(x, why?): November 2021

Tuesday, November 30, 2021

Geometry Problems of the Day (Geometry Regents, August 2011)

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

Geometry Regents, August 2011

Part I: Each correct answer will receive 2 credits.

6. A straightedge and compass were used to create the construction below. Arc EF was drawn from point B, and arcs with equal radii were drawn from E and F.

Which statement is false?

1) m∠ABD = m∠DBC
2) 1/2 (m∠ABC) = m∠ABD
3) 2(m∠DBC) = m∠ABC
4) 2(m∠ABC) = m∠CBD

Answer: 4) 2(m∠ABC) = m∠CBD

This is a construction of an angle bisector. So ADB = DBC, ABD is 1/2 of ABC, and ABC is twice DBC. But ABC is the large angle, so it would equal twice CBD, not the other way around.

Choice (4) is false.

7. What is the length of the line segment whose endpoints are (1,−4) and (9,2)?

1) 5
2) 2 √(17)
3) 10
4) 2 √(26)

Answer: 3) 10

Use the distance formula or Pythagorean Theorem. You may need to simplify a radical.

d = SQRT( (9-1)² + (-4-2)² )
= SQRT ( (8)² + (-6)² )

At this point, you should recognize that this will be the hypotenuse of a 6-8-10 triangle. If you didn't, continue with the calculations. You'll get 10.

8. What is the image of the point (2,−3) after the transformation r_y-axis?

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

Answer: 2) (-2,-3)

The original point is in Quadrant IV. If you flip it over the y-axis, it will end up in Quadrant III. The signs will both be negative but the magnitude of the numbers will remain the same.

In other words, the image will be at (-2,-3), which is Choice (2)?

9. In the diagram below, lines n and m are cut by transversals p and q.

Which value of x would make lines n and m parallel?

1) 110
2) 80
3) 70
4) 50

Answer: 3) 70

For the lines to be parallel, x would have to be congruent to the corresponding angle, which is the unmarked one between the 30 and the 80 degree angles.

Those three angles make a straight line so 30 + 80 + x = 180. Therefore, x = 70.

10. What is an equation of the circle with a radius of 5 and center at (1,−4)?

1) (x + 1)² + (y − 4)² = 5
2) (x + 1)² + (y − 4)² = 5
3) (x + 1)² + (y − 4)² = 25
4) (x − 1)² + (y + 4)² = 25

Answer: 4) (x − 1)² + (y + 4)² = 25

The equation for a circle is (x − h)² + (y - k)² = r², where (h,k) is the center of the circle and r is the radius. There are minus signs in the formula, so the signs on the points are flipped.

Since the radius is 5, 5² = 25. Eliminate Choices (1) and (2).

Since the signs get flipped, the correct choice is (4).

More to come. Comments and questions welcome.

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Algebra Problems of the Day (Integrated Algebra Regents, August 2011)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones. The Integrated Algebra Regents covered most of the same material as the current Algebra Regents, with a few differences.

More Regents problems.

Integrated Algebra Regents, August 2011

Part I: Each correct answer will receive 2 credits.

6. Based on the box-and-whisker plot below, which statement is false?

1) The median is 7.
2) The range is 12.
3) The first quartile is 4.
4) The third quartile is 11.

Answer: 2) The range is 12..

What we can identify from the graph is that the minimum value is 2, the first quartile is 4, the median is 7, the third quartile is 11, and the maximum value is 12. We can calculate that the range is 12 - 2 = 10 and the interquartile range (IQR) is 11 - 4 = 7.

Choice (2) is the only statement that is false.

7. The ninth grade class at a local high school needs to purchase a park permit for $250.00 for their upcoming class picnic. Each ninth grader attending the picnic pays $0.75. Each guest pays $1.25. If 200 ninth graders attend the picnic, which inequality can be used to determine the number of guests, x, needed to cover the cost of the permit?

