Tuesday, June 30, 2015

The Water Tower Problem and the Perils of Rounding

Question 34, the first Part IV question of the New York State Geometry (Common Core) Regents exam for June 2015, read as follows:

The water tower in the picture below is modeled by the two-dimensional figure beside it. The water tower is composed of a hemisphere, a cylinder, and a cone. Let C be the center of the hemisphere and let D be the center of the base of the cone.

If AC = 8.5 feet, BF = 25 feet, and m∠EFD = 47°, determine and state, to the nearest cubic foot, the volume of the water tower.

The water tower was constructed to hold a maximum of 400,000 pounds of water. If water weighs 62.4 pounds per cubic foot, can the water tower be filled to 85% of its volume and not exceed the weight limit? Justify your answer.

If you are a regular reader of this blog, you may have seen the discussion generated by this problem between me and a fellow educator when I previously posted the questions in another thread with questions and answers from the exam.

Most of it stemmed from the fact that I made an error in calculating the problem, but I had no way of knowing how significant the error was or what rubric the state would use to grade the mistake. Discussion followed speculating what they would and should do, based on what has happened in the past (recent past, not your parents' past) and what they said they wanted to do now.

Cutting to the chase for those who don't want to read that entire thread, I made a rounding error, and my Volume was off by 1 as a result. Looking at it now, it was likely a one-point deduction. However, before we get there, let's review the problem.

Breaking it into parts

Part of the reason that this question was overwhelming was the irregular nature of the shape. (Another part was that the question was divided into three sections spread across two pages, including the two illustrations above.) This has been done before on exams, but rarely with three sections, and never (to my knowledge) with 3-D objects. It's usually an area or perimeter question.

The water tower consists of a hemisphere (half a sphere), a cylinder and a cone. The radius of the all three parts is the same, and it is given. The height of the hemisphere is its radius. The height of the cylinder is given. The height of the cone needs to be calculated using trigonometric ratios. We have one angle and the adjacent leg of the right triangle, and we want to know the height, which is opposite, so we need to use tangent.

tan(47) = x/8.5, so x = 8.5(tan(47) = 9.11513403521

And here is the problem. Rounding.

One of the biggest problems on Regents exams is rounding. If the answer isn't rounded correctly, to the correct number of places (as specified in the problem), the student will lose 1 point -- which is a hefty price to pay for a two-point problem.

This, however, is only part of the problem, and not a final answer. No number of decimal places for height is specified because it's still the middle of the problem. In the past, a test might have asked for the height (if only to guide the students toward the answer) and then used that height to find the Volume, but this test did not do that, so rounding is NOT appropriate at this point.

I have warned my students in recent years about the dangers of using 3.14 for pi in problems, particularly when large numbers are involved. Current-day exams assume (demand, actually) that the student is using a scientific or graphing calculator and therefore has access to a function key with the value of Pi to 8 or more decimals, and this should be used. This is fine, considering that you can use the pi symbol in your calculations. In the case of this question, you need to carry all those decimals around with you.

All of them? No, not really. But how many? That's the tricky part. The more the better.

The state releases sample answers (not from actual students' exams). For this question, the sample correct answer had rounded to FIVE decimal places. A second response only used FOUR places, but also used 3.14. It lost a point and was faulted for the latter mistake, but not the former one.

So the Volume of the entire water tower needed to be calculated as

V = 1/3(pi)(8.5)2(9.11513) + (pi)(8.5)2(25)+(1/2)(4/3)(pi)(8.5)3

which yields 7650.373... etc., which rounds to 7650.

Some careful experimenting yields the following results: Using both 9.1151 and 9.115 give answers that still round to 7650.37 and still give the correct answer. However, using only 9.12 gives a Volume of 7650.74, which is rounded "correctly" gives an incorrect answer of 7651. Had it been truncated instead of rounded, the correct final result would have been obtained but whether or not the two errors (which cancelled out) were caught by the scorer is questionable. (They are grading a lot of papers.) Likewise, if 9.11 (an incorrect amount) were used, the final answer would've been 7549.98, which rounds to the correct answer. Again, would it have been caught and marked incorrect? I couldn't tell you.

Finishing the Problem

Whatever answer you obtained (even one far off from the ones mentioned above), that answer has to be carried forward into the last part of the problem.

7650(62.4) = 477,360.
.85(477,360)= 405,756 pounds, which is more than 400,000 pounds, so
No, the water tower cannot be filled to 85% of its Volume.

If you didn't completely answer this part of the question, you would have lost a point or, possibly, two. You cannot state "No" without the calculations and justification. There is never credit for a 50-50 guess.

Conclusion

Rounding in the middle of a problem is hazardous to your grade, as is not rounding correctly at the end of the problem.
Generally speaking, no, you don't need eight decimal places for simple problems. However, when problems get more complex and require more significant digits, each decimal places grows in importance. When in doubt keep the extra digits.

Monday, June 29, 2015

Summer To-Do Diagram

(Click on the comic if you can't see the full image.)
(C)Copyright 2015, C. Burke.

It's oddly accurate while leaving out a whole lot of stuff.

But definitely the proofreading!

There are new curriculums to be learned, and I need to refresh in upper-level classes in case I'm called to teach those. And now is a good time to assess what worked and what didn't -- particularly, with the new material which I hadn't taught in the same manner before.

So I'm not really off.

... except that I am. Who am I kidding?




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Wednesday, June 24, 2015

New York State Geometry Regents, June 2015 Part 1: Multiple Choice

Here is Part 1 of the New York Geometry Regents (non-Common Core) exam, the multiple choice.
Part 2 was posted here. And Parts 3 and 4 were posted here.

Illustrations and diagrams will be added when available.

Part 1

1. Quadrilateral ABCD undergoes a transformation, producing a quadrilateral A'B'C'D'. For which transformation would the area of A'B'C'D' not be equals to the area of ABCD?

(3) a dilation by a scale factor of 2. A dilation of 2 would increase the area by a factor of 4 (2 x 2).

2. The diameter of a sphere is 12 inches. What is the volume of the sphere to the nearest cubic inch?

(3) 905. Remember that the radius is only 6. Use the formula from the reference sheet: 4/3 (3.141592...)(6)2 = 905, to the nearest integer.

3. A right rectangular prism is shown in the diagram below.


Which line segments are coplanar?

(4) EA and GC are both in a vertical plane. The other pairs of lines are skew.

4. What are the coordinates of the image of point A(2, -7) under the translation (x,y) -> (x - 3, y + 5)?

(1) (-1, -2). (2 - 3 = -1, -7 + 5 = -2)

5. Point M is the midpoint of AB. If the coordinates of M are (2, 8) and the coordinates of A are (10, 12), what are the coordinates of B?

(2) (-6, 4). 10 - 2 = 8; 2 - 8 = -6. 12 - 8 = 4; 8 - 4 = 4.

6. In the diagram below, QM is an altitude of right triangle PQR, PM = 8, and RM = 18.

What is the length of QM?

(3) 12. Right-Triangle Altitude Theorem. Use the proportion 8/x = x/18. Cross multiply: x2 = 144. x = 12.

