Wednesday, June 29, 2016

June 2016 Geometry Regents (Common Core), Parts 3 and 4

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

Part II is posted here.

June 2016 Geometry Regents, Part III

32. A barrel of fuel oil is a right circular cylinder where the inside measurements of the barrel are a diameter of 22.5 inches and a height of 33.5 inches. There are 231 cubic inches in a liquid gallon. Determine and state, to the nearest tenth, the gallons of fuel that are in a barrel of fuel oil.

The Volume of a cylinder is pi*r2*h. So V = pi(11.25)2(33.5) = 13319.9
To convert cubic inches into gallons, divide by 231: 13319.9 / 231 = 57.66...
Answer: 57.7 gallons.

33. Given: Parallelogram ABCD, EFG, and diagonal DFB


Prove: Triangle DEF ~ triangle BGF

You could write a two-column proof or a paragraph proof. For this blog, paragraph is a little easier.

Angle DFE is congruent to Angle BFG because they are vertical angles. AD is parallel to BC because opposite sides of a parallelogram are parallel. BD is a transversal. Angle ADB is congruent to CBD because they are alternate interior angles. Therefore Triangle DEF ~ triangle BGF because of AA (Angle-Angle Theorem).

34. In the diagram below, Triangle A'B'C' is the image of Triangle ABC after a transformation.


Describe the transformation that was performed.
Explain why Triangle A'B'C' ~ Triangle ABC.

The transformation was a Dilation of scale factor 2.5 centered on (0, 0). Point A(-2, 4) -> A'(-5, 10). Point B(-2, -4) -> B'(-5, -10). Point C(4, -4) -> C'(-10,-10).
-5/-2 = 2.5. 10/4 = 2.5. -10/-4 = 2.5. 10/4 = 2.5

Dilations preserve shape so the angles are the same size. Therefor the triangles are similar.

June 2016 Geometry Regents, Part IV

35. Given: Quadrilateral ABCD with diagonals AC and BD that bisect each other, and < = <2

Prove: Triangle ACD is an isosceles triangle and triangle AEB is a right triangle

Make sure you restate the Given information. Make sure that you restate what you want to prove as your final statement. (In this case, you are proving two things, so one of them will be in the middle, right before you start proving the second half.) Do NOT use what you are trying to prove and a reason why something must be true.

Statement Reason
1. AC and BD bisect each other. Given
2. <1 = <2 Given
2.5 ABCD is a parallelogram a quadrilateral with diagonals that bisect each other is a parallelogram
(edited to add missing step)
3. AB || CD Opposite sides of a parallelogram are parallel.
4. Angle DCA = Angle 1 Alternate interior angles
5. Angle DCA = Angle 2 Transitive Property of Congruence
6. Triangle ACD is isosceles If the base angles of a triangle are congruent, then the triangle is Isosceles
7. AD = CD The sides opposite congruent base angles of an isosceles triangle are congruent
8. ABCD is a rhombus A parallelogram with consecutive sides congruent is a rhombus
9. Angle AEB is a right angle. Diagonals of a rhombus are perpendicular.
10. Triangle AEB is a right triangle. A triangle with a right angles is a right triangle.

36. A water glass can be modeled by a truncated right cone (a cone which is cut parallel to its base) as shown below.


The diameter of the top of the glass is 3 inches, the diameter at the bottom of the glass is 2 inches, and the height of the glass is 5 inches.
The base with a diameter of 2 inches must be parallel to the base with a diameter of 3 inches in order to find the height of the cone. Explain why.
Determine and state, in inches, the height of the larger cone.
Determine and state, to the nearest tenth of a cubic inch, the volume of the water glass.

To find the height, you need to have similar triangles. To have similar triangles, the bases must be parallel so that the base angles are congruent.

The radius of the large circle is 1.5. The radius of the smaller circle is 1. The height to the smaller circle is x. The height to the larger circle is x + 5.

