Wednesday, August 22, 2018

Blnx

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(C)Copyright 2018, C. Burke.

Don't try to steal it! It's already taken!

So this happened:

A Twitter colleague, Taylor Grant, @teachbarefoot, announced his new blog

Captain’s ln(x): A Fresh Start
https://teachbarefoot.wordpress.com/2018/08/21/captains-lnx-a-fresh-start/

To which I replied, Now that is a great name for a bln(x)!

Taylor liked my pun better, and re-christened his bln(x) to include the "b".

Obviously, I'm not stealing his idea because a) I suggested the "b", and b) I skipped the parentheses to save space.

Enjoy his blog. My blog will remain the usual comics, test questions and answers, and the occasional mathematical insights, which are hopefully my own and not someone else's restated in (mostly) my own words.




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Tuesday, August 21, 2018

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

June 2017, Part I

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.


13. A student studying public policy created a model for the population of Detroit, where the population decreased 25% over a decade. He used the model P = 714(0.75)d, where P is the population, in thousands, d decades after 2010. Another student, Suzanne, wants to use a model that would predict the population after y years. Suzanne’s model is best represented by
Which explanation is appropriate for Miles and his dad to make?
1) P = 714(0.6500)y
2) P = 714(0.8500)y
3) P = 714(0.9716)y
4) P = 714(0.9750)y

Answer: 3) P = 714(0.9716)y
There are 10 years to 1 decade, so d = 10y
So P = 714(0.75)d = 714(0.75)10y = 714(0.7510)y
And P = 714(0.9716)y





14. The probability that Gary and Jane have a child with blue eyes is 0.25, and the probability that they have a child with blond hair is 0.5. The probability that they have a child with both blue eyes and blond hair is 0.125. Given this information, the events blue eyes and blond hair are

I: dependent
II: independent
III: mutually exclusive

1) I, only
2) II, only
3) I and III
4) II and III

Answer: (2) II, only
They are independent because P(A and B) = P(A) * P(B). That is 0.125 = 0.5 * 0.25.
This eliminates choices 1 and 3.
The events are not mutually exclusive, because P(A or B) = P(A) + P(B) - P(A and B) = 0.25 + 0.5 - 0.125 = 0.625, but 0.625 =/= 0.5 + 0.25.





15. Based on climate data that have been collected in Bar Harbor, Maine, the average monthly temperature, in degrees F, can be modeled by the equation B(x) = 23.914sin(0.508x - 2.116) + 55.300. The same governmental agency collected average monthly temperature data for Phoenix, Arizona, and found the temperatures could be modeled by the equation P(x) = 20.238sin(0.525x - 2.148) + 86.729.
Which statement can not be concluded based on the average monthly temperature models x months after starting data collection?
1) The average monthly temperature variation is more in Bar Harbor than in Phoenix.
2) The midline average monthly temperature for Bar Harbor is lower than the midline temperature for Phoenix.
3) The maximum average monthly temperature for Bar Harbor is 79° F, to the nearest degree.
4) The minimum average monthly temperature for Phoenix is 20° F, to the nearest degree.

Answer: 4) The minimum average monthly temperature for Phoenix is 20° F, to the nearest degree.
If you graph both of these functions you will find the following information:
For Bar Harbor: the minimum value is 31.39, the midline 55.3, the maximum is 79.21 and the range is 47.83.
For Phoenix, min is 66.49, mid is 86.73, max is 106.97 and range is 40.48.
Choices 1, 2, and 3 can be seen in the data. Choice 4 is incorrect, the minimum average monthly temperature for Phoenix is approximately 66° F, to the nearest degree.



Comments and questions welcome.

More Algebra 2 problems.

Monday, August 20, 2018

Palindrome Week

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(C)Copyright 2018, C. Burke.

Dammit, I'm mad.

I guess that was, you know, "palindrome weak".




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Saturday, August 18, 2018

Dec Savage: The Plane of Terror

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(C)Copyright 2018, C. Burke.

