathes in and out, can be modeled by a sine graph. A scientist is studying the differences in this volume for people at rest compared to people told to take a deep breath. When examining the graphs, should the scientist focus on the amplitude, period, or midline? Explain your choice.

The scientist should focus on the amplitude because the height of the graph corresponds to the Volume of air in the lungs.

26. Explain how (31/5)2 can be written as the equivalent radical expression "fifth root of 9".

Look at the illustration below:

The question says "explain" so you MUST include an explanation for full credit. You can't just write an equation.
The exponents can be be switched so you have 3 square to the one-fifth power. Three squared is 9. The one-fifth power is the same as taking the fifth root of the number. So (31/5)2 is the same as the fifth root of 9.

27. Simplify xi(i - 7i)2, where i is the imaginary unit.

Subtract i - 7i to get -6i. Square (-6i) to get 36i2. Replace i2 with -1, so you have xi(-36). Rewrite as -36xi.

Illustrated below.

to be continued.

Any questions?

Comments, corrections welcome.

Thursday, September 22, 2016

August 2016 Common Core Geometry Regents, Parts 3 and 4

What follows is a portion of the Common Core Geometry exam. Other parts will be posted on other days. Illustrations will be added at a later time when they become available.

Part 1 is was posted here.
Part 2 is was posted here.

August 2016 Geometry Regents, Part III

32. Using a compass and straightedge, construct and label triangle A'B'C', the image of triangle ABC after a dilation with a scale factor of 2 and centered at B. [Leave all construction marks.]
Describe the relationship between the lengths of AC and A'C'.

Constructions are not easy to show on this blog, but look at the illustration.
The scale factor is two, so you need to double the lengths of the sides. Start at B, open the compass to point C and make an arc. Now move to C without changing the compass and make another arc. Using the straightedge, draw the line from B through C to the new arc. Label that point C'.
Repeat the process for A to create A'. Then use the straightedge to draw A'C'. Finally, at point B, label it B' as well. That point does not move.
The relationship between AC and A'C' is that A'C' = 2AC or AC = 1/2 A'C'. The image is twice the length of the original line.

33. The grid below shows triangle ABC and triangle DEF.
Let trianlge A'B'C' be the image of triangle ABC after a rotation about point A. Determine and state location B' if the location of C' is (8, -3). Explain your answer.

Is triangle DEF congruent to triangle A'B'C'? Explain your answer.

If you plot C'(8, -3), you will see that the triangle has rotated 90 degrees counterclockwise. In a rotation, distance is preserved. C was six units away from A, C' is six units away from A. Point B is one unit above C and four units to the right. After a 90 degree rotation, B' will be one unit to the left of C' and four units above it. B' is at (7, 1)

Triangle DEF is congruent to triangle A'B'C'. This can be shown by finding the lengths of the sides or by doing a series of rigid motions. If A'B'C' is reflected over the y-axis, the image will have points A"(-2, -3), B"(-7, 1), C"(-8, -3). If A"B"C" is translated by (0, -2), then the image will coincide will DEF.

Likewise, you could reflect A'B'C' over the line y = -1 to have it coincide in a single step.

34. As modeled below, a movie is projected onto a large outdoor screen. The bottom of the 60-foot-tall screen is 12 feet off the ground. The projector sits on the ground at a horizontal distance of 75 feet from the screen.


Determine and state, to the nearest tenth of a degree, the measure of , the projection angle

You have two right angles. The first has legs 12 and 75, and the second has legs of 72 and 75. There legs are Opposite and Adjacent to angles with the projector. You can use the Tangent function to find the angles and then find the difference of the two to get theta.

Tan x = 12 / 75
x = tan-1(12/75) = 9.09
Tan x = 72 / 75
x = tan-1(12/75) = 43.83
43.83 - 9.09 = 34.74 = 34.7 degrees.

August 2016 Geometry Regents, Part IV

35. Given: Circle O, chords AB and CD intersect at E

Theorem: If two chords intersect in a circle, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

Prove this theorem by proving (AE) • (EB) = (CE) • (ED)

The procedure to show that theis theorem is true requires you to create two triangles, by drawing extra chords, and then showing that the two triangles are similar in shape. This will make the sides proportional, and by cross-multiplying, you will get the equation in the theorem.

