Geometry Problems of the Day (Geometry Regents, August 2013)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Geometry Regents, August 2013

Part IV: A correct answer will receive 6 credits. Partial credit is available.


38.Quadrilateral ABCD with vertices A(-7,4), B(-3,6), C(3,0), and D(1,-8) is graphed on the set of axes below. Quadrilateral MNPQ is formed by joining M, N, P, and Q, the midpoints of AB, BC, CD, and AD, respectively.

Prove that quadrilateral MNPQ is a parallelogram.

Prove that quadrilateral MNPQ is not a rhombus

Answer:

Draw quadrilateral MNPQ. It will be help to see it visually even though you can do it algebraically.

The midpoints are (average of 2 x-coodinates, average of 2 y-coordinates).

They are M((-7-3)/2, (4+6)/2), N((-3+3)/2, (6+0)/2), P((3+1)/2, (0-8)/2), and Q((1-7)/2, (-8+4)/2),
Or M(-5, 5), N(0, 3), P(2, -4), and Q(-3, -2).
Again, you could've counted out the boxes on the graph if you hate formulas and averages or dealing with signed numbers in fractions.

To prove it's a parallelogram, you need to show that the opposite slopes are equal, which means that the opposite sides are parallel.

To prove that it NOT a rhombus, you can EITHER show that any two sides are not the same length, OR that the diagonals are not perpendicular to each other.

Find the Slopes of MN, NP, PQ, and QM:

MN: (3 - 5) / (0 - -5) = -2 / 5 = -2/5
NP: (-4 - 3) / (2 - 0) = -7 / 2 = -7/2
PQ: (-2 - -4) / (-3 - 2) = 2 / -5 = -2/5
QM: (5 - -2) / (-5 - -3) = 7 / -2 = -7/2

The opposite sides have the SAME slope, so the opposite sides are PARALLEL. Therefore, MNPQ is a Parallelogram.

Find the Slopes of diagonals MP and NQ:

MP: (-4 - 5) / (2 - -5) = -9 / 7
NQ: (-2 - 3) / (-3 - 0) = -5 / -3 = 5 / 3

The product of (-9/7)(5/3) =/= -1, so the diagonals are not perpendicular. Therefore, MNPQ is NOT a Rhombus.

Find the Lengths of sides:

MN: SQRT ((5)² + (2)²) = SQRT(29)
NP: SQRT((2² + (7)²) = SQRT(51)

The sides of the parallelogram are NOT congruent. Therefore, MNPQ is NOT a rhombus.

End of Exam.

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Algebra 2/Trigonometry Regents, January 2014

Part IV: A correct answer will receive 6 credits. Partial credit is available.


39. Solve algebraically for all values of x:

log_{(x + 3)}(2x + 3) + log_{(x + 3)}(x + 5) = 2

Answer:

Have I ever mentioned how much I hated logs as a student. It's likely the notation and trying to memorize a bunch of rules that actually make plenty of sense will you think about the rules for exponents but didn't make a lot of sense for me at the time.

log_{(x + 3)}(2x + 3) + log_{(x + 3)}(x + 5) = 2
log_{(x + 3)}((2x + 3)(x + 5)) = 2
log_{(x + 3)}(2x² + 13x + 15) = 2
(x + 3)² = 2x² + 13x + 15
x² + 6x + 9 = 2x² + 13x + 15
x² + 7x + 6 = 0
(x + 6)(x + 1) = 0
x + 6 = 0 or x + 1 = 0
x = -6 or x = -1

We reject x = -6 because the base x + 3 would be (-6) + 3 = -3, which is not allowed.

So x = -1 only.

End of exam.

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

9/11 Twenty Years Later

Twenty years have passed since that fateful morning that changed the world, and I wanted to be sure that I made a note of the date on the blog. I did a check, and I've done exactly one comic, ten years ago for the tenth anniversary, but that was all that was needed. I wasn't going to harp on it and I couldn't make math problems out of it. I didn't want to trivialize it or be morbid about it. So a handful of times, I had a simple entry like this one to wish everyone a Somber September 11th.

What follows below are the comic from 2011, and a post from a few years later, which was actually written in 2005 about my day walking home in Brooklyn with a bunch of strangers, none of whom I ever met again, but we were important to each other in that moment.