Are the odds stacked against Dec?

I'm still decided on the format. I'll probably use the pulp order (or publication order) but parody the Bantam covers.

I fiddled with the logo, and I made the image taller so I could include a "cover blurb", which makes up for the lack of dialogue. Also, the blurb means people who see the image away from the blog or the comic page will see that text and not miss the joke. Without context, it's just an odd image.




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Friday, August 17, 2018

Pleased as Punch

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(C)Copyright 2018, C. Burke.

Historically, the expression is *pleased as Punch & Judy*.

And Ken is the puppet master.

I had more exposition, but it was just a recap of last week's comic and blog notes. Who needs that, right? what we really need is a wacky Wikipedia link!




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Wednesday, August 15, 2018

(x, why?) School Life #6: The Next Level

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(C)Copyright 2018, C. Burke.

When you beat bosses together, it's getting serious.

My intent had been to have more updates this week, but this one took a lot longer than anticipated, even in black and white.




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Monday, August 13, 2018

(x, why?) School Life #5: At The Beach

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(C)Copyright 2018, C. Burke.

Well, someone's got to get these two together, right?

Debated whether or not to do "School Life" in the summer, and then whether or not this strip would fall into that category. (For one thing, it's in color, not black and white.) That's what I get for starting so many sub-series within the main series. Why am I even numbering these things?

How should things work out for Vanessa and Sven? Or Missy and Vaughn? Or the green-haired kid whom I don't think I created a name for...




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Saturday, August 11, 2018

Dec Savage: The Shape of Bronze

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(C)Copyright 2018, C. Burke.

''Doc? Are we going to Widow's Peak?'' ''No!''

Something old becomes something new. I had a bit of debate with myself about how to do this. Just the covers? Do I need dialogue? If so, where to add it? Maybe underneath, or just the "alt" text. I don't know.

If it's just the covers, with no dialogue, then I spent too much time -- including a Twitter poll -- deciding on the supporting characters. We'll see as time moves on.

Also, if anyone knows how to *easily* remove the shadows from WordArt, or knows a similar program to produce the titles, that would be great.

By the way, the alternate title was going to be Dec Visage, to remove it further from the original, but I thought that that might be too far. "Visage", of course, is a word refering to a face, which 3-D objects, like prisms and pyramids, have.




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Friday, August 10, 2018

Logical Progression

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(C)Copyright 2018, C. Burke.

ObMath: The mean age of the guys should equal the mean age of the girls. And other mean stuff going on.

The dialogue for this comic was originally wordier, trying to get it where it needed to go (and even included, sort of, a reference to the above fact, giving their relative ages). Judy and Chuck have been together almost as long as Ken and Michele (possibly longer, but I'm only going by first mentions).

Stranger fact: even though Chuck has been mentioned a few times, I don't think he's appeared since comic #321 on the Fourth of July in 2009. NINE YEARS! He was supposed to be shown at the wedding, but I simplified that strip as much as I could, so none of the +1's were shown.




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Wednesday, August 08, 2018

(x, why?) Mini: Viscous

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(C)Copyright 2018, C. Burke.

Stand back. You don't to get into the thick of it!




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Monday, August 06, 2018

(x, why?) Mini: Changin' Sines

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(C)Copyright 2018, C. Burke.

Don't stand in the amplitude. Don't block the wave.

Yeah, that's about all I had. That's why this became a "mini". For what it's worth, this looks like the 100th Mini I've done.
I say "looks like" because my count has been off once before due to mislabeling a file. But this is the 100th entry that has the Mini tag.




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Saturday, August 04, 2018

The Casebook of Sherlock Pi: The Adventures of the Cardboard Box

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(C)Copyright 2018, C. Burke.

Doctor Woo Hoo and his Companion. Or Doctor Cutey and her Companion.




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Wednesday, August 01, 2018

Angles in the Night

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(C)Copyright 2018, C. Burke.

They were complementary, and it turned out so right.