StatementReason
1. Circle O, chords AB and CDGiven
2. AD and BC are drawnAuxiliary lines can be drawn
3. <1 = <2Vertical angles
4. <A = <CInscribed angles that intercept the same arc are congruent
5. Triangle ADE ~ triangle CBEAA
6. AE/CE = ED/EBCorresponding sides of similar triangles are proportional
7. (AE)•(EB) = (CE)•(ED)if a proportion the products of the means equals the product of the extremes

36. A snow cone consists of a paper cone completely filled with shaved ice and topped with a hemisphere of shaved ice, as shown in the diagram below. The inside diameter of both the cone and the hemisphere is 8.3 centimeters. The height of the cone is 10.2 centimeters.

The desired density of the shaved ice is 0.697 g/cm3, and the cost, per kilogram, of ice is $3.83.
Determine and state the cost of the ice needed to make 50 snow cones.

Density is mass divided by the Volume. We know the density. We can calculate the volume. Then we can find the mass. Once we have the mass of 1 snow cones, we can multiply to find the mass of 50 snow cones and then calculate the cost of 50 snow cones. Note: Use the PI key on your calculator. Do NOT use 3.14 -- it doesn't have enough decimal places. Also, do NOT round in the middle of this problem.

Volume of a cone = (1/3) (pi)(r)2h
V = (1/3)(3.141592...)(4.15)2(10.2)=183.96067022

Volume of a hemisphere = (1/2)(4/3)(pi)(r)3
Volume = (1/2)(4/3)(3.141592...)(4.15)3 = 149.693486552
Total Volume = 183.96067022 + 149.693486552 = 333.654156772

d = m / V, m = Vd = (333.654156772)(0.697) = 232.55694727g in 1 snow cone.
Multiply by 50 to 11627.8473635g in 50 snow cones.
Convert to kilograms by dividing by 1000: 11,6278473635kg of ice needed.
Multiply by $3.83: 44.5346554022 = $44.53 is the cost

Yes, this is an insane amount of work for one question, but none is it involves anything really crazy. You just have to stick with it to the end. Remember, in questions like this one, one little error will throw off the rest of the numbers. HOWEVER, if you are consistent all the way to the end, and give an answer that matches the work you did, you may only lose 1 point for a "little" error early on.

END OF PART IV

How did you do? Any questions? (I appreciate pointing out any "typos" in my problems. Thank you.)

Friday, September 16, 2016

August 2016 Common Core Geometry Regents, Part 2

What follows is a portion of the Common Core Geometry exam. Other parts will be posted on other days. Illustrations will be added at a later time when they become available.

Part 1 is was posted here.

August 2016 Geometry Regents, Part II

25. 5 Lines AE and BD are tangent to circles O and P at A, E, B, and D, as shown in the diagram below. If AC:CE = 5:3, and BD = 56, determine and state the length of CD

Because they are tangent lines, AC = BC and CD = CE. So if AC:CE = 5:3, then BC:CD = 5:3 also.
So 5x + 3x = 56
8x = 56
x = 7
CD = 3x = 3(7) = 21.

26. 6 In the diagram below, triangle ABC has coordinates A(1,1), B(4,1), and C(4,5). Graph and label triangle A" B" C" , the image of ABC after the translation five units to the right and two units up followed by the reflection over the line y 0.

First, remember that the line y = 0 is the x-axis, not the y-axis.
Move each point 5 units to the right and 2 units up. Point A (1, 1) will move to A'(6, 3), etc.
Next, flip the points over the x-axis, so point A'(6, 3) becomes A"(6, -3).
You're final image will look like this.

27. A regular hexagon is rotated in a counterclockwise direction about its center. Determine and state the minimum number of degrees in the rotation such that the hexagon will coincide with itself.

A regular hexagon will coincide with itself every 1/6 of a revolution. (1/6)(360) = 60 degrees.

28. In the diagram of ABC shown below, use a compass and straightedge to construct the median to AB. [Leave all construction marks.]

Construction marks are difficult for me to show using the tools I'm currently using, so I will describe the process.
The median is a line drawn from point C to the midpoint of AB, so you need to find the midpoint of AB. The way to do this is to construct the perpendicular bisector of AB, label the midpoint and then draw a line from C to this new point.

The red lines are created by opening the compass so that it is wider than half the distance between A and B. If you make the arcs to small, then they won't intersect. You will have to make the arc bigger and try again. Remember that both arcs MUST be the same size when you center on A and when you center it on B.

Draw the perpendicular bisector, which is the line through the two points where the arcs meet. This line will cross AB at its midpoint.

