Saturday, June 24, 2017

June 2017: Common Core Geometry Regents, Part 3

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

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

The answers to Part II can be found here.

June 2017, Geometry (Common Core), Part III

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

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

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

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

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

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


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

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

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



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

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

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

End of Part III

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

Friday, June 23, 2017

June 2017: Common Core Geometry Regents, Part 2

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

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

June 2017, Geometry (Common Core), Part II

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

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

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




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

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

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

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


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

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

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

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

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

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

Those were some misunderstandings I came across speaking to students.

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

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

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


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

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




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

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

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

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

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


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

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

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


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

Explain your answer.

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

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

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


End of Part II

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

End-of-Year Review 2017

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(C)Copyright 2017, C. Burke.

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




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Thursday, June 22, 2017

June 2017: Common Core Algebra Regents, Part 4

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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




End of Part IV

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

Wednesday, June 21, 2017

June 2017: Common Core Algebra Regents, Part 3

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

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

The answers to Part II can be found here.

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

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

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

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


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

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

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

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

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

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


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

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

The Quadratic Formula is used in the image below:

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.






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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.




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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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Monday, May 29, 2017

Memorial Day BBQ Math

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(C)Copyright 2017, C. Burke.

So how many beers is 1 steak, 2 hot dogs and a cheeseburger and a half?




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Wednesday, May 24, 2017

Orbit Memory

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(C)Copyright 2017, C. Burke.

Cost a lot of Bucks ... and was gone in a Flash!

If the worksheets on the flip side of the original sketches are any indication, this pun has been waiting to happen since about 2009.

I didn't have the skills then (and, frankly, I don't really have them now, but I try more).

Whatever the scientist's name was supposed to have been is lost to time (and my own faulty Memory). Victor Vargos was the name of the scientist is a schlock satire I wrote shortly after college. He was a good guy and saved the world, but I felt I needed a name here.

And I was already days late, and switched up my comic at the late moment in the process.




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Wednesday, May 17, 2017

Center, Pt. 3

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(C)Copyright 2017, C. Burke.

It was almost the satyr of the centaur circle, which would have been a lot of work to throw out at the last minute!

I think I'm done beating this dead ... well, you know.

See also, Center, Pt. 1 and Center, Pt. 2.




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Monday, May 15, 2017

Another Grim Venn

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(C)Copyright 2017, C. Burke.

Perhaps this is what you were expecting the last time?

Cross reference: A Grim Venn




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Wednesday, May 10, 2017

Village Squares

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(C)Copyright 2017, C. Burke.

"If you know just two sides
There's a rule that provides
How to find the space confined.
If the angle that's included
Has a value that computed
You can use the Law of Sines!"




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Monday, May 08, 2017

A Grim Venn

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(C)Copyright 2017, C. Burke.

Yes, it's grim, but it's also fantastic!




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Thursday, May 04, 2017

Center, Pt. 2

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(C)Copyright 2017, C. Burke.

Yes, I've said this in class. As any fun teacher would. Amirite?

And for the geeks out there: May the Fourth be with you. I didn't forget, but I didn't have time to make a special comic. I do Star Wars stuff throughout the year when the mood strikes.




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Tuesday, May 02, 2017

Center, Pt. 1

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(C)Copyright 2017, C. Burke.

Nice to know things are fine.

Warning you now, that there's a part two because I couldn't decide on which joke to use, so I'm using them both.
And I won't even try to hide it by doing the other one two or three weeks from now.




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Wednesday, April 26, 2017

Apothem

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(C)Copyright 2017, C. Burke.

I tried a rousing rendition of the Nation Apothem, but it didn't help.




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Saturday, April 22, 2017

(x, why?) Mini: Scatter!

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(C)Copyright 2017, C. Burke.

Bad shapes! Bad shapes! What you gonna do...?

They're my keystones to comedy.




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Tuesday, April 18, 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

19. Parallelogram ABCD has coordinates A(0,7) and C(2,1). Which statement would prove that ABCD is a rhombus?

(3) The slope of BD is 1/3..

The diagonals of a rhombus are perpendicular. That means that the slopes of the diagonals are inverse reciprocals.
The slope of AC is (1 - 7) / (2 - 0) = -6 / 2 = -3.
Therefore, the slope of BD must be +1/3.

The midpoint of AC is (1, 4), which would be true regardless of the shape of the parallelogram. The length of diagonal BD is not restricted by the location of points A and C; the length of one diagonal does not affect the other. Finally, the slope of AC is NOT 1/3.

