## Monday, September 20, 2021

### Factor Song

(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?).

A true artist is never appreciated in his own time. Or class.

Folks who follow the math problem calender on Twitter might've gotten a sneak peek last week at this little ditty. The plan was to use it here as well, but I wasn't exactly sure how I was going to use it. Or when, since the new school year keeps me busy until I get things settled into a routine.

The problem written in the top panel is how it appeared on the calendar. In the bottom panel, I rewrote it in a way more commonly seen in math class when covering combinations.

### 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, June 2013)

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, June 2013

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

32. Find, algebraically, the measure of the obtuse angle, to the nearest degree, that satisfies the equation 5 csc θ = 8.

Remember that csc = 1/sin.

5 csc θ = 8
5 = 8 sin θ
sin θ = 5/8
θ = sin-1 (5/8) = 38.6...

Since 39 degrees would be an acute angle, we want 180 - 39 degrees = 141 degrees.

33. If g(x) = (ax * SQRT(1 - x))2, express g(10) in simplest form.

Subsitute and evaluate.

g(10) = (a(10) * SQRT(1 - (10)))2
= (10a * SQRT(-9))2
= (10a * 3i)2
= (30ai)2
= -900a

34. Express (cot x sin x) / sec x as a single trigonometric function, in simplest form, for all values of x for which it is defined.

Replace cot with cos / sin and sec with 1 / cos.

(cot x sin x) / sec x
= (cos x / sin x ) * (sin x) * 1 / (1 / cos x)
= (cos x) * (cos x)
= cos2 x

35. On a multiple-choice test, Abby randomly guesses on all seven questions. Each question has four choices. Find the probability, to the nearest thousandth, that Abby gets exactly three questions correct.

There is a 1/4 chance that each question is correct, and a 3/4 chance that each question is wrong.

The probability that Q1,2,3 are correct and Q4,5,6,7 are incorrect is (1/4)(1/4)(1/4)(3/4)(3/4)(3/4)(3/4).

This number needs to be multiplied by the number of ways that exactly three correct answers can be chosen, which is 7C3.

So 7C3 (1/4)3 (3/4)4 = 0.173

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, June 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, June 2013

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

32. On the ray drawn below, using a compass and straightedge, construct an equilateral triangle with a vertex at R. The length of a side of the triangle must be equal to a length of the diagonal of rectangle ABCD.

Use the straightedge to draw diagonal AC. This isn't essential but it's a good visual. Put the compass at point A and open it to the length of AC. Make a little arc through point C.

Next, without changing the compass, move it to point R and make and arc from the ray counterclockwise above the ray. Then move the compass to the point where the arc intercepted the ray and make another arc above the ray that intersects the other one. The point of intersection is the third point of the equilateral triangle.

Use the straightedge to complete the triangle.

33. On the set of axes below, graph the locus of points 4 units from the x-axis and equidistant from the points whose coordinates are (-2,0) and (8,0). Mark with an X all points that satisfy both conditions.

The locus of points 4 units from the x-axis are the two horizontal parallel lines y = 4 and y = -4. The points equidistant from (-2,0) and (8,0) are sit on the perpendicular bisector of a line segment joining those points.

First draw y = 4 and y = 4.

Next, both (-2, 0) and (8, 0) are on the x-axis, which is a horizontal line. The midpoint between those two points is ( (8-2)/2, 0), or (3, 0). Draw a vertical line at x = 3.

Place an X at point (3, 4) and (3, -4).

34. The coordinates of two vertices of square ABCD are A(2,1) and B(4,4). Determine the slope of side BC.

Squares have right angles, so the sides are perpendicular. Find the slope of AB. The slope of BC is the

The slope of AB is (4 - 1) / (4 - 2) = 3 / 2.

The slope of BC is -2/3 because that is the inverse reciprocal, and (3/2)(-2/3) = -1.

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.

## Sunday, September 19, 2021

### Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, June 2013)

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, June 2013

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

28. Determine the sum and the product of the roots of the equation 12x2 + x - 6 = 0.

You do not need to find what the roots are to get the sum and product of the roots.

The product of the roots is given by the formula c / a. The sum is given by the formula -b / a.

The sum is -1 / 12 = -1/12

The product is -6 / 12 = -1/2

If you wanted to find the roots:

12x2 + x - 6 = 0, a = 12, b = 1, c = -6.
Find the factors of ac that add up to b.
The factors of -72 that add up to 1 are 9 and -8.
12x2 + 9x - 8x - 6 = 0
3x(4x + 3) - 2(4x + 3) = 0
(3x - 2)(4x + 3) = 0
3x - 2 = 0 or 4x + 3 = 0
3x = 2 or 4x = -3
x = 2/3 or x = -3/4.

