Regents Exam Questions and Answers Will Be Delayed

Today I got a phone call from my former AP at my old school. I just assumed he was calling about summer school or tutoring or something.

Actually, he was a little confused and passing on a message. He had received a call from someone working for the New York State Regents because of my blog. Apparently, they traced me to his school. (I haven't worked there since Fall 2014.)

They had a problem with the answers being posted so soon. Since he's just the middle man in this conversation, I didn't protest at all that no answers were posted until after the time when students with conflicts and test modifications would still have finished it. Nor complain that questions and answers get posted to social media by students and other teachers, although probably not as organized as I was.

So what this means is that there will be no answers posted here.

However, if anyone has a specific question about a specific problem that they had trouble with and they want to talk about it in email, we could try that.

June 2016 Common Core Algebra 1 Regents, Part 2

What follows is a portion of the Common Core Integrate Algebra exam. Other parts will be posted on other days. Illustrations will be added at a later time when they become available.

June 2016 Algebra Regents, Part II

25. Given that f(x) = 2x + 1, find g(x) if g(x) = 2[f(x)]² - 1.

Take the expression for f(x) and square it. Then double that trinomial. Then subtract 1 from the expression.
Multiplying (2x + 1)(2x + 1) gives you 4x² + 4x + 1.
Double it to: 8x² + 8x + 2
Subtract 1: 8x² + 8x + 1, which is the final answer.

26. Determine is the product of 3*SQRT(2) and 8*SQRT(18) is rational or irrational. Explain your answer.

You can put it in your calculator or do it by hand. Either way, you will get 24 * 6 = 144, which is a rational number.
144 can be written as a fraction.
144 is a Whole number and all whole numbers are Rational.
There are many similar things you can right as an explanation.

27. On the set of axes below, draw the graph of y = x² - 4x - 1.
State the equation of the axis of symmetry. (image omitted)

Put the equation into your graphing calculator and not the points in the table of values. Plot the points and draw the graph.
By looking at your graph, you will be able to see that the Axis of Symmetry is x = 2

Note that they asked for an equation for the axis of symmetry. If you just state "2" without the "x = ", you will not get the point!

28. Amy solved the equation 2x² + 5x - 42 = 0. She stated that the solutions to the equation were 7/2 and -6. Do you agree with Amy's solution? Explain why or why not.

First thing to notice: they didn't ask you to solve it like Amy did. Just whether or not you agree and could back it up. Second: don't forget to answer the question: "Yes, I agree with Amy because ..." (if you disagreed, I have some bad news for you.)
The algebraic solution will be added later. If you did it algebraically, you should have agreed with Amy.

If you put the equation y = 2x² + 5x - 42 into the graphing calculator, you will see that -6 is one solution and the other solution is between 3 and 4. You can use the function of the calculator that finds the zeroes to show that the other point is 7/2 (or 3.5).

Likewise, you can check her work, showing that 2(-6)² + 5(-6) - 42 = 0 and 2(7/2)² + 5(7/2) - 42 = 0. I would add a note that says that quadratic equations can have up to 2 solutions, just to exclude the possibility that there are other answers besides these two.

29. Sue and Kathy were doing their algebra homework. They were asked to write the equation of the line that passes through the points (-3, 4) and (6, 1). Sue wrote y - 4 = -1/3(x + 3) and Kathy wrote y = -1/3(x) + 3. Justify why both students are correct.

Sue wrote her answer in point-slope form and Kathy wrote her answer in slope-intercept form. However, you still have to prove that they both have the correct answer. You can calculate that the slope is (4 - 1) / (-3 - 6) = 3 / (-9) = -1/3, which is in both equations.

Sue used the slope -3 and the point (-3, 4) to write her equation.
If you distribute the (-1/3) on the right side, and then add 4 to both sides to isolate the y, you will get Kathy's equation.

If this didn't occur to you, but you checked both points in both equations and found out that they all worked, you should get full credit.

30. During a recent snowstorm in Red Hook, NY, Jaime noted that there were 4 inches of snow on the ground at 3:00 pm and there were 6 inches of snow on the ground at 7:00 pm.

If she were to graph these data, what does the shape of the line connecting these two points represent in the context of this problem?

Note that they never asked you to graph the data or find the slope, only to explain the slope.

The slope of the line would be the number of inches that fell every hour.

If you wanted to know how much that was: difference in snow divided by the difference in time = 2 inches / 4 hours = 1/2 inch/hour.

