Thursday, October 21, 2021

Algebra 2 Problems of the Day (Algebra 2 Regents, June 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, June 2012

Part I: Each correct answer will receive 2 credits.


11. Which graph represents the function log 2 x = y? position?


Answer: 1) (see image)


The log2 x function is the inverse of the exponential function 2x, a reflection across the line y = x.

When x = 1 then y = 0 because when 0 is the exponent of 2 then the result is 1.





12. A circle is drawn to represent a pizza with a 12 inch diameter. The circle is cut into eight congruent pieces. What is the length of the outer edge of any one piece of this circle?

1) 3π/4
2) π
3) 3π/2
4) 3π

Answer: 3) 3π/2


The circumference of the pizza is 12π. One eighth of that is 12π/8, which simplifies to 3π/2.





13. What is the solution set for the equation √(5x + 29) = x + 3?

1) {4}
2) {-5}
3) {4, 5}
4) {-5, 4}

Answer: 1) {4}


You can work this out (see below) or you can check if 4, -5 or 5 work.

√(5x + 29) ?= x + 3
√(5(4) + 29) ?= (4) + 3
√(49) ?= 7
True, x = 4 is a solution. Eliminate Choice (2).

√(5x + 29) ?= x + 3
√(5(5) + 29) ?= (5) + 3
√(54) ?= 8
False, x = 5 is NOT a solution. Eliminate Choice (3).

√(5x + 29) ?= x + 3
√(5(-5) + 29) ?= (-5) + 3
√(4) ?= -2
False, x = 2 is NOT a solution. Eliminate Choice (4).

Choice (1) is the solution.

Alternatively, you could solve the equation:

√(5x + 29) = x + 3
5x + 29 = (x + 3)2
5x + 29 = x2 + 6x + 9
x2 + x - 20 = 0
(x + 5)(x - 4) = 0
x + 5 = 0 or x - 4 = 0
x = -5 or x = 4

Next check for extraneous values. As shown above x = -5 is NOT a solution. Discard it. Then x = 4 is the only solution.





14. When factored completely, x3 + 3x2 - 4x - 12 equals

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

Answer: 2) (x + 2)(x - 2)(x + 3)


It says "factored completely" and (x2 - 4) is not factored completely. It can be broken down further. Eliminate Choices (3) and (4).

If you factor the polynomial by grouping:

x3 + 3x2 - 4x - 12
= x3 - 4x + 3x2 - 12
= x(x2 - 4) + 3(x - 4)
= (x2 - 4)(x + 3)
= (x + 2)(x - 2)(x + 3)





15. What is the middle term in the expansion of (x/2 - 2y)6?

1) 20x3y3
2) -15/4 x4y2
3) -20x3y3
4) 15/4 x4y2

Answer: 3) -20x3y3


The expansion has 7 terms, starting with the x6y0 term and ending with the x0y6 term. The middle term would be the fourth one and it would be in the form x3y3, so eliminate Choices (2) and (4).

The coefficient of the fourth term would be 6C3 * (1/2)3 * (-2)3 = (20)(1/8)(-1/8) = -20.




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 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, June 2012

Part I: Each correct answer will receive 2 credits.


11. In the diagram of triagnle ABC shown below, D is the midpoint of AB, E is the midpoint of BC, and F is the midpoint of AC.


If AB = 20, BC = 12, and AC = 16, what is the perimeter of trapezoid ABEF?

1) 24
2) 36
3) 40
4) 44

Answer: 4) 44


Add all of AB to half of AC and half of BC. Then add half of AB, which is the length of EF. So 20 + 6 + 8 + 10 = 44

Choice (1) is the size of triangle DEF which connects the midpoints.

Choices (2) and (3) are the perimeters of the other two trapezoids.





12. In the diagram below, LMO is isosceles with LO = MO.


If m∠L = 55 and m∠NOM = 28, what is m∠N?

1) 27
2) 28
3) 42
4) 70

Answer: 1) 27


Since m∠L = 55 then m∠LMO = 55. This makes m∠OMN = 125 becuase it is supplementary.

Triangle MNO has 180 degrees (like all triangles), so m∠N = 180 - 28 - 125 = 27

Choice (2) is there in case you thought MNO was isosceles. There is nothing that suggests that it is (because it is NOT).





13. If AB is contained in plane P, and AB is perpendicular to plane R, which statement is true?

1) AB is paralel to plane R.
2) Plane P is parallel to plane R.
3) AB is perpendicular to plane P.
4) Plane P is perpendicular to plane R.