1) 0.75x − (1.25)(200) ≥ 250.00
2) 0.75x + (1.25)(200) ≥ 250.00
3) 0.75x + (1.25)(200) ≥ 250.00
4) 0.75x + (1.25)(200) ≥ 250.00

Answer: 4) 0.75x + (1.25)(200) ≥ 250.00

Obviously, the sum has to be greater than $250.00, so eliminate the two subtraction answers, Choices (1) and (3).

If 200 ninth graders are paying $0.75 each then (200)(0.75) needs to appear in the solution. That's Choice (4). The x number of guests will each pay 1.25, so 1.25x is added to the (200)(0.75) already collected.

8. Which equation represents the line that passes through the point (1,5) and has a slope of −2?

1) y = -2x + 7
2) y = -2x + 11
3) y = 2x - 9
4) y = 3x + 3

Answer: 1) y = -2x + 7

A slope of -2 means that the equation is y = -2x + b, and you have to find b. Eliminate Choices (3) and (4).

You can solve for b, or you can substitute 1 for x and see which equation gives you y = 5.

y = -2(1) + 7 = 5. Choice (1) is your answer.

y = -2(1) + 11 = 9. Not the equation we're looking for.

9. What is the solution of the system of equations 2x − 5y = 11 and −2x + 3y = −9

1) (−3,−1)
2) (−1,3)
3) (3,−1)
4) (3,−1)

Answer: 3) (3,−1)

You could solve algebraically, or you could work backward from the choices.

Algebraically, if you add the two equations together, you get -2y = 2, so y = -1. That eliminates Choices (2) and (4).

Again, you could solve for x, or you could substitute x = 3 and see if it works:

2(3) - 5(-1) ?= 11
6 + 5 = 11. Choice (3) is correct.

2(-3) - 5(-1) ?= 11
-6 + 5 != 11. Choice (1) was incorrect.

10. Which algebraic expression represents 15 less than x divided by 9?

1) x/9 - 15
2) 9x - 15
3) 15 - x/9
4) 15 - 9x

Answer: 1) x/9 - 15

Only two of the choices have division in them. The other two have multiplication and subtraction. Eliminate choices (2) and (4).

If it is "15 less than", then you are taking 15 away from x/9, so it is choice (1).

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Monday, November 29, 2021

The Battle of Two Points

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Doesn't everyone?

I realize now, several hours after creating the image, seeing anything may prove difficult since neither stick figure has eyes (that we can see). But I liked how the bandana and the facial hair came out. The swords wasn't too bad either.

Suggested when I politely corrected someone who was trying to self-correct but ended on "fo-KYE".

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Wednesday, November 24, 2021

Thanksgiving Geometry

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A reflex angle, of course.

A day early because I wanted to use it as a classroom problem, and there's no school tomorrow. YAY! Something to be thankful for!

If I don't post again tomorrow, Have a Happy Thanksgiving! Get stuffed!

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Geometry Problems of the Day (Geometry Regents, August 2011)

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

EDIT: I don't know what exam I started below, but it doesn't appear to be August 2011. It will be corrected soon.

More Regents problems.

Geometry Regents, August 2011

Part I: Each correct answer will receive 2 credits.

1. In the diagram below, AB, BC, and AC are tangents to circle O at points F, E, and D, respectively, AF = 6, CD = 5, and BE = 4.

What is the perimeter of △ABC?

1) 15
2) 25
3) 30
4) 60

Answer: 3) 30

When two tangents of a circle intersect, the segments between the point of intersection and the circle will be congruent. So AD ≅ AF; BE ≅ BF; CD ≅ CE.

So the perimeter is 6 + 6 + 5 + 5 + 4 + 4 = 30, or 2 * (6 + 5 + 4) = 30.

2. Quadrilateral MNOP is a trapezoid with MN || OP. If M′N′O′P′ is the image of MNOP after a reflection over the x-axis, which two sides of quadrilateral M′N′O′P′ are parallel?

1) M'N' and O'P'
2) M'N' and N'O'
3) P'M' and O'P'
4) P'M' and N'O'

Answer: 1) M'N' and O'P'

A reflection doesn't change the shape of the object. Lines that were parallel will still be parallel in the image.