31. The formula for the sum of the degree measures of the interior angles of a polygon is S = 180(n - 2). Solve for n, the number of sides of the polygon, in terms of S.

Isolate n, and move everything else to the other side using Inverse Operations.
Divide by 180: S / 180 = n - 2.
Add 2 to both sides: (S / 180) + 2 = n.

32. In the diagram below, f(x) = x³ + 2x² is graphed. Also graphed is g(x), the result of a translation of f(x).
Determine an equation of g(x). Explain your reasoning.

It doesn't matter if you never graphed a cubic function all year or if you don't remember what one looks like. All you need to remember is that this is an example of a translation. How do the points on the two lines differ?

All the points on g(x) are 4 places lower. That means that all the y-values are 4 less for g(x) than for f(x). So g(x) = f(x) - 4.
Therefore g(x) = x³ + 2x² - 4

Is there another way to do this? Sure. You could have, for example, put some of the points into lists in the graphing calculator and did a Cubic regression to get the equation of the line. This is a totally valid approach as long as you explain what you did and where you're answer came from.

END OF PART II

How did you do? Any questions?

My original scratch writings are shown below.

Daily Regents: Percentages and Statistics (January 2016) #Algebra

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams. At least, that is the plan.

Common Core Algebra, January 2015, Question 26

26. The school newspaper surveyed the student body for an article about club membership. The table below shows the number of students in each grade level who belong to one or more clubs.
If there are 180 students in ninth grade, what percentage of the ninth grade students belong to more than one club?

Very straightforward question. The number of ninth graders in more than one club is 33 + 12 = 45. The total number of ninth graders is 180. Divide 45/180 = .25 or 25%.

You would have lost a point if you had done "1 or more" clubs, or picked a different year. You likely would have lost a point if you left the answer as a decimal when it asked for a percentage.

Any questions?

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If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note.

Daily Regents: Transversals and Parallel Lines (June 2015) Geometry

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams. At least, that is the plan.

Common Core Geometry, January 2015, Question 32

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

Angle HIG is supplementary to DIG, which is 115 degrees. This makes HIG 65 degrees.
Angles HIG and IGH are congruent because they are base angles of an isosceles triangle (GH = IH). This means that IGH is also 65 degrees.
Angle GHI is 50 because a triangle has 180 degrees, and 180 - 65 - 65 = 50 degrees
Corresponding angles GHI and EGB are congruent so AB || CD.

Look at the image below:

Any questions?

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If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note.

Daily Regents: Right Triangle Altitude Theorem (January 2016)

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams next week. At least, that is the plan.

I had a question about this problem on the thread where I posted all the multiple-choice answers: http://mrburkemath.blogspot.com/2016/02/january-2016-new-york-geometry-common.html

Common Core Geometry, January 2016, Question 22

30. 2 In the diagram below, CD is the altitude drawn to the hypotenuse AB of right triangle ABC.
Which lengths would not produce an altitude that measures 6√2?

The Right Triangle Altitude Theorem tells us that (AD)(DB) = (CD)²
The square of the altitude is (6√2)(6√2) = (36)(2) = 72.
So which choice has lengths of AD and DB that have a product of something other than 72?

Careful! Two of the options give you AB instead of DB. Subtract AB - AD to get DB.

Look at the four choices:

1. - 2 * 36 = 72. Not the answer
2. - 3 * (24 - 3) = 3 * 21 = 63, not 72. This is the correct answer.
3. - 6 * 12 = 72. Not the answer
4. - 8 * (17 - 8) = 8 * 9 = 72. Not the answer.

The correct answer is choice (2).

Any questions?

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If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents: Rigid Motions and Similar Triangles (June 2015) Geometry

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams next week. At least, that is the plan.

Common Core Geometry, June 2015, Questions 30

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'.

A reflection is a rigid motion that preserves distance. Since the distance between the points in the same AB = A'B', BC = B'C' and CA = C'A'. The sides are congruent, so the two triangles are congruent.

Common Core Geometry, June 2015, Questions 31

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.

Tim and his shadow form a triangle that is similar to the flagpole and its shadow. The ratios of the shadows to the heights are proportional.

Be careful! Tim is standing 12.45 meters from the flagpole. That means his shadow is only 16.60 - 12.45 = 4.15 meters long. Don't use the incorrect number.