7. What is an equation of the line that passes through the point (2, 4) and is perpendicular to the line whose equation is 3y = 6x + 3?

(1) y = -1/2 x + 5. The slope of the given line is 6/3 = 2, so the perpendicular line has a slope of -1/2. (Eliminate choices 3 and 4.) Plug in 2 for x and 4 for y and you get the first equation.

8. In all isosceles triangles, the exterior angle of a base angle must always be

(3) an obtuse angle. The base angles of an isosceles triangle must be acute, so the exterior angle must be obtuse.

9. If triangle W'X'Y' is the image of WXY after the transformation R90, which statement is false?

(2) WX || W'X'. If it's rotated 90 degrees, it can't be parallel to the original. The sides, angles, and the entire triangle will be congruent.

10. Which equation represents the circle shown in the graph below?

(1) (x - 2)2 + y2 = 9. Use the formula. Flip the sign, square the radius.

11. In quadrilateral ABCD, each diagonal bisects opposite angles. If m<DAB = 70, then ABCD must be a

(3) rhombus. Bisecting the opposite angles is true of rhombuses and squares (which are rhombuses), but the angle is not 90 degrees.

12. Which diagram illustrates a correct construction of an altitude of triangle ABC?

(2). It's the only construction that shows an altitude. Choice (1) shows a median. Choice (3) shows an angle bisector. Choice (4) shows a perpendicular bisector. The perpendicular bisector will come in handy in Part 2!

13. From external point A, two tangents to circle O are drawn. The points of tangency are B and C. Chord BC is drawn to form triangle ABC. If m<ABC = 66, what is m<A?

(2) 48. ABC is an isosceles triangle with base angles of 66. 66 + 66 = 132. 180 - 132 = 48.

14. Point A lies on plane P. How many distinct lines passing through point A are perpendicular to plane P?

(1) 1. By definition, really. Not much to say.

15. Students made four statements about a circle.
A: The coordinates of its center are (4, -3).
B: The coordinates of its center are (-4, 3).
C: The length of the radius is 5(2)^.5.
D: The length of the radius is 25. If the equation of the circle (x + 4)2 + (y - 3)2 = 50, which statements are corrct?

(3) B and C. Another formula of a circle question, asked in a totally different way. Flip the signs, square root of r2.

16. Points A, B, C, and D are located on circle O, forming trapezoid ABCD with AB || DC. Which statement must be true?

(2) AD = BC. The intercepted arcs of to parallel chords are congruent.

17. If triangle ABC ~ LMN, which statement is not always true?

(1) m<A = m<N. When presented with two triangles that are stated as similar or congruent, proper notation is that angle A corresponds to angle L, angle B corresponds to angle M, and angle C corresponds to angle N. Pick two letters for a side, and the corresponding two letters indicate the corresponding side. I point this out because some students are a little careless about naming conventions when they write, so they might have forgotten this, if they were ever aware of it. Some teachers might even be too lax at enforcing this. Here is one of the times where it is important to know that the test writers following the rules.

18. The equations representing lines k, m, and n are given below.
k: 3y + 6 = 2x
m: 3y + 2x + 6 = 0
n: 2y = 3x + 6
Which statement is true?

(4) m is perpendicular to n. The slope of k is 2/3. The slope of m is -2/3. The slope of n is 3/2.

19. A regular polygon with an exterior angle of 40o is a

(3) nonagon. 360/x = 40. x = 360/40 = 9. Nine sides is a nonagon.

20. In triangle ABC shown below, L is the midpoint of BC, M is the midpoint of AB, and N is the midpoint of AC.


If MN = 8, ML = 5, and NL = 6, the perimeter of trapezoid BMNC is

(4) 35. Midpoints are connected by midsegments, which are half the length of the third side and also parallel to it. The perimeter is 5 + 8 + 6 + 8 + 8 = 35.

21. The sum of the interior angles of a regular polygon is 720 degrees. How many sides does the polygon have?

(2) 6. This is such a common question that you should have known it, but ...

180(n - 2) = 720
n - 2 = 4
n = 6

22. In the prism shown below, AD is perpendicular to AE and AD is perpendicular to AB.
Which plane is perpendicular to AD? (3) EAB. The bottom and top of the prism are perpendicular to AD. The bottom is listed as a choice.

23. In triangle ABC, m<A = 65 and m<B is greater than m<A. The lengths of the sides of triangle ABC in order from smallest to largest are

(1) AB, BC, AC. If B > A, then A + B > 130 and C < 50. Therefore C < A < B, and AB < BC < AC.

24. Which equation represents a circle whose center is the origin and that passes through the point (-4, 0)?

(2) x2 + y2 = 16. Square the radius: 42 = 16.

25. The lengths of two sides of a triangle are 7 and 11. Which inequality represents all possible values for x, the length of the third side of the triangle?

(4) 4 < x < 18. Triangle Inequality Theorem: 11 - 7 < x < 11 + 7.

26. Which statement is the inverse of "If x + 3 = 7, then x = 4"?

(3) If x + 3 =/= 7, then x =/= 4. Inverse adds "NOT" (or removes it if it was already there).

27. In the diagram below of triangle MAR, medians MN, AT, and RH intersect at O.


If TO = 10, what is the length of TA?

(1) 30. AO is twice the length of TO, and TA is three times the length of TO.

28. What is an equation of the line that passes though the point (4, 5) and is parallel to the line whose equation is y = 2/3 x - 4?

(4) 3y - 2x = 7. Choices (1) and (2) have a slope of -3/2, so eliminate them. Calculate 3(5) - 2(4) = 7, which is choice (4).

That's it. It's up. Everyone happy?

Tuesday, June 23, 2015

New York State Geometry Regents, June 2015 Parts 3 and 4

Here are Parts 3 and 4 of the New York Geometry Regents (non-Common Core) exam. Part 2 was posted here.

Part 3

35. Solve the following system of equations graphically. State the coordinates of all points in the solution.

y + 4x = x2 + 5
x + y = 5
(0, 5) and (3, 2) were the solutions and they had to be stated. If you made a graphing error, then whatever points of intersection you had needed to be stated. If you graphed the system so that neither line touched, you would have needed to state that there was No Solution. Otherwise, you would have lost more points on top of your graphing error.
Note that arrows at the ends of the lines were not required. Labeling the lines wasn't required. I'm assuming because it was a Geometry test and not an Algebra test. I still recommend you do these things anyway, in case the standards change again.

36.


Two points for x and y, with work. Three points if you got as far as 18 and 34, or if you came up with 52 (without solving 18 and 34 separately). With all those calculations, many students lost a point for a computation error, such as -1 + 15 = 16.

37. Point P is 5 units from line j. Sketch the locus of points that are 3 units from line j and also sketch the locus of points that are 8 units from P. Label with an X all points that satisfy both conditions.

The circle will touch the bottom line in only one place because the 5 + 3 = 8. (The addition wasn't required, but it's a good reference for yourself while sketching.)
Note that it didn't have to look pretty. In a "sketch", you don't need to measure with a ruler or draw a circle with a compass. (Neither of which I had on hand, obviously.)