Answer: 4) Plane P is perpendicular to plane R.


AB is in P. AB is ⊥ R. Therefore P ⊥ R.

It's that simple.

If it were more complicated than that, you would need a lot more information to make a decision. But it isn't and you don't.





14. In the diagram below of triangle ABC, AE ≅ BE, AF ≅ CF, and CD ≅ BD.


Point P must be the

1) centroid
2) circumcenter
3) incenter
4) ortocenter

Answer: 1) centroid


Points D, E, and F are all midpoints. That makes AB, BC, and CA medians. Medians meet at the centroid.

The circumcenter is the concurrence point for three perpendicular bisectors of a triangle. AD, BF, and CE are NOT shown to be perpendicular to the lines they bisect.

The incenter is the concurrence point for three angle bisectors of a triangle.

The orthocenter is the concurrence point for three altitudes of a triangle.





15. What is the equation of the line that passes through the point (-9,6) and is perpendicular to the line y = 3x - 5?

1) y = 3x + 21
2) y = -1/3 x - 3
3) y = 3x + 33
4) y = -1/3 x + 3

Answer: 4) y = -1/3 x + 3


Perpendicular lines have slopes that are inverse reciprocals. Parallel lines have the same slopes.

The given line has a slope of 3. Choices (1) and (3) are lines with a slope of 3, so they are parallel to the given line. Eliminate them.

The slope of the perpendicular line must be -1/3. Check (-9,6) to see if it is a point on either line.

y = -1/3(-9) - 3 = 3 - 3 = 0. (-9, 6) is not on this line.

y = -1/3(-9) + 3 = 3 + 3 = 6. (-9, 6) IS a point on this line.




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, October 20, 2021

Algebra 2 Problems of the Day (Algebra 2 Regents, June 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, June 2012

Part I: Each correct answer will receive 2 credits.


6. If m∠θ = -50, which diagram represents θ drawn in standard position?


Answer: 4) (see image)


If θ has a negative value, you drop down below the x-axis, in a clockwise direction, starting from the right side of axis.

Choice (3) is 50 degrees. Choices (1) and (2) are not standard positions because they start on the left.





7. If logb x = 3 log b p - (2 log b t + 1/2 log b r), then the value of x is


Answer: 4) See image


Sometimes I think the Regents people just want to give me a headache.





8. Which equation has roots with the sum equal to 9/4 and the product equal to 3/4?

1) 4x2 + 9x + 3 = 0
2) 4x2 + 9x - 3 = 0
3) 4x2 - 9x + 3 = 0
4) 4x2 - 9x - 3 = 0

Answer: 3) 4x2 - 9x + 3 = 0


The rule for the Sum of the Roots is −b/a, and for the Product of the Roots is c/a. Note that the Sum has a minus sign and the Product does not.

If -b/a = +9/4 then either a or b must be negative. Since a = 4 in all four choices, b must be -9. Eliminate Choices (1) and (2).

If c/a = +3/4 then either a and c are both positive or they are both negative. Since a must be +4, then c must be +3. Eliminate Choice (4).





9. Which graph represents the solution set of | (4x - 5) / 3 | > 1?

Answer: 3) see image


You have to remove the absolute value symbols and then solve TWO inequalities: one for > 1 and one for < -1

(4x - 5) / 3 > 1
4x - 5 > 3
4x > 8
x > 2
(4x - 5) / 3 < -1
4x - 5 < -3
4x < 2
x < 1/2





10. Which expression is equivalent to (x-1y4) / (3x-5y-1)

1) x4y5 / 3
2) x5y4 / 3
3) 3x4y5
4) y4 / (3x5)

Answer: 1) x4y5 / 3


When you have the same variable in the numerator and the denominator, you can SUBTRACT the exponents.

x-1 / x-5 = x-1 - -5 = x4

y4 / y-1 = y4 - -1 = y5

The 3 in the denominator doesn't move. It remains 1/3 because there is no exponent affecting it.

The final result is 1/3 x4 y5




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 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, June 2012

Part I: Each correct answer will receive 2 credits.


6. In the diagram below of ABCD, AC ≅ BD.


Using this information, it could be proven that

1) BC = AB
2) AB = CD
3) AD - BC = CD
4) AB + CD = AD

Answer: 2) AB = CD


Since BC ≅ BC, it could be removed from AC and BD, which are congruent, and the remaining segments will still be congruent because of the Subtraction Property of Congruency.

In Choice (1), there is nothing to suggest that B is a midpoint of AC.

