Wednesday, April 25, 2018

Algebra 2 Problems of the Day (open-ended)

Continuing with daily Algebra 2 questions and answers.

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January 2018, Part II

Questions in Part II are worth 2 credits. All work must be shown or explained for full credit. A correct numerical answer without work is only worth 1 credit.


31.The zeros of a quartic polynomial function h are 1, +2, and 3.
Sketch a graph of y = h(x) on the grid below.

Answer:
They are asking for a "sketch", not an exact graph, because they did not give you the actual function. There are an infinite number of quartic (fourth power) graphs that have these zeroes.
You can enter the following into your graphing calculator to see an example:

y = (x + 2)(x + 1)(x - 2)(x - 3)

You can also multiply that by any coefficient, positive or negative. As always, a positive multiplier means that the graph opens upward, and a negative means that it will open downward. Either is acceptable.
Here is one possible answer:

Note that you needed to have the zeroes in the correct places for full credit. If you flipped the signs, for example, you would lose a credit.



32. 2 Explain why 813/4 equals 27.

Answer:

The fourth root of 81 is 3 and 3 to the third power is 27.
You can write it as 81 = (3 * 3 * 3 * 3)3/4 = (3 * 3 * 3) = 27 and add an explanation.

Reminder "Explain" means that you have to write words to "explain" (sorry about the repetition, except that I'm not). "Justify" means that you can just write equations as proof, but "explain" means at least a sentence.



Comments and questions welcome.

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Tuesday, April 24, 2018

Algebra 2 Problems of the Day (open-ended)

Continuing with daily Algebra 2 questions and answers.

More Algebra 2 problems.

January 2018, Part II

Questions in Part II are worth 2 credits. All work must be shown or explained for full credit. A correct numerical answer without work is only worth 1 credit.


29. Researchers in a local area found that the population of rabbits with an initial population of 20 grew continuously at the rate of 5% per month. The fox population had an initial value of 30 and grew continuously at the rate of 3% per month.
Find, to the nearest tenth of a month, how long it takes for these populations to be equal.

Answer:
The equation for rabbit population is y = 20e.05x.
The equation for fox population is y = 30e.03x.
Set them equal to each other 20e.05x = 30e.03x
Divide by 30: (20/30)e.05x = e.03x
Divide by the e term: (2/3) = (e.03x)/(e.05x)
Simplify the exponent: 2/3 = e-.02x
ln (2/3) / -.02 = -.02x / -.02
20.273... = x
20.3 months.

You also could have graphed the two equations, found the point of intersection, record that point of intersection on your exam paper as an ordered pair, and then state the answer of 20.3 months based on the x coordinate.



30. Consider the function h(x) = 2sin(3x) + 1 and the function q represented in the table below.


Determine which function has the smaller minimum value for the domain [-2,2]. Justify your answer.

Answer:
The minimum for q(x) is -8, according to the table. You don't need to work out what the equation is.
The minimum for h(x) is -1. You don't need to explain how you know this as it would likely be assumed that you got it from the calculator, or you just knew that y = 2 sin(3x) has a minimum of -2 and a maximum of 2, so if you add 1 to those numbers, the minimum is -1.
So q(x) has the smaller minimum. Don't forget to mention this. It isn't enough to just state the two minimums. But you DO need to state the two minimums.



Comments and questions welcome.

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Monday, April 23, 2018

Boyfriend

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(C)Copyright 2018, C. Burke.

Because it is a high school, and these things happen, even when I can't work math into it.




Come back often for more funny math and geeky comics.




Algebra 2 Problems of the Day (open-ended)

Continuing with daily Algebra 2 questions and answers.

More Algebra 2 problems.

January 2018, Part II

Questions in Part II are worth 2 credits. All work must be shown or explained for full credit. A correct numerical answer without work is only worth 1 credit.


27.A formula for work problems involving two people is shown below.