20. Point Q is on such that MQ:QN = 2:3. If M has coordinates (3,5) and N has coordinates (8,-5), the coordinates of Q are

(1) (5, 1)
2 + 3 = 5, so Q is 2/5 of the way from M to N.
(2/5)(8 - 3) = +2 and (2/5)(-5 -5) = -4.
The coordinates of Q are (3 + 2, 5 - 4), or Q(5, 1).




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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

19. Which statement regarding the graphs of the functions below is untrue?

f(x) = 3 sin 2x, from -π < x < π
h(x) = log2x
g(x) = (x - 0.5)(x + 4)(x - 2)
j(x) = -|4x - 2| + 3 (1)

(2) f(x), h(x), and j(x) have one y-intercept.

All four are functions, so they have at most one y-intercept. However, the log function has a domain of positive numbers, x > 0, so it will not have a y-intercept.


20. When g(x) is divided by x + 4, the remainder is 0. Given g(x) = x4 + 3x3 - 6x2 - 6x + 8, which conclusion about g(x) is true? (1

(2) g(-4) = 0

The polynomial can be divided by x + 4 without a remainder, meaning (x + 4) is a factor. This makes -4 a zero of the function.




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Monday, April 17, 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

17. Which rotation about its center will carry a regular decagon onto itself?

(4) 252°.

A full rotation is 360°. One-tenth of a rotation is 36°, and each one-tenth of a rotation will map a regular decagon onto itself.
Of the choices given, only 252° is a multiple of 36°.

18. The equation of a circle is x2 + y2 - 6y + 1 = 0. What are the coordinates of the center and the length of the radius of this circle?

(1) center (0,3) and radius 2(2)1/2
First, convert x2 + y2 - 6y + 1 = 0 to standard form by completing the square.
Half of -6 is -3, and (-3)2 = 9, so add 8 to both sides of the equation to increase 1 to 9.
So x2 + y2 - 6y + 9 = 8
Then x2 + (y - 3)2 = 8.
The coordinates of the vertex are (0, 3) and the radius is SQRT(8), which is 2(2)1/2




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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

17. What is the solution, if any, of the equation

2 / (x + 3) - 3 / (4 - x) = (2x - 2) / (x2 - x - 12)

(1) -1

Follow along with the image below.
Because x2 - x - 12 is the product of (x + 3)(x - 4), we can multiply the entire equation by (x + 3)(x - 4).
To get (x - 4) in the denominator of the middle fraction, multiply the top and bottom of the fraction by -1.
Multiplying removes all the denominators and sets up an equation that solved in a couple of steps.
The solution is x = -1.

Because it was multiple choice, you could also use the graphing calculator to solve the problem:
As shown below, put the left side of the equation into y1 and the right side into y2 and graph.
Check the values for -1 and -5. You can see that -1 has the same values but -5 does not. Likewise, all real numbers and no real solutions are obviously incorrect.




18. In 2013, approximately 1.6 million students took the Critical Reading portion of the SAT exam. The mean score, the modal score, and the standard deviation were calculated to be 496, 430, and 115, respectively.
Which interval reflects 95% of the Critical Reading scores?

(4) 496 + 230

The standard deviation uses the mean, which is 496 in this example. About 95% of the data falls within TWO standard deviations from the mean. Double 115 and you get 230.




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Sunday, April 16, 2017

Happy Easter 2017!

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(C)Copyright 2017, C. Burke.

Hope your hunts are fun! (But hiding bacon is just mean, whatever the doctors say.)




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Saturday, April 15, 2017

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

15. The loudness of sound is measured in units called decibels (dB).
These units are measured by first assigning an intensity I0 to a very softsound that is called the threshold sound. The sound to be measured is assigned an intensity, I, and the decibel rating, d, of this sound is found using d = 10 log (I / I0).
The threshold sound audible to the average person is 1.0 X 10-12 W/m2 (watts per square meter).
Consider the following sound level classifications:

How would a sound with intensity 6.3 X 10-3 W/m2 be classified?
Moderate 45-69 dB
Loud 70-89 dB
Very Loud 90-109 dB
Defeaning > 110 dB

(3) very loud

Don't let all the logs and subscripts bother you. At the heart of this question is simple substitution and evaluation. Plug in the values and let the calculator do the rest.
The formula you are given is: d = 10 log (I / I0).
I0 = 1.0 X 10-12, and I = 6.3 X 10-3
So calculate (using enough parentheses!) the following: 10 log ( (6.3 X 10-3) / (1.0 X 10-12) )
You should get 97.99341..., which is between 90 and 109, or "Very Loud".