So 1.65 / 4.15 = x / 16.60
Cross multiply: 4.15x = (1.65)(16.60)
Divide: x = (1.65)(16.60)/4.15 = 6.6 meters

The flagpole is 6.6 meters high.

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents: Area of a Sector of a Circle (June 2015) Geometry

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams next week. At least, that is the plan.

Common Core Geometry, June 2015, Questions 29

29. In the diagram below of circle O, the area of the shaded sector AOC is 12*pi in²and the length of OA is 6 inches. Determine and state m<AOC.

As shown in the illustration below, the ratio of the area of the sector of a circle to the whole circle is proportional to the measure of the central angle to 360 degrees.

In other words, whatever fraction of the area is shaded in, multiply that same fraction times 360 degrees to get the measure of the central angle of the sector.

Since area is pi*r², the area of the circle is (6)²pi or 36pi. The sector is 12pi. (12 pi) / (36 pi) = 1/3. The shaded sector is 1/3 of the circle.
Multiply 1/3 times 360, and the central angle is 120 degrees.

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents:Trigonometric Ratios (June 2015)

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams next week. At least, that is the plan.

Geometry, June 2015, Questions 28

27. 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.

A trigonometric ratio problem. The height of the platform is opposite the angle. The ramp is the hypotenuse. This means we have the sine ratio for the angle.

sin x = (4.5 / 11.75)

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents: Coordinate Geometry (June 2015)

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams next week. At least, that is the plan.

Geometry, June 2015, Questions 27

27. The coordinates 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.]

You could solve this by graphing or algebraically. (Algebra shows up in Geometry all the time!)

A ratio of 2:3 means 2/5ths and 3/5ths of the line.
The difference in the x-coordinates is 10. Two-fifths of 10 is 4, and -6 + 4 is -2, the x coordinate for P.
The difference in the y-coordinates is 5. Two-fifths of 5 is 2, and -5 + 2 is -3, the y coordinate for P.

P is at (-2, -3).

See image below for details.

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents: Parallelograms (June 2015) Geometry

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams next week. At least, that is the plan.

Geometry, June 2015, Questions 26

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

The opposite angles of a parallelogram are congruent. Therefore, m<O = 118^o.
Triangle LNO has 180 degrees, so m<NLO + m<LON + m<LNO = 180
m<NLO + 118 + 22 = 180
m<NLO + 140 = 180
m<NLO = 40 degrees.

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents: Systems of Inequalities (August 2014)

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams next week. At least, that is the plan.

August 2014, Questions 37

37. Edith babysits for x hours a week after school at a job that pays $4 an hour. She has accepted a job that pays $8 an hour as a library assistant working y hours a week. She will work both jobs. She is able to work no more than 15 hours a week, due to school commitments. Edith wants to earn at least $80 a week, working a combination of both jobs.
Write a system of inequalities that can be used to represent the situation. Graph these inequalities on the set of axes below.

Determine and state one combination of hours that will allow Edith to earn at least $80 per week while working no more than 15 hours.

If one job pays $4 per hour and she works x hours, she makes 4x dollars. If the other job pays $8 per hour and she works y hours, then she makes 5y dollars. The total is 4x + 8y, which must be greater than or equal to $80, so
4x + 8y > 80

If the total number of hours worked about both jobs must be less than or equal to 15 hours, then
x + y < 15

That is the system of inequalities to graph. Both lines will be solid. The one for her pay will be shaded above. The one for her hours will be shaded below.

You can graph these by finding the x- and y-intercepts, or by re-writing them in y-intercept form and putting them in the graphing calculator.

4x + 8(0) = 80 4x = 80 x = 20, (20, 0)

4(0) + 8y = 80 8y = 80 y = 10, (0, 10)

x + (0) = 15 x = 15, (15, 0)

(0) + y = 15 y = 15, (0, 15)

The answer to the second part varies. You can pick any part in the double-shaded region, S. Since the lines are solid, those boundary points are good as well.
So 0 hours babysitting and 10 hours at the library, or 3 hours babysitting and 9 hours at the library.

See image below.

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents: Rational Plus Irrational (January 2015)

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams begin next month. At least, that is the plan.

January 2015, Questions 25

36. Ms. Fox asked her class “Is the sum of 4.2 and SQRT(2) rational or irrational?” Patrick answered that the sum would be irrational.
State whether Patrick is correct or incorrect. Justify your reasoning.