If MN || OP in the pre-image, then M'N' || O'P' in the image.

3. In the diagram below of △ABC, D is the midpoint of AB, and E is the midpoint of BC.

If AC = 4x + 10, which expression represents DE?

1) x + 2.5
2) 2x + 5
3) 2x + 10
4) 8x + 20

Answer: 2) 2x + 5

The midsegment is half of the length of the side of the triangle that it is parallel to.

Half of 4x + 10 is 2x + 5.

4. Which statement is true about every parallelogram?

1) All four sides are congruent.
2) The interior angles are all congruent.
3) Two pairs of opposite sides are congruent.
4) The diagonals are perpendicular to each other.

Answer: 3) Two pairs of opposite sides are congruent

The opposite sides of a parallelogram are always congruent.

All four sides will be congruent only in a rhombus (including squares). The interior angles are all congruent only in rectangles (including squares). The diagonals are perpendicular only in rhombuses.

5. The diagram below shows a rectangular prism.
Which pair of edges are segments of lines that are coplanar ?

1) AB and DH
2) AE and DC
3) BC and EH
4) CG and EF

Answer: 3) BC and EH

The edges of a prism that are coplanar will be pointed in the same direction. If they aren't pointed in the same direction, they will be skew.

AB goes front to back, but DH goes up and down. Eliminate Choice (1).

AE goes up and down, but DC goes front to back. Eliminate Choice (2).

BC goes left to right, and EH goes left to right. This is the correct answer.

CG goes up and down, EF goes front to back. Eliminate Choice (4).

at long before the end.

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Algebra Problems of the Day (Integrated Algebra Regents, August 2011)

More Regents problems.

Integrated Algebra Regents, August 2011

Part I: Each correct answer will receive 2 credits.

1. The number of calories burned while jogging varies directly with the number of minutes spent jogging. If George burns 150 calories by jogging for 20 minutes, how many calories does he burn by jogging for 30 minutes?

1) 100
2) 180
3) 200
4) 225

Answer: 4) 225

If it varies directly, then the amounts will be proportional. So write a proportion.

150 / 20 = x / 30

20x = 4500

x = 225

2.The scatter plot below represents the relationship between the number of peanuts a student eats and the student’s bowling score.

Which conclusion about the scatter plot is valid?

1) There is almost no relationship between eating peanuts and bowling score.
2) Students who eat more peanuts have higher bowling scores.
3) Students who eat more peanuts have lower bowling scores.
4) No bowlers eat peanuts.

Answer: 1) There is almost no relationship between eating peanuts and bowling score.

The graph doesn't show any positive or negative trends. The points appear to be randomly scattered about the graph. So eating peanuts doesn't raise or lower bowling scores.

Choice (4) is the most obviously incorrect. Almost all of the bowlers are peanuts.

3. If the universal set is {pennies, nickels, dimes, quarters}, what is the complement of the set {nickels}?

1) {}
2) {pennies, quarters}
3) {pennies, dimes, quarters}
4) {pennies, nickels, dimes, quarters}

Answer: 3) {pennies, dimes, quarters}

The complement of a set is everything that is in the universal set that is NOT in the set.

So the complement of {nickels} is {pennies, dimes, quarters}

4. Which situation does not describe a causal relationship?

1) The higher the volume on a radio, the louder the sound will be.
2) The faster a student types a research paper, the more pages the paper will have.
3) The shorter the distance driven, the less gasoline that will be used.
4) The slower the pace of a runner, the longer it will take the runner to finish the race.

Answer: 2) The faster a student types a research paper, the more pages the paper will have.

First of all, the word is "causal", as in one thing causes another. It is not "casual", which is commonly misread.

In Choice (1), turning up the volume increases the sound. That is cause and effect. Eliminate Choice (1).

In Choice (2), the speed the student types doesn't affect the length of the paper. It may affect how soon the paper is completed, but not the length. Choice (2) is the answer.

In Choice (3), if you don't drive as far, you won't need as much gas as you would for a longer drive. Eliminate Choice (3).