The product of the roots is (2/3)(-3/4) = -6/12 = -1/2

The sum of the roots is 2/3 - 3/4 = 8/12 - 9/12 = -1/12.

Make sure these are labeled.

29. Solve algebraically for x:

log27 (2x - 1) = 4/3

Rewrite the equation using exponents:
27(4/3) = 2x - 1
34 = 2x - 1
81 = 2x - 1
82 = 2x
x = 41

30. Find the number of possible different 10-letter arrangements using the letters of the word “STATISTICS.”

The number of 10-letter arrangements of 10 letters is 10! (that is, 10 factorial). However, there are duplicate letters that need to be accounted for because of the repetition: 3 Ss, 3 Ts, and 2 Is.

10!
3! * 3! * 2!

= 10 * 9 * 8 * 7 *6 * 5 * 4 * 3 * 2 * 1
3 * 2 * 1 * 3 * 2 * 1 * 3 * 2 * 1

= 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
3 * 2 * 1 * 3 * 2 * 1 * 2 * 1

= 10 * 7! = 10 * 5040 = 50,400

You could have entered the equation into your calculator at any point after writing the factorial notation, if your calculator handles it.

I simplified to 7! just because I knew that 7! = 5040, which is basically where I have the factorials memorized to. For 8!, I usually do 8 * 5040 in my head. The numbers aren't so bad.

31. Express the product of cos 30° and sin 45° in simplest radical form.

cos 30° = SQRT(3) / 2 and sin 45° = SQRT (2) / 2

The product of SQRT (3) / 2 * SQRT (2) / 2 = SQRT (6) / 4. This cannot be reduced.

Looking at the answer key, they also accepted SQRT (3) / (2 SQRT (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, June 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, June 2013

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

29. A right circular cylinder has a height of 7 inches and the base has a diameter of 6 inches. Determine the lateral area, in square inches, of the cylinder in terms of π.

The lateral area is the surface area not including the top and bottom. If the cylinder were a soup can, it would be the area of the rectangular label on it.

The length of the rectangle is the circumference of the circle. The width is the height of the cylinder.

Therefore, A = (6π)(7) = 42π

30. Determine, in degrees, the measure of each interior angle of a regular octagon.

The formula for the sum of the interior angles for a polygon is (n-2)* 180. To find the size of each individual angle in a regular polygon, you have to divide the total by the number of angles: (n-2)*180/n.

An octagon has 8 sides, so (n-2)*180/n = (8-2)(180)/(8) = (6)(180)/(8) = 135.

Alternatively, the sum of the exterior angles of ANY polygon is 360. Each individual exterior angle is a regular polygon is 360/n. This angle is supplementary to the interior angle.

So 180 - 360/8 = 180 - 45 = 135.

31. Triangle ABC has vertices at A(3,0), B(9,-5), and C(7,-8). Find the length of AC in simplest radical form.

Note that point B is irrelevant to the problem. You can make a sketch of the triangle in the graph paper in the back of the booklet, but you still need to show work on the question page.

You can use the Distance Formula or Pythagorean Theorem to solve this.

d = SQRT ( (7 - 3)2 + (-8 - 0)2 )
= SQRT ( 42 + (-8)2 )
= SQRT ( 16 + 64 )
= SQRT (80)
= SQRT ( 2 * 2 * 2 * 2 * 5)
= 2 * 2 * SQRT(5)
= 4 SQRT(5)

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.

## Saturday, September 18, 2021

### Geometry Problems of the Day (Geometry Regents, June 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, June 2013

25. Which graph represents a circle whose equation is x2 + (y - 1)2 = 9?

The center of the circle is (0, 1) and the radius is 3. That is Choice (1).

The equation for a circle is (x - h)2 + (y - 1)2 = r2, where (h,k) is the center and r is the radius.

26. What is the perimeter of a rhombus whose diagonals are 16 and 30?

1) 92
2) 68
3) 60
4) 17

The diagonals of a rhombus will split the quadrilateral into 4 congruent right triangles. Since the diagonals are 16 and 30, then the legs of the right triangle will be 8 and 15.

Using Pythagorean Theorem, the hypotenuse, which is the side of the rhombus, will be 17. (You should know your Pythagorean Triples.)

The perimeter is 4 * 17 = 68.

27. In right triangle ABC shown in the diagram below, altitude BD is drawn to hypotenuse AC, CD = 12, and AD = 3.

What is the length of AB?

1) 5 SQRT(3)
2) 6
3) 3 SQRT(5)
4) 9

According to the Right Triangle Altitude Theorem, (AD)(DC) = (DB)2.