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Monday, July 30, 2018

Old Blue Eyes

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(C)Copyright 2018, C. Burke.

Don't be the negative one here. That's his job.

I can't believe I haven't done this before. And there may be more of this to come.




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Saturday, July 28, 2018

The Casebook of Sherlock Pi: The Noble Bachelor

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(C)Copyright 2018, C. Burke.

He is happy just as He is. Say what you want, you won't get a reaction.




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Friday, July 27, 2018

Togetherness

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(C)Copyright 2018, C. Burke.

Mr. & Mrs. Wayne will pop up any time now.

I wanted a Ken-and-Michele-are-back comic, and then I left them out of it.

I also left out the math -- probability that two people would capsize the chair was approximately 1.0.

I also figured that they would have gone out with Judy and Chuck first, but where's the math in that? Other than figuring the tip? (Chuck might be the least mature of the group, but I didn't want him to be a dweeb or anything.)




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Thursday, July 26, 2018

Parallel vs. Perpendicular vs. Skew: Getting Shiggy edition

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(C)Copyright 2018, C. Burke.

It's only skew if the dancer *misses* the pole. But that wouldn't be as funny!
Don't try this at home. Or on the road.

Don't get ''shiggy'' with it.






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Wednesday, July 25, 2018

(x, why?) Mini: 10 to the -6th Aggressions

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(C)Copyright 2018, C. Burke.

I had a couple different ideas, but thise was the quicker one to do after my day off.






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Monday, July 23, 2018

June 2018 Common Core Geometry Regents, Parts III and IV

The following are some of the multiple questions from the recent June 2018 New York State Common Core Geometry Regents exam.
The answers to Part I can be found here
The answers to Part II can be found here

June 2018 Geometry, Part III

Each correct answer is worth up to 4 credits. Partial credit can be given. Work must be shown or explained.


32. Triangle ABC has vertices with coordinates A(-1,-1), B(4,0), and C(0,4). Prove that ABC is an isosceles triangle but not an equilateral triangle. [The use of the set of axes below is optional.]

Answer:
If it is isosceles, then at least two legs have the same length. If it is not equilateral, then the third leg will have a different length.
Looking at the coordinates of the points, it should be obvious that AB and AC are congruent because you'll be using the same numbers in the calculations.
AB = SQRT ( (-1 - 4)2 + (-1 - 0)2 ) = SQRT(26)
AC = SQRT ( (-1 - 0)2 + (-1 - 4)2 ) = SQRT(26)
BC = SQRT ( (4 - 0)2 + (0 - 4)2 ) = SQRT(32)
There is no need to simplify because you're only looking for equality or inequality.
AB is the same length as AC but not the same as BC. Only two sides are congruent, so ABC is isosceles but not equilateral.


33. The map of a campground is shown below. Campsite C, first aid station F, and supply station S lie along a straight path. The path from the supply station to the tower, T, is perpendicular to the path from the supply station to the campsite. The length of path FS is 400 feet. The angle formed by path TF and path FS is 72°. The angle formed by path TC and path CS is 55°.


Determine and state, to the nearest foot, the distance from the campsite to the tower.

Answer:
We can find the length of TS by using the tangent function with triangle TSF.
Once we know TS, we can use the sine function with triangle TSC to find the length of CT, the distance from the campsite to the tower.
Make sure your calculator is in Degree mode.

Tan 72 = x / 400
x = 400 * tan 72 = 1231.07...

Sin 55 = 1231.07 / y
y = 1231.07 / sin 55 = 1502.85897... = 1503 feet.
The distance from the campsite to the tower is 1503 feet.

Note that you could have skipped the intermediary skip and used (x / 400) in the last equation. Finding the length of TS was not required for the problem.


34. Shae has recently begun kickboxing and purchased training equipment as modeled in the diagram below. The total weight of the bag, pole, and unfilled base is 270 pounds. The cylindrical base is 18 inches tall with a diameter of 20 inches. The dry sand used to fill the base weighs 95.46 lbs per cubic foot.