(C)Copyright 2011, C. Burke. All rights reserved. Real-world examples shouldn't be overlooked nor forgotten.

---

As requested (and as promised):

This was originally posted (I think) in 2005. It's been an annual repost since then.

At about 8:40am on that morning, I was walking into the Court building on Adams Street (actually, the Court St. entrance) in Downtown Brooklyn to start serving jury duty. As luck would have it, I had been halfway to the train station before I’d realized that I’d left my Walkman on the kitchen table, so I didn’t bother going back for it. I figured I’d just get a newspaper. Besides, I didn’t know how well I’d be able to pick up AM inside the building anyway.

I sat in a dark room watching a video on How to Be a Good Juror, oblivious to what was going on right across the river. We were told to relax in the room that they have, and I looked out the window at the Marriot Hotel. Traffic on Adams St was snarled, not moving. Must’ve been an accident on the Brooklyn Bridge, I thought. (It was a block away.)

People were standing around outside the hotel. Must be waiting for a tour bus or something. What did I know.

Fire trucks and ambulances started flying by on the wrong side of Adams Street, which had no traffic. Okay, traffic doesn’t come into Brooklyn much in the morning, but something was odd here. I had been facing 180 degrees from where I needed to be looking.

Finally, they had made an announcement. America was at war, under attack. The World Trade Center and the Pentagon had been destroyed. They were trying to get coverage on the TV sets in the jury rooms. I don’t think they succeeded. Even if they wanted to, only CBS would be available because it still broadcast from the Empire State Building.

People were beside themselves, many broke down, everyone was rushing for the payphones. I met a woman who had been listening to her radio. She let me share her earbuds. She was shaken and needed a cigarette. I don’t smoke, but I walked her to the smoking room. (There was one on the floor. Quite a few people were there.)

We were dismissed. Not much was going to get done in the Court building that week. Groups of people huddled outside with questions, comments, gossip and hearsay. Some of the lawyers said they saw it happen. What kind of plane was it?

Does anyone know if the trains are running? What about the buses? No trains. No LIRR. A few buses and they’re all packed. It was time to start walking and no one wanted to walk alone. We walked in groups.

Dust was falling from the sky in downtown Brooklyn like a dirty snow that was covering the cars. Papers fell too. We started walking up Atlantic Avenue. People were wandering around with their cell phones out trying to get a signal. no luck.

We took a turn down Third Avenue. I needed to. I wanted to stop at my mother’s house. It was a good resting point for me. The group I’d tagged along with decided to join me. One guy stopped in a hardware store for masks and passed them out.

When we passed Third St and reached the Gowanus Canal, we had our first real look. It was like a scene out of a bad movie. The skyline was there. But the Towers were missing. Just a terrible column of smoke and a cloud drifting our way.

We didn’t stay long. We kept walking. I made it to my mother’s house and said good-bye to the others. Some were walking all the way to Staten Island. One who had joined our group had walked over the Brooklyn Bridge — after having walked down 50 floors of Tower 1. God was looking out for him.

I watched some of the coverage until the trains were running again. I took one that left me about a mile or so from my inlaws, the meeting place for the rest of the family. I stopped in at St. Athanasius on the way. I hadn’t been there since a wedding about 15-20 years earlier. I stayed for a little while and walked the rest of the way.

Thankfully, my wife, who worked at the foot of the Brooklyn Bridge on the Manhattan side had evacuated immediately before the trains had stopped running.

It’s not a walk that I’ll ever forget.

---

Mom isn't there anymore and my "resting place" has since been bulldozed, sadly. Just more things to remember about that day.

Geometry Problems of the Day (Geometry Regents, August 2013)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Geometry Regents, August 2013

Part III: Each correct answer will receive 4 credits. Partial credit is available.


35. Given: Triangle ABC, BD bisects angle ABC, BD ⊥ AC

Prove: AB ≅ CB

Answer:

Prove that the triangles are congruent, which means that the corresponding parts are congruent. Since BD is perpendicular to AC, the two smaller triangles are both right triangles.

Make you state the Given and that your last line is what you want to prove.

Statement:	Reason:
1. BD bisects angle ABC, BD ⊥ AC	Given.
2. Angles ADB and CDB are right angles	Definition of Perpendicular
3. Angle ABD ≅ Angle CDB	All right angles are congruent.
4. Angle CBD ≅ angle ABD	Definition of angle bisector.
5. BD ≅ BD	Reflexive Property.
6. Triangle ABD ≅ triangle CBD.	ASA
7. AB ≅ CB	CPCTC (Corresponding Parts of Congruent Triangles are Congruent.