The sum of any rational and irrational number is always irrational.
An irrational number has a non-repeating, non-terminating decimal. Adding .2 to the decimal would not make it repeat or terminate.

4.2 + 1.4142135623746.... = 5.6142135623746....

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

June 2016 Integrated Algebra Regents, Part IV

On June 2, 2016, New York State gave a special Integrated Algebra Regents exam, which only seniors and "super seniors" were eligible to take. Those were the students who entered high school prior to the implementation of Common Core. This test follows the older curriculum.

I am currently scoring exams, and while I cannot talk about that specifically, I do have a copy of the questions. The work you see below is mine, written on photocopies of the pages that they gave me for scoring purposes.

A cleaner, fuller (and possibly more clear) explanation of the problems may come at a future date. However, this is likely the last time that this test will be administered. Then again, they have said that before.

Part IV

All questions in this section were open-ended and worth 4 points. A computational, graphing or rounding error cost 1 point. A conceptual error (e.g., using the wrong formula) cost 2 points.

37. On the set of axes below, solve the following system of equations graphically for all values of x and y. State the coordinates of all solutions.

y = x² - 4x - 5
y + 3x = 1

You can put the first equation in your graphing calculator to get the table of values for the parabola.
To put the linear equation into the calculator, you first had to subtract 3x from both sides of the equation. This gives you y = -3x + 1.

The graph is shown below. The points of intersection are (-2, 7) and (3, -8).

See image below:

38. Express in simplest form: (x² + 5x + 6)/(x² - x - 12) : (x² + x - 6)/(2x - 10)

First, flip over the second fraction and change the problem to a multiplication problem.

I worked out the rest on the test paper. This is too complicated for markup language.
Either answer is acceptable: the factored answer or the multiplied answer.

See image below:

39. The length of a rectangle is (3*SQRT(8) + 2) and the width is (2*SQRT(2) + 3).
Express the perimeter of the rectangle in simplest radical form.
Express the area of the rectangle in simplest radical form.

A few notes, and then I'll leave the work to the image.
Perimeter is 2L + 2W or 2(L + W).
Area is L * W.
SQRT(8) is simplest radical form is SQRT(4*2) = SQRT(4)*SQRT(2) = 2*SQRT(2).
Terms with SQRT(2) in them can be added just as terms with 'x' could.

See image below:

How did you do?

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

June 2016 Integrated Algebra Regents, Part III

On June 2, 2016, New York State gave a special Integrated Algebra Regents exam, which only seniors and "super seniors" were eligible to take. Those were the students who entered high school prior to the implementation of Common Core. This test follows the older curriculum.

I am currently scoring exams, and while I cannot talk about that specifically, I do have a copy of the questions. The work you see below is mine, written on photocopies of the pages that they gave me for scoring purposes.

A cleaner, fuller (and possibly more clear) explanation of the problems may come at a future date. However, this is likely the last time that this test will be administered. Then again, they have said that before.

Part III

All questions in this section were open-ended and worth 3 points. A computational or rounding error cost 1 point. A conceptual error (e.g., using the wrong formula) cost 2 points. One of each cost all 3 points, and the answer -- despite the amount of work -- is worth 0. (There are exceptions, where the rubric states a point is awarded for finding a specific checkpoint in the middle of the problem and then not having any correct work afterward.)

34. Ryan bought three bags of mixed tulip bulbs at a local garden store. The first bag contained 7 yellow bulbs, 8 red bulbs, and 5 white bulbs. The second bag contained 3 yellow bulbs, 11 red bulbs, and 6 white bulbs. The third bag contained 13 yellow bulbs, 2 red bulbs and 5 white bulbs. Ryan combined the contents of these three bags into a single container. He randomly selected one bulb, planted it, and then randomly selected another and planted that one. Determine if it is more likely that Ryan planted a red bulb and then another red bulb, or planted a yellow bulb and then a white bulb. Justify your answer.

You needed to find the compound probabilities of two dependent event (without replacement) and compare the two.
There was no guideline in the rubric, and it didn't come up in our discussion, if you only found the number of outcomes and compared the two with the proper justification.

When added together, there were 23 yellow, 21 red and 16 white for a total of 60 bulbs. There would only be 59 when drawing the second bulb.

P(R then R) = (21 / 60) * (20 / 59) = 420/3540
P(Y then W) = (23 / 60) * (16 / 59) = 368/3540
Therefore, it is more likely that he would get red then red.