Draw the median by using the straightedge and marking the line between C and the midpoint.

29. Triangle MNP is the image of triangle JKL after a 120° counterclockwise rotation about point Q. If the measure of angle L is 47° and the measure of angle N is 57°, determine the measure of angle M. Explain how you arrived at your answer.

When the triangles rotate, M is the image of J, N is the image of K and P is the image of L. Size and shape are preserved in a rotation so <M = <J, <N = <K annd <P = <L.
&LtM + &LtN + <P = 180
<M + 47 + 57 = 180
<M + 104 = 180
<M = 76

30. A circle has a center at (1,–2) and radius of 4. Does the point (3.4,1.2) lie on the circle? Justify your answer.

If (3.4, 1.2) is on the circle then the following equation must be true.
(3.4 - 1)2 + (1.2 + 2)2 = 42 ?
(2.4)2 + (3.2)2 = 16 ?
5.76 + 10.24 = 16 ?
16 = 16. Check
(3.4, 1.2) lies on the circle

31. In the diagram below, a window of a house is 15 feet above the ground. A ladder is placed against the house with its base at an angle of 75° with the ground. Determine and state the length of the ladder to the nearest tenth of a foot.

You have a right triangle. You have an angle. You have the opposite side. You want to know the hypotenuse. O-H means you need to use the sine ratio. (Make sure your calculator is in DEGREE mode.
sin 75 = 15 / x or as a proportion: sin 75 / 1 = 15 / x.
cross-multiply: (sin 75) x = 15
x = 15 / sin 75 = 15.5291427062
The ladder is 15.5 feet to the nearest tenth.

END OF PART II

How did you do? Any questions? (I appreciate pointing out any "typos" in my problems. Thank you.)

Friday, September 09, 2016

August 2016 Common Core Geometry Regents, Part 1

What follows is a portion of the Common Core Geometry exam. Other parts will be posted on other days. Illustrations will be added at a later time when they become available.

August 2016 Geometry Regents, Part I

1. In the diagram below, lines l, m, n and p intersect line r.


Which statement is true?

(2) l || p. If two exterior angles on the same side of a transversal are supplementary, then the lines are parallel. The sum of 112 and 68 is 180 degrees.

2. Which transformation would not always proudce an image that would be congruent to the original figure?

(2) dilation. A dilation will change the size of an image, so it will not be congruent. The others only change the orientation.

3. If an equilateral triangle is continuously rotated around ones of its medians, which 3-dimensional object is generated?

(1) cone. One point will stay in play and the other two will form a circle.

4. In the diagram below, m<BDC = 100, m<A = 50, and m<DBC = 30.


Which statement is true?

(2) Triangle ABC is isosceles. BDC = 100, DBC = 30, so <C = 50, because 100 + 30 + 50 = 180. Angle A is also 50. Beacuse angle A = angle C, it is an isosceles triangle.

5. Which point shown in the graph below is the image of point P after a counterclockwise rotation of 90o about the angle?

(1) A. Counterclockwise from Quadrant IV brings you to Quadrant I. Point P is close to the x-axis, so the image will be close to the y-axis, which is where A is. Point B looks like a reflection across the x-axis.

6. In triangle ABC, where <C is a right angle, cos A = SQRT(21) / 5. What is sin B?

(1) SQRT(21) / 5. Because Sin B = Cos A

7. Quadrilateral ABCD with diagonals AC and BD is shown in the diagram below.


Which information is not enough to prove ABCD is a parallelogram?

(3) AB = CD and BC || AD. Either two pairs of parallel sides are needed, or two pair of congruent sides, or one pair that is both parallel and congruent would prove it. However, one pair of parallel sides and a different pair congruent is not sufficient. That information could be true is ABCD was, for example, an isosceles trapezoid.

8. An equilateral triangle has sides of length 20. To the nearest tenth, what is the height of the equilateral triangle?

(3) 17.3. You could guess this one. Draw a triangle and an altitude. Choice (1) 10 is the length of the base of the right triangle you just created. Choice (4) 23.1 is longer than the hypotenuse. It's either (2) or (3), but you know that it will be much bigger than the base. In fact, it will be the base TIMES SQRT (3), which is approximately 1.732. So 10(1.732) = 17.32.

Using the Pythagorean Theorem: x2 + 102 = 202
so x2 + 100 = 400
so x2 = 300
x = 17.3205..., which is 17.3 to the nearest tenth.