Choice (3) is an incorrect equation. AD - BC = CD PLUS AB.

Likewise, Choice (4) is an incorrect equation. AB + CD = AD MINUS BC.





7. The diameter of a sphere is 15 inches. What is the volume of the sphere, to the nearest tenth of a cubic inch?

1) 706.9
2) 1767.1
3) 2827.4
4) 14,137.2

Answer: 2) 4 is not an odd integer; true


If the diameter is 15 then the RADIUS is 7.5, and you need the radius for the formula, which is 4/3 π r3

(4/3) π (7.5)3 = 1767.14..., or 1767.1, which is Choice (2).

If you used the diameter of 15 by mistake, you would have gotten Choice (4) and wouldn't have known you got it wrong.

Choice (1) shows the SURFACE AREA of a sphere with a 15 inch diameter.

Choice (3) showns the SURFACE AREA of a sphere with a 15 inch RADIUS.





8. The diagram below shows the construction of AB through point P parallel to CD.


Which theorem justifies this method of construction?

1) If two lines in a plane are perpendicular to a transversal at different points, then the lines are parallel.
2) If two lines in a plane are cut by a transversal to form congruent corresponding angles, then the lines are parallel.
3) If two lines in a plane are cut by a transversal to form congruent alternate interior angles, then the lines are parallel.
4) If two lines in a plane are cut by a transversal to form congruent alternate exterior angles, then the lines are parallel.

Answer: 2) If two lines in a plane are cut by a transversal to form congruent corresponding angles, then the lines are parallel.


Knowing nothing else about construction, you can deduce that the answer is Choice (2) because the angles that are marked off are corresponding angles.

The angles were not constructed to be perpendicular. The are neither alternate interior nor exterior. Eliminate Choices (1), (3) and (4).

The construction shown is how you can draw a line parallel to a given line through a point, P, not on that line. The theorem about corresponding angles justifies this.





9. Parallelogram ABCD has coordinates A(1,5), B(6,3), C(3,-1), and D(-2,1). What are the coordinates of E, the intersection of diagonals AC and BD?

1) (2,2)
2) (4.5,1)
3) (3.5,2)
4) (-1,3)

Answer: 1) (2,2)


THe diagonals of a parallelogram bisect each other. You need to find the midpoint of AC or of BD. Or you can check your work by finding both of them, which MUST BE the same point.

To find the midpoint, take the average of the two x-coordinates and the average of the two y-coordinates.

E = ( (1+3)/2, (5-1)/2 ) = (2,2)

E = ( (6-2)/2, (3+1)/2 ) = (2,2)

Choice (3) is the midpoint of AB. I don't see the "logic" behind the other two incorrect choices.





10. What is the equation of a circle whose center is 4 units above the origin in the coordinate plane and whose radius is 6?

1) x2 + (y - 6)2 = 16
2) (x - 6)2 + y2 = 16
3) x2 + (y - 4)2 = 36
4) (x - 4)2 + y2 = 36

Answer: 3) x2 + (y - 4)2 = 36


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

Note that this comes up so often that I cut and paste the above sentence from blog post to blog post.

The radius is 6, so r2 = 36. Eliminate choices (1) and (2).

Four units above the origin is the point (0,4), which is the center. That means that the correct equation is Choice (3).

Choice (4) would have been correct if the center was four units to the RIGHT of the origin, at (4,0).




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, October 19, 2021

Geometry Problems of the Day (Geometry Regents, June 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, June 2012

Part I: Each correct answer will receive 2 credits.


1. Triangle ABC is graphed on the set of axes below.


Which transformation produces an image that is similar to, but not congruent to, triangle ABC?

1) T2,3
2) D2
3) ry = x
4) R90

Answer: 2) D2


A Dilation does not preserve distance, or size, by defintion. The shape will be conserved, meaning that all of the angles in the image are congruent to the original, which is the definition of similar.

Translations, reflections and rotations are produce images that are congruent to the original pre-image.





2. A student wrote the sentence “4 is an odd integer.” What is the negation of this sentence and the truth value of the negation?

1) 3 is an odd integer; true
2) 4 is not an odd integer; true
3) 4 is not an even integer; false
4) 4 is an even integer; false

Answer: 2) 4 is not an odd integer; true


The negation of "is" is "is not", and the rest of the sentence doesn't matter.

The negation of this sentence is "4 is not an odd integer" and that negation is true.

Negative does not make 4 into 3. Why 3? Why not any other number? This makes no sense. A negation is very specific.