Part 4

Part 4 was worth 6 credits. Please don't ask me individual questions about your proof. This is A proof. Yours might've looked similar or not. It might've had more statements. It might have skipped a statement. As long as the important stuff is there, you did okay. Looking at the sample proofs given, this one would have received full credit.

38.


There are many possible variations, which is why I try not to grade the proofs if I don't have to. I would go insane applying Regents rubrics/grades to them.

How'd everyone do?

New York State Geometry Regents, June 2015 Part 2

Here is Part 2 of the New York Geometry Regents (non-Common Core) exam. Part 1 will be posted when it is typed up. (I'm only one man. With a job. Working a lot of Overtime.) These were scanned in, obviously, which made it a little easier. Parts 3 and 4 are here.

Part 2

29. The measures of the angles of a triangle are in the ratio 5:6:7. Determine the measure, in degrees, of the smallest angle of the triangle.

30.


According to the rubric I read, work wasn't required because it's something that could be done in your head (flipping two numbers). But don't make a mistake in what you did write.

31.


I hope you remembered to use the slant height and not the height.

32. Using a compass and a straightedge, locate the midpoint of AB by construction.

The perpendicular bisector was not required so long as there were marks to indicate how the midpoint was arrived at. There should be a midpoint if there is no bisector.

33.

34.


An incenter is formed from three angle bisectors. That fact is necessary in solving the problem.

Monday, June 22, 2015

New York State Algebra I (Common Core) Regents, June 2015 Parts 3 and 4

Here is Part 3 of the Algebra 1 Common Core Regents exam. Here are the links to Part 1 and Part 2.

Part 3

33. Albert says that the two systems of equations shown below have the same solutions.

8x + 9y = 48
12x + 5y = 21
8x + 9y = 48
-8.5y = -51

Determine and state whether you agree with Albert. Justify your answer.

Solve the second system first. If -8.5y = -51, then divide and get y = 6. Then 8x + 9(6) = 48.

8x + 54 = 48
8x = -6
x = -0.75

Substitute these numbers into the first system: 12(-0.75) + 5(6) = 21 ?

-9 + 30 = 21 (?)
21 = 21 (check)

Albert was correct.
Don't forget to say that Albert was correct even though there is NO CREDIT for saying he is correct without the justification.

34. The equation to determine the weekly earnings of an employee at the Hamburger Shack is given by w(x), where x is the number of hours worked.

Determine the difference in salary, in dollars, for an employee who works 52 hours versus one who works 38 hours.
Determine the number of hours an employee must work in order to earn $445. Explain how you arrived at this answer.

This is how overtime generally works in the real world.
The 38 hour employee earned $380. (38 * 10)
The 52 hour employee earned $400 + 12 * 15 = $580. The difference is $200.

In order to earn $445, they need to work 40 hours, earning $400, and then another $45, which at $15/hour is three more hours, for a total of 43 hours.

35. An online electronics store must sell at least $2500 worth of printers and computers per day. Each printer costs $50 and each computer costs $500. The store can ship a maximum of 15 items per day. On the set of axes below, graph a system of inequalities that models these constraints.

Determine a combination of printers and computers that would allow the electronics store to meet all the constraints. Explain how you obtained you answer.

First, the inequalities you need to graph are:

p + c < 15
50p + 500c > 2500

The graph looks like this:

Possible combinations that would work would be, for example, 0 printers and 5 computers, or 10 printers and 4 computers.

36. An application developer released a new app to be downloaded. The table below gives the number of downloads for the first four weeks after the launch of the app.

Write an exponential equation that models these data.
Use this model to predict how many download the developer would expect in the 26th week if this trend continues. Round your answer to the nearest download.
Exponential regression, which can be done in the calculator.
Since I don't have one handy, I can see that it increases by 60, 90, 135 ... which increasing by half, so an equation could be
y = 120(1.5)^(w-1)

Using that formula, on the 26th week, there would be 3030140 downloads (rounded).

Using that formula, on the 52nd week, there would be 1.15 X 10^11 downloads, which is not a reasonable number because it is more than the number of people on the planet to begin with, all of whom, by this point, would have already downloaded it.

I'm curious if anyone is going to explain that apps die out after a few months and wouldn't keep growing.

Part 4

This question is worth 6 credits.

37. A football player attempts to kick a football over a goal post. The path of the football can be modeled by the function h(x) = -1/225 x2 + 2/3 x, where x is the horizontal distance from the kick, and h(x) is the height of the football above the ground, when both are measured in feet.

On the set of axes below, graph the function y = h(x) over the interval 0 < x < 150.

Determine the vertex of y = h(x). Interpret the meaning of this vertex in the context of the problem.

The goal post is 10 feet high and 45 yards away from the kick. Will the ball be high enough to pass over the goal post? Justify your answer.

The graph looks like the one below. Put the equation in your calculator and check the Table of Values. Note that your graphing calculator defaults to showing from -10 to +10, not 1 to 150, so you won't see it unless you change the display settings. Also, be careful about entering the equation. I entered "255" instead of "225" the first time and you can see the erasures. I didn't realize my mistake until I noticed that 75 wasn't the vertex, which I reasoned it should be.

The vertex of y=h(x) is (75, 25), which means that at 75 feet away from the kicker, the ball will reach a maximum height of 25 feet and start descending.

At 45 yards, or 135 feet, the ball will be 9 feet of the ground (135, 9), so it will not pass over the goal post, which is 10 feet high.

That's it for this test. Finally complete. Except for some illustrations, which will be added when they are available.

Completing the Square

(Click on the comic if you can't see the full image.)
(C)Copyright 2015, C. Burke.

After 1000 comics, sometimes I surprise myself when I realize that there are puns that I still haven't used.

Also, 1017 is not a perfect square. You need to add + 7 to complete the square.




Come back often for more funny math and geeky comics.




Sunday, June 21, 2015

Fathers Day 2015

(Click on the comic if you can't see the full image.)
(C)Copyright 2015, C. Burke.

I wonder if Fudgie the Whale is still around...

Remember, if you're the Only, you are, by definition, the Greatest!

HAPPY FATHERS DAY TO ALL!

2015 -- give or take 6000 years.




Come back often for more funny math and geeky comics.




Saturday, June 20, 2015

New York State Algebra I (Common Core) Regents, June 2015 Part 2

I'll be working all weekend (yes, even on Fathers Day) grading exams, so getting these answers up will take some time. I'll edit the answers when I have time to make illustrations.

Part 2

25. Graph the function y = |x - 3| on the set of axes below.

Explain how the graph of y = |x - 3| has changed from the related graph y = |x|.

This absolute value graph is V-shaped, with a slope of -1 on the left side and 1 on the right side. It's vertex is at (3, 0).
The change from y = |x| is that the graph has shifted three units to the right.

26. Alex is selling tickets to a school play. an adult ticket costs $6.50 and a student ticket costs $4.00. Alex sells x adult tickets and 12 student tickets. Write a function f(x), to represent how much money Alex collected from selling tickets.

f(x) = 6.50x + 4.00(12) or
f(x) = 6.50x + 48.00

27. John and Sarah are each saving money for a car. The total amount of money John will save is given by the function F(x) = 60 + 5x. The total amount of money Sarah will save is given by the function g(x) = x2 + 46. After how many weeks, x, will they have the same amount of money? Explain how you arrived at your answer.