1/t1 + 1/t2 = 1/tb
t1 = the time taken by the first person to complete the job
t2 = the time taken by the second person to complete the job
tb = the time it takes for them working together to complete the job

Fred and Barney are carpenters who build the same model desk. It takes Fred eight hours to build the desk while it only takes Barney six hours. Write an equation that can be used to find the time it would take both carpenters working together to build a desk.
Determine, to the nearest tenth of an hour, how long it would take Fred and Barney working together to build a desk.

Answer:
The equation needed to solve this would be

1/8 + 1/6 = 1/tb or 1/8 + 1/6 = 1/x

Once you have this, you can solve for x (or tb).

3/24 + 4/26 = 1/x
7/24 = 1/x
x = 24/7 = 3.428571... = 3.4 hours

Alternatively, multiply the entire equation by (8)(6)(x) to eliminate the fractions:

(8)(6)(x)(1/8) + (8)(6)(x)(1/6) = (8)(6)(x)(1/x)
6x + 8x = 48
14x = 48
x = 48 / 14 = 3.4 hours.

Also of note, whoever wrote this questions obviously likes The Flintstones.



28.Completely factor the following expression:

x2 + 3xy + 3x3 + y

Answer:
First, rewrite the expression as 3x3 + x2 + 3xy + y
Factor by grouping: x2(3x + 1) + y(3x + 1)
Factor again: (x2 + y)(3x + 1)



Comments and questions welcome.

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Sunday, April 22, 2018

Algebra 2 Problems of the Day (open-ended)

Continuing with daily Algebra 2 questions and answers.

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January 2018, Part II

Questions in Part II are worth 2 credits. All work must be shown or explained for full credit. A correct numerical answer without work is only worth 1 credit.
Elizabeth tried to find the product of (2 + 4i) and (3 - i), and her work is shown below.


(2 + 4i)(3 - i)
= 6 - 2i + 12i - 4i2
= 6 + 10i - 4i2
= 6 + 10i - 4(1)
= 6 + 10i - 4
= 2 + 10i

Identify the error in the process shown and determine the correct product of (2 + 4i) and (3 - i).

Answer:
Elizabeth replaced i2 with 1 instead of -1.
The correct answer is:


= 6 + 10i - 4(-1)
= 6 + 10i + 4
= 10 + 10i




26.A runner is using a nine-week training app to prepare for a “fun run.” The table below represents the amount of the program completed, A, and the distance covered in a session, D, in miles.

Based on these data, write an exponential regression equation, rounded to the nearest thousandth, to model the distance the runner is able to complete in a session as she continues through the nine-week program.

Answer:

Enter the data into two lists (L1 and L2, most likely). Check for errors.
Go to STAT, CALC and select ExpReg.
You should get the following output:
y = a*b^x
a = 1.223034549
b = 2.652024589
Round these numbers to the nearest thousandth. (You will lose a point if you do not round correctly.)
y = 1.223(2.652)x.
You could have used A for x and D for y in your answer.



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Saturday, April 21, 2018

Algebra 2 Problems of the Day

Continuing with daily Algebra 2 questions and answers.

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January 2018
23. If the function g(x) = acx represents exponential growth, which statement about g(x) is false?

(1) a > 0 and b > 1
(2) The y-intercept is (0, a).
(3) The asymptote is y = 0.
(4) The x-intercept is (b, 0).

Answer: (4) The x-intercept is (b, 0).
The function has no x-intercept. And when x = b, then g(x) = abb, not 0.
Note that choice (3) and (4) are mutually exclusive, so one of them has to be false.
Since it is exponential growth, a > 0 and b > 1. And when x = 0, g(x) = ab0 = a.



24.At her job, Pat earns $25,000 the first year and receives a raise of $1000 each year. The explicit formula for the nth term of this sequence is an = 25,000 + (n - 1)1000. Which rule best represents the equivalent recursive formula?
(1) an = 24,000 + 1000n
(2) an = 25,000 + 1000n
(3) a1 = 25,000, an - 1 + 1000
(4) a1 = 25,000, an + 1 + 1000

Answer: (3) a1 = 25,000, an - 1 + 1000
In the recursive formula, each term is the sum of the term before it (an-1) plus 1000, which is choice (3).
Note that choices (1) and (2) are not recursive formulas.