Set up a proportion: x / 1 = (x + 5) / 1. 5
1.5x = x + 5
.5x = 5
x = 10
The larger cone is 15 inches.

The Volume of a cone = (1/3)(pi)r2h
The Volume of the water glass = (1/3)(pi)r12h1 - (1/3)(pi)r22h2
= (1/3)pi(1.5)2(15) - (1/3)pi(1)2(10)
= 35.343 - 10.472 = 24.871 = 24.9 cubic inches

Saturday, June 25, 2016

Common Core Geometry, Part 2, June 2016

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

June 2016 Geometry Regents, Part II

25. Describe a sequence of transformations that will map triangle ABC onto triangle DEF as shown below.

There are multiple possible answers. One possibility is a reflection over the x-axis, followed by a translation 4 units to the right. (The reverse order would work, too.)

26. Point P is on segment AB such that AP:PB is 4:5. If A has coordinates (4,2), and B has coordinates (22,2), determine and state the coordinates of P.

The distance between the two x-coordinates is 18 units. If AP:PB is 4:5, then the two segments are 4x and 5x in length, and 4x + 5x = 18. So 9x = 18, and x = 2.
4(2) = 8, 4 + 8 = 12. P is at (12, 2).

27. In triangle CED as shown below, points A and B are located on sides CE and ED, respectively. Line segment AB is drawn such that AE = 3.75, AC = 5, EB = 4.5, and BD = 6.
Explain why AB is parallel to CD.

As shown in the illustration below, if you can show that the sides of the smaller triangle and the larger triangle are proportional, then the triangles are similar. If they are similar, then the corresponding angles of the two triangles are congruent.

Angle EAB is congruent to angle ECD and they are corresponding angles on a transversal. Therefore, AB is parallel to CD.

28. Find the value of R that will make the equation sin 73° = cos R true when 0° < R < 90°.
Explain your answer.

The sine of an angle is equal to the cosine of the complementary angle. 90 - 73 = 17.
R = 17.
EDIT Arithmetic error corrected. Minus 1 for me.

29. In the diagram below, Circle 1 has radius 4, while Circle 2 has radius 6.5. Angle A intercepts an arc of length pi, and angle B intercepts an arc of length 13*pi / 8.


Dominic thinks that angles A and B have the same radian measure. State whether Dominic is correct or not. Explain why.

The measure of angle A is arclength / radius, which is pi / 4 radians.
The measure of angle B is 13(pi)/8 / 6.5, which is (13 pi) / (8*6.5) = (13 pi) / 52 = pi / 4.
Angle A and B have the same measure.

Another approach, which is a little more old-school, is to use circumference to find arclength.

The circumference of a circle is pi*d or 2*pi*r.
The length of an arc of a circle is the circumference times the size of the angle/360.

In the first circle, (A / 360) (2) (4) (pi) = pi
So (A / 360) (2) (4) = 1
Use inverse operations to isolate A, and A = 45.

In the second circle, (B / 360) (2) (6.5) (pi) = 13(pi) / 8
So (B / 360) (2) (6.5) = 13 / 8
Use inverse operations to isolate B, and B = 45.

So A and B have the same measure and that is 45.

30. A ladder leans against a building. The top of the ladder touches the building 10 feet above the ground. The foot of the ladder is 4 feet from the building. Find, to the nearest degree, the angle that the ladder makes with the level ground.

The ladder makes a right triangle. The wall is the opposite side. The ground is the adjacent side. The ladder is the hypotenuse, but we don't know or need to find the length of the ladder. So we have opposite and adjacent, so we are using tan. More specifically, since we are looking for the size of the angle, we need tan-1.

tan(x) = 10/4, so x = tan-1 (10/4) = 68 degrees.

31. In the diagram below, radius OA is drawn in circle 0. Using a compass and a straightedge, construct a line tangent to circle 0 at point A. [Leave all construction marks.]

Sorry, but I'm still not good with handling constructions electronically.