One way: Enter y1 = 60 + 5x and y2 = x2 + 46 into your graphing calculator and look at the Table of Values. Copy the Table of Values onto your paper (and explain that this is what you did).

WeekJohnSarah
165 47
270 50
375 55
480 62
585 71
690 82
795 95
8100 110
In seven weeks the amounts will be the same.

Second way: Solve the following quadratic equation: x2 + 46 = 60 + 5x

x2 + 46 - 60 - 5x = 0
x2 - 5x - 14 = 0
(x - 7)(x + 2) = 0
x = 7 or x = -2
Throw out -2 as irrelevant. In seven weeks the amounts will be the same.

28. If the difference (3x2 - 2x + 5) - (x2 +3x - 2) is multiplied by 1/2X2, what is the result, written in standard form.

The subtraction gives you 2x2 - 5x + 7. Times 1/2x2 gives x4 - 2.5x3 + 3.5x2.

29. Dylan invested $600 in a savings account at 1.6% annual interest rate. He made no deposits or withdrawals on the account for 2 years. The interest was compounded annually. Find, to the nearest cent, the balance in the account after 2 years.

Because it is only 2 years, you have your choice of using the compound interest formula (exponential growth) or using the simple interest formula, adding the interest and using the simple interest formula a second time on the new amount (the principal). The second method takes longer, but is good if you forgot the first formula.

The formula given in the reference table is overly complex -- it's like that because you'll need that one in the future. On the other hand, the example in question 17 in Part 1 is a better example to use.

(600)(1.016)2= 619.3536 = $619.35
Or (600)(1.016) = 609.60 and (609.60)(1.016) = 619.3536 = $619.35

30. Determine the smallest integer that makes -3x + 7 - 5x < 15 true.

-8x + 7 < 15
-8x < 8
x > -1
The smallest integer that makes this statement true is 0 (because it has to be greater than -1).

31. The residual plots from two different sets of bivariate data afe graphed below.

Explain, using evidence from graph A and graph B, which graph indicates that the model for the data is a good fit.

FOR MORE INFORMATION ABOUT RESIDUALS, CHECK OUT MY PREVIOUS POST ABOUT THE QUESTION FROM THE AUGUST 2014 REGENTS EXAM.

Basically, Graph A is a better model because the residuals are scattered about and there is no pattern to them. Graph B is not a good fit because the residuals form a pattern, which is sort of a curve. This can happen, for example, when a linear model (linear regression) is used but a quadratic regression would be better. (You don't need the last sentence -- only the part about why A is good and B is bad.)

32. A landscaper is creating a rectangular flower bed such that the width is half of the length. The area of the flower bed is 34 square feet. Write and solve an equation to determine the width of the flower bed, to the nearest tenth of a foot.

A = L * W = 34. L = 2W
(2W)(W) = 34
2W2 = 34
W2 = 17
W = SQRT(17) = 4.123105626 = 4.1 feet.

That's it for Part II. Hopefully, I'll have Part III up Sunday or Monday.
Happy Fathers Day to all your Fathers (and to you, too, if you are a Father).

Thursday, June 18, 2015

New York State Algebra I (Common Core) Regents, June 2015 Part 1

Sorry for the rush job, but there were TWO Algebra tests back-to-back, one day after the other. Here is Part 1 for the Common Core exam. As always, images will be added as time permits.

Part 1

1. The cost of airing a commercial on television is modeled by the function C(n) = 110n + 900, where n is the number of times the commercial is aired. Based on this model, which statement is true?

(3) The commercial costs $900 to produce and $110 each time it is aired. None of the other choices fit the function.

2. The graph below represents a jogger's speed during her 20-minute jog around her neighborhood.

Which statement best describes what the jogger was doing during the 9-12 minute interval of her jog?

(4) She was jogging at a constant rate. The y-axis is speed, and it doesn't change during that interval -- it's flat.

3. If the area of a rectangle is expressed as x4 - 9y2, then the product of the legnth and the width of the rectange could be expressed as

(2) (x2 - 3y)(x2 + 3y) is the only choice with the correct product. Use the Difference of Squares Rule.

4. Which table represents a function?

(3) Only table three doesn't repeat any x values. No x value can have multiple f(x) values associated with it.

5. Which inequality is represented in the graph below?

(1) y > -3x + 4. The graph is shaded above the line, and the y-intercept is 4. That is enough information to give you the correct choice.

6. Mo's farm stand sold a total of 165 pounds of apples and peaches. She sold apples for $1.75 per pound and peaches for $2.50 per pound. If she made $33750, how many pounds of peaches did she sell?

(3) 65. If you set up a system of equations and solve, you will see that she sold 100 pounds of apples and 65 pounds of peaches.

7. Morgan can start wrestling at age 5 in Division 1. He remains in that division until he next odd birthday when he is required to move up to the next division level. Which graph correctly represents this information.

(1) The step graph with the solid endpoint on the left and the open endpoint on the right. Note that choices (2) and (4) don't make any sense.

8. Which statement is not always true?

(2) The product of two irrational numbers is rational. The square root of 2 times the square root of 3 is the square root of 6, which is NOT rational. Not true.

9. The graph of the function f(x) = (x + 4)(1/2) is shown below:

The domain of the function is

(4) {x | x > -4}. Only values of x from -4 and above make any sense in that function. Lower numbers would not have real number results.

10. What are the zeroes of the function f(x) = x2 - 13x - 30?

4. 15 and -2. Set the function equal to 0 and factor.
x2 - 13x - 30 = 0
(x - 15) (x + 2) = 0
x - 15 = 0 or x + 2 = 0
x = 15 or x = -2

11.Joey enlarged a 3-inch by 5-inch photograph on a copy machine. he enlarged it four times. The table below shows the area of the photograph after each enlargement.

What is the average rate of change of the area from the original photograph to the fourth enlargement, to the nearest tenth?

(3) 5.4. Calculate the rate of change: (36.6 - 15) / (4 - 0) = 21.6 / 4 = 5.4

12. Which equation(s) represent the graph below?
I. y = (x + 2)(x2 - 4x - 12)
II. y = (x - 3)(x2 + x - 2)
III. y = (x - 1)(x2 - 5x - 6)

(2). II, only. The zeroes are at -2, 1 and 3. Each equation shows one of those zeros. You have to factor the trinomials to find out if they will reveal the other two zeroes. If you factor you get
I. (x + 2)(x - 6)(x + 2)
II. (x - 3)(x + 2)(x - 1)
III. (x - 1)(x - 6)(x + 1)
Only choice II fits the graph.
You also could have put the three equations in your graphing calculator and graphed them or checked the Tables of Values.

13. A laboratory technician studied the population growth of a colony of bacteria. He recorded the number of bacteria every other day, as shown in the partial table below.

Which function would accurately model the technician's data?

(2) f(t) = 25t + 1. It is exponential, so choices (3) and (4) are out. For choice (1), if t = 0, 250 = 1, not 25.