(DB)2 = (3)(12) = 36
DB = 6
But this isn't the side we are looking for.

Using Pythagorean Theorem, 32 + 62 = (AB)2
9 + 36 = (AB)2
45 = (AB)2
AB = SQRT(45) = SQRT(9 * 5) = SQRT(9)SQRT(5) = 3 SQRT(5)

More directly, you could have written the proportion:

AB / AC = AD / AB
AB / 15 = 3 / AB

... which would have led to the same answer.

28. Secants JKL and JMN are drawn to circle O from an external point, J. If JK = 8, LK = 4, and JM = 6, what is the length of JN?

1) 16
2) 12
3) 10
4) 8

The Secant Secant Theorem tells us that the following products are equal:
(JK)(JL) = (JM)(JN)

If LK = 4 then JL = 8 + 4 = 12.

(8)(12) = (6)(JN)
96 = 6(JN)
JN = 16

End of Part I.

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, June 2013)

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, June 2013

25. Expressed with a rational denominator and in simplest form,

x / (x - SQRT(x) )
is:

1) ( x2 + x SQRT(x) ) / ( x2 - x )
2) - SQRT(x)
3) ( x + SQRT(x) ) / (1 - x)
4) ( x + SQRT(x) ) / (x - 1)

Answer: 4) ( x + SQRT(x) ) / (x - 1)

To rationalize the denominator, multiply by the conjugate. Note that it says simplest form, so after you multiply, you might be able to simplify.

x / ( x - SQRT(x) ) * ( x + SQRT(x) ) / ( x + SQRT(x) )
= x ( x + SQRT(x) ) / ( ( x - SQRT(x) ) ( x + SQRT(x) ) )
= x ( x + SQRT(x) ) / ( x2 - x )
= x ( x + SQRT(x) ) / ( x (x - 1) )
= ( x + SQRT(x) ) / (x - 1)

26. What is the common ratio of the sequence

1/64 a5b3, -3/32 a3b4, 9/16 ab5, ... ?

1) -3b/(2 a2)
2) -6b/(a2)
3) -3a2/b
4) -6a2/b

Look at what is happening in each of the terms:

The numerator is being multiplied by -3, and the denominator is divided by 2, which means it is multipled by 1/2. So each term is multiplied by -3 / (1/2), which is -6.

Each a term has its exponent reduced by 2, so that's a-2 or 1/a2.

Each b term has its exponent increased by 1, so that's b1 or just b.

That makes the common ratio -6b / a2

27. In triangle KLM, KL = 20, LM = 13, and m∠K = 40. The measure of ∠M

1) must be between 0° and 90°
2) must equal 90°
3) must be between 90° and 180°
4) is ambiguous

Answer: 1) must be between 0° and 90°

Angle M is opposite side KL. Angle K is opposite LM.

THe Law of Sines says that sin M / 20 = sin 40 / 13

So sin M = 20 * sin 40 / 13 = 0.9889

M = 81.45 degrees

According to the Regents, the answer is (4) ambiguous, but I have to assume that's based on just looking at the size of the angles. Angle M MUST BE bigger than Angle K. Without doing any work, you don't know if it is over 90, 90 or under 90. HOWEVER, with the Law of Sines, we have enough information to calculate the answer. We KNOW FOR A FACT that it is about 81 degrees.

Whether the Regents ever corrected this mistake, I have no idea.

If your teacher is using these old questions for test prep and marks you incorrect, bring this to his or her attention, politely.

End of Part I

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.

## Friday, September 17, 2021

### Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, June 2013)

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, June 2013

21. The expression log 4m2 is equivalent to

1) 2(log 4 + log m)
2) 2 log 4 + log m
3) log 4 + 2 log m
4) log 16 + 2 log m

Answer: 3) log 4 + 2 log m

When you multiply to expressions with the same base you add the exponents. If you have a log of two things being multiplied, you can add the logs. When you raise a power to another power, you multiply the exponents. If you have a log of a variable with an exponent, you can multiply the log by the exponent.

In other words:

log 4m2
= log 4 + log m2
= log 4 + 2 log m

22. In triangle PQR, p equals

1) r sin P / sin Q
2) r sin P / sin R
3) r sin R / sin P
4) q sin R / sin Q

Answer: 2) r sin P / sin R

Use the Law of Sines:
sin P / p = sin Q / q = sin R / r

Split this into sin P / p = sin Q / q or sin P / p = sin R / r

Solving for p gives either p = q sin P / sin Q or p = r sin P / sin R.

23. If tan ( Arc cos Sqrt(3)/k ) = Sqrt(3) / 3, then k is

1) 1
2) 2
3) SQRT (2)
4) 3 SQRT (2)

In you take Tan -1 SQRT(3)/3, you get π/6.