Use the straightedge to extend the radius. From the point A, measure off two points on other side. Then construct a perpendicular bisector. Let's call these first two marks B and C. Go to B and make an arc above and below the line. Do the same at C so that you make two X's. Draw the perpendicular bisector, it will be tangent to the circle.


End of Part II

Any questions?

Friday, June 24, 2016

June 2016 Common Core Algebra 1 Regents, Parts 3 and 4

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.

Part I is posted here.
Part II is posted here.

June 2016 Algebra Regents, Part III

33. The height, H, in feet, of an object dropped from the top of a building after t seconds is given by H(t) = -16t2 + 144.
How many feet did the object fall between one and two seconds after it was dropped?
Determine, algebraically, how many seconds it will take for the object to reach the ground.

h(1) = -16(1)2 + 144 = 128
h(2) = -16(2)2 + 144 = 80
h(2) - h(1) = 128 - 80 = 48 feet between the 1st and 2nd second.

Solve for h(t) = 0.

-16t2 + 144 = 0
-16(t2 - 9) = 0
-16(t + 3)(t - 3) = 0
t = -3 or t = 3

It takes 3 seconds to reach the ground.

34. The sum of two numbers, x and y, is more than 8. When you double x and add it to y, the sum is less than 14.
Graph the inequalities that represent this scenario on the set of axes below.
Kai says that the point (6,2) is a solution to this system. Determine if he is correct and explain your reasoning.

The first inequality you need to graph is x + y > 8. The second one is 2x + y < 16.
The graph looks like this:

(graph will be uploaded later)

Looking at the graph, Kai is incorrect because (6, 2) is on a broken line which is not part of the solution set.

Note that if you drew the graph with solid lines, you lost a point for that. However, Kai would have been correct according to that mistaken graph. You have to be consistent.

35. An airplane leaves New York City and heads toward Los Angeles. As it climbs, the plane gradually increases its speed until it reaches cruising altitude, at which time it maintains a constant speed for several hours as long as it stays at cruising altitude. After flying for 32 minutes, the plane reaches cruising altitude and has flown 192 miles. After flying for a total of 92 minutes, the plane has flown a total of 762 miles.
Determine the speed of the plane, at cruising altitude, in miles per minute.
Write an equation to represent the number of miles the plane has flown, y, during x minutes at cruising altitude, only.
Assuming that the plane maintains its speed at cruising altitude, determine the total number of miles the plane has flown 2 hours into the flight.

There was much discussion over this question, but in the end, there was no arguing with the rubric for scoring this question. If you feel it's an unfair question, appeal to the state. In the meantime...

To find the speed at cruising altitude, use the two points given (32 minutes, 192 miles) and (92 minutes, 762 miles).
Speed in Miles per minute = changes in distance (miles) / change in time (minutes)
(762 - 192) / (92 - 32) = 9.5
If you showed your work, you have 1 point already.

The second part was where many students got caught up.
The word only was meant to apply to both the x and the y values, not just the x. In other words, you did not need to account for the distance traveled before reaching cruising altitude.
Because of this, the correct equation was y = 9.5x.
If you included "+ 192", you didn't get credit.

For the last part, you not only need to remember the initial 192 miles, but also the first 32 minutes of the flight. The question states that it is 2 hours into the flight, not 2 hours at cruising altitude. Also remember that you are dealing with miles per minute, so you need to convert.

2 hours = 120 minutes
120 - 32 = 88 minutes at cruising altitude
y = 9.5(88) + 192 = 836 + 192 = 1028 miles

36. On the set of axes below, graph:


How many values of x satisfy the equation f(x) = g(x)? Explain your answer, using evidence from your graphs.