9. Given: Triangle AEC, triangle DEF, and FE perpendicular to CE


What is a correct sequence of similarity transformations that shows triangle AEC ~ triangle DEF?

(4) a counterclockwise rotation of 90 degrees about point E followed by a dilation with a scale factor 2 centered at point E. You can see that the figure rotated only 90 degrees, not 180 -- it would be upside down then -- so eliminate choices (1) and (3). Next, a translation would move DEF away from ACE, which didn't occur. However, the size did change, which indicates a dilation.

10. In the diagram of right triangle ABC, CD intersects hypotenuse AB at D.

If AD = 4 and DB = 6, which length of AC makes CD perpendicular to AB?

(1) 2*SQRT(6). If CD is perpendicular to AB, then CD is an altitude. The Right Triangle Altitude Theorem says that the product of AD times DB must equal the length of CD2. Since 4 * 6 = 24, CD must be SQRT(24), which reduces to 2*SQRT(6) in simplest form.

11. Segment CD is the perpendicular bisector of AB at E. Which pair of segments does not have to be congruent?

(4) DE, CE. AB is not said to bisect CD, so you cannot assume that DE and CE are congruent. AE and BE are congruent by definition. The other points would form congruent right triangles, with the corresponding hypotenuses being congruent.

12. In triangle CHR, O is on HR, and D is on CR so that <H = <RDO


If RD = 4, RO = 6 and OH = 4, what is the length of CD?

(3) 11. It may not seem obvious at first that the two triangles are similar because DO is not parallel to CH. However, you are given one pair of angles congruent, and both triangles contain angle R. Therefore, they are similar.

The proper proportion to set up is: RD / HR = RO / CR. And HR = 4 + 6 = 10.
So 4 / 6 = 10 / CR
4*CR = 60
CR = 15
CD = CR - DR = 15 - 4 = 11.
Don't forget to subtract the length of RD from the side of the triangle because they are looking for CD.

13. The cross section of a regular pyramid contains the altitude of the pyramid. The shape of this cross section is a (3) triangle. If the cross section contains the altitude, then imagine a vertical sheet of paper slicing through the top of a pyramid. No matter how you line it up, the sides of the pyramid will form a triangle on the paper.

14. The diagonals of rhombus TEAM intersect at P(2, 10). If the equation of the line that contains diagonal TA is y = -x + 3, what is the equation of a line that contains diagonal EM?

(1) y = x - 1. The lines must be perpendicular, and that means the slope of EM must be 1 because the slope of TA is -1. This eliminates choices (3) and (4). Given P(2, 1), substitute 2 for x in the first equation and you get y = 1, which is correct.

15. The coordinates of vertices A and B of triangle ABC are A(3, 4) and B(3, 12). If the area of triangle ABC is 24 square units, what could be the coordinates of point C?

(3) (-3, 8) Area of a triangle is 1/2 b h. The base has a length of 12 - 4 = 8.
(1/2)(8)h = 24, so 4h = 24 and h = 6.
The third coordinate has to have an x value that is six more or six less than 3. 3 - 6 = -3, so choice (3) works.

16. What are the coordinates of the center and the length of the radius of the circle represented by the equation x2 + y2 - 4x + 8y + 11 = 0?

(1) center (2, -4) and the radius 3. They are making these more difficult. You need to rewrite this in standard form (x - h)2 + (y - k)2 = r2.
To do this, we need to complete the square ... twice.

x2 + y2 - 4x + 8y + 11 = 0
regroup: x2 -4x + y2 + 8y + 11 = 0
complete the squares: x2 -4x + 4 + y2 + 8y + 16 + 11 = 0 + 4 + 16
simplify: (x - 2)2 + (y + 4)2 + 11 = 20
simplify: (x - 2)2 + (y + 4)2 = 9
So the center is (2, -4) and the radius is 3.

If you didn't remember how to complete the square, you could have started by writing an equation for choice (1), expanding it, and seeing if you got the same equation. If it was wrong, it would have been a matter of signs, so if you started with an incorrect choice, you could have deduced the correct one.

17. The density of the American white oak tree is 752 kilograms per cubic meter. If the trunk of an American white oak tree has a circumference of 4.5 meters and the height of the trunk is 8 meters, what is the approximate number of kilograms of the trunk?