Cos &pi/6 = SQRT(3)/2, so k = 2.

24. Which expression is equivalent to 2x-2y-2 / (4 y-5)

1) y3 / (2 x2)
2) 2y3 / (x2)
3) 2x2 / (y3)
4) x2 / (2 y3)

Answer: 1) y3 / (2 x2)

A variable in the numerator with a negative exponent gets moved to the denominator and its exponent is switched to positive. If it's in the demoninator, it comes up to the numerator.

2x-2y-2 / (4 y-5)
= 2y5 / (4 x2 y2)

Now simplify the factors:

2y5 / (4 x2 y2)
= 1 y3 / (2 x2)

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, June 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, June 2013

21. In triangle ABC, m∠A = 60, m∠B = 80, and m∠C = 40. Which inequality is true?
AB > BC
2) AC > BC
3) AC < BA
4) BC < BA

It may not look obvious, but Choices (1) and (4) are the same thing. Either they are both true or they are both false, but they cannot both be true.

For any triangle the longest side is opposite the biggest angle, and the smallest side is opposite the smallest angle. If two angles are congruent, then the opposite sides are congruent.

If ∠B is the largest angle then side AC is the longest side, which means it's bigger than BC, so Choice (2) is correct.

22. Circle O with ∠AOC and ∠ABC is shown in the diagram below.

What is the ratio of m∠AOC to m∠ABC?

1) 1:1
2) 2:1
3) 3:1
4) 1:2

∠AOC is a central angle and ∠ABC is an inscribed angle. The relationship between the two is that a central angle is twice the size of an inscribed angle that intercepts the same arc (in this case, arc AC).

So the ratio of ∠AOC to ∠ABC is 2:1. Note that the order in the question is important. It is NOT 1:2. That would be the ratio of ∠ABC to ∠AOC.

23. A rectangular prism has a base with a length of 25, a width of 9, and a height of 12. A second prism has a square base with a side of 15. If the volumes of the two prisms are equal, what is the height of the second prism?

1) 6
2) 8
3) 12
4) 15

Calculate the Volume of the first prism, and then solve for the height of the second one.

V = (25)(9)(12) = 2700

V = (15)(15)H = 2700
H = 2700 / 225 = 12

The slopes are not the same and they are not inverse reciprocials (that is, they do not have a product of -1). Therefore, they intersect but are not perpendicular.

24. In triangles ABC and DEF, AB = 4, AC = 5, DE = 8, DF = 10, and ∠A ≅ ∠D. Which method could be used to prove triangle ABC ∼ triangle DEF?

1) AA
2) SAS
3) SSS
4) ASA angle.

When you see similarity, your first thought is to go to AA, but you are only given one angle here.

However, you are given two pairs of sides and the included angle, which is SAS. The corresponding sides are proportional. so the triangles are similar.

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.

## Thursday, September 16, 2021

### (x, why?) Mini: Betweenness

(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?).

Between the laughs and the math lies obsession.

If B came between point A and point A, then Point A and Point B would have to be the same point. However, contrary to the large representation in the comic for the benefit of your eyesight (and especially mine), since points are dimenionalness spots in space, the only way for there not to be any other dimensional location in spot in between A and B would be for A and B to be the same point.

### 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, June 2013)

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, June 2013

16. Which value of r represents data with a strong positive linear correlation between two variables?

1) 0.89
2) 0.34
3) 1.04
4) 0.01

A strong positive correlation means that the value of r will be close to 1 WITHOUT GOING OVER 1. This is because the number 1 represents a 100% correlation, which would be the data is perfectly linear.

Choice (1) is Strong. Choice (2) is weak. Choice (3) is impossible. Choice (4) would suggest that there is no correlation between the two variables at all.

Which problem involves evaluating 6P4?

1) How many different four-digit ID numbers can be formed using 1, 2, 3, 4, 5, and 6 without repetition?
2) How many different subcommittees of four can be chosen from a committee having six members?
3) How many different outfits can be made using six shirts and four pairs of pants?
4) How many different ways can one boy and one girl be selected from a group of four boys and six girls?

Answer: 1) How many different four-digit ID numbers can be formed using 1, 2, 3, 4, 5, and 6 without repetition?

The permuation 6P4 refers to selecting four objects from a group of six when the order they are selected is important.

Choice (1) matches that description. This is the answer.

Choice (2) is a Combination problem because the order the members are selected is not important. (If the first person were, say, head of the committee, and the second one the assitant, etc, then it would be different.)

Choice (3) is an example of the Counting Principle, where 6 * 4 = 24 outfits can be made.