(Graph will be posted later)

It is important that you had a break in the line at x = -1. The linear portion ends, the quadratic portion needed to have an open circle.
Most of the mistakes I saw fit into these categories:

  • Draw three equations across the entire plane
  • Connecting the two parts of the piecewise function
  • Forgetting the open circle
  • Shading the graph like it was an inequality
  • Graphing a broken line for the quadratic because of the greater than symbol
  • Saying that there were no solutions because the three lines didn't intersect at a single point (they didn't have to)

There was one solution because f(x) and g(x) only intersect one time.
You did not have to give the coordinates of the solution, and the solution was NOT a proper explanation. Seriously. You had to reference the fact that the lines only cross/intersect one time so there is one solution.

Also, if you had a graphing error, your final answer had to match the graph you drew. If, for example, you graph g(x) = -1/2x + 1, that would be one graphing error, but that line would intersect f(x) two times. Your answer had to match your graph.

June 2016 Algebra Regents, Part IV

37. Franco and Caryl went to a bakery to buy desserts. Franco bought 3 packages of cupcakes and 2 packages of brownies for $19. Caryl bought 2 packages of cupcakes and 4 packages of brownies for $24. Let x equal the price of one package of cupcakes and y equal the price of one package of brownies.
Write a system of equations that describes the given situation.
On the set of axes below, graph the system of equations.
Determine the exact cost of one package of cupcakes and the exact cost of one package of brownies in dollars and cents. Justify your solution.

The two equations were

3x + 2y = 19
2x + 4y = 24

Note: if you used c and b instead of x and y, you lost 1 point because the instructions said to use x and y.

(The graph will be loaded later.)

You could solve the system of equations using elimination. You could also solve them using the functions on the graphing calculator, but you needed to explain how you got your answer. A correct pair of answers without an explanation or procedure was only 1 point instead of 2.

Multiply the first equation by 2 and subtract

6x + 4y = 38
2x + 4y = 24
4x = 14
x = 3.50

2(3.50) + 4y = 24
7.00 + 4y = 24
4y = 17
y = 4.25

Cupcakes were 3.50 and brownies were 4.25.

June 2016 Common Core Algebra 1 Regents, Part 1

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.

Part II is posted here.

June 2016 Algebra Regents, Part I

1. The expression x4 - 16 is equivalent to

(3) (x2 + 4)(x2 - 4). Difference of Squares Rule
Additionally, a +4 times -4 gives you - 16, and there's only one choice.
If you were really stuck, you could have multiplied all the choices or put them in your graphing calculator and looked for a match

2. An expression of the fifth degree is written with a leading coefficient of seven and a constant of six. Which expression is correctly written for these conditions?

(4) 7x5 + 2x2 + 6. "Fifth degree is written with a leading coefficient of seven" means 7x5.
Also, only one choice had a constant of 6.

3. The table below shows the year and the number of households in a building that had high-speed broadband internet access. (table omitted)
For which interval of time was the average rate of change the smallest?

(1) 2002-2004. Average rate of change is (difference is number of households) / (difference in years).
In each case, the number of years is 2, so you only need to find the smallest numerator.
23 - 11 = 12. 33 - 16 = 17. 42 - 23 = 19. 47 - 33 = 14.
Choice 1 is smallest.

4. The scatterplot beow compares the number of bags of popcorn and the number of sodas sold at each performance of the circus over one week. (image omitted)
Which conclusion can be drawn from the scatterplot?

(2) There is a positive correlation between popcorn sales and soda sales. As one rises, so does the other.

5. The Celluloid Cinema sold 150 tickets to a movie. Some of these were child tickets and the rest were adult tickets. A child ticket cost $7.75 and an adult ticket cost $10.25. If the cinema sold $1470 worth of tickets, which system of equation could be used to determine how many adult tickets, a, and how many child tickets, c, were sold?

(1) a + c = 150; 10.25a + 7.75c = 1470. The number of adult (a) + the number of child (c) tickets is (=) 150.
The amount of money from adult tickets is 10.25 times a. The amount of money from child tickets is 7.75 times c.

6. The tables below (image omitted) show the values of four different functions for given values of x.
Which table represent a linear function?