36. Quadrilateral MATH has coordinates M(-6,-3), A(-1,-3), T(-2,-1), and H(-4,-1). The image of quadrilateral MATH after the composition r_x-axis ° T_7,5 is quadrilateral M"A"T"H". State and label the coordinates of M"A"T"H". [The use of the set of axes below is optional.]

Answer:

The composition of transformations should be read as the Reflection OF THE Translation. In other words, you must do the Translation first (getting M'A'T'H') before you can reflect. You will lose half credit if you do it the other way around.

M(-6,-3) --> M'(-6 + 7, -3 + 5), M'(1, 2)
A(-1,-3) --> A'(-1 + 7, -3 + 5), A'(6, 2)
T(-2,-1) --> T'(-2 + 7, -1 + 5), T'(5, 4)
H(-4,-1) --> H'(-4 + 7, -1 + 5), H'(3, 4)

Reflecting over the x-axis will flip the sign of the y-coordinate and leave the x-coodinate alone.

M'(1, 2) --> M"(1, -2)
A'(6, 2) --> A"(6, -2)
T'(5, 4) -->T"(5, -4)
H'(3, 4) --> H"(3, -4)

37. Trapezoid TRAP, with median MQ, is shown in the diagram below. Solve algebraically for x and y.

Answer:

The median is average of the two bases, so it is half the size of the sum of the bases.

Angles T and P are supplementary. Angle A is superfluous, it doesn't help with anything. It's only there to throw you off.

2(16y + 1) = 12y + 1 + 18y + 6
32y + 2 = 30y + 7
2y = 5
y = 2.5

12x - 4 + 7x + 13 = 180
19x + 9 = 180
19x = 171
x = 9

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Algebra 2/Trigonometry Regents, January 2014

Part III: Each correct answer will receive 4 credits. Partial credit is available.


36. Solve algebraically for all exact values of x in the interval 0 < x < 2π:

2 sin²x + 5 sin x = 3

Answer:

Start by setting it up like any other quadratic equal to 0.

2 sin²x + 5 sin x = 3
2 sin²x + 5 sin x - 3 = 0
(2 sin x - 1)(sin x + 3) = 0
2 sinx x - 1 = 0 or sin x + 3 = 0
2 sin x = 1 or sin x = -3 (discard)
sin x = 1/2

In the interval 0 < x < 2π, sin x = 1/2 at π/6 and 5π/6.

If you showed the above work, but answered 30 and 150, you would've lost a point.

37. Because Sam’s backyard gets very little sunlight, the probability that a geranium planted there will flower is 0.28. Sam planted five geraniums. Determine the probability, to the nearest thousandth, that at least four geraniums will flower.

Answer:

The probabilty that at least four will flower is the probabilty that four will flower plus the probability that five will flower. Or the probability that only 1 or none will not flower.

P(exactly 4 will flower) = (₅C₄)(0.28)⁴(0.72) = 0.02212...
P(5 will flower) = (0.28)⁵ = 0.00172...
P(4 or 5) = 0.02384 = 0.024

38. Two sides of a parallelogram measure 27 cm and 32 cm. The included angle measures 48°. Find the length of the longer diagonal of the parallelogram, to the nearest centimeter.

Answer:

If the included angle is 48 degrees, then the supplementary angle in the parallelogram will be 180 - 48 = 132 degrees. The diagonal opposite this angle will be the longer of the two.

Use the Law of Cosines (from the back of the booklet): a² = b² + c² - 2bc cos A, where a will be the diagonal, b and c are the given sides, and A is the included angle, in this case 132.

a² = b² + c² - 2bc cos A
a² = (27)² + (32)² - 2(27)(32)cos(132)
a² = 2909.257
a = 53.937... = 54 cm.

End of Part III.

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

School Life #25: Declaring Independence

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

(C)Copyright 2021, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

Where you go each day depends on how independent you are.

Sometimes being the dependent variable is better than being independent.