Choice (4) is an example of the Counting Principle, where 4 * 6 = 24 pairs of one boy and one girl can be selected.

18. Which equation is represented by the graph below?

1) (x - 3)2 + (y + 1)2 = 5
2) (x + 3)2 + (y - 1)2 = 5
3) (x - 1)2 + (y + 3)2 = 13
4) (x + 3)2 + (y - 1)2 = 13

Answer: 4) (x + 3)2 + (y - 1)2 = 13

The equation of a circle is given by the formula: (x - 2)2 + (y - 5)2 = r2, where (h, k) is the center and r is the radius. Notice that there are MINUS signs in the formula for (h, k).

The radius isn't obvious, and you could figure it out using Distance formula or Pythagorean Theorem. Or you could see that it is somewhere between 3 and 4, so r2 must be between 9 and 16. It could be 13, but it couldn't be 5.

Aside from this, the four choices each show a different center, so you only need to know the center to find the correct equation. The center is at (-3, 1). When you flip the signs you get (x + 3)2 + (y - 1)2.

19. If x = 3i, y = 2i, and z = m + i, the expression xy2z equals

1) -12 - 12mi
2) -6 - 6mi
3) 12 - 12mi
4) 6 - 6mi

Multiply the expressions, use the Distributive Property, and don't forget that i2 = -1.

xy2z
= (3i)(2i)2(m + i)
= (3i)(4i2)(m + i)
= (3i)(-4)(m + i)
= -12i(m + i)
= -12mi - 12i2
= -12mi + 12
= 12 - 12mi

20. An angle, P, drawn in standard position, terminates in Quadrant II if

1) cos P < 0 and csc P < 0
2) sin P > 0 and cos P > 0
3) csc P > 0 and cot P < 0
4) tan P < 0 and sec P > 0

Answer: 3) csc P > 0 and cot P < 0

In Quadrant II, x is negative (x < 0) and y is positive (y > 0).

Remember that cos is x, so it should be negative, and sin is y, so it should be positive. Also, sec, is 1/cos, so it is negative, csc is 1/sin so it should be positive, tan is sin/cos, so it is negative, and cot is cos/sin, so it is negative.

You might want to make a little chart next to the question that you can refer to.

Choice (1) has the first part right and the second part wrong.

Choice (2) has the first part right and the second part wrong.

Choice (3) has both parts correct.

Choice (4) has the first part right and the second part wrong.

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, June 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, June 2013

16. What is the equation of a line passing through the point (6,1) and parallel to the line whose equation is 3x = 2y + 4?

1) y = -2/3 x + 5
2) y = -2/3 x - 3
3) y = 3/2 x - 8
4) y = 3/2 x - 5

Answer: 3) y = 3/2 x - 8

The point (6, 1) has to be a solution to the linear equation, and the linear equation has to have a slope that is equal to the slope of the line it is parallel to.

Rewrite 3x = 2y + 4 as 2y = 3x - 4 and then y = 3/2 - 2. The slope of the line is 3/2, so eliminate Choices (1) and (2). Those lines are perpendicular to the given line.

Substitute x = 6 into the other equations, 3/2 (6) - 8 = 9 - 8 = 1, which is the y-coordinate we are looking for. Choice (3) is the answer.

Check: 3/2 (6) - 5 = 9 - 5 = 4, which is not the point we are looking for.

17. The volume of a sphere is approximately 44.6022 cubic centimeters. What is the radius of the sphere, to the nearest tenth of a centimeter?

1) 2.2
2) 3.3
3) 4.4
4) 4.7

The Volume of a sphere, using the formula in the back of the booklet, is V = 4/3 π r3.

Write an equation using the given Volume:

4/3 π r3 = 44.6022
r3 = 44.6022 / ( (4/3)( π) )
r3 = 10.647...
r = CBRT(10.647...) = 2.199..., or about 2.2.

Check: V = (4/3)(3.141592)(2.2)3 = 44.6022...

If you don't know how to find the cube root function on your calculator, you can raise the number to the power of (1/3). Use parentheses around the fraction.

18. Points A(5,3) and B(7,6) lie on AB. Points C(6,4) and D(9,0) lie on CD. Which statement is true?

1) AB || CD
2) AB ⊥ CD
3) AB and CD are the same line.
4) AB and CD instersect, but are not perpendicular.

Answer: 4) AB and CD instersect, but are not perpendicular.

First find to two slopes to find out if either (2) or (4) are true. If the slopes are the same, check if they are the same line.

Slope of AB: (6 - 3) / (7 - 5) = 3/2

Slope of CD: (0 - 4) / (9 - 6) = -4/3

The slopes are not the same and they are not inverse reciprocials (that is, they do not have a product of -1). Therefore, they intersect but are not perpendicular.