To the nearest pound, determine and state the total weight of the training equipment if the base is filled to 85% of its capacity.

Answer:
Find the Volume of the Base in cubic feet (not cubic inches). Multiply that by .85 to find 85% of the Volume. Multiply that by 95.46 to find the weight of the sand in the base. Then add 270 pounds for the bag, pole and unfilled base.
Remember that the radius is half the diameter: 20 / 2 = 10 inches, which is 10/12 of a foot. The height is 18 inches, which is 18/12 feet.
V = pi * r2 * h = (3.141592...)(10/12)2 * 18/12 = 3.27249
.85V = .85 * 3.27249 = 2.7816 cubic feet.
The weight of the sand = 2.7816 * 95.46 = 265.53 pounds.
Total weight = 265.53 + 270 = 535.53 or 536 pounds.


End of Part III

Part IV

A correct answer is worth up to 6 credits. Partial credit can be given. Work must be shown or explained.


35. Parallelogram ABCD, BF ⊥ AFD, and DE ⊥ BEC.


Prove: BEDF is a rectangle

Answer:
BEDF is a rectangle if it is a parallelogram as has a right angle. You need not prove the length of the opposite sides are congruent.
1. ABCD is a parallelogram, BF ⊥ AFD, DE ⊥ BEC: Given
2. BC || AD: Opposite sides of a parallelogram are parallel.
3. BE || DE: Parts of parallel lines are parallel.
4. BF || DE: Two lines that are perpendicular to the same line are parallel.
5. BEDF is a parallelogram: A quadrilateral with two pairs of parallel sides is a parallelogram.
6. Angle DEB is a right angle: Perpendicular lines form right angles.
Note: Line 6 wasn't actually *Given*, even though the boxes for right angles are shown in the diagram. Whatever reason you give, the fact that it is a right angle is important and must be stated!
7. BEDF is a rectangle: A parallelogram with a right angles is a rectangle.

End of Part IV

How did you do?

Questions, comments and corrections welcome.

Saturday, July 21, 2018

Trigonometry Jones and the i of Vector

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(C)Copyright 2018, C. Burke.

It uses the Right-Hand Rule because everyone lose that Vector's left-hand was lost and became a powerful artifact!






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Friday, July 20, 2018

(x, why?) Mini: Altitude

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(C)Copyright 2018, C. Burke.

For instance, finding the area, altitude is key!






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Wednesday, July 18, 2018

The Young Pioneers

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(C)Copyright 2018, C. Burke.

Episode 2: Building a Log Cabin






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Tuesday, July 17, 2018

June 2018 Common Core Geometry Regents, Part II

The following are some of the multiple questions from the recent June 2018 New York State Common Core Geometry Regents exam.
The answers to Part I can be found here

June 2018 Geometry, Part II

Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.


25. Triangle A'B'C' is the image of triangle ABC after a translation of 2 units to the right and 3 units up. Is triangle ABC congruent to triangle A'B'C'? Explain why

Answer:
Yes, because a translation is a rigid motion, which preserves shape and size.
The corresponing angles are congruent, and the corresponding sides are congruent, so the triangles are congruent.


26. Triangle ABC and point D(1,2) are graphed on the set of axes below.

See image below.

Graph and label triangle A'B'C', the image of triangle ABC, after a dilation of scale factor 2 centered at point D.

Answer:
See image below. No explanation is needed. Just the graph, with labels.
Each image is twice as far away from D as the original point was.
To get from D to A, go down 1 unit and 3 left. To get from A to A', do this again. A' is (-5, 0)
To get from D to B, go up 3 units and 1 left. To get from B to B', do this again. B' is (-1, 8)
To get from D to C, go down 3 units and 3 right. To get from C to C', do this again. C' is (7, -4)




27. Quadrilaterals BIKE and GOLF are graphed on the set of axes below.


Describe a sequence of transformations that maps quadrilateral BIKE onto quadrilateral GOLF.