Since the blog comics aren't numbered, you might not be aware that this is comic number 1776. Were I the planning type, I might've tried to get this to coincide with The Battle of Brooklyn, which was fought August 27, 1776 not far from the house where I grew up. And I'm not talking in terms of miles -- when I say there were soldiers buried in my backyard, I'm literally refering to the area immediately behind that house, or the yards really close to it. You should've seen my father's garden, with roses that bloomed nearly eleven months out of the year. (Okay, that part might be exaggeration. Might've only been 10 1/2 months.)

I also write Fiction! You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon.

Come back often for more funny math and geeky comics.

Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Algebra 2/Trigonometry Regents, January 2014

Part II: Each correct answer will receive 2 credits. Partial credit is available.


32. Solve |-4x + 5| < 13 aglebraically for x.

Answer:

You need to split the inequatily into two, one < 13 and one > -13.

-4x + 5 < 13
-4x < 8
x > -2

-4x + 5 > -13
-4x > -18
x < 4.5

x > -2 OR x < 4.5
or -2 < x < 4.5

33. Express 4xi + 5yi⁸ + 6xi³ + 2yi⁴ in simplest a + bi form.

Answer:

Since i² = -1, then i³ = -i, and i⁴ = 1. Since i⁴ * i⁴ = i⁸, then i⁸ = 1 * 1 = 1.

4xi + 5yi⁸ + 6xi³ + 2yi⁴
= 4xi + 5y(1) + 6x(-i) + 2y(1)
= 4xi + 5y - 6xi + 2y
= 7y - 2xi

34. In an arithmetic sequence, a₄ = 19 and a_{7 = 31. Determine a formula for an, the nth term of this sequence.}

Answer:

Finding the common difference of an aritmetic sequence is like finding the slope of a linear equation.

d = (31 - 19) / (7 - 4) = 12 / 3 = 4

Substitute this into the formula for an arithmetic sequence, and solve for a₁.

a₄ = (4)(4 - 1) + a₁
19 = 12 + a₁
a₁ = 7

a_n = 4(n - 1) + 7

35. 5 Circle O shown below has a radius of 12 centimeters. To the nearest tenth of a centimeter, determine the length of the arc, x, subtended by an angle of 83°50′.

Answer:

The circumference of the circle is 2πr, so x is a fraction of that amount and that fraction will be (83°50′)/360°

(83 5/6) / 360 = (503/6) / 360 = 503 / 2160

x = (2)(3.141592)(12)(503/2160) = 17.558... or 17.6 cm.

End of Part II.

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

Geometry Problems of the Day (Geometry Regents, August 2013)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Geometry Regents, August 2013

Part II: Each correct answer will receive 2 credits. Partial credit is available.


32.In triangle ABC, the measure of angle A is fifteen less than twice the measure of angle B. The measure of angle C equals the sum of the measures of angle A and angle B. Determine the measure of angle B.

Answer:

The sum of the three angles is 180 degrees. Write an equation and solve it.

A = 2B + 15
C = A + B
A + B + C = 180
A + B + A + B = 180
2(A + B) = 180
2(2B + 15 + B) = 180
2(3B + 15) = 180
3B + 15 = 90
3B = 75
B = 25

This would make A equal to 2(25) + 15, or 65, and C = 25 + 65 = 90.

30. A circle has the equation (x - 3)² + (y + 4)² = 10. Find the coordinates of the center of the circle and the length of the circle’s radius.

Answer:

The center of the circle is (3, -4) and the radius is SQRT(10).

There isn't any work to show, and each answer is worth a point. It's important to remember to flip the signs in the equation to get the coordinates of the center.

34. Two intersecting lines are shown in the diagram below. Sketch the locus of points that are equidistant from the two lines. Sketch the locus of points that are a given distance, d, from the point of intersection of the given lines. State the number of points that satisfy both conditions.\

Answer:

A construction question and a locus of points question in the same Part II? (See yesterday's post.) Wow!

The points equidistant from these two lines would be two more lines going through the center of the X. Sketch a vertical and a horizontal line through the central point.

The locus of points of a distance, d, is a circle with a radius of d centered on the point of intersection of the lines. Draw a circle. It doesn't matter how big because it's only a sketch and you aren't told the size of d.

State the number of points that satisfiy both conditions: 4, the points on the circle that intersect the vertical and horizontal lines. (NOT the points on the diagonal lines.)

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

Geometry Problems of the Day (Geometry Regents, August 2013)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Geometry Regents, August 2013

Part I: Each correct answer will receive 2 credits. Partial credit is available.


