Thursday, June 22, 2017

June 2017: Common Core Algebra Regents, Part 4

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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




End of Part IV

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

Wednesday, June 21, 2017

June 2017: Common Core Algebra Regents, Part 3

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

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

The answers to Part II can be found here.

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

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

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

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


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

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

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

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

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

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


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

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

The Quadratic Formula is used in the image below:

Your reasons could have been anything like these:
I solved by factoring because I thought it was easy enough to factor and find the zeroes if I used "borrow and payback" (not likely in this example).
I completed the square because I like that approach when the answer may be a fraction or irrational.
I always use the quadratic formula when the equation looks complicated because it always works, even if the answer is irrational.
Or anything like these, so long as it matches the answer you gave.

End of Part III

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

Tuesday, June 20, 2017

June 2017: Common Core Algebra Regents, Part 2

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

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

Answers to Part III can be found here.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

See graph below.

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

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

y = -2(x - 4) - 4

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

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

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

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

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

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

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

End of Part II

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

a || b

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(C)Copyright 2017, C. Burke.

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




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Thursday, June 15, 2017

Geometry Regents Questions? Post them here

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

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

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

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

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

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

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

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

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

Does this help/answer your question?

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

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

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

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

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

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

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

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

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

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

Does this help/answer your question?

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

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

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

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

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

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

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

Monday, June 12, 2017

RIP Adam "Batman" West

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(C)Copyright 2017, C. Burke.

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

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

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




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Friday, June 09, 2017

Failure

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(C)Copyright 2017, C. Burke.

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




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Wednesday, June 07, 2017

The 0'Factor, Episode 17

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(C)Copyright 2017, C. Burke.

And things might get a little improper before that happens!

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

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




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Sunday, June 04, 2017

Geometry Problems of the Day

The following problems were taken from the GEOMETRY (COMMON CORE) Regents Exam given on Thursday, January 26, 2017.
Previous problems can be found here.

Part 1

23. A plane intersects a hexagonal prism. The plane is perpendicular to the base of the prism. Which two-dimensional figure is the cross section of the plane intersecting the prism?

(4) rectangle.
A hexagonal prism has a six-sided hexagon on its "bottom" and "top". Imagine a hexagonal building. The walls holding up the roof would be shaped like rectangles, going straight up, perpendicular to the ground.
Each of these walls part of planes that would be perpendicular to the base. So the answer is rectangle.



24. A water cup in the shape of a cone has a height of 4 inches and a maximum diameter of 3 inches. What is the volume of the water in the cup, to the nearest tenth of a cubic inch, when the cup is filled to half its height?

(1) 1.2
The equation for Volume of a cone is V = 1/3 π r2h, however, in this case, we only want 1/2 of the height. There are TWO problems with the radius. First, we're given the diameter of the top of the cone, not the radius. The radius of the top of the cone is 1.5, not 3. However, that's NOT the radius that we want. We need the radius of the circle that is halfway down the cone.
Luckily, the smaller cone and the larger cone are similar (have the same shape), so the radius is proportional. At half the height, the radius is also half, or 0.75.
Plug in these values and you have V = 1/3 (3.141592...) (0.75)2 (2) = 1.178097..., which rounds to 1.2.

Did you get tripped up by that one?

That's the end of Part I. I hope you all did well.




Continue to the next problems.

Algebra 2 Problems of the Day

The following problems were taken from the ALGEBRA II (Common Core) Regents Exam given on Friday, January 27, 2017.
Previous problems can be found here

Part 1

21. Joelle has a credit card that has a 19.2% annual interest rate computations. compounded monthly. She owes a total balance of B dollars after m months. Assuming she makes no payments on her account, the table below illustrates the balance she owes after m months.

Over which interval of time is her average rate of change for the balance on her credit card account the greatest?

(4) month 60 to month 73
Look at the image below. Find the average rate of change by calculation (y1 - y2) / (x1 - x2).




22. Which graph represents a cosine function with no horizontal shift, an amplitude of 2, and a period of 2/3 π ?

(3) See below
Choices (2) and (4) are out because they start at 0 (sine graphs). Graph (1) shows a function with a period of 2/9 π, as it repeats three 3 in the space of 2/3 π. Choice (3) shows a function that repeats 3 times in the space of 2π, so it has a period of 2/3π.






Continue to the next problems

Thursday, June 01, 2017

Geometry Problems of the Day

The following problems were taken from the GEOMETRY (COMMON CORE) Regents Exam given on Thursday, January 26, 2017.
Previous problems can be found here.

Part 1

21. In the diagram below of circle O, GO 8 and m∠GOJ = 60°. What is the area, in terms of π, of the shaded region?

(4) 160π / 3.
Since 60° is 1/6th of the 360° degree in the complete circle, then the unshaded region of the circle is (1/6) πr2 = (1/6) π82
and the shaded portion would be (5/6) πr2 = (5/6) π82 = (5 * 64π) / 6 = (5 * 32π) / 3 = 160π/ 3.



22. A circle whose center is the origin passes through the point (5,12).
Which point also lies on this circle?

(3) (11, 2 sqrt(12))
The equation of the circle is x2 + y2 = r2. We can find r using the Distance Formula or Pythagorean Theorem: 52 + 122 = r2.
25 + 144 = 169 = r2
r = 13 (which you really should have known. Look up Pythagorean triples.)

Which of the other points creates a right triangle with a hypotenuse of 13?
(10, 3) definitely do not -- they don't even create a triangle with a side of 13. (-12, 13) can't because the hypotenuse is longer than the legs (plus it would have to be -12 and either 5 or -5).
Check 112 + (2sqrt(12))2 = 121 + 4(12) = 169 = 132. Looks good.
Check (-8)2 + (5sqrt(21))2 = 64 + 25(21) = way too much. No good.




Continue to the next problems.

Bubblegum

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(C)Copyright 2017, C. Burke.

These things have a way of blowing up in your face. Kind of like bubblegum.




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