Answer:
Notice that letters of BIKE go counterclockwise and GOLF go clockwise. A simple translation will not be good enough. A translation of T5,8 will map BIKE over GOLF; however, B doesn't map to G, and K doesn't map to L.
You need to include a reflection in your transformations. The directions ask for a sequence.
There are many possibilities, but the simplest (to me) would be a reflection over the y-axis followed by a translation of 5 units up.


28. In the diagram below, secants RST and RQP, drawn from point R, intersect circle O at S, T, Q, and P.


If RS = 6, ST = 4, and RP = 15, what is the length of RQ?

Answer:
(RS)(RT) = (RQ)(RP)
(6)(6 + 4) = (RQ)(15)
60 = 15(RQ)
RQ = 4.


29. Using a compass and straightedge, construct the median to side AC in ABC below.
[Leave all construction marks.]

Answer:
See image.
To construct a median, you need to find the midpoint of AC. You can find the midpoint by constructing a perpendicular bisector of AC. Label the midpoint D. Then use the straightedge to draw BD.




30.
Skye says that the two triangles below are congruent. Margaret says that the two triangles are similar.
Are Skye and Margaret both correct? Explain why.

Answer:
Skye and Margaret are both correct. Using the Pythagorean Theorem, 52 + 122 = 169, which is 132, so the length of the hypotenuse of the first triangle is 13. By the Hypotenuse-Leg Theorem, the two triangles are congruent.
If the triangles are congruent, then their corresponding angles are congruent, and that makes them similar triangles as well.


31. Randy’s basketball is in the shape of a sphere with a maximum circumference of 29.5 inches. Determine and state the volume of the basketball, to the nearest cubic inch.

Answer:
The circumference is 2*pi*r = 29.5
So the radius, r = 29.5 / (2 * pi) = 4.695...
Volume of a sphere = (4/3) * pi * r3 = (4/3) * pi * (4.695)3 = 433.506...
To the nearest cubic inch, the Volume is 434 cubic inches.

End of Part II

How did you do?

Questions, comments and corrections welcome.

Monday, July 16, 2018

June 2018 Common Core Geometry Regents, Part I (multiple choice)

The following are some of the multiple questions from the recent June 2018 New York State Common Core Geometry Regents exam.

June 2018 Geometry, Part I

Each correct answer is worth up to 2 credits. No partial credit. Work need not be shown.


1. After a counterclockwise rotation about point X, scalene triangle ABC maps onto triangle RST, as shown in the diagram below.


Which statement must be true?

Answer: (1) ∠A = ∠R
A rigid motion, such as a rotation, does not change size, so the corresponding angles and sides remain the same.
Angle A corresponds to Angle R. Angle B corresponds to Angle S. Angle C corresponds to Angle T.
Choice 2 does not have corresponding angles. Choices 3 and 4 do not have corresponding sides.ity symbol.


2. In the diagram below, AB || DEF, AE and BD intersect at C, m∠B = 43°, and m∠CEF = 152°.


Which statement is true?

Answer: (3) m∠ACD = 71°
Because of alternate interior angles, angle D = 43 degrees. Angle CED is supplementary to 152 degrees, so it must be 180 - 152 = 28 degrees.
By the Exterior Angle Theorem, Angle ACD must be the sum of the two remote angles, D and CED. So 43 + 28 = 71 degrees.


3. In the diagram below, line m is parallel to line n. Figure 2 is the image of Figure 1 after a reflection over line m. Figure 3 is the image of Figure 2 after a reflection over line n.


Which single transformation would carry Figure 1 onto Figure 3?

Answer: (4) a translation
A translation will move Figure 1 directly to Figure 3. Reflecting the figure switched the orientation, but reflecting it a second time switched it back to its original orientation.
A dilation would have increased its size. A rotation would have changed its orientation, as would a single reflection.


4. In the diagram below, AF and DB intersect at C, and AD and FBE are drawn such that m∠D = 65°, m∠CBE = 115°, DC = 7.2, AC = 9.6, and FC = 21.6.