Note: some students did write these numbers are decimals or percents. Their math was checked for accuracy (no rounding needed), and then the comparison.

See image below:

35. A particular jewelry box is in the shape of a rectangular prism. The box is advertised as having an interior length of 20.3 centimeters, an interior width of 12.7 centimeters, and an interior height of 10.2 centimeters. However, when a customer measures the interior of the box, she finds that the interior height is actually 6.3 centimeters. Upon further examination, she discovers that the bottom of the interior of the box lifts up to reveal a hidden compartment. Find the volume of this hidden compartment to the nearest cubic centimeter.

Rounding matters. The right formula matters.

I was ready to go off on a rant (and I did, on Twitter) because so many students lost 2 of the 3 points because they used the formula for Surface Area. This meant that they had to do a lot more work -- and do it perfectly -- to get a single point. A rounding error and it would be zero.

Why did so many students do Surface Area? My guess: the formula is in the back of the book. Volume = Length X Width X Height is not, likely because it's so easy that every student should know it. But for whatever reasons, these students didn't.

The height of the hidden compartment was 10.2 - 6.3 = 3.9. Volume = 12.7 X 20.3 X 3.9 = 1005.459 = 1005 cm³.
Rounding before you multiplied was a 1-point error.

Alternatively, you could have calculated Volume = 12.7 X 20.3 X 10.2 = 2629.662 and Volume = 12.7 X 20.3 X 6.3 = 1624.283 and then subtracted to get the same answer. More work, but not incorrect.

See image below:

36. Solve algebraically for all values of x that satisfy the equation: x / (x + 4) = 3 / (x + 2)

Cross-multiply the numerators and denominators. Use the Distributive Property.
This gives you x² + 2x = 3x + 12
Subtract everything from the right side to get: x² - x - 12 = 0. (This was good enough for 1 point.)
Factor: (x - 4)(x + 3) = 0
x - 4 = 0 or x + 3 = 0
x = 4 or x = -3

Note: If you made a computational error, you probably had a problem factoring. However, if you used the Quadratic Formula and followed through to get an answer, however complicated it might have been, it would only be a 1-point error. We had a couple of these.

See image below:

How did you do?

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents: Quadratic Equations with Area (August 2016)

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams begin next month. At least, that is the plan.

August 2014, Questions 36

36. A school is building a rectangular soccer field that has an area of 6000 square yards. The soccer field must be 40 yards longer than its width. Determine algebraically the dimensions of the soccer field, in yards.

Area of a rectangle is L * W. This is important: this formula will NOT be given to you because they assume you learned it in or before middle school. Seriously: it's A = L* W.

The width of the field can be called W. That means that the length is 40 more than the width, so L = W + 40.
The Area = L * W = W(W + 40) = 6000.

Distribute the W, and you get W² + 40W = 6000
Subtract 6000 from both sides: W² + 40W - 6000 = 0. You have a quadratic equation that needs to be factored.

The factors of -6000 with a sum of +40 are 100 and -60.
So (W + 100)(W - 60) = 0
Then W + 100 = 0 or W - 60 = 0
And W = -100 or W = 60.

Because we are looking for width, we can discard the negative value, and use the positive value. The width is 60. The length is 60 + 40 = 100.

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

June 2016 Integrated Algebra Regents, Part II

On June 2, 2016, New York State gave a special Integrated Algebra Regents exam, which only seniors and "super seniors" were eligible to take. Those were the students who entered high school prior to the implementation of Common Core. This test follows the older curriculum.

I am currently scoring exams, and while I cannot talk about that specifically, I do have a copy of the questions. The work you see below is mine, written on photocopies of the pages that they gave me for scoring purposes.

A cleaner, fuller (and possibly more clear) explanation of the problems may come at a future date. However, this is likely the last time that this test will be administered. Then again, they have said that before.

Part II

All questions in this section were open-ended and worth 2 points.

31. Jim calculated the area of a triangle to be 51.75 cm². The actual area of the triangle is 53.24 cm². Find the relative error in Jim's calculation of the area to the nearest thousandth.

Relative error is defined as the difference between the two amounts divided by the actual amount. Write the answer as a decimal. Do NOT convert to a percent.

53.24 - 51.75 = 1.49.
1.49 / 53.24 = 0.028 to the nearest thousandth. (Not rounding correctly will cost you one of the two points.)