19. Which set of equations represents two circles that have the same center?

1) x2 + (y + 4)2 = 16 and (x + 4)2 + y2 = 16
2) (x + 3)2 + (y - 3)2 = 16 and (x - 3)2 + (y + 3)2 = 25
3) (x - 7)2 + (y - 2)2 = 16 and (x + 7)2 + (y + 2)2 = 25
4) (x - 2)2 + (y - 5)2 = 16 and (x - 2)2 + (y - 5)2 = 25 angle.

Answer: 4) (x - 2)2 + (y - 5)2 = 16 and (x - 2)2 + (y - 5)2 = 25

If two circles are concentric, have the same center, then their equations will be the same excpt for the size of the radius squared, which comes are the equal sign.

Choice (1) has centers at (0, -4) and (-4, 0).

Choice (2) has centers at (-3, 3) and (3, -3).

Choice (3) has centers at (7, 2) and (-7, -2).

Choice (4) has centers at (2, 5) only.

20. Transversal EF intersects AB and CD, as shown in the diagram below.

Which statement could always be used to prove AB || CD?

1) ∠2 ≅ ∠4
2) ∠7 ≅ ∠8
3) ∠3 and ∠6 are supplementary
4) ∠1 and ∠5 are supplementary

Answer: 3) ∠3 and ∠6 are supplementary

If two lines are parallel and crossed by a transversal, then there are several statements that are true regarding corresponding, congruent and supplementary angles.

Choice (1) is true because they are vertical, but it doesn't prove anything about line CD.

Choice (2) is false. They are supplementary. Also they have nothing to do with line AB.

Choice (3) is true. Same-side interior angles will be supplementary. If they are supplementary, then the lines are parallel.

Choice (4) is false. The corresponding angles are congruent, not supplementary.

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.

## Wednesday, September 15, 2021

### Geometry Problems of the Day (Geometry Regents, June 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, June 2013

11. In ABC, D is the midpoint of AB and E is the midpoint of BC. If AC = 3x - 15 and DE = 6, what is the value of x?

1) 6
2) 7
3) 9
4) 12

If DE is the midsegment and DE = 6 then AC = 12 because the side of the triangle parallel to the midsegment is double the size of the midsegment. (And the midsegment is half the size of that side.)

So 3x - 15 = 12
and 3x = 27
then x = 9.

12. What are the coordinates of the center of a circle if the endpoints of its diameter are A(8,-4) and B(-3,2)

1) (2.5,1)
2) (2.5,-1)
3) (5.5,-3)
4) (5.5,3)

If you are given the endpoints of a diameter then the center of the circle is the midpoint. Average the two x-coordinates and the two y-coordinates.

Point M( (8 + -3)/2, (-4 + 2)/2 ) = (2.5, -1)

13. Which graph could be used to find the solution to the following system of equations?

y = (x + 3)2 - 1
x + y = 2

The vertex of the parabola is (-3, -1). You don't have to worry about flipping the signs because all of the choices showing the vertex at x = -3. You DO NOT flip the sign of the y-coordinate. Subtracting 1 lowers the graph. Eliminate Choices (3) and (4).

The line x + y = 2 can be rewritten as y = -x + 2, which has a slope of -1, which goes DOWN as it moves to the right.

14. What is the converse of “If an angle measures 90 degrees, then it is a right angle”

1) If an angle is a right angle, then it measures 90 degrees.
2) An angle is a right angle if it measures 90 degrees.
3) If an angle is not a right angle, then it does not measure 90 degrees.
4) If an angle does not measure 90 degrees, then it is not a right angle.

Answer: 1) If an angle is a right angle, then it measures 90 degrees.

For Converse, think of sneakers, and walking around. The converse is when the IF walks over and becomes the THEN and vice versa. INVERSE is the one with the "Nots" (or the Nots removed).

Choice (1) has the two clause reversed. This is the answer.

Choice (2) is a closed statement of fact, which is true. But it is not a conditional statement, so it is not the Converse of the original statement.

Choice (3) is the Contrapositive. The clauses are switched but they are also negated.

Choice (4) is the Inverse. The clauses are negated, not switched.

15. As shown in the diagram below, a right pyramid has a square base, computations. ABCD, and EF is the slant height.

Which statement is not true?

1) EA ≅ EC
2) EB ≅ EF
3) triangle AEB ≅ triangle BEC
4) triangle CED is isosceles

It's a right pyramid with a square base.

Choice (1) is always true. These lines will always be congruent.