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Friday, April 20, 2018

Algebra 2 Problems of the Day

Continuing with daily Algebra 2 questions and answers.

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January 2018
21. What is the inverse of f(x) = -6(x - 2)?

(1) f-1(x) = -2 - x/6
(2) f-1(x) = 2 - x/6
(3) f-1(x) = 1 / (-6(x - 2))
(4) f-1(x) = 6(x - 2)

Answer: (2) f-1(x) = 2 - x/6
Inverse operations. Divide by negative six, then add two.
x = -6(f-1(x) - 2)
x / (-6) = f-1(x) - 2
2 - x/6 = f-1(x).



22. Brian deposited 1 cent into an empty non-interest bearing bank account on the first day of the month. He then additionally deposited 3 cents on the second day, 9 cents on the third day, and 27 cents on the fourth day. What would be the total amount of money in the account at the end of the 20th day if the pattern continued?
(1) $11,622,614.67
(2) $17,433,922.00
(3) $116,226,146.80
(4) $1,743,392,200.00

Answer: (2) $17,433,922.00
Do not answer the 20th term in the geometric sequence. They are looking for the sum of the first 20 terms.
The formula for finding the sum of the first n terms in a geometric sequence is

Sn = (a1(1 - rn)) / (1 - r),

where n is the number of terms, r is the common ratio, and a1 is the initial term.
In this question, the common ratio is 3, because the sequence goes 1, 3, 9, 27 ...
So the sum is (1 * (1 - 320)/(1 - 3) = 1743392200, which is the number of cents. Divide this by 100 to convert it to dollars, or $17,433,922.00.
Alternatively, you could have used .01 for the initial term in the formula, which would have given you the answer immediately.



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Thursday, April 19, 2018

Algebra 2 Problems of the Day

Continuing with daily Algebra 2 questions and answers.

More Algebra 2 problems.

January 2018
19. If p(x) = 2x3 - 3x + 5, what is the remainder of p(x) : (x - 5)?

(1) -230
(2) 0
(3) 30
(4) 240

Answer: (4) 240
The Polynomial Remainder Theorem tells us that is p(x) is divided by (x - r), then the remainder, R, can be found by evaluating p(r).
If (x - 5) is a factor of p(x), then when x = 5, p(x) would = 0. If it is not a factor, then the value of p(5) will be the remainder when you divide the polynomials.
If you calculate p(5), you will get 2(5)3 - 3(5) + 5 = 240, which is the remainder.
Alternatively, if you forgot this, you can do the polynomial division. This will give you 240 as a remainder. See the image below:





20. The results of simulating tossing a coin 10 times, recording the number of heads, and repeating this 50 times are shown in the graph below.

Based on the results of the simulation, which statement is false?
(1) Five heads occurred most often, which is consistent with the theoretical probability of obtaining a heads.
(2) Eight heads is unusual, as it falls outside the middle 95% of the data.
(3) Obtaining three heads or fewer occurred 28% of the time.
(4) Seven heads is not unusual, as it falls within the middle 95% of the data.

Answer: (2) Eight heads is unusual, as it falls outside the middle 95% of the data.
Eight does not fall outside the middle 95% of the data. There are 50 data points, so 47.5 pieces of data are in the middle, leaving 2.5 / 2 = 1.25 pieces of data more than two standard deviations above and below the mean. But there are two results greater than 8, so it's not outside of the middle 95%.



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Wednesday, April 18, 2018

Algebra 2 Problems of the Day

Continuing with daily Algebra 2 questions and answers.

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January 2018
17. The function below models the average price of gas in a small town computations. since January 1st.
G(t) = -0.0049t4 + 0.0923t3 - 0.56t2 + 1.166t + 3.23, where 0 ≤ t ≤ 10.