See image below:

32. A 12-foot ladder is placed against a wall. The ladder makes an angle of 73^o with the floor. Determine, to the nearest tenth of a foot, how high up the wall the ladder will reach.

You are given the hypotenuse and you are looking for the side opposite the given angle. That means that you need to use sine, which is opposite / hypotenuse.

sin 73 = x / 12
So 12 * sin 73 = x. You can put this in your calculator, which must be in degree mode.

x = 11.5, to the nearest tenth. Again, correct rounding is important.

Using the wrong trig function will cost you half credit (one point). Two mistakes mean 0 credits.

See image below:

33. On the sets of axes below, draw the graph of the function y = 3^x. Include the interval -2 < x < 2.

You can make a table of values or just plot points to show your work: (2, 9), (1, 3), (0, 1), (-1, 1/3), (-2, 1/9). The last two would be approximate to graph. This is to be expected. You might want to label them, but the rubric didn't say it was required.

Draw the curve through the points.

See image below:

How did you do?

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents: Quad-Linear Systems (August 2014)

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams begin next month. At least, that is the plan.

August 2014, Questions 35

35. Let f(x)= -2x² and g(x) = 2x - 4. On the set of axes below, draw the graphs of y = f(x) and y = g(x).

Using this graph, determine and state all values of x for which f(x) =g(x).

Using your graphing calculator, find the points from each function.
The linear function, g(x) has a y-intercept of -4 and a slope of +2.
The quadratic function, f(x) is a parabola with the origin as its vertex and it will open downward.

The two lines will intersect twice: at (-2, -8) and at (1, -2). They do NOT ask for the points of intersection. You are asked only for the x values. The x values are -2 and 1.

Believe it or not, if you give coordinates, you could actually lose a point for not answering the question that was asked.

Any questions?

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If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents: Literal Equations (August 2014)

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams begin next month. At least, that is the plan.

August 2014, Questions 34

34. The formula for the area of a trapezoid is A = (1/2)h(b₁ + b₂).Express b₁ in terms of A, h, and b₂.

The area of a trapezoid is 60 square feet, its height is 6 ft, and one base is 12 ft. Find the number of feet in the other base.

You want to isolate the variable we want to solve for, in this case b₁:

A = (1/2)h(b₁ + b₂)
Multiply by 2: 2A = h(b₁ + b₂)
Divide by h: 2A/h = b₁ + b₂
Subtract b₂ from both sides: (2A/h) - b₂ = b₁
b₁ = (2A/h) - b₂

To find the answer to the second part, substitute the values you have into the equation you just created, and evaluate.

b₁ = (2A/h) - b₂ = (2(60)/6) - 12 = 120/6 - 12 = 20 - 12 = 8
The other base is 8 ft.

Any questions?

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If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents: Graphing and Translating Absolute Value (August 2014)

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams begin next month. At least, that is the plan.

August 2014, Questions 33

33. On the axes below, graph f(x) = |3x|.
If g(x) = f(x) - 2, how is the graph of f(x) translated to form the graph of g(x)?
If h(x) = f(x - 4), how is the graph of f(x) translated to form the graph of h(x)?

You can use the graphing calculator to see what the function looks like. Press the MATH key, and then arrow over to the NUM menu and abs( will be the first option. Depending on your calculator's operating system, you will either get abs( written as a function, or you will see the symbols: | |

If you have the symbols, put the variable between them. If you have the function, add the variable and then close the parentheses! This will be important later in the problem.

g(x) will translate the function 2 spaces down. If you subtract two from a number after taking the square root, the value will always decrease by 2. Example: f(0) = 0, g(0) = 0 - 2 = -2.

h(x) will translate the function 4 spaces to the right. Taking the absolute value after subtracting means that any negative results will become positive. Example f(0) = 0, h(0) = f(0-4) = 4, but f(4) = 4, h(4) = f(4-4) = 0. The vertex of the function has shifted four places to the right.

You can look at the graphs of g(x) and h(x) in the calculator to compare.
Remember: if your calculator's operating system shows abs( for absolute value, you must close the parentheses:
y = abs(x) - 2
y = abs(x - 4)

Any questions?

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If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents Review: Pinning the Posts

Due to family obligations, it looks like there won't be a Sunday update.

However, in order to keep the "Daily" tag going, I pinned the earlier posts on Pinterest:

https://www.pinterest.com/cjburke23/algebra-regents-review/

Bookmark it, and you will be able to find all the questions later on.