Choice (2) is never true. EF is the slant height and EB is an edge. EFB forms a right triangle and EB is the hypotenuse, which means it must be longer that EF.

Choice (3) is always true. The four triangular faces will always be congruent.

Choice (4) is always true. The four triangular faces will always be isosceles.

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, June 2013)

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, June 2013

11.In the right triangle shown below, what is the measure of angle S, to the nearest minute?

1) 28°1'
2) 28°4'
3) 61°56'
4) 61°93'

A minute is 1/60 of a degree, the same way that a minute is 1/60 of an hour. It is a "minute" (pronounce this last one as "my NEWT") portion.

Since there are only 60 minutes in a degrees, you can eliminate choice (4).

You are given the opposite side from S and the hypotenuse, so use Sine.

Sin x = 8/17
x = sin-1 (8/17) = 28.072..., but that's a decimal you have to convert the decimal to Sexigesmal:
0.072 * 60 = 4.32

So angle S has a measure of 28°4'.

Note that if you'd done cosine by mistake, you would've gotten choice (3).

12. Which ordered pair is in the solution set of the system of equations shown below?

y2 - x2 + 32 = 0
3y - x = 0

1) (2,6)
2) (3,1)
3) (-1,-3)
4) (-6,-2)

Eliminate Choice (1) and (3) immediately. The x coordinate should be three times the y coordinate, not the other way around.

From there, you can check choices (2) and (4), but it should be obvious that 3 and 1 are too small to add 32 and make zero.

(-2)2 - (-6)2 + 32 = 4 - 36 + 32 = 0

13. Susie invests \$500 in an account that is compounded continuously computations. at an annual interest rate of 5%, according to the formula A = Pert where A is the amount accrued, P is the principal, r is the rate of interest, and t is the time, in years. Approximately how many years will it take for Susie’s money to double?

1) 1.4
2) 6.0
3) 13.9
4) 14.7

Substitute what you are given into the equation A = Pert, which gives you 1000 = (500)e(.05)t.

You can take try the choices or you can solve it algebraically.

(500)e(.05)(1.4) = 536.25
(500)e(.05)(6.0) = 674.93
(500)e(.05)(13.9) = 1001.85
(500)e(.05)(14.7) = 1042.74

You want all the y values that are less than or equal to 0. There are no y values greater than 0.

Algebraically, solve for t:

1000 = (500)e(.05)t
2 = e.05t
ln 2 = ln e.05t
ln 2 = .05t
t = (ln 2) / .05 = 13.86...

14. If n is a negative integer, then which statement is always true?

1) 6n-2 < 4n-1
2) n/4 > -6n-1
3) 6n-1 < 4n-1
4) 4n-1 > (6n)-1

Let's look at the cases one at a time. Notice that the negative exponents only go with the variable, except in the last instance where it is next to parentheses.

6n-2 < 4n-1 means 6/n2 < 4/n.
Since n is a negative integer, the square of n will be be positive. Therefore 6/n2 will be positive, but 4/n will be negative. So the statement can never be true.

n/4 > -6n-1 means n/4 > -6/n.
n/4 will always be negative but -6/n will always be positive. So this statement can never be true.

6n-1 < 4n-1 means that 6/n < 4/n.
Since n is negative, this is always true.

4n-1 > (6n)-1 means 4/n > 1/6n, or 4/n > 4/24n.
Since n is negative, 4/24n would be much closer to 0 than 4/n. So this statement can never be true.

15. The expression 4 + (k=2)∑(5), 3(k-x) is equal to

1) 58 - 4x
2) 46 - 4x
3) 58 - 12x
4) 46 - 12x

Write out the summation:

4 + 3(2 - x) + 3(3 - x) + 3(4 - x) + 3(5 - x)

This first thing you should see is that there will be 12x, not 4x, so eliminate Choices (1) and (2).

4 + 6 - 3x + 9 - 3x + 12 - 3x + 15 - 3x = 4 + 6 + 9 + 12 + 15 - 12x = 46 - 12x

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.

## Tuesday, September 14, 2021

### Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, June 2013)

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, June 2013

6. Which equation represents the graph below?

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

Answer: 3) y = - 2 cos 2x

Cosine (0) = 1 and Sine (0) = 0. The function is cosine, so eliminate Choices (1) and (2).

Between 0 and 2π the graph makes two cycles instead of one, so this is cos 2x.

2. What is the graph of the solution set of |2x - 1| > 5?

First of all, substitute 0 for x. |2(0) - 1| > 5 is NOT a true statement. Zero is NOT a solution, so you can eliminate Choices (2) and (4).