29.Triangle ABC has vertices A(6,6), B(9,0), and C(3, -3). State and label the coordinates of triangle A'B'C', the image of triangle ABC after a dilation of D_1/3.

Answer:

A dilation of 1/3 centered on the origin -- no center of dilation was stated, so the default is the origin -- means that all points in the image will have coordinates that are 1/3 of the pre-image.

Make sure you write something. Don't do it all in your head!

A(6,6) * 1/3 --> A'(2,2)

B(9,0) * 1/3 --> B'(3,0)

C(3,-3) * 1/3 --> C'(1,-1)

30. Using a compass and straightedge, construct the bisector of ∠MJH. [Leave all construction marks.]

Answer:

Make an arc about point J. Mark off the points on JM and JH. Next, make an arc about the point on JM. Make an arc of the SAME size about the point on JH. Draw a line from J through the point where the two new arcs intersect. This is your angle bisector.

31. Find, in simplest radical form, the length of the line segment with endpoints whose coordinates are (-1,4) and (3,-2).

Answer:

Use the distance formula, or make a sketch on graph paper and use the Pythagorean Theorem.

d = SQRT( (-1 - 3)² + (4 - (-2))²) )
= SQRT ( (4)² + (6)² )
= SQRT ( 16 + 36 )
= SQRT (52)
= SQRT (2 * 2 * 13)
= 2 * SQRT (13)

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Algebra 2/Trigonometry Regents, January 2014

Part II: Each correct answer will receive 2 credits. Partial credit is available.


28. Show that sec θ sin θ cot θ = 1 is an identity.

Answer:

Remember that sec θ = 1 / cos θ and cot θ = 1 / tan θ = cos θ / sin θ.

So

sec θ sin θ cot θ = 1
1 / cos θ * sin θ * cos θ / sin θ = 1
(sin θ * cos θ) / (cos θ * sin θ) = 1
1 / 1 = 1

29. Find, to the nearest tenth of a square foot, the area of a rhombus that has a side of 6 feet and an angle of 50°.

Answer:

A rhombus can be divided into four congruent right triangles. The side of the rhombus will be the hypotenuse of the triangle. The given angle will be bisected by the diagonal, meaning that the angles in the triangle will be 25, 90 and 65 degrees. You can use sine and cosine to find the lengths of the base and the height to find the Area of one of the triangles and multiply that by four.

The quicker way is that the area of a rhombus can be found with the formula A = b² * Sin(a)

A = (6)² * sin(50) = 27.57... = 27.6 to the nearest tenth of a square foot.

The LONG way:

6 * Sin (25) = 6(.422...) = 2.535...
6 * cos (25) = 6(.906...) = 5.437...
A = 4 * (1/2) * (2.535) * (5.437) = 27.56..., which rounds to 27.6.
Note the extra accuracy used in the intermediate steps to prevent rounding errors.

30. The following is a list of the individual points scored by all twelve members of the Webster High School basketball team at a recent game:

2 2 3 4 6 7 9 10 10 11 12 14
Find the interquartile range for this set of data

Answer:

The Interquartile Range (IQR) is the difference between the Upper Quartile (Q3) and the Lower Quartile (Q1). The data are in order, so finding these is easy. Mark up the page to show your work.

There are twelve pieces of data, so the median is the average of the 6th and the 7th. Q1 is the middle of the lower six values, which is the average of the 3rd and 4th, or 3.5. Q3 is the middle of the upper six values, which is the average of the ninth and tenth, or 10.5

The IQR = 10.5 - 3.5 = 7.

31. Determine algebraically the x-coordinate of all points where the graphs of xy = 10 and y = x + 3 intersect

Answer:

Use substitution to solve the system of equations.

xy = 10 and y = x + 3
x(x + 3) = 10
x² + 3x = 10
x² + 3x - 10 = 0
(x + 5)(x - 2) = 0
x + 5 = 0 or x - 2 = 0
x = -5 or x = 2

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

Geometry Problems of the Day (Geometry Regents, August 2013)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Geometry Regents, August 2013

Part I: Each correct answer will receive 2 credits.


25. In the diagram below, AC and AD are tangent to circle B at points C and D, respectively, and BC, BD, and BA are drawn.

If AC = 12 and AB = 15, what is the length of BD?