14. Which quadratic function has the largest maximum?

(3) k(x) = -5x2 - 12x + 4

Again, if you graph (1) and (3), you will find out that that maximum value is more than 11. It's actually 11.2 when x = -1.2. You can find x using the formula for the axis of symmetry if you don't know how to find the maximum with your calculator. This is obviously bigger than (4). In choice (2), the maximum is more than 9, but it's much less than 11. (It's 9.5.) If you are really worried, you can do a quadratic regression to find the equation, which is f(x) = -2x2 + 6x + 5.

15. If f(x) = 3x and g(x) = 2x + 5, at which value of x is f(x) < g(x)?

(1) -1. Plug the values in or enter the equations in your calculator. Or try this trick on the calculator
3^{-3, -1, 2, 4} [ENTER]
2*{-3, -1, 2, 4}+5 [ENTER]
Compare the results. Make sure you type the choices in the same order, so you can compare them easily.

16. Beverly did a study this past spring using data she collected from a cafeteria. She recorded data weekly for ice cream sales and soda sales. Beverly found the line of best fit and the correlation coefficient as shown in the diagram below.

Given the information, which statement(s) can correctly be concluded?
I. Eating more ice cream causes a person to become thirsty.
II. Drinking more soda causes a person to become hungry.
III. There is a strong correlation between ice cream sales and soda sales.

(2) III, only. The correlation constant is .96, which shows a strong connection. There's nothing to indicate increased hungry or thirst. (When I'm hungry, I don't think "ice cream!", for example.) This is one of those cases that fits the expression "correlation does not prove causation". Something else (like rising or falling temperatures) could be causing sales of each to rise or fall at the same time.

17. The function V(t) = 1350(1.017)t represents the value V(t), in dollars, of a comic book t years after its purchase. The yearly rate of appreciation of the comic book is

(4) 0.017%.
(2) 1.7%. This is a straight definition question. Keep this question in mind in PART 2 when you have to write a compound interest formula. It's a good example with numbers, as opposed to the formal definition in the reference table. Move the decimal two places over to get the percent.

18. When directed to solve a quadratic equation by completing the square, Sam arrived at the equation (x - 5/2)2 = 13/4. Which equation could have been the original equation given to Sam?

(4) x2 - 5x + 3 = 0. The term - 5/2 comes from being half of -5, so choices (1) and (2) are out. Squaring 5/2 yields 25/4, which had to be added to both sides of the equation. Subtract 13/4 - 25/4 = -12/4 = -3. Which means that the original equation ended in + 3.

19. The distance a free falling object has traveled can be modeled by the equation d = 1/2 at2, where a is acceleration due to gravity and t is the amount of time the object has fallen. What is t in terms of a and d?

(2) t = the square root of (2d / a). As shown, multiply both sides by 2, divide by a, and take the square root.

20. The table below shows the annual salaries for the 24 member of a professional sports team in terms of millions of dollars. [table omitted] The team signs an additional player to a contract worth 10 million dollars per year. Which statement about the median and the mean is true?

(3) Only the mean will increase. DO NOT CALCULATE THE MEAN!! Seriously. Look at the table. The mean must be between 0.5 and 7.2. The number 10 MUST be higher than the mean, so the new average will go up.

On the other hand, check the median. With 24 players, the median in the average of the 12th and 13th, both of which are 1.4, so the median is 1.4. When you add a 25th player, the median is just the 13th player, which is still 1.4. So, in this case, the median does not change.

21. A student is asked to solve the equation 4(3x - 1)2 - 17 = 83
The student's solution to the problem starts as

4(3x - 1)2 = 100
(3x - 1)2 = 25

A correct next step in the solution of the problem is

(1) 3x - 1 = +5. Take the square root of both sides. You could also expand the binomial by squaring it, but the only choice to do this, (4), inexplicably rooted the other side of the equation. If (4) said 25 instead of 5, it would have been correct.

22. A pattern of blocks is shown below.

If the pattern continues, which formula(s) could be used to determine the number of blocks in the nth term?

(3) II and III. II shows the recursive formula and III shows the explicit formula. The first formula makes no sense -- it increases by 1 instead of 4.

23. What are the solutions to the equation x2 - 8x = 24?

(1) x = 4 +2(10)^(.5) You have two choices. The easiest is to complete the square again. If you forgot how, then you can use the Quadratic formula, which is on the reference sheet. Obviously, from the choices, straight factoring is not an option.

24. Natasha is planning a school celebration and wants to have live music and food for everyone who attends. She has found a band that will charge her $750 and a caterer who will provide snacks and drinks for $2.25 per person. If her goal is to keep the average cost per person between $2.75 and $3.25, how many people, p, must attend?

(4) 750 < p < 1500. See the image below.

UPDATE: Part 2 is now online.

Math Regents Questions

Until I get a chance to type up the Regents questions, if anyone has a specific question about a specific problem, leave it in the comments and I'll get to it as soon as possible.

Be sure I know which test you're talking about. There were TWO Algebra exams.

Selfie

(Click on the comic if you can't see the full image.)
(C)Copyright 2015, C. Burke.

There will come a day when we will not be able to tell the difference between pictures of ourselves and those of our mirror image counterparts!




Come back often for more funny math and geeky comics.




Monday, June 15, 2015

Manifest Destiny

(Click on the comic if you can't see the full image.)
(C)Copyright 2015, C. Burke.

They walked the New Frontier not realizing 160-so years later, men would Walk The Moon.

I had another rhyme about "Canyon so Grand" and "mapping", but, unfortunately, that wasn't their expedition. I didn't think it was, but I checked anyway. (Thank you, Internet, for once more saving me from looking foolish when I venture outside my comfort zone.)

If you'd like to know why Lewis & Clark are drawn that way, you'll have to venture back in time -- not that far, just May 2009. There are four not-quite-sequential comics on the subject:

Also part of history today: The Magna Carta is 800 Years Old today.




Come back often for more funny math and geeky comics.




Friday, June 12, 2015

Game of Homophones

(Click on the comic if you can't see the full image.)
(C)Copyright 2015, C. Burke.

When you're playing your game of homophones, ewe one or ewe dye.

So watch it.




Come back often for more funny math and geeky comics.




Wednesday, June 10, 2015

Mean Girls (with Belle & Mr. Whiskers)

(Click on the comic if you can't see the full image.)
(C)Copyright 2015, C. Burke.

As the saying goes, they're so mean, they have no standard deviation!

This is the seventh appearance for Belle & Mr. Whiskers over the years.

Add a few more characters, and I'll have enough for a children's series ... featuring material that no one in the target demographic would understand. Sigh.




Come back often for more funny math and geeky comics.




Monday, June 08, 2015

Distance

(Click on the comic if you can't see the full image.)
(C)Copyright 2015, C. Burke.

Actually, any great speed I imagine is usually in a rocket of some kind, so DiRT isn't as much of a problem.




Come back often for more funny math and geeky comics.




Saturday, June 06, 2015

New York Regents Geometry (Common Core) June 2015, Parts 3 and 4

Feeling better now. Let's try to finish this off.