In Choice (4), running slower will take longer to finish than running faster. Unlike Choice (2), this one doesn't mention the length of the course being run. Eliminate Choice (1).

5. A cylinder has a diameter of 10 inches and a height of 2.3 inches. What is the volume of this cylinder, to the nearest tenth of a cubic inch?

1) 72.3
2) 83.1
3) 180.6
4) 722.6

Answer: 3) 180.6

The volume of a cylinder is given by the formula V = πr²h, and the radius is half of the diameter.

So V = (3.141592)(5)²(2.3) = 180.64154, which is about 180.6.

More to come. Comments and questions welcome.

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Tuesday, November 23, 2021

Geometry Problems of the Day (Geometry Regents, January 2012)

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, January 2012

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

38. In the diagram below of quadrilateral ABCD,AD ≅ BC and ∠DAE ≅ ∠BCE.

Line segments AC, DB, and FG intersect at E.

Prove: Triangle AEF ≅ triangle CEG

Answer:

There is enough here to show that ABCD is a parallelogram, if that becomes important. There are properties that could be used, the most important of which is probably that the diagonals will bisect each other.

You are given a pair of congruent angles. You have a pair of vertical angles. You can show that the contained sides are parallel. So you can prove congruency through ASA.

This is not an obvious one. But then again, it's a Part IV problem.

Statement	Reason
1. AD ≅ BC, ∠DAE ≅ ∠BCE	Given
2. AD \|\| BC	Alternate interior angles are congruent
3. ABCD is a parallelogram	A quadrilateral with one pair of sides that are both parallel and congruent is a parallelogram
4. AE ≅ CE	The diagonals of a parallelogram bisect each other
5. ∠AEF ≅ &∠CEG	Vertical angles are congruent
5. Triangle AEF ≅ triangle CEG	ASA

End of Exam.

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Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2012)

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 2012

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

36. Perform the indicated operations and simplify completely:

(x³ - 3x² + 6x - 18) / (x² - 4x) * (2x = 4) / (x⁴ - 3x³) ÷ (x² + 2x - 8) / (16 - x²)

Answer:

First, change the division to multiplication and replace the last fraction with its inverse (i.e., "flip it", replace it with its reciprocal). Then, factor everything that can be factored. Finally, cancel out the common factors.

In particular, the first numerator can be factored by grouping. The last numerator (after flipping) is the Differenc of Two Perfect Squares. However, (16 - x²) can be reversed by multiplying by -1, becoming -1(x² - 16), which will be easier to work with.

What remains will be -2(x² + 6) / x⁴. You can multiply the two factors in the numerator, or leave them separate. Both are acceptable.

End of Exam.

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Monday, November 22, 2021

x = 2 sec y

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What a freq!

The construction of "sec y" is older than this blog, so I usually dismiss it. This is the first time that I've thought about it that a creative use for the pun came to mind.

Out of curiosity, I wondered what y = 2 sec xy might like. The graph below was created using Desmos.com. I love how it looks, and some of the finer details are missing.

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Friday, November 19, 2021

Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2012)

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 2012

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

36. The diagram below shows the plans for a cell phone tower. A guy wire attached to the top of the tower makes an angle of 65 degrees with the ground. From a point on the ground 100 feet from the end of the guy wire, the angle of elevation to the top of the tower is 32 degrees. Find the height of the tower, to the nearest foot.

Answer:

We are concerned with the two smaller triangles: the one with the guy wire and the 62-degree angle, and the one with the guy wire and the tower. You need to use the Law of Sines to find the length of the guy wire. You can then use that as the hypotenuse of the right triangle to find the height of the tower.

The Exterior Angle Theorem tells us that the top angle in the left side triangle is 65 - 32 = 33 degrees. So:

sin 32 / x = sin 33 / 100

x = 100 * sin 32 / sin 33 = 97.297 ...

We have the hypotenuse of the right side triangle and we need the opposite side, so use sine.

sin 65 = x / 162.116

x = 97.297 * sin 65 = 88.18

The height of the tower is 88 feet.