What is the length of CB?

Answer: (3) 16.2
If angle DBE = 115 degrees, then angle DBF = 65 degrees. This makes angles DBF and D congruent. Angles ACD and BCF are congruent because they are vertical angles. Therefore, the two triangles are similar. (Of course, if these triangles weren't similar to begin with, this would've been a much harder problem to deal with!)
That means that the corresponding sides are proportional.
7.2 / 9.6 = x / 21.6
9.6x = (7.2)(21.6)
x = (7.2)(21.6)/9.6 = 16.2


5. Given square RSTV, where RS = 9 cm. If square RSTV is dilated by a scale factor of 3 about a given center, what is the perimeter, in centimeters, of the image of RSTV after the dilation?

Answer: (4) 108
If RS = 9, then the perimeter is 9 * 4 = 36. If the square is dilated by a factor of 3, then the perimeter will be 3 times as much. So 36 * 3 = 108.
Alternatively, 9 * 3 = 27 cm as the length of the side of the image, and 27 * 4 = 108.


6. In right triangle ABC, hypotenuse AB has a length of 26 cm, and side BC has a length of 17.6 cm. What is the measure of angle B, to the nearest degree?

Answer: (2) 47°
cos B = 17.6 / 26
B = cos-1 (17.6/26) = 47.396... or 47 degrees.


7. The greenhouse pictured below can be modeled as a rectangular prism with a half-cylinder on top. The rectangular prism is 20 feet wide, 12 feet high, and 45 feet long. The half-cylinder has a diameter of 20 feet.


To the nearest cubic foot, what is the volume of the greenhouse?

Answer: (1) 17,869
The Volume of the rectangular prism is Length x Width x Height. The volume of the half-cylinder is 1/2 pi x radius2 x height. Don't confuse to two things that are labeled "height".
Vp = (12)(20)(45) = 10800
Vc = (1/2)(10)2(pi)(45) = 7068.583...
For a total of approximately 17,869


8. In a right triangle, the acute angles have the relationship sin (2x + 4) = cos (46). What is the value of x?

Answer: (1) 20
In a right triangle, the relationship between sin and cos of the acute angles is as follows:

sin (O) = cos (90 - O)
So if sin(2x + 4) = cos(46) then 2x + 4 = 90 - 46
2x + 4 = 44
2x = 40
x = 20


9. In the diagram below, AB||DFC, EDA||CBG, and EFB and AG are drawn.


Which statement is always true?

Answer: (4) Triangle DEF ~ Triangle AEB
Because AB || EDF then Angle A = Angle EDF, and angle E is congruent to itself by the Reflexive Property. By AA, triangles DEF and AEB are similar.
Of the incorrect answers, choice (1) would have been correct had it said "similar to" and not "congruent to", because of alternate interior angles and vertical angles. However, there is no indication that F is the midpoint of EB or CD, or that BC is congruent to DE.


10. The base of a pyramid is a rectangle with a width of 4.6 cm and a length of 9 cm. What is the height, in centimeters, of the pyramid if its volume is 82.8 cm3?

Answer: (1) 6
The formula for the Volume of a rectangular-based pyramid is V = 1/3 L * W * H.
So 82.8 = (1/3) * (4.6) * (9) * H.
And H = 82.8 / ((1/3) * (4.6) * (9))
H = 6 cm.


11. In the diagram below of right triangle AED, BC || DE.


Which statement is always true?

Answer: (2) AB/AD = BC/DE
Short hypotenuse / Long hoptenuse = Short altitude / Long altitude
The incorrect choices:
(1) Short base / Long base =/= Long altitude / Short altitude -- incorrect order
(3) Short base / *segment* of long base =/= Short altitude / Long altitude -- not corresponding parts
(4) Long altitude / Short altitude =/= *segment* of long hypotenuse / Short hypotenuse -- not corresponding parts


12. What is an equation of the line that passes through the point (6,8) and is perpendicular to a line with equation y = 3/2 x + 5?