16. Pedro and Bobby each own an ant farm. Pedro starts with 100 ants and says his farm is growing exponentially at a rate of 15% per month. Bobby starts with 350 ants and says his farm is steadily decreasing by 5 ants per month.
Assuming both boys are accurate in describing the population of their ant farms, after how many months will they both have approximately the same number of ants?

(2) 8

The equation for Pedro's farm is y = 100(1.15)x. The equation for Bobby's farm is y = 350 - 5x. (Note that Bobby is losing 5 ants, not 5%.)
Graph these on your calculator.
The two lines will intersect at a fraction past 8 months, where both will have a little less than 310 ants.
Use the CALC function (2nd and TRACE) and choose Intersection from the menu. Hit ENTER three times to get the answer.




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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

15. In circle O, secants ADB and AEC are drawn from external point A such that points D, B, E, and C are on circle O. If AD = 8, AE = 6, and EC is 12 more than BD, the length of BD is

(2) 22.

I sketched the image before solving so it isn't drawn to scale. This is a problem that you might want to work backward from the answers, plugging numbers into the equations.


The rule to remember is that (AD)(AB) = (AE)(AC)
or (8)(x + 8) = (6)(x + 12 + 6)
Simplify: 8x + 64 = 6x + 108
2x = 44
x = 22

Check: (8)(22 + 8) = (6)(22 + 12 + 6)
(8)(30) = (6)(40)
240 = 240

16. A parallelogram is always a rectangle if

(1) the diagonals are congruent
Congruent diagonals is true of rectangles and isosceles trapezoids (which aren't parallelograms, so they don't count here).
Opposite angles congruent and diagonals bisecting each other are true of all parallelograms, not just rectangles.
Diagonals intersecting at right angles proves that the quadrilateral is a rhombus, not a rectangle.




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Friday, April 14, 2017

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

13. The price of a postage stamp in the years since the end of World War I is shown in the scatterplot below.

The equation that best models the price, in cents, of a postage stamp based on these data is

(3) y = 1.43(1.04)x

Looking at the graph, you can see that it has an exponential shape, starting somewhat flat and then increasing more quickly. This eliminates choices (1) and (4). (And, really, a sine graph would be a silly choice here.) If you put the other two into your graphing calculator and expand the size of the window, you will see that choice (2) increases much more quickly than the graph in the question. Choice (3) is a better fit.
If you have a problem expanding the window size, you can also check the Tables of Values to see that (2) shoots up to quickly.

14. The eighth and tenth terms of a sequence are 64 and 100. If the computations. sequence is either arithmetic or geometric, the ninth term can not be

(1) -82

In an arithmetic sequence, the common difference can be found be subtracting the 8th term from the 10th term and dividing by 2 (which is 10 - 8). If we do that, we get (100-64)/2, which is 18.
64 + 18 = 82, which eliminates choice (4). (At this point, you might suspect that -82 is likely the answer.)
In a geometric sequence, the common ratio can be found by dividing the 10th term by the 8th term and then taking the square root (because the ratio is applied twice to get from the 8th term to the 10th term).
So (100/64)^.5 = +1.25, which are the common ratios. It is important to remember that the square root may positive or negative.
Then 64 * 1.25 = 80, and 80 * 1.25 = 100
And 64 * (-1.25) = -80 and -80 * 1.25 = 100.
This eliminates choices (2) and (3).




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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

13. On the graph below, point A(3,4) and with coordinates B(4,3) and C(2,1) are graphed. What are the coordinates of B' and C' after undergoes a dilation centered at point A with a scale factor of 2?

(1) B'(5,2) and C'(1,-2) .
A dilation with scale factor 2 centered at A means that everything will be twice as far from A as it currently is.
Point B is 1 unit right and 1 unit down from A, so B' will be 2 units right and 2 units down from A, or (5, 2).
Point C is 1 unit left and 3 units down from A, so C' will be 2 units left and 6 units down from A, or (1, -2).

14. In the diagram of right triangle ADE below, BC || DE.

Which ratio is always equivalent to the sine of ∠A?

(3) BC / AB
Sine = Opposite / Hypotenuse. There are two right triangles containing Angle A. The sides opposite angle A are BC in the smaller triangle and DE in the bigger triangle. Note that because BC || DE, Angle ACB must be a right angle because it corresponds to Angle E. The hypotenuse of the smaller triangle is AB, and the hypotenuse of the bigger triangle is AD.
The two possible answers to this question, therefore, are BC / AB, and DE / AD. The answer is choice (3).




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