1) 6
2) 9
3) 24
4) 36

Answer: 2) 9

BD is a radius of the circle, so is BC. Therefore, BD = BC. BC is the shorter leg of a right triangle because BC is a radius and AC is a tangent, so angle ACB is a right angle.

If you don't recognize the Pythagorean Triple, use the Pythagorean Theorem to solve this.

12² + x² = 15²
144 + x² = 225
x² = 81
x = 9

26. 6 Triangle ABC shown below is a right triangle with altitude AD drawn to the hypotenuse BC.

If BD = 2 and DC = 10, what is the length of AB?

1) 2 * SQRT(2)
2) 2 * SQRT(5)
3) 2 * SQRT(6)
4) 2 * SQRT(30)

Answer: 3) 2 * SQRT(6)

When you draw an altitude in a Right Triangle from the right angle to the hypotenuse, the result is three similar right triangles. That means that the corresponding sides will be proportional.

AB is the shorter leg of BAC. It is also the hypotenuse of BDA. So you can set up a proportion of shorter leg to hypotenuse:

x / 12 = 2 / x
x² = 24
x = SQRT(24)
x = SQRT(2 * 2 * 2 * 3)
x = 2 * SQRT(2 * 3)
x = 2 * SQRT(6)

27. Triangle ABC has vertices A(0,0), B(6,8), and C(8,4). Which equation represents the perpendicular bisector of BC?

1) y = 2x - 6
2) y = -2x + 4
3) y = 1/2 x + 5/2
4) y = -1/2 x + 19/2

Answer: 3) y = 1/2 x + 5/2

The perpedicular bisector of BC will have a slope that is the negative reciprocal of the slope of BC and go through the midpoint of BC.

Note 1: Point A is irrelevant to this problem.

Note 2: All four choices have different slopes, so you don't need to find the midpoint of BC.

The slope of BC is (4 - 8) / (8 - 6) = -4 / 2 = -2. Therefore, the slope of the perpendicular bisector must be +1/2.

28. Chords AB and CD intersect at point E in a circle with center at O. If AE = 8, AB = 20, and DE = 16, what is the length of CE?

1) 6
2) 9
3) 10
4) 12

Answer: (1) 6

If AB intersects CD at point E then the product of (AE)(BE) must be equal to the product of (CE)(DE). Also, if AB = 20 and AE = 8, then BE = 20 - 8 = 12

(AE)(BE) = (CE)(DE)
(8)(12) = (CE)(16)
96 = 16(CE)
6 = (CE)

Just for giggles, another way to solve the above problem:

16(CE) = (8)(12) = (2)(2)(2)(2)(2)(3) = (2)(2)(2)(2) * (2)(3) = (16)(6)
So CE = 6
You can always use those Properties of Real Numbers that you learned in Algebra class.

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Algebra 2/Trigonometry Regents, January 2014

Part I: Each correct answer will receive 2 credits.


25. Which equation represents a graph that has a period of 4π?

1) y = 3 sin 1/2 x
2) y = 3 sin 2x
3) y = 3 sin 1/4 x
4) y = 3 sin 4x

Answer: 1) y = 3 sin 1/2 x

The graph of y = sin x has a period of 2π. When the argument of the sine function is halved, the period is doubled, which is a period of 4π.

22. The expression x²(x + 2) - (x + 2) is equivalent to

1) x²
2) x² - 1
3) x³ + 2x² - x + 2
4) (x + 1)(x - 1)(x + 2)

Answer: 4) (x + 1)(x - 1)(x + 2)

Before you try to multiply everything, notice that there are two terms, and each one has (x + 2) in it. This can be factored out.

x²(x + 2) - (x + 2) is the same as (x + 2)(x² - 1)

Moreover, (x² - 1) is the Difference of Squares, which can be factored into the conjugates (x - 1)(x + 1).

So x²(x + 2) - (x + 2) is the same as (x + 2)(x - 1)(x + 1), which is Choice (4).

27. Approximately how many degrees does five radians equal?

1) 286
2) 900
3) π/36
4) 5π

Answer: 1) 286

To convert, remember that 360 degrees is 2π radians, and 180 degrees is π radians.

Set up a proportion:

180 / π = x / 5
πx = (5)(180)
x = (5)(180)/π
x = 286.478...

Five radians is approximately 286 degrees.

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.