(1) f(x). A linear function will have a constant rate of change, slope, common difference, etc. 19-12 = 7, 26-19 = 7, 33-26 = 7 (x increases by 1 in each case).

7. The acidity in a swimming pool is considered normal if the average of three pH readings, p, is defined such that 7.0 < p < 7.8. If the first two reading are 7.2 and 7.6, which value for the third reading will result in an overall rating of normal?

(2) 7.3. Here is why Number Sense is important. If all three numbers are between 7.0 and 7.8, then the average must be as well. Common sense says that 7.3 will result in an average between those two extremes.

That being said, an average is found by adding the three numbers and dividing by three (the number of data items).
The averages are 7.0, 7.4, 7.8, 7.9. Only the second one is within the compound inequality.

Calculator trick: 7.2 + 7.6 = 14.8, right? Try this in your graphing calculator:

(14.8 + {6.2,7.3,8.6,8.8})/3 [ENTER]

The parentheses and the brackets are important. No spaces after the commas.

8. Dan took 12.5 seconds to run the 100-meter dash. He calculated the time to be approximately
(1) 0.2083 minute. The 100 is irrelevant. They didn't ask about his speed, just his time. To convert seconds to minutes, divide by 60.
12.5 seconds is less than one minute so 750 minutes is silly (and really slow: 6 1/2 hours to finish the race!)
Also, .2 and .5 hours are well over one minute, so they can be crossed out.

9. When 3x + 2 < 5(x - 4) is solved for x, the solution is

(4) x > 11.

3x + 2 < 5(x - 4)
3x + 2 < 5x - 20
2 < 2x - 20
22 < 2x 11 < x
x > 11

If you graphed y = 3x + 2 and y = 5(x - 4), you would have found the intersection point at x = 11, which, in this case, would have eliminated 3 choices.

10. The expression 3(x2 - 1) - (x2 - 7x + 10) is equivalent to

(2) 2x2 + 7x - 13. 3x2 - x2 = 2x2, which is in all the choices.
0 - (-7x) = +7x, which eliminates (1) and (3).
3(-1) - 10 = -13. Choice (2).

11. The range of the function f(x) = x2 + 2x - 8 is all real numbers

(2) greater than or equal to -9. Range is the y values on the graph. You can put this in your graphing calculator and see the answer.
Or you can find the vertex. X = -b / (2a) = -2/(2*1) = -1, and f(-1) = (-1)2 + 2(-1) - 8 = -9.
The parabola opens up, so the vertex is a minimun and all other f(x) must be higher than that.

12. The zeroes of the function f(x) = x2 - 5x - 6 are

(1) -1 and 6. Again, put it in your calculator and look at the graph and the Table of Values. Or factor it.

x2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6 or x = -1

13. In a sequence, the first term is 4 and the common difference is 3. The fifth term of this sequence is

(3) 16. Either write f(n) = 3(n - 1) + 4, f(5) = 3(5-1) + 4 = 16, or just write the sequence: 4, 7, 10, 13, 16.

14. The growth of a certain organism can be modeled by C(t) = 10(1.029)24x, where C(t) is the total number of cells after t hours. Which function is approximately equivalent to C(t)?

C(t) = 10(1.986)t.
First of all, the graph is exponential growth. Two of the choices have a base of .083, which is exponential decay. So choices (1) and (2) are out.
At this point, you can put the original equation and choices (3) and (4) into your graphing calculator.

Looking at choice (3), the only differences are that the base has changed and the exponent is t instead of 24t. That means that if 1.029 to the 24th power is 1.986, then this is the correct answer. If you put that in your calculator, you see that it is the same.

15. A public opinion poll was taken to explore the relationship between age and support for a candidate in an election. The results of the poll are summarized in the table below. (Table omitted).
What percent of the 21-40 age group was for the candidate?

(4) 60. The entire age group is 30 + 12 + 8 = 50 people. "For" was 30 out of 50, and 30/50 = .6 = 60%.