Answer: (2) y - 8 = - 2/3 (x - 6)
Perpendicular means that the slope of the new line is -2/3, so eliminate choices (1) and (3), which are parallel.
Point-slope form is y - y0 = m(x - x0). Note the signs.
Plug in 6 for x0 and 8 for y0


13. The diagram below shows parallelogram ABCD with diagonals and intersecting at E.


What additional information is sufficient to prove that parallelogram ABCD is also a rhombus?

Answer: (4) AC is perpendicular to BD
The diagonals of a rhombus are perpendicular to each other.
Choices (1), (2), and (3) are true for all parallelograms, not just rhombuses.


14. Directed line segment DE has endpoints D(-4,-2) and E(1,8). Point F divides such that DF:FE is 2:3. What are the coordinates of F?

Answer: (1) (-3,0)
2 + 3 = 5, so point F is 2/5 of the way from D to E.
1 - (-4) = 5, and (2/5)(5) = 2. Add 2 to -4 and you get -2, which must be the x-coordinate for F. So Choice (1).
8 - (-2) = 10, and (2/5)(10) = 4. Add 4 to -2 and you get 2, which must be the y-coordinate for F. Still Choice (1).
Notice that point F must be on the line BETWEEN D and E, so Choice (4) makes no sense at all.


15. Triangle DAN is graphed on the set of axes below. The vertices of DAN have coordinates D(-6,-1), A(6,3), and N(-3,10).


What is the area of triangle DAN?

Answer: (1) 60
Reason this one out. It's a multiple-choice question, so you don't have to go the long way.
You might notice that the slope of DA is 1/3. That means that the altitude of the triangle can be found by starting at N(-3, 10) and finding points that are 3 down and 1 to the right, because the perpendicular slope is -3. This gives you the point (0, 1). Call it P.
You could then find the length of AD and NP, and then use those vales as base and height in the formula (1/2) b h.
HOWEVER, there is another way. Make a big rectangle around the triangle, so that D, A, and N are each touching a side of the rectangle.
This rectangle with have dimensions 11 X 12, or 132 square units.
Now remove the "negative" space -- that is, the three triangles outside of DAN.
First thing you will realize is that all the bases and heights are whole numbers, so their areas are rational. That means that choices (3) and (4) cannot be correct.
At this point, knowing how the Regents operates, you can be pretty sure that the answer is 60, and that 120 is there in case you forget to multiply by 1/2, a common error.

Doing it this way, you have 132 - (1/2)(11)(3) - (1/2)(9)(7) - (1/2)(12)(4) = 60.

The long way: the length of AD is the square root of ((-6 - 6)2 + (-1 - 3)2)) = SQRT(144 + 16) = SQRT(160)
The length of NP is the square root of ((-3 - 0)2 + (10 - 1)2)) = SQRT(9 + 81) = SQRT(90)
A = (1/2) * SQRT(160) * SQRT(90) = (1/2) * SQRT(14,440) = (1/2)(120) = 60.


16. Triangle ABC, with vertices at A(0,0), B(3,5), and C(0,5), is graphed on the set of axes shown below.

Which figure is formed when ABC is rotated continuously about BC?

Answer: (3) (see image)


When you rotate the triangle, you will get a cone. The axis BC that it is rotated around, will be the altitude of the cone. The base of the triangle will become the radius of the cone.
So the height is 3, the radius is 5 and the diameter is 10. Choice (3).
Note that Choices (1) and (2) are rotated about the wrong line.


17. In the diagram below of circle O, chords AB and CD intersect at E.
If mAC = 72° and m∠AEC = 58°, how many degrees are in mDB ?

Answer: (3) 44°
58 is the average of 72 and mDB.
58 = (72 + x)/2
2(58) = 72 + x
116 - 72 = x
44 = x.
Alternatively, 72 - 58 = 14, and 58 - 14 = 44.