Changing "is an odd" to "is not an even" requires changing TWO things: adding the "not" and switching "odd" to "even". This is not how a negation is done.

With Choice (4), it could be argued argued that "not an odd integer" is the same as "an even integer". However, this ignores the case that it might not be an integer at all, which would mean that the number is neither odd nor even. There are more than two possibilities. In any event, the statment "4 is an even integer" is true, not false, so the Choice is incorrect regardless of any argument.





3. As shown in the diagram below, EF intersects planes P , Q , and R/


If EF is perpendicular to planes P and R , which statement must be true?

1) Plane P is perpendicular to plane Q.
2) Plane R is perpendicular to plane P.
3) Plane P is parallel to plane Q.
4) Plane R is parallel to plane P.

Answer: 4) Plane R is parallel to plane P.


If you picture EF as a horizontal line, it would be perpendicular to vertical planes. That would make planes P and R parallel to each other.

Plane Q has no special relationship with either planes P or R, nor with line EF. Plane Q is shown to intersect P and R, so it cannot be parallel to them. If it were perpendicular to P and R, then plane Q would either have to be parallel to EF, which is can't because they intersect, or it would have to contain the line EF in its entirety. It isn't shown to do that, and the word "intersect" would suggest that this isn't the case.





4. In the diagram below, LATE is an isosceles trapezoid with LE ≅ AT, LA = 24, ET = 40, and AT = 10. Altitudes LF and AG are drawn.


What is the length of LF?

1) 6
2) 8
3) 3
4) 4

Answer: 1) 6


Since it is an isosceles trapezoid, you know that triangles LEF and AGT are congruent (by AAS, if you need to work it out). That means that EF ≅ GT. Call each of those x.

Also, FG ≅ LA.

So x + 24 + x = 40, meaing 2x + 24 = 40. Then x = 8.

So EF = 8, LE = 10 and LFE is a right triangle.

A quick use of Pythagorean Theorem tells you that LF = 6. And you should seriously know 6-8-10 is a Pythagorean Triple without needing to use the theorem.

remain parallel.





5. In the diagram below of circle O, diameter AB is parallel to chord CD.


If mCD = 70, what is mAC?

1) 110
2) 70
3) 55
4) 35

Answer: 3) 55


AC = BD and AC + CD + BD = 180

So 2AC + 70 = 180
2AC = 110
AC = 55




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 Regents, June 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, June 2012

Part I: Each correct answer will receive 2 credits.


1. What is the product of (2/5 x - 3/4 y2) and (2/5 x + 3/4 y2)?

1) 4/25 x2 - 9/16 y4
2) 4/25 x - 9/16 y2
3) 2/25 x2 - 3/4 y4
4) 4/5 x

Answer: 1) 4/25 x2 - 9/16 y4


The multiplication of two conjugates (the same binomial except one has a plus and the other a minus) will have a product that is the Difference of Two Perfect Squares.

In general: (ax + by) * (ax - by) = a2x2 - b2y2.

Replace a with 2/5 and b with 3/4 and you get:

(2/5 x + 3/4 y) * (2/5 x - 3/4 y) = (2/5)2x2 - (3/4)2y2
= 4/25 x2 - 9/16 y2





2. What is the domain of the function shown below?



1) -1 < x < 6
2) -1 < y < 6
3) 2 < x < 5
4) 2 < y < 5

Answer: 1) -1 < x < 6


Domain refers to the x values, so eliminate Choices (2) and (4) which refer to y.

The only values of x that have points are from -1 through 6, inclusive. That is Choice (1).

Note that Choice (4) is the range of the function.





3. What is the solution set for 2cosθ - 1 = 0 in the interval 0° < θ < 360°

1) {30°, 150°}
2) {60°, 120°}
3) {30°, 330°}
4) {60°, 300°}

Answer: 4) {60°, 300°}


Solve for cosθ, and then find the values of θ between 0 and 360 that will give the correct answer.

2cosθ - 1 = 0
2cosθ = 1
cosθ = 1/2

The values on the Unit Circle when cosθ = 1/2 is at 60 degrees above and below the x-axis. However, since the stated interval was 0 to 360, we want 60 and 300, not 60 and -60.

-5 - -3 = -2 and 2 - 6 = -4

Point B is (-5 - 2, 2 - 4), or (-7, -2)





4. The expression ∛(64a16) is equivalent to

1) 8a4
2) 8a8
3) 4a5 ∛(a)
4) 4a5 ∛(a5)

Answer: 3) length of sides


The cube root of 64 is 4 because 43 = 64.