37. If log ₄ x = 2.5 and log _y 125 = -3/2, find the numerical value of x/y, in simplest form.

Answer:

Use inverse operations.

log ₄ x = 2.5

4^2.5 = x

4^5/2 = x

2⁵ = x

x = 32

log _y 125 = -3/2

y^-3/2 = 125

1/y^3/2 = 125

y^3/2 = 1/125

y^1/2 = 1/5

y = 1/25

Substitute: x/y = 32 / (1/25) = 800

38. A population of single-celled organisms was grown in a Petri dish over a period of 16 hours. The number of organisms at a given time is recorded in the table below.

Determine the exponential regression equation model for these data, rounding all values to the nearest ten-thousandth.

Using this equation, predict the number of single-celled organisms, to the nearest whole number, at the end of the 18th hour.

Answer:

Put the data into lists in your graphing calculator, and run an exponential regression.

The nearest ten-thousandth means four decimal places. You should have gotten a = 27.2025 and b = 1.1509, so the equation is y = 27.2025(1.1509)^x.

To predict the number of organisms, substitute 18 for x and calculate.

y = 27.2025(1.1509)¹⁸ = 341.41... = 341.

The prediction is 341 organisms.

End of Part III.

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Geometry Problems of the Day (Geometry Regents, January 2012)

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, January 2012

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

35. Triangle ABC has coordinates A(2,-2), B(2,1), and C(4,-2). Triangle A'B'C' is the image of triangle ABC under T_5,-2.

On the set of axes below, graph and label triangle ABC and its image, triangle A'B'C'.

Determine the relationship between the area of triangle ABC and the area of tirangle A'B'C'.

Justify your response.

Answer:

Here is the graph of ABC and its image, which is five units to the right and two units down.

The area of the triangle and its image is the same because a translation is a rigid motion which preserves size, shape and area.

Alternatively, since it the question doesn't say how to justify your response, you could find the area of both triangles:

Area of ABC = 1/2(3)(2) = 3

Area of A'B'C' = 1/2(3)(2) = 3

Area of ABC = Area of A'B'C'.

36. A paint can is in the shape of a right circular cylinder. The volume of the paint can is 600π cubic inches and its altitude is 12 inches.

Find the radius, in inches, of the base of the paint can. Express the answer in simplest radical form.

Find, to the nearest tenth of a square inch, the lateral area of the paint can.

Answer:

The lateral area of a cylinder is the surface area without the top and bottom, like a tube, or a label on a soup can. If you open up the cylinder, it's a rectangle with a length equal to the circumference of the circles on top and bottom (and width equal to the height of the cylinder).

V = π r² h

600π = π r² (12)

r² = 600 / 12 = 50

r = √(50) = √(25*2) = 5 √(2)

The radius is 5 √(2).

LA = 2 π r h

LA = 2 π 5√(2) (12) = 533.14...

The lateral area is 533.1.

37. Triangle HKL has vertices H(-7,2), K(3,-4), and L(5,4). The midpoint of HL is M and the midpoint of LK is N.

Determine and state the coordinates of points M and N.

Justify the statement: MN is parallel to HK.

[The use of the set of axes below is optional.]

Answer:

Normally, I wouldn't bother with the graph and do everything algebraically. But when it comes to proveing that two lines are parallel because they have the same slope, it's nice to it visually. Also, when you see the boxes in front of you, it's harder to make a sign error that dives you the wrong coordinate.

MN is parallel to HK because they have the same slope as shown in this graph:

Algebraically:

Midpoint of HL is M( (-7+5)/2, (2+4)/2 ) = M(-1,3)

Midpoint of LK is N( (3+5)/2, (-4+4)/2 ) = N(4,0)

To show that MN is parallel to HK, find the slopes of each line:

Slope of MN = (3 - 0) / (-1 - 4) = -3/5.

Slope of HK = (-4 - 2) / (3 - -7) = -6/10 = -3/5.

HK and MN have the same slope, so they are parallel.

End of Part III.

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Thursday, November 18, 2021

Algebra Problems of the Day (Integrated Algebra Regents, January 2012)

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

More Regents problems.