If G(t) is the average price of gas in dollars and t represents the number of months since January 1st, the absolute maximum G(t) reaches over the given domain is about

(1) $1.60
(2) $3.92
(3) $4.01
(4) $7.73

Answer: (3) $4.01
Graph the function and use "maximum" to find the highest value, which you should see is just above $4.00.
See the graph below:

At approximately t = 1.6, G(t) = 4.01, approximately.



18. Written in simplest form, (c2 - d2) / (d2 + cd - 2c2), where c =/= d, is equivalent to
(1) (c + d) / (d + 2c)
(2) (c - d) / (d + 2c)
(3) (-c - d) / (d + 2c)
(4) (-c + d) / (d + 2c)

Answer: (3) (-c - d) / (d + 2c)
The numerator, (c2 - d2), is the difference of two perfect squares, and factors into the conjugates, (c + d)(c - d).
Note that all four choices have (d + 2c) as the denominator, which makes factoring (d2 + cd - 2c2) that much easier into (d + 2c)(d - c).
(c - d) / (d - c) = -1, which reduces the fraction to (-1)(c + d) / (d + 2c).
Distribute the -1, and you get choice (3).



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Tuesday, April 17, 2018

E-mote

(Click on the comic if you can't see the full image.)

(C)Copyright 2018, C. Burke.

They're irrational, you know.

I remember when they were just ''smileys''. Then ''emoticons'' (emote icons). Finally, ''emoji''. Like ''Gojira'' instead of ''Godzilla''.




Come back often for more funny math and geeky comics.




Algebra 2 Problems of the Day

Algebra 2 is not my usual subject, but I do get asked about the problems occasionally. So I've decided to run a couple of Regents problems daily for a while. If there's a positive reaction (or at least, a lack of negative reaction), I may continue it.

More Algebra 2 problems.

January 2018
15. The terminal side of θ, an angle in standard position, intersects the unit circle at P(-1/3, -sqrt(8)/3). What is the value of sec θ?

(1) -3
(2) -3*sqrt(8)/8
(3) -1/3
(4) -sqrt(8)/3

Answer: (1) -3
The coordinates of P are (cos θ, sin θ)
sec θ = 1 / cos θ
cos θ = -1/3
sec θ = 1 / (-1/3) = -3





16. What is the equation of the directrix for the parabola -8(y - 3) = (x + 4)2?

(1) y = 5
(2) y = 1
(3) y = -2
(4) y = -6

Answer: (1) y = 5
When the parabola is written in this form -- (x − p)2=±4a(y−q) -- then (p,q) will be the vertex and a is the focus length. In other words, the distance in one direction from the vertex will be the focus, and in the other direction will be the directrix.

The vertex is (-4, 3) and the focal length is 2. The negative tells us that the parabola is opening down, so the directrix is 2 units above the vertex, which is y = 5.



Comments and questions welcome.

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Monday, April 16, 2018

Algebra 2 Problems of the Day

Algebra 2 is not my usual subject, but I do get asked about the problems occasionally. So I've decided to run a couple of Regents problems daily for a while. If there's a positive reaction (or at least, a lack of negative reaction), I may continue it.

More Algebra 2 problems.

January 2018
13. If aebt = c, where a, b, and c are positive, then t equals

Answer: (3) ln(c/a) / b
You start with: aebt = c
Divide both sides by a: ebt = c/a
Take the natural log: ln(ebt) = ln(c/a)
which gives you: bt = ln(c/a)
Divide by b: t = ln(c/a) / b.



14. For which values of x, rounded to the nearest hundredth, will |x2 - 9| - 3 = log3x?

(1) 2.29 and 3.63
(2) 2.37 and 3.54
(3) 2.84 and 3.17
(4) 2.92 and 3.06

Answer: (1) 2.29 and 3.63
If you graph the system: y = |x2 - 9| - 3 and y = log(x)/log(3), you can use the intersection function the points of intersection (2.29, 0.754) and (3.63, 1.173).
Use (2nd)(CALC), option (5)Intersect and hit ENTER three times.
Or you can graph log(x)/log(3) - |x2 - 9| + 3, and look for the zeroes.
Use (2nd)(CALC), option (2)Zero.