To find the cube root of a16, break it into a5 * a5 * a5 * a1. So the cube root is a5 * ∛(a)

So the final answer is 4 * a5 * ∛(a)

A dilation will retain the orientation and the shape of the original. Since the shape is similar, the size angles will be the same and lines that were parallel will remain parallel.





5. Which summation represents 5 + 7 + 9 + 11 + ... + 43?

Answer: 2) (see image)


The summations are like little loops (FOR/NEXT, or DO UNTIL, etc, for programmers)

In Choice (1), all the integers from 5 to 43 are being added, but we only want the odd numbers. Eliminate Choice (1).

In Choice (2), the first number is 2(1) + 3 = 5 and the last is 2(20) + 3 = 43. The sequence is counting by 2s. This is the solution.

In Choice (3), the first number is 2(4) - 3 = 5 but the last is 2(24) - 3 = 45, not 43. Had the top number been 23 instead of 24, this would have worked. Eliminate Choice (3).

In Choice (4), the sequence is increasing by 3 and not 2. Eliminate Choice (4).

Once you see that the series is increasing by 2, Choices (1) and (4) should have been eliminated immediately.




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.



Monday, October 18, 2021

Math Horror Movies: Red

(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?).

In Space, No One Can Hear You Get Mad Dept.

That time of year, again, when I need more shlocky Halloween-type horror B-movie titles.

I love 'em and hate 'em at the same time.



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.





Come back often for more funny math and geeky comics.



Geometry Problems of the Day (Geometry Regents, August 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, August 2012

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


38. Chords AB and CD intersect at E in circle O, as shown in the diagram below. Secant FDA and tangent FB are drawn to circle O from external point F and chord AC is drawn. The mDA = 56, mDB = 112, and the ratio of mAC : mCD = 3:1.

Determine m∠CEB.

Determine m∠F.

Determine m∠DAC.

Answer:


Rules to remember: an inscribed angle is half the size of the arc it intercepts. Where two chords intersect, the vertical angles created will be equal to the average of the two arcs the chords intercept on either side. Finally, when two secants, or two tangents, or a tangent and a secant intercept a circle, the size of the angle of the two secants (or whatever) will be one-half of the difference between the size of the FAR (the bigger, more remote) arc minus the NEAR (the smaller, closer) arc.

In the circle given, you have measures for arcs AD and DB, 56 and 112, respectively, which total 168 degrees. There are 360 degrees total in the circle, so what remains, 360 - 168 = 192 is the sum of AC and CB. Note that arc ACB is a major arc, greater than a semicircle, so the number 192 looks reasonable.

We also know that AC:CB = 3:1, so AC = 3/4(192) = 144, and CB = 1/4(192) = 48.

m∠CEB = 1/2 (m CB + m DA) = 1/2(48 + 56) = 52 degrees.

m∠F = 1/2 (m ACD - m DB) = 1/2(192 - 112) = 40 degrees.

Angle DAC intercepts arc DBC, which in the sum of arcs DB and BC. Intercepted angles are half the size of the arc they intercept. So m∠DAC = 1/2 (112 + 48) = 80 degrees.

Can check your work here because the angles of triangle FAB must add up to 180. Angle F = 40, angle DAB is 1/2(112) = 56, and angle ABF is 1/2(168) = 84, and 40 + 56 + 84 = 180.




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.



Sunday, October 17, 2021

Geometry Problems of the Day (Geometry Regents, August 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, August 2012

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


35. As shown in the diagram below, the diagonals of parallelogram QRST intersect at E. If QE = x2 + 6x, SE = x + 14, and TE = 6x - 1, determine TE algebraically.


Answer:


Since the diagonals of the parallelogram bisect each other, we know that QE = ES. Set up and solve this quadratic equation. Then use the value of x to find the length of TE.

x2 + 6x = x + 14
x2 + 5x - 14 = 0
(x + 7)(x - 2) = 0
x + 7 = 0 or x - 2 = 0
x = -7 or x = 2
Discard x = -7

Note that x = -7 is a valid answer for both QE and ES but creates an invalid answer for TE.

TE = 6(2) - 1 = 11.





36. The vertices of triangle RST are R(-6,5), S(-7,-2), and T(1,4).
The image of triangle RST after the composition T–2,3 ° ry = x is triangle R"S"T".
State the coordinates of triangle R"S"T".
[The use of the set of axes below is optional.]

Answer:


Remember with compositions, you do the second transformation first. You want to show the Translation OF THE Reflection, not a Translation followed by a Reflection.

