Saturday, March 25, 2017

Algebra 2 Regents Problems of the Day

The following problems were taken from the ALGEBRA II (Common Core) Regents Exam given on Friday, January 27, 2017.
Previous problems can be found here

Part 1

5. What is the solution to the system of equations y = 3x - 2 and y = g(x) where g(x) is defined by the function below?

(4) {(1,1),(6,16)}
First of all, this is an odd question. Three of the answers make no sense at all. It is plain to see that neither (0, -2) nor (1, 6) are points on g(x), the parabola shown.
Second, it is quite plain that (1, 1) is on that line, and that (1) = 3(1) - 2, so it fits the other equation in the system as well. The choice is (4).
Just to be thorough, how do we know that (6, 16) is a solution to g(x)?
You can see that g(x) has points (0, 4), (1, 1) and (2, 0), which is the vertex. This makes (3, 1) and (5, 4) points by reflection over the line of symmetry, if you couldn't tell from the graph. The parent function is x2 and it has been shifted two spaces to the right.
This means that g(x) = (x - 2)2.
Then g(6) = (6 - 2)2 = 16 and (16) = 3(6) - 2.

6. Which statement about statistical analysis is false?

(3) Observational studies can determine cause and effect relationships.
Experiments do that, not observational studies.




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Geometry Regents Problems of the Day

The following problems were taken from the GEOMETRY (COMMON CORE) Regents Exam given on Thursday, January 26, 2017.
Previous problems can be found here.

Part 1

5. In the diagram below, if triangle ABE = triangle and AEFC is drawn, then it could be proven that quadrilateral ABCD is a

(4) Parallelogram.
First: common sense. If the figure could be proved to be a rectangle or a rhombus, it would have to be a parallelogram as well. You can't have multiple answers. If it were a square, then all four would be true.
Second: figures are not necessarily drawn to scale (even if they don't tell you), so you shouldn't assume that it isn't a square based solely on the picture. Looking at an image isn't proof.

If the triangles are congruent, then by CPCTC (Corresponding parts of congruent triangles are congruent) we know that AB = CD. We also know, for the same reason, that angle BAE = DCF. Those two angles are alternate interior angles along the transversal. That makes AB || CD.
If two sides of a quadrilateral are both parallel and congruent, the shape is a parallelogram.
We do not have any additional information to prove (nor assume!) that the shape is a rhombus.

6. Under which transformation would triangle A'B'C', the image of triangle ABC, not be congruent to ABC?

(4) dilation with a scale factor of 2 centered at the origin.
A dilation increased the size, so it will no longer be congruent.




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Friday, March 24, 2017

(x, why?) Mini: Irregular

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(C)Copyright 2017, C. Burke.

To be fair, 3/5ths of the pentagon feels ''right''.

On the other hand, I won't share this with my class because they'll understand the Geometry portion (I hope), but not the doctor on TV part. I'm not explaining it.




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Thursday, March 23, 2017

Radical

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(C)Copyright 2017, C. Burke.

He's definitely lower than 3, but he's still more than 2.




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Wednesday, March 22, 2017

Algebra 2 Regents Problems of the Day

The following problems were taken from the ALGEBRA II (Common Core) Regents Exam given on Friday, January 27, 2017.
Previous problems can be found here

Part 1

3. When factored completely, m5 + m3 - 6m is equivalent to

(4) m(m2 + 3)(m2 - 2)
Start with: m5 + m3 - 6m
Each term has m, factor it: (m)(m4 + m2 - 6)
Factor into two binomials: What are two factors of -6 that add up to +1: +3 and -2
So it factors into m(m2 + 3)(m2 - 2).

4. If sin2(32°) + cos2(M) = 1, then M equals

(1) 32°.
The rule is: sin2x + cos2x = 1.
In this case x = 32°, so M must be 32° as well.




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Geometry Regents Problems of the Day

The following problems were taken from the GEOMETRY (COMMON CORE) Regents Exam given on Thursday, January 26, 2017.
Previous problems can be found here.

Part 1

3. Given triangle ABC = triangle DEF, which statement is not always true?(

(1) BC = DF.
BC and DF are not corresponding sides. BC corresponds to EF and DF corresponds to AC. They would only be congruent if the triangles were isosceles or equilateral. We don't know whether they are or not.

4. In the diagram below, DE, DF and EF are midsegments of triangle ABC.


The perimeter of quadrilateral ADEF is equivalent to

(4) AB + AC.
FE = DB and DE = FC. So AD + DE + EF + FA = AD + FC + DB + FA.
Rearrange the terms: AD + DB + AF + FC = AB + AC.




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Sunday, March 19, 2017

Algebra 2 Regents Problems of the Day

The following problems were taken from the ALGEBRA II (Common Core) Regents Exam given on Friday, January 27, 2017.

Part 1

1. Relative to the graph of y = 3sin x, what is the shift of the graph of y = 3sin(x + π/3)?

(2) π/3 left.
When the value within the parentheses is added to x, the graph shifts to the left. When the value within the parentheses is subtracted from x, the graph shifts to the right.
To move up or down, the addition or subtraction would have to take place outside the parantheses.

2. A rabbit population doubles every 4 weeks. There are currently five rabbits in a restricted area. If t represents the time, in weeks, and P(t) is the population of rabbits with respect to time, about how many rabbits will there be in 98 days?

(1) 56.
The starting value is 5. The base is 2. The exponent is 98 days / 4 weeks, which is 98 / 28 after converting weeks to days.
That gives a value of approximately 56.5685424949..., which is approximately 56.
(Yes, that would round up to 57, but that is not one of the choices, leaving 56 as the closest.)




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Geometry Regents Problems of the Day

The following problems were taken from the GEOMETRY (COMMON CORE) Regents Exam given on Thursday, January 26, 2017.

Part 1

1. Which equation represents the line that passes through the point (-2,2) and is parallel to y = 1/2 x + 8?

(3) y = 1/2 x + 3.
A parallel line will have the same slope, 1/2, so y = 1/2 x + b.
Next, substitute the coordinates (-2, 2) in place of x and y, giving you

2 = 1/2 (-2) + b
2 = -1 + b
3 = b
so y = 1/2 x + 3

2. In the diagram below, triangle ADE is the image of triangle ABC after a reflection over the line AC followed by a dilation of scale factor (AE/AC) centered at point A.

Which statement must be true?

(2) m∠ABC = m∠ADE.
Neither reflection nor dilation changes the shape of the figure, nor the sizes of the angles, and ∠ABC and ∠ADE are corresponding angles in the two triangles.




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Friday, March 17, 2017

Happy St. Patrick's Day 2017

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(C)Copyright 2017, C. Burke.

Tora Lora Torus just doesn't sound right. But now I might get Pearl Harbor comments

This St. Paddy's Day, may all your non-Irish friends be Green with envy!




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Wednesday, March 15, 2017

Sides

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(C)Copyright 2017, C. Burke.

Followed by the sound of a dagger striking . . .




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Tuesday, March 14, 2017

Pi in the Sky

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(C)Copyright 2017, C. Burke.

Next Up: Pi on Sports!

I almost went with a family trip on an airline instead. Maybe I should make a note of that for the future.




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Thursday, March 09, 2017

(x, why?) Mini: Irreplaceable

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(C)Copyright 2017, C. Burke.

Speaks for itself. I hope.




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