18. In triangle SRK below, medians SC, KE, and RL intersect at M.
Which statement must always be true?

Answer: (1) 3(MC) = SC
The three medians of a triangle meet at a centroid, which is point M. The distance from the angle to the centroid is ALWAYS twice as much as from the centroid to the midpoint of the opposite side. That always means that the distance from the centroid to the midpoint is one-third the length of the entire median. So 3(MC) = SC is always true.
Incorrect choices: (2) MC is 1/3 of SC, not 1/3 of SM; (3) RM is 2 times ML, not MC, which is unrelated to RM; (4) SM and KM could be the same size but there is no reason that they must be.


19. MThe regular polygon below is rotated about its center.
Which angle of rotation will carry the figure onto itself?

Answer: (3) 216°
One full rotation is 360 degrees. A regular pentagon will carry onto itself every 1/5th of a rotation.
360 / 5 = 72 degrees. So this will happen at any multiple of 72 degrees.
72 * 3 = 216.

Incorrect choices: (1) 60 degrees is for a hexagon; (2) 108 degrees is the measure of one interior angle of a regular pentagon; (4) 540 degrees is the total number of degrees of the interior angles.


20. What is an equation of circle O shown in the graph below?

Answer: (2) x2 - 10x + y2 - 4x = -13
The standard form for a circle is (x - h)2 + (y - k)2 = r2, where (h,k) is the center.
The center is (5,2) and the radius is 4, so (x - 5)2 + (y - 2)2 = 42.
The choices are not in standard form, so you have to square the binomials.
x2 - 10x + 25 + y2 - 4x + 4 = 16
x2 - 10x + y2 - 4x = 16 - 25 - 4
x2 - 10x + y2 - 4x = -13


21. In the diagram below of PQR, ST is drawn parallel to PR, PS = 2, SQ = 5, and TR = 5.


What is the length of QR?

Answer: (4) 17.5
The ratio between the side lengths is the same. So if PS = 2 and RT = 5, then QR is 2.5 times the length of PQ. And QR is 2.5 times as long as QP, so 2.5 * 7 = 17.5
Or you can write a proportion to find QT.
5 / 7 = x / (x + 5)
7x = 5(x + 5)
7x = 5x + 25
2x = 25
x = 12.5
If QT = 12.5 and TR = 5, then QR = 12.5 + 5 = 17.5


22. The diagram below shows circle O with radii OA and OB. The measure of angle AOB is 120°, and the length of a radius is 6 inches.


Which expression represents the length of arc AB, in inches?

Answer: (4) (1/3)(12π)
The length of an arc of the circle is the Circumference (2πr) times the fraction of the central angle divided by 360.
(120/360)(2π)(6) = (120/360)(12π) = (1/3)(12π)


23. Line segment CD is the altitude drawn to hypotenuse EF in right triangle ECF. If EC = 10 and EF = 24, then, to the nearest tenth, ED is

Answer: (1) 4.2
If it helps, draw a diagram.


ED is the shorter leg of right triangle EDC, and EC is the shorter leg of triangle ECF.
EC is the hypotenuse of right triangle EDC, and EF is the hypotenuse of triangle ECF.
Write a proportion:
x / 10 = 10 / 24
24x = (10)(10)
x = 100 / 24 = 4.2

If you draw a diagram, you can see that ED had to be shorter than EC, so choices (3) and (4) make no sense. Eliminate them.


24. Line MN is dilated by a scale factor of 2 centered at the point (0,6). If is MN represented by y = -3x + 6, which equation can represent M'N', the image of MN?

Answer: (2) y = -3x + 6
A dilation would not change the slope of a line, so eliminate Choices (3) and (4).
A dilation would only change the y-intercept UNLESS the line goes through the center of dilation -- that is, if the center of dilation is a point on the line. In that case, the line remains unchanged because the image of every point will still fall on the same line.
Check for this: 6 = -3(0) + 6
6 = 6, which is true, so the center of dilation is on the line, and the image is the same line.

End of Part I

How did you do?

Questions, comments and corrections welcome.