To refect over the line y = x, switch the x-coordinate with the y-coordinate.

R(-6,5) --> R'(5, -6)
S(-7,-2) --> S'(-2,-7)
T(1,4) --> T'(4,1)

A Translation of (-2,3) means subtract 2 from each x-coordinate and add 3 to every y-coordinate.

R'(5, -6) --> R"(3,-3)
S'(-2,-7) --> S"(-4,-4)
T'(4,1) --> T"(2,4)

If you used the grid, make sure you show both steps and Label every point and state the final coordinates. Make sure your final answer is stated clearly.





37. On the set of axes below, solve the following system of equations graphically and state the coordinates of all points in the solution.

(x + 3)2 + (y - 2)2 = 25
2y + 4 = -x

Answer:


The first equation is a circle with its center at (-3, 2) and a radius of 5.

The second is a linear equation that should be rewritten in slope-intercept form for quicker graphing:

2y + 4 = -x
2y = -x - 4
y = -1/2 x - 2

The slope of the line is -1/2 and the y-intercept is -2.

The graph looks like this:




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.



Saturday, October 16, 2021

Geometry Problems of the Day (Geometry Regents, August 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, August 2012

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


32. The coordinates of the endpoints of FG are (-4,3) and (2,5). Find the length of FG in simplest radical form.

Answer:


Uset the distance formula or pythagorean theorem.

d = SQRT( (-4 - 2)2 + (3 - 5)2 )
= SQRT( (-6)2 + (-2)2 )
= SQRT( 36 + 4 )
= SQRT(40)
= SQRT(4) * SQRT(10)
= 2 SQRT(10)





33. Using a compass and straightedge, construct a line perpendicular to AB through point P.
[Leave all construction marks.]

Answer:


We can use the procedure to make a perpendicular bisector that goes through point P. Before we can do that, we need to have a point between A and P that is the same distance away from P as B is. We can measure the distance of PB and make another point on the other side with a little arc.

Next, make an arc above and below P from this new point. Then do the same thing from point B. Finally, connect those two new points, which should line up with P. If they don't line up, then a mistake was made. If this occurs, correct your work. Don't erase. Leave all the marks.





34. The graph below shows the locus of points equidistant from the x-axis and y-axis. On the same set of axes, graph the locus of points 3 units from the line x = 0. Label with an X all points that satisfy both conditions.


Answer:


The line x = 0 is the y-axis. The locus of points 3 units away would be two vertical lines on either side of the y-axis. That means that there will be four spots that satisfy both conditions.




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.



Friday, October 15, 2021

(x, why?) Mini: Clock

(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?).

The size of the interior angles is left as an exercise to the reader.

Probably the oddest thought to come out of this, after deciding to put clock faces on the shapes, was what times I should display. Originally, they all had 3 o'clock from Clarke's Third Law, but that was boring. Then the first one was midnight, and then the last one was midnight.

For the record, the last shape is a 24-gon, although it doesn't look like it after reducing its size.



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.



Geometry Problems of the Day (Geometry Regents, August 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, August 2012

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


29. Write the negation of the statement “2 is a prime number,” and determine the truth value of the negation

Answer:


"2 is not a prime number."

This statement is false.

Negation means, basically, adding a "not" or removing a "not" that is already there. Negating a statement means it will have the opposite truth value from the original statement.

Since 2 is a prime number, the negation is false.





30. The coordinates of the vertices of triangle ABC are A(1,2), B(-4,3), and C(-3,-5). State the coordinates of triangle A'B'C', the image of triangle ABC after a rotation of 90° about the origin.
[The use of the set of axes below is optional.].

Answer:


If the direction of the rotation is not specified, then the rotation is counter-clockwise. Think of the numbering of the quadrants. Quadrant I becomes Quadrant II, etc.

The rule for a rotation of 90 degrees counterclockwise is P(x, y) --> P'(-y, x). That is, switch the order of the x-coordinate and the y-coordinate and then switch the sign of the new x-coordinate.

A(1,2) --> A'(-2,1)
B(-4,3) --> B'(-3,-4)
C(-3,-5) --> C'(5,-3)





31.A cylinder has a height of 7 cm and a base with a diameter of 10 cm. Determine the volume, in cubic centimeters, of the cylinder in terms of π.

Answer:


Remember to use the RADIUS and not the Diameter. Divide 10/2 to get Radius = 5.

V = π r2 h
= π (5)2 (7)
= 175 π




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.