Here are the questions and answers to Parts 3 and 4 of Tuesday's exam. There was quite a bit of discussion about the test at last night's meet up.
Links to Part 1 and Part 2 .

Part 3

There are 3 problems, each worth 4 credits.

32. In the diagram below, EF intersects AB and CD at G and H, respectively, and GI is drawn such that GH = IH.


If m<EGB = 50o and m<DIG = 115o, explain why AB || CD.

You could write this as proof, but it just says "explain", so writing your answer as a paragraph is fine.
Angle DIG is 115 degrees. This means that angle HIG is 65 degrees because it is supplementary to DIG. Because GH is congruent to HI, GHI is an isosceles triangle with HIG and HGI as the base angles. Therefore, HGI has a measure of 65 degrees. The sum of the angles in triangle GHI is 180 degrees, so angle GHI is 180 - 65 - 65 = 50. Angle GHI is congruent to angle EGB, so AB || CD because the corresponding angles are congruent.

33. Given: Quadrilateral ABCD is a parallelogram with diagonals AC and BD intersecting at E. Prove triangle AEB is congruent to triangle CEB.
Describe a single rigid motion that maps AED onto GEB.

Due to formatting, I can't write a proof right now. I'll update it later as an image. But, basically ...
AE = EC and BE = DE because the diagonals of parallelograms bisect each other. Angles AED and CEB are congruent because vertical angles are congruent. Triangle AED is congruent to CEB because of SAS.

A rigid motion that would map AED on CEB is a 180 degree rotation about point E.

34. In the diagram below, the line of sight from the park ranger station, P, to the lifeguard chair, L, on the beach of a lake is perpendicular to the path joining the campground, C, and the first aid station, F. The campground is 0.25 mile from the lifeguard chair. The straight paths from both the campground and first aid station to the park ranger station are perpendicular.


If the path from the park ranger station to the campground is 0.55 mile, determine and state, to the nearest hundredth of a mile, the distance between the park ranger station and the lifeguard chair.
Gerald believes the distance from the first and station to the campground is at least 1.5 miles. Is Gerald correct? Justify your answer.

The length of PL is simple Pythagorean Theorem: x2 + 0.252 = 0.552. So x2 = .3025 - .0625 = .24, and x = .489... or .49 to the nearest hundredth.

To answer the second part, you have your choice of ratios. Either you can compare the base and hypotenuse of PLC (.25/.55) to the base and hypotenuse of PFC (.55/x) or you can use the Right Triangle Altitude Theorem to find FL and then add FL + LC to get FC. (I would recommend the first, but given the way the diagram is drawn, some will be drawn to the second method even if it is longer because it's more obvious to them.)

Using the first method: .25/.55 = .55 / x means that x = .55 * .55 / .25 = 1.21 miles, for Gerald is incorrect. It is shorter than 1.5 miles.

Using the second method: .25 / .49 = .49 / y means that y = .49 * .49 / .25 = 0.96. Add 0.96 + 0.25 = 1.21 miles. Gerald is still incorrect.

Part 4

There is 1 problem, worth 6 credits. And what a doozy of a problem. I have warned my students about area problems where you have to break the irregular figure down into parts, and that's essentially what this question was, but in three dimensions.

35. The water tower in the picture below is modeled by the two-dimensional figure beside it. The water tower is composed of a hemisphere, a cylinder and a cone. Let C be the center of the hemisphere and let D be the center of the base of the cone.


If AC = 8.5 feet, BF = 25 feet and m<EFD = 47o, determine and state, to the nearest cubic foot, the volume of the water tower.
The water tower was constructed to hold a maximum of 400,000 pounds of water. If water weighs 62.4 pounds per cubic foot, can the water tower be filled to 85% of its volume and not exceed the weight limit? Justify your answer.

You have the Volume formulas for the cylinder and cone in the reference table. Also, you have the formula for a sphere, which you will take half of.
If AC is 8.5, then BC and DF are also 8.5, so we have our radii for the formulas.
Cylinder: V = pi*r2h = (3.141592...)(8.5)2(25) = 5674.5017...
Hemisphere: V = (1/2)(4/3)pi*r3 = (2/3)(3.141592...)(8.5)3 = 1286.2204...
Cone: V = (1/3) pi*r2h -- except we need to find the height! Trigonometry, again.

The side adjacent to the 47o angle is 8.5. The height, which we are looking for, is the side opposite the angle. That means that we need to use Tangent to find the height.
Tan(47) = x/8.5, so x = 8.5 * tan(47), which is 9.12. Soooooooo ....
Cone: V = (1/3) pi*r2h = (1/3)(3.141592...)(8.5)2(9.12) = 690.0194

The total volume of the water tower is 5674.5017 + 1286.2204 + 690.0194 = 7650.7415, which rounds to 7651.

And that's just part of the answer! If you got that far, you probably got a lot of points. Loss of points for computational errors. (I wonder how many I made.)

Second part: .85 * 7651 * 62.4 = 405,809 pounds. Therefore, the water tower cannot be filled to 85% of its volume and not exceed 400,000 pounds.
Note: If you got an incorrect answer to the first part of the question, you can still receive credit for this portion if you mistake is consistent.

36. In the coordinate plane, the vertices of triangle RST are R(6, -1), S(1, -4) and T(-5, 6). Prove that RST is a right triangle.
State the coordinates of point P such that quadrilateral RSTP is a rectangle.
Prove that your quadrilateral RSTP is a rectangle.
[The use of the set of axes is optional.]

You can prove that a triangle is a right triangle by showing that two sides form a right angle. You can prove two sides form a right angle by showing that they are perpendicular, which will be true if their slopes are negative reciprocals, that is, they have a product of -1.

Slope of RS: (-4 - -1)/(1 - 6) = 3/5.
Slope of ST: (6 - -4)/(-5 - 1) = -5/3.
Slope of TR: (-1 - 6)/(6 - -5) = -7/11.
The slopes of RS and ST are negative reciprocals, so RS is perpendicular to ST. Therefore <S is a right angle and triangle RST is a right triangle.

Finding point P: They gave you one hint. The rectangle has to be RSTP, so P has to be above R and to the right, and will be opposite from S.
The properties of rectangles are that the opposite sides are the same length and have the same slope. To get from point S to point R, you have to go up 5 and to the right 3 units. Or to use a rigid motion expression, if you S is moved be a translation of (x + 5, y + 3), it will be mapped onto R.
The image of T(-5, 6) after a translation of T+5,+3 is P(0, 9).

Proving that it's a rectangle: First, show that it's a parallelogram by showing that the opposite sides are parallel.
Slope of RP: (9 - -1)/(0 - 6) = -5/3.
Slope of TP: (9 - 6)/(0 - -5) = 3/5.
RP || ST because they have the same slope.
TP || RS because they have the same slope. Therefore, RSTP is a parallelogram.
Angle S is right angle. Therefore, RSTP is a rectangle.


...


And that's that. The rest of the Regents Exams are over a week away. Answers are farther away than that. :)

Friday, June 05, 2015

New York Regents Geometry (Common Core) June 2015, Part 2

My apologies for the delay. I was literally away from the keyboard all day yesterday, between meetings, professional development and the #NYCMathTweetup last night.