This you only have to check if x = -3 is a solution or not to know the answer.

| 2(-3) - 1 | = | -6 - 1 | = | -7 | = 7, which IS greater than 5, so x = -3 IS a solution. This means that choice (3) is eliminated, and the answer is (1).

Algebraically, break up the equation:

2x - 1 > 5
2x > 6
x > 3

2x - 1 < -5
2x < -4
x < -2

The algebraic solution gives us graph number 1.

8. What is the range of the function shown below?

1) x < 0
2) x > 0
3) y < 0
4) y > 0

3) y < 0

Range refers to the y-values, so eliminate choices (1) and (2).

You want all the y values that are less than or equal to 0. There are no y values greater than 0.

9. The expression sin(θ + 90)° is equivalent to

1) -sin θ
2) -cos θ
3) sin θ
4) cos θ

If you recall from rotations in Geometry, a counterclockwise rotation of 90 degrees moved point A(x, y) to point A'(-y, x).

Similarly, a point B(cos θ, sin θ) will have an image B'(-sin θ, cos θ)

So sin θ becomes cos θ

10. The points (2,3), (4, 3/4), and (6,d) lie on the graph of a function.
If y is inversely proportional to the square of x, what is the value of d?

1) 1
2) 1/3
3) 3
4) 27

If y is inversely proportional to x then as x gets bigger y will get smaller. This means that d must be smaller than 3/4. THere is only one possibility.

(2)2(3) = 12

(4)2(3/4) = 12

(6)2d = 12
36d = 12
d = 1/3

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, June 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, June 2013

6. Which equation represents the circle whose center is (-5,3) and that passes through the point (-1,3)?

1) (x + 1)2 + (y - 3)2 = 16
2) (x - 1)2 + (y + 3)2 = 16
3) (x + 5)2 + (y - 3)2 = 16
4) (x - 5)2 + (y + 3)2 = 16

Answer: 3) (x + 5)2 + (y - 3)2 = 16

The equation of a circle is given by the formula (x - h)2 + (y - k)2 = r2, where (h, k) is the center of the circle and r is the radius. Note that there are MINUS signs in the formula, so if h or k is negative, then a plus will be indicated.

The center is (-5, 3), so eliminate choices (1) and (2) immediately. Since the x coordinate is negative there is a plus sign in the equation (x + 5). That is Choice (3).

7. As shown in the diagram below, when right triangle DAB is reflected over the x-axis, its image is triangle DCB.

Which statement justifies why AB ≅ CB?

1) Distance is preserved under reflection.
2) Orientation is preserved under reflection.
3) Points on the line of reflection remain invariant.
4) Right angles remain congruent under reflection.

Answer: 1) Distance is preserved under reflection.

The two lines remain congruent because the distance (length) is preserved.

Choice (2) says orientation is preserved in a reflection. This is not true.

Choice (3) is a true statement but doesn't have anything to do with lines AB and CB, which are not on the line of reflection.

Choice (4) is a true statement but doesn't have anything to do with lines AB and CB.

8. In ABC, m∠A = 3x + 1, m∠B = 4x - 17, and m∠C = 5x - 20. Which type of triangle is ABC?

1) right
2) scalene
3) isosceles
4) equilateral

You can immediately eliminate right because if the triangle is right, it still have to be one of the other three.

You have to find the size of each of the angles to determine how many are congruent, 0, 2, or 3.

3x + 1 + 4x - 17 + 5x - 20 = 180
12x - 36 = 180
12x = 216
x = 18

m∠A = 3x + 1 = 3(18) + 1 = 55, m∠B = 4x - 17 = 4(18) - 17 = 55, and m∠C = 5x - 20 = 5(18) - 20 = 70.

The triangle is isosceles.

9. What is the equation for circle O shown in the graph below?

1) (x - 3)2 + (y + 1)2 = 6
2) (x + 3)2 + (y - 1)2 = 6
3) (x - 3)2 + (y + 1)2 = 9
4) (x + 3)2 + (y - 1)2 = 9

Answer: 3) (x - 3)2 + (y + 1)2 = 9

The equation of a circle is given by the formula (x - h)2 + (y - k)2 = r2, where (h, k) is the center of the circle and r is the radius. Note that there are MINUS signs in the formula, so if h or k is negative, then a plus will be indicated.

The radius is 3, and 32 = 9. Eliminate (1) and (2).

The center is (3, -1). You FLIP the signs in the equation, so the answer is (3).

10. Point A is on line m. How many distinct planes will be perpendicular computations. to line m and pass through point A?

1) one
2) two
3) zero
4) infinte

While an infinitie number of planes can be perpendicular to a given line -- think of an elevator in a videogame that keeps going up or down forever -- only ONE will be perpendicular to a line at a specific point.

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.