(2) 9694. d = m / V or d = m / (pi * r2 * h). C = 2*pi*r so r = C / (2 * pi) = 0.7162
752 = m / (pi*0.716122*8)
m = 752 * (pi*0.716122*8) = 9692.355
which is approximately 9694, with errors for rounding in the middle of the problem.

Yes, this was an unnecessarily complicated problem for a multiple-choice question. They could have give a diameter instead of the circumference, or even just given the radius itself.

(1) 150(0.85)m. The hot chocolate is getting cooler, and the temperature is getting lower. Therefore, the base must be less than one (exponential decay), so choices (2) and (4) are out. If m = 0, then 150(0.85)0 = 150(1) = 150, which fits the table. The correct choice is (1). In choice (3), m - 1 would give an exponent of -1, which would actually increase the temperature at m = 0.

18. Point P is on the directed line segment from point X(-6, -2) to point Y(6, 7) and divides the segment in the ratio 1:5. What are the coordinates of point P?

(4) (-4, -1/2). The ratio 1:5 represents 1x + 5x which is 6x. If the calculate the distance between the two x-coordinates, the x-coordinate of P is 1/6th to the right of the x coordinate of point X. If the calculate the distance between the two y-coordinates, the y-coordinate of P is 1/6th to the above of the y coordinate of point X (not point Y).

From -6 to 6 is 12 spaces, and 1/6(12) = 2; -6 + 2 = -4. The answer is choice (4). You can verify with y.
From -2 to 7 is 9 spaces, and 1/6(9) = 1.5; -2 + 1.5 = -1/2, which is choice (4).

19. Due to a typographical error in the test booklet, there is no correct answer for question 19. I'm repeating it here just in case a teacher accidentally assigns it and a student puts it into a search engine.

In circle O, diameter AB, chord BC and radius OC are drawn, and the measure of arc BC is 108o. Some students wrote these formulas to find the area of sector COB (omitted)
Which students wrote correct formulas?

No answer.

20. Tennis balls are sold in cylindrical cans with the balls stacked one on top of the other. A tennis ball has a diameter of 6.7 cm. To the nearest cubic centimeter, what is the minimum volume of the can that holds a stack of 4 tennis balls?

(4) 945. Volume = (pi)r2h, r = 6.7/2 = 3.35, h = 4(6.7) = 26.8
V = (pi)(3.35)^2(26.8) = 944.87... = 945

21. Line segment A'B', whose endpoints are (4, -2) and (16, 14) is the image of AB after a dilation of 1/2 centered at the origin. What is the length of AB?

(4) 40. You could either find the length and double it (the image is 1/2 the original) or you can double the x- and y-values and then find the length. Your choice. The distance between the x values is 16 - 4 = 12. The distance between the y-values is 14 - (-2) = 16. Using either the distance formula, or Pythagorean theorem, you will find that the distance between the two is 20. (You know that 12-16-20 is just a 3-4-5 right triangle times 4, right?)
Since the image is half the length of the original, the length of AB is 40.

22. Given: Triangle ABE and triangle CBD shown in the diagram below with DB = BE


Which statement is needed to prove triangle ABE = ACBD using only SAS = SAS?

(3) AD = CE. You are given DB = BE. Angle B is congruent to itself by the Reflexive Property. You need to know that AB = BC, but since you already know that DB = BE, then if AD = CE, you can conclude that AB = BC by the Addition Postulate. That will give you SAS = SAS.

23. In the diagram below, BC is the diameter of circle A.


Point D, which is unique from points D and C, is plotted on circle A. Which statement must always be true?

(1) Triangle BCD is a right triangle. Angle D will be an inscribed angle and the arc it intercepts is a semi-circle, which is 180 degrees. Therefore, angle D must be half of that, or 90 degrees. So BCD must be a right triangle, always.

24. In the diagram below, ABCD is a parallelogram, AB is extended through B to E, and CE is drawn.

If CE = BE and m<D = 112o, what is m<E?

(1) 44o. If angle D = 112, then angle A is supplementary, or 180 - 112 = 68. Angle CBE is congruent to A, so it is also 68 degrees. Because BE = CE, triangle BCE is an isosceles triangle and the base angles are congruent, so angle BCE is also 68 degrees. Angle E = 180 - 68 - 68 = 112 - 68 = 44 degrees.

END OF PART I

How did you do? Any questions? (I appreciate pointing out any "typos" in my problems. Thank you.)

More Geometry problems.

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