Given that this is a multiple choice question, you could also use a list of information and enter that last equation into the calculator to see which gives you zero -- or very close to zero, because we have approximate answers. Be careful, though, because there's a lot of information to enter and typos happen.



Comments and questions welcome.

Sunday, April 15, 2018

Algebra 2 Problems of the Day

Algebra 2 is not my usual subject, but I do get asked about the problems occasionally. So I've decided to run a couple of Regents problems daily for a while. If there's a positive reaction (or at least, a lack of negative reaction), I may continue it.

More Algebra 2 problems.

January 2018
11. If n = sqrt(a5) and m = a, where a > 0, an expression n/m could be

(1) a5/2
(2) a4
(3) (a2)1/3
(4) (a3)1/2
See image below

Answer: (4) (a3)1/2
The square root of a value is the same as raising it to a power of 1/2, so n can be expressed as a5/2.
Also, m can be expressed as a1.
This means that n/m is the same as (a5/2)/a1.
When dividing, keep the base, subtract the exponents: a(5/2 - 1) = a(3/2)
A fractional exponent of 3/2 would mean take the square root of the third power, which is choice (4).



12. The solutions to x + 3 - (4 / (x - 1) ) = 5 are


Answer: (1) 3/2 + sqrt(17)/2
Follow the logic in the image:


Subtract 5 from each side, then add (4 / (x - 1)) to each side.
This will set up a rational equation. Multiply both sides by (x - 1) -- or "cross-multiply", if you prefer.
Multiply the binomials, then set up the quadratic equation to solve.
Use the Quadratic Formula to find the roots. Note that you have a positive discriminant, so the roots are real, and there is no "i" in the answer.
If you split the file fraction, you get choice (1).



Comments and questions welcome.

Saturday, April 14, 2018

Algebra 2 Problems of the Day

Algebra 2 is not my usual subject, but I do get asked about the problems occasionally. So I've decided to run a couple of Regents problems daily for a while. If there's a positive reaction (or at least, a lack of negative reaction), I may continue it.

More Algebra 2 problems.

January 2018
9. What is the quotient when 10x3 - 3x2 - 7x + 3 is divided by 2x - 1?

(1) 5x2 + x + 3
(2) 5x2 - x + 3
(3) 5x2 - x - 3
(4) 5x2 + x - 3

Answer: (4) 5x2 + x - 3
First of all, if you divide +3 by -1, the result must be -3, so we can eliminate choices (1) and (2).
Since it's multiple choice, it might be easier just to multiply the two remaining choices by 2x - 1 to see which one works. As they only differ by one sign, it should be quick to do, as shown in the image below:


As you can see, +x gives you +2x2, which when added to the -5x2 in the first column, makes a total of -3x2. So the correct answer is choice (4).

If you wanted to divide (after eliminating the two bad choices), 2x - 1 goes into (10x3 - 3x2), 5x2 times.
(10x3 - 3x2) - (10x3 - 5x2) = 2x2
Bring down the next term, -7x. At this point, you will notice that 2x goes into (2x2), +x times, not -x times. You now have enough information to answer the question.



10. Judith puts $5000 into an investment account with interest compounded continuously. Which approximate annual rate is needed for the account to grow to $9110 after 30 years?

(1) 2%
(2) 2.2%
(3) 0.02%
(4) 0.022%

Answer: (1) 2%
You can check each rate to see which gives you $9110 after 30 years, or you can work backward to solve it.
Use the Continuously Compounded Interest formula A = Pert
9100 = 5000e30r
(9100/5000) = e30r
ln(9100/5000) = ln(e30r)
ln(9100/5000) = 30r
ln(9100/5000)/30 = r
r = 0.0199978..., which is approximately 0.02, or 2%.



Comments and questions welcome.

Friday, April 13, 2018

Algebra 2 Problems of the Day

Algebra 2 is not my usual subject, but I do get asked about the problems occasionally. So I've decided to run a couple of Regents problems daily for a while. If there's a positive reaction (or at least, a lack of negative reaction), I may continue it.