Integrated Algebra Regents, January 2012

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

37. The sum of three consecutive odd integers is 18 less than five times the middle number. Find the three integers. [Only an algebraic solution can receive full credit.]

Answer:

Consecutive integers are one after another, like n, n+1, n+2... Consecutive odd integers are each two more than the previous one, like n, n+2, n+4, where n is an odd integer.

Change the sentence in the question into an equation:

n + n + 2 + n + 4 = 5(n + 2) - 18

3n + 6 = 5n + 10 - 18

3n + 6 = 5n - 8

14 = 2n

7 = n

The numbers are 7, 9 and 11.

You could also have used n - 2, n and n + 2 and save yourself some work:

n - 2 + n + 2 = 5n - 18

3n = 5n - 18

-2n = - 18

n = 9

The numbers are still, 7, 9, and 11 becuase you used n-2, n and n+2.

End of Exam.

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Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2012)

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 2012

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

32. A blood bank needs twenty people to help with a blood drive. Twenty-five people have volunteered. Find how many different groups of twenty can be formed from the twenty-five volunteers.

Answer:

Since the order is not important, this is a combination question, not a permutation question. With combinations, the number of ways that 20 could be chosen out of 25 is the same as the number of ways that 5 could NOT be chosen out of 25.

Both of these numbers can be found in the 25th row of Pascal's triangle, but you wouldn't see a triangle that large -- the numbers in the center would be huge!

₂₅C₂₀ = ₂₅C₅ =

25 * 24 * 23 * 22 * 21
5 * 4 * 3 * 2 * 1

= 53130

33. On the axes below, for -2 < x < 2, graph y = 2^{x + 1} - 3.

Answer:

Graph the exponential function. DO NOT go beyond -2 or 2. Only graph the domain that was asked for. No arrows.

You can use your graphing calculator to find the points to plot: (-2,-2.5) (-1,2), (0,-1), (1,1), (2, 5)

34. Write an equation of the circle shown in the diagram below.

Answer:

Use the distance formula or Pythagorean Theorem to find r². Note that you don't need to find the exact radius, so you don't have to take any square roots or simplify any radicals.

r² = (-8 - -5)² + (4 - 2)² = (-3)² + (2)² = 9 + 4 = 13

The equation for a circle is (x - h)² + (y - k)² = r², where (h,k) is the center. Note that there are minus signs in the formula so the signs of the coordinates are flipped.

So the equation is (x + 5)² + (y - 2)² = 13.

35. Express the exact value of csc 60°, with a rational denominator.

Answer:

Coseceant is 1/sine and sine 60 degrees is √(3)/2.

csc = 1/sin 60 = 1/(√(3)/2) = 2/√(3) = 2√(3) / 3

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction!

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

(x, why?) Mini: Discriminant

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

Not real roots? He must be imagining it!

I thought about adding a bee square, but I didn't think about it for very long.

The English language is good for allowing the use of ants for puns.

I also write Fiction!

If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

Come back often for more funny math and geeky comics.

Tuesday, November 30, 2021

Geometry Regents, August 2011

I also write Fiction!

Integrated Algebra Regents, August 2011

I also write Fiction!

Monday, November 29, 2021

I also write Fiction!

Wednesday, November 24, 2021

I also write Fiction!

Geometry Regents, August 2011

I also write Fiction!

Integrated Algebra Regents, August 2011

I also write Fiction!

Tuesday, November 23, 2021

Geometry Regents, January 2012

I also write Fiction!

Algebra 2/Trigonometry Regents, January 2012

I also write Fiction!

Monday, November 22, 2021

I also write Fiction!

Friday, November 19, 2021

Algebra 2/Trigonometry Regents, January 2012

I also write Fiction!

Geometry Regents, January 2012

I also write Fiction!

Thursday, November 18, 2021

Integrated Algebra Regents, January 2012

I also write Fiction!

Algebra 2/Trigonometry Regents, January 2012

I also write Fiction!

I also write Fiction!

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by Christopher J. Burke