Here are the questions and answers to Part 2 of Tuesday's exam. There was quite a bit of discussion about the test at last night's meet up.
Link to Part 1.

Part 2

There were seven open-ended questions in Part II, each worth 2 points. You had to show your work to get full credit.

25. Use a compass and straightedge to construct an inscribed square in circle T shown bleow. [Leave all construction marks.]

Using the straightedge, draw a diameter. Now draw a perpendicular bisector using the two points on the circle at either end of the diameter. This will create a second diameter (a secant, actually) perpendicular to the first one. Finally, using the straightedge, draw the four sides of the square by connecting the consecutive points on the circle. (I really wish I had an image here ... visually, it's a little more obvious.)

26. The diagram below shows parallelogram LMNO with diagonal LN, m<M = 118 degrees, and m<LNO = 22.
Explain why m<NLO = 40 degrees.

You have a couple of choices. First, because the figure is a parallelogram, angle O is congruent to angle M, so <O = 118o. Triangle LON has 180 degrees, so 180 = 118 + 22 + x. Solve for x, NLO = 40 degrees.

Second, using basiocally the same math as above, the consecutive angles of a parallelogram as supplementary, so 180 = 118 + MNL + 22. Angle MNL is 40 degrees. Angle NLO and MNL are alternate interior angles and are, therefore, congruent.

27.The coordinats of the endpoints of AB are A(-6, -5) and B(4, 0). Point P is on AB. Determine and state the coordinates of point P, such that AP:PB is 2:3. [The use of the set of axes below is optional.]

The ratio 2:3 means that you can divide AB into five congruent segments and that point P would be at the end of the second segment. P is located at a point that is two-fifths of the distance from A to B. You will notice that the difference in the x-coordiantes is 4 - (-6) = 10 and the difference of the y-coordinates is 0 - (-5) = 5. Two-fifths of 10 is 4; two-fifths of 5 is 2. Point P is at (-6 + 4, -5 + 2), which is P(-2, -3).

You can also graph the points and see that the slope of the line is 1/2. If you draw the triangles formed by going up 1, right 2 five times, the second triangle will bring you to point P.

28. The diagram below shows a ramp connecting the ground to a loading platform 4.5 feet above the ground. The ramp measures 11.75 feet from the ground to the top of the loading platform.


Determine and state, to the nearest degree, the angle of elevation formed by the ramp and the ground.

Trigonometry. You have the opposite side and the hypotenuse, so you need to use the sine ratio.
sin (x) = 4.5 / 11.75. Take the inverse and you get
x = sin-1 (4.5 / 11.75).
Put that in your calculator and you get 22.52 degrees, which rounds to 23 degrees.

29. In the diagram below of circle O, the area of the shaded sector AOC is 12(pi) in2 and the length of OA is 6 inches. Determine and state m<AOC.

The measure of the central angle can be found using a proportion: central angle / 360 = sector area / circle area.
The area of the entire circle is A = (pi)r2 = (pi)(6)2 = 36(pi).

Cross multiply: x / 360 = 12 / 36 and you get x = 120 degrees.
Likewise 12(pi)/36(pi) = 1/3. One-third of 360 degrees is 120.

30. After a reflection over a line triangle A'B'C' is the image of triangle ABC. Explain why triangle ABC is congruent to triangle A'B'C'.

In a reflection. the image retains the size and shape of the original. This means that AB = A'B', BC = B'C' and CA = C'A', so by SSS the two triangles are congruent.

I'm guessing that Definition of a Reflection is NOT a sufficient answer. State something that you know.

31. A flagpole casts a shadow 16.60 meters long. Tim stands at a distance of 12.45 meters from the base of the flagpole, such that the end of Tim's shadow meets the end of the flagpole's shadow. If Tim is 1.65 meters tall, determine and state the height of the flagpole to the nearest tenth of a meter.

Sketch a diagram. You will have two similar triangles, such that one is contained within the other, with two parallel, vertical lines. You can set up a proportion height : shadow = height : shadow. Notice that if Tim is 12.45 meters from the pole, that means that his shadow is 16.6 - 12.45 = 4.15 meters long.
So x / 16.60 = 1.65 /4.15. Cross-multiply, x = 6.6 meters.

Another good reason to sketch a diagram is so that you won't make the mistake of saying that Tim's shadow is 12.45 meters long because it's written in the problem and you're too busy punching keys to visual it.

That's it for Part II. Still working on the rest.

Wednesday, June 03, 2015

New York Regents Geometry (Common Core) June 2015, Part 1

Yesterday was the first-ever Common Core Geometry Regents exam. I could not get a copy of it until this morning. What follows are the questions from Part 1. (If the images are omitted, come back soon -- they'll be added at some point.) Come back over the next few days for the rest of the exam.

Part 1 only had 24 questions, instead of 28, and they were a little more challenging that those found on the old exam. One thing that struck me was that there was only one Eqaution of a Circle question, not 3 or 4, but they made it tougher. I hope you paid attention in Common Core Algebra last year.

Part I -- Multiple Choice

1. What object is formed when right triangle RST is rotated around leg RS?

(4) A cone. Notice that it doesn't mention an image. This isn't a transformation. An object is being created, and it will be created in three dimensions. That object will be a cone as every point on the hypotenuse spins about in a circle.

2. The vertices of triangle JKL have coordinates J(5, 1), K(-2, -3) and L(-4, 1). Under which transformation is the image triangle J'K'L' not congruent to JKL?

(4) A dilation with a scale factor of 2 and centered on the origin. Dilations produce images that are different sizes from the original, and therefore are NOT congruent. (They are similar.)

3. The center of circle Q has coordinates (3, -2). If circle Q passes through R(7, 1), what is the length of its diameter?

(3) 10. QR is a radius and it has length 5. The difference of the x coordinates is 4; the difference of the y coordinates is 3. The square root of the sum of 4 squared plus 3 squared is 5. The diameter is twice the radius, which is 10.

4. In the image below, congruent figures 1, 2, and 3 are shown.

Which sequence of transformations maps figure 1 onto figure 2 and then figure 2 onto figure 3?

(4) A translation followed by a rotation. Neither move is a reflection, and the translation happens first.

5. As shown in the diagram below (omitted), the angle of elevation from a point on the groun to the top on the tree is 34 degrees.

If the point is 20 feet from the base of the tree, what is the height of the tree, to the nearest tenth of a foot?

(3) 13.5. Trigonometry. The height is opposite the angle. The ground is adjacent to the angle. Opposite over adjacent is Tangent. Tan (34) = x/20, so x = 20 * Tan (34). Put that in your calculator and you get 13.490..., which rounds to 13.5.

6. Which figure can have the same cross section as a sphere?

(2). A cone. The cross section of a sphere will always be a circle, no matter how you slice it. It is possible to slice a cone so that the cross-section is also a circle. The others are prisms and pyramids with polygonal (not circular) bases.

7. A shipping container is in the shape of a right rectangular prism with a length of 12 feet, a width of 8.5 feet, and a height of 4 feet. The container is completely filled with contents that weigh, on average, 0.25 pound per cubic foot. What is the wight, in pounds, of the contents in the container?