16. Which equation and ordered pair represent the correct vertex from and vertex for f(x) = x2 - 12x + 7?

(3) f(x) = (x - 6)2 - 29. First, (x - 6) means that h, the x-coordinate of the vertex, is 6, so eliminate choices (2) and (4).
Next, either complete the square or put the equations in the graphing calculator! Seriously.
To complete the square do the following:

f(x) = x2 - 12x + 7
f(x) + 36 = x2 - 12x + 36 + 7
f(x) + 36 = (x - 6)2 + 7
f(x) = (x - 6)2 - 29

17. A student invests $500 for 3 years in a savings account that earns 4% interest per year. No further deposits or withdrawals are made during this time. Which statement does not yield the correct balance in the account at the end of 3 years?

(2) 500(1 - .04)3. It should be obvious that choice (2) shows decay and not growth so it cannot be the answer.
Choice (1) is the standard way to write it. Choice (3) is what you get if you expand the exponent of 3 -- three factors of (1 + .04). Choice (4) is what you get if you use the simple interest formula and add the interest year after year.

18. The line represent by the equation 4y + 2x = 33.6 shares a solution point with the line represent by the table below. (Table omitted)
The solution for this system is

(4) (6.0, 5.4). First off, eliminate choices (2) and (3). The y values 5.0 and 4.6 are in the table with different x values, so neither can be the intersection point.
Next, you need to find the equation of the line. At this point, AGAIN, I will remind you that you can use your calculator to give you the equation. Put two sets of points into Lists and do a Linear Regression to get the equation of the line. y = .2x + 4.2.

Check .2(-14) + 4.2 = 1.4, no good, and .2(6) + 4.2 = 5.4, which checks.

If you didn't do a linear regression, first find the slope of the line:
Using (2, 4.6) and (4, 5): (5 - 4.6) / (4 - 2) = .4 / 2 = .2, the slope of the line is .2, so y = .2x + b.
Next find b: 5 = .2(4) + b
5 = .8 + b
b = 4.2
So y = .2x + 4.2
Now check which choices work. If one choice works, double check that it works for 4y + 2x = 33.6.

19. What is the solution of the equation 2(x + 2)2 - 4 = 28?

(3) 2 and -6. Test-taking Strategy: Try 6 first: Put 2(6 + 2)2 - 4 in the caculator. It does not equal 28. Cross out (1) and (4). You know now that "2" has to be a solution, so you only need to check (-6).
2(-6 + 2)2 - 4 = 28. Check.

Or use the graphing calculator! Subtract 28 from both sides and you have 2(x + 2)2 - 4 - 28 = 0.
Graph y = 2(x + 2)2 - 32 and look for the zeroes!

20. The dot plot below represents the number of pets owned by students in the class (image omitted)
Which statement about the data is true?

(3) The mean is 3. If you analyze the data, there are 20 dots. The 10th and 11th piece of information are both 3, so the median is 3. Split the data and find Q1 and Q3 (which are, basically, the median of the lower half and the median of the upper half). Q1 = 2 and Q3 = 4, so the interquartile range is 4 - 2 = 2. Choices (1) and (2) are eliminated.
An outlier is a piece of data so far away from the rest of the data that it is discounted (maybe it was a mistake in data entry?) so that it doesn't sway (or "skew") the results. There are no points way off (say, at 10 or 11). So choice (4) is eliminated. That leaves choice (3).

If you add up the values represented by each of the 20 dots and then divide by 20, you will get 2.75, not 3.

Once again, you can use the calculator. If you use list L1, you can enter the 20 pieces of data: 0, 0, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, and do 1-Var Stats. You win get the mean, Q1, median and Q3. (You can then calculate IQR.) And you can see the box-and-whisker plot, if you wanted to, and Trace the points.

21. What is the largest integer, x, for which the value of f(x) = 5x4 + 30x2 + 9 will be greater than the value of g(x) = 3x?