(3) 102. The weight of the contents is the weight of one cubic foot times the number of cubic feet. The number of cubic feet is the Volume, which is length times width times height. So Weight = 12 X 8.5 X 4 X 0.25, which is 102 pounds. If you noticed, 4 x 0.25 = 1, so you could have canceled those two numbers.

8. In the diagram of circle A shown below (omitted), chords CD and EF intersect at G, and chords CE and FD are drawn.

Which statement is NOT always true?

(1) CG is congruent to FG. There is no reason that those segments must be congruent. On the other hand, inscribed angles C and F intercept the same arc, so they are congruent. Moreover, inscribed angles E and D intercept the same arc, and angle CGE and DGF are vertical angles, so the two triangles are similar. The proportion in choice (3) is true because the triangles are similar.

9. Which equation represents a line that is perpendicular to the line represented by 2x - y = 7?

(1) y = -1/2 x + 6. The slope of the given line is 2, which you can find be re-writing the equation in slope-intercept form, or recognizing that in Standard form, Ax + By = C, the slope of the line is -A/B, which is -2/-1 = 2. The slope of the perpendicular line is the negative reciprocal, which is -1/2.

10. Which regular polygon has a minimum rotation of 45 degrees to carry the polygon onto itself?

(1) Octagon. To carry the polygon onto itself means that as you rotate the polygon, you will at some point get an image that is the same as the original. A rectangle would have to turn 180 degrees. An equilateral triangle would have to rotate 120 degrees. A square would rotate 90 degrees. Etc. Divide 360 / 45 = 8. A polygon with 8 sides will carry onto itself every 45 degrees. Note that 45 degrees is also the size of the exterior angles of an octagon, a figure which is found by dividing 360 / 8. The two concepts are related.

11. In the diagram of triangle ADC below, EB || DC, AE = 9, ED = 5, and AB = 9.2


What is the length of AC, to the nearest tenth?

(3) 14.3. Create a proportion 9/9.2 = 5/x, and cross-multiply. 9x = 46, and x = 5.111..., or 5.1. The length of AC is 9.2 + 5.1 = 14.3.

12. In scalene triangle AbC shown in the diagram below, m<C = 90.


Which equation is always true?

(4) sin A = cos B. This is a basic Trig identity. What is opposite A is adjacent to B. Choices (1) and (2) are only true if angles A and B are congruent (e.g., 45 degrees). Choice (3) is silly as you can't take the sine of the right angle.

13. Quadrilateral ABCD has diagonals AC and BD. Which information is NOT sufficient to prove ABCD is a parallelogram?

(4) AB = CD and BC || AD. This choice could also be an isosceles trapezoid. The other choices show properties of parallelograms.

14. The equation of a circle is x2 + y2 + 6y = 7. What are the coordinates of the center and the length of the radius of the circle?

(2) Center (0, -3) and radius 4. This is not the standard form for the equation of a circle. To fix it, you have to complete the square.


x2 + y2 + 6y + 9 = 7 + 9
x2 + (y + 3)2 = 16

The center is (0, -3) and the radius is 4.

15. Triangles ABC and DEF are drawn below


If AB = 9, BC = 15, DE = 6, EF = 10 and <B = <E, which statement is true?

(3) Triangle ABC ~ Triangle DEF. The two pairs of sides are proportional and the included angle is congruent, so by SAS the two triangles are similar. The other statements do NOT match corresponding parts, so they are NOT true.

16. If triangle ABC is dilated by a scale factor of 3, which statement is true of the image A'B'C'?

(2)B'C' = 3BC. The sides of the new triangle will be three times as long.

17. Steve drew line segments ABCD, EFG, BF, and CF as shown in the diagram below.

Triangle BFC is formed.
Which statement will allow Steve to prove ABCD || EFG?

(1) <CFG = <FCB. These two angles are Alternate Interior Angles. None of the other pairs are alternate interior angles (or corresponding for that matter).

18. In the diagram below, CD is the image of AB after a dilation of scale factor k with center E.


Which ratio is equal to the scale factor k of the dilation?

(1) EC/EA. A scale factor of 2 moves all the points to a distance twice as far from the center of dilation. So the ratio of the distances, in that case, would be 2. Likewise, EC/EA will give the scale factor k of the problem.

19. A gallon of paint will cover approximately 450 square feet. An artist wants to paint all the outside surfaces of a cube measuring 12 feet on each edge. What is the least number of gallons of paint he must buy to paint the cube?

(2) 2. The amount of paint needed is 6 X 12 X 12 = 864. Divide that by 450 and you get a number between 1 and 2, so you need 2 cans of paint to complete the task.

20. In circle O shown below, AC is perpendicular to CD at point C, and chords AB, BC, AE and CE are drawn.


Which statement is NOT always true?

(1) <ACB = <BCD. Choice (1) implies that chord CB bisects angle ACD (which is a right angle), but there is nothing stated or shown that proves this. As for the other choices: Angles ABC and ACD are both right angles because the first is an inscribed angle that intercepts a semicircle and the latter is a formed by a diameter and a tangent. Angles BAC and DCB intercept the same arc, and both are half the size of that arc. Angles CBA and AEC are right angles, which intercept a semicircle.

21. In the diagram below, triangle ABC ~ DEC.


If AC = 12, DC = 7, DE = 5 and the perimeter of triangle ABC is 30, what is the perimeter of triangle DEC?

(4) 17.5. In the smaller triangle, DC corresponds to AC in the larger triangle. That gives a ratio of 7/12. The smaller triangle's perimeter is unknown, but the larger triangle has a perimeter of 30, so the ratio is x/30. That gives the proportional 7/12 = x/30.
Cross-multiply and you will find that x = 17.5.

22. The line 3y = -2x + 8 is transformed by a dilation centered at the origin. Which linear equation could be its image?

(1) 2x + 3y = 5. A dilation will not change an image's orientation, which means that the slope of the line is preserved. You had to find the slope of the line, but you didn't need to convert anything into Slope-Intercept form. All four choices are in Standard form, so rewrite the given equation in Standard form by adding 2x to both sides of the equation. That gives you 2x + 3y = 8. Only one choice has 2x + 3y in it, and that is Choice (1).

23. A circle with a radius of 5 was divided into 24 congruent sectors. The sectors were then rearranged, as shown in the diagram below.


To the nearest integer, the value of x is

(2) 16. The sectors are arranged into what looks almost like a parallelogram. As you divide the circle into more and more sectors, the base of the parallelogram gets closer and closer to one-half the circumference (with the other base being the other half of the circumference). The radius is 5. Circumference is 2 X pi X r, and half of that is just pi X r, or (3.14)(5) = 15.70, or roughly 16.

24. Which statement is sufficient evidence that triangle DEF is congruent to triangle ABC?

(3) There is a sequence of rigid motions that maps AB onto DE, BC onto EF, and AC onto DF. Side-side-side. Choice (1) only gives two sides. Choice (2) proves that they are similar, but not that they are congruent. Choice (4) does not have enough information.


Part 2 when I can get to it. Hopefully, soon.