(3) 9. Calculate the values of f(x) and g(x) for 7, 8, 9 and 10.
Or put the equations in the graphing calculator and see which is the last time f(x) is bigger by checking the Table of values. (The graph is really big. You won't see it by looking at the graph itself.

22. The graph of the functions f(x) = |x - 3| + 1 and g(x) = 2x + 1 are shown. Which statement about these functions is true?

(2) The solution to f(x) = g(x) is 1. When x = 1, f(x) = 3 and g(x) = 3. The solution is the value of x that makes it happen, not the function values.
You may have noticed that choices (1) and (4) essentially say the same thing.

23. A store sells self-serve yogurt sundaes. The function C(w) represents the cost, in dollars, of a sundae weighing w ounces. An appropriate domain for the function would be

(4) nonnegative rational numbers. Negative ounces do not make sense. (Zero ounces cost zero, so that's okay.) You don't have to buy an integer value of yogurt in your sundae. It would be difficult to imagine how they could scoop it out exactly.

24. Sara was asked to solve this word problem: "The product of two consecutive integers is 156. What are the integers?"
What type of equation should she create to solve this problem?

(2) quadratic. "The product of two consecutive numbers" would be represented by (N)(N + 1), which will create a quadratic equation.


End of Part I

Who did you do? Any questions?


Thursday, June 23, 2016

Coffee For Jenny

(Click on the comic if you can't see the full image.)
(C)Copyright 2016, C. Burke.

Like anyone remembers phone numbers any more.




Come back often for more funny math and geeky comics.




Friday, June 17, 2016

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.

Thursday, June 16, 2016

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.

Wednesday, June 15, 2016

Silence of the Pupils

(Click on the comic if you can't see the full image.)
(C)Copyright 2016, C. Burke.

Let's be real: when are the pupils ever silent? They had an absolute party yesterday while they were (supposed to be) taking a state exam!

Sometimes people ask me about the inspiration for certain comics. For this one: I haven't a clue. It was just in my head. On the other hand, I have been to get them comfortable using the graphing calculator for anything they need it to do.




Come back often for more funny math and geeky comics.




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?


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


Tuesday, June 14, 2016

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 = 50o and m<DIG = 115o, 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?


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


Monday, June 13, 2016

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?


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.


Negative Times Negative

(Click on the comic if you can't see the full image.)
(C)Copyright 2016, C. Burke.

It's not true that there's no homework.

The first time I said this in class was probably in my first month of teaching. I find a way to use it every year.

Funny thing: I started plotting this one out a month ago. (You might've noticed the lack of comics lately.) And I realized that I needed to give the teacher a last name so that they students could address her properly. And I decided on one -- and consulted a person about it.

And then I finished the dialogue boxes and realized that I'd forgotten to use her name.

Well, it'll come up sometime in the future -- soon -- and I'll give you the story. Bonus: why it's Barbara in this comic and not Annie.




Come back often for more funny math and geeky comics.




Sunday, June 12, 2016

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.


Saturday, June 11, 2016

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.


Friday, June 10, 2016

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.


Thursday, June 09, 2016

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.


Wednesday, June 08, 2016

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 = 118o, and m<LNO = 22o.
Explain why m<NLO is 40 degrees.

The opposite angles of a parallelogram are congruent. Therefore, m<O = 118o.
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.


Tuesday, June 07, 2016

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.


Monday, June 06, 2016

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.


Sunday, June 05, 2016

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.


Saturday, June 04, 2016

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.


Friday, June 03, 2016

Yoda, Math Teacher

(Click on the comic if you can't see the full image.)
(C)Copyright 2016, C. Burke.

Yoda Man! Yoda Real MVP! Yoda -- except that wouldn't work in Yoda-speak. Sigh.

It's not y.




Come back often for more funny math and geeky comics.




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.


Thursday, June 02, 2016

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 73o 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 = 3x. 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.


Wednesday, June 01, 2016

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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