January 2018
7. 7 There are 440 students at Thomas Paine High School enrolled in U.S. History. On the April report card, the students’ grades are approximately normally distributed with a mean of 79 and a standard deviation of 7. Students who earn a grade less than or equal to 64.9 must attend summer school. The number of students who must attend summer school for U.S. History is closest to

(1) 3
(2) 5
(3) 10
(4) 22

Answer: (3) 10.
If the standard deviation is 7, then 72 is one standard deviation from the mean, and 65 is two standard deviations from the mean. So anyone who scored 64.9 or lower had a score more than 2 standard deviations away from the mean. That would be approximately 2.3% of the 440 students, which is .023 * 440 = 10.12, or about 10.



8. For a given time, x, in seconds, an electric current, y, can be represented by
y = 2.5(1 - 2.7-10x). Which equation is not equivalent?


Answer: (4)
In each of the four choices, the 25 has been distributed. The Distributive Property gives you choice (1), so eliminate that. In choice (2), you would multiply the exponents (2) and (-.05x) because you have a power of a power, and that gives you -.10x again, so this can be eliminated. Choice (3) removes the negative by inverting the number into its reciprocal, which is what a negative exponent does, so this is eliminated. Choice (4), the only one remaining, shows the product of two expressions with the same base, but the rule would be the add the exponents, not multiply them.

Adding the exponents in choice (4) would yield (-2 + .05x), not (-10x). Choice (4) is not equivalent.



Comments and questions welcome.

Thursday, April 12, 2018

Algebra 2 Problems of the Day

Algebra 2 is not my usual subject, but I do get asked about the problems occasionally. So I've decided to run a couple of Regents problems daily for a while. If there's a positive reaction (or at least, a lack of negative reaction), I may continue it.

More Algebra 2 problems.

January 2018
5. A certain pain reliever is taken in 220 mg dosages and has a half-life of 12 hours. The function A = 220(1/2)t/12 can be used to model this situation, where A is the amount of pain reliever in milligrams remaining in the body after t hours.
According to this function, which statement is true?

(1) Every hour, the amount of pain reliever remaining is cut in half.
(2) In 12 hours, there is no pain reliever remaining in the body.
(3) In 24 hours, there is no pain reliever remaining in the body.
(4) In 12 hours, 110 mg of pain reliever is remaining.

Answer: (4) In 12 hours, 110 mg of pain reliever is remaining.
Substitute 12 for t, and the exponent becomes (12/12), which is 1. 220 times (1/2) is 110 mg. Even without doing any math, you are told in the question that the pain reliever has a "half-life of 12 hours" -- meaning that in 12 hours, there will be half as much, which is 110.
Choice (1) is incorrect because of the fraction in the exponent. Had the exponent been simply t, then Choice (1) would have been correct. Choices (3) and (4) can be eliminated because exponential decay will not go to zero.



6. The expression (x + a)(x + b) can not be written as

(1) a(x + b) + x(x + b)
(2) x2 + abx + ab
(3) x2 + (a + b)x + ab
(4) x(x + a)+ b(x + a)

Answer: (2) x2 + abx + ab
The coefficient for the middle term will be the sum of a and b (as in Choice (3)), not their product.
If you use the Distributive Property on choices (1), (3) and (4), you will get x2 + ax + bx + ab in each case.



Comments and questions welcome.

Twisted System

(Click on the comic if you can't see the full image.)

(C)Copyright 2018, C. Burke.

Each individual line is asymptoted. It's more of a protest song.




Come back often for more funny math and geeky comics.




Wednesday, April 11, 2018

Algebra 2 Problems of the Day

Algebra 2 is not my usual subject, but I do get asked about the problems occasionally. So I've decided to run a couple of Regents problems daily for a while. If there's a positive reaction (or at least, a lack of negative reaction), I may continue it.

More Algebra 2 problems.

January 2018
3. For the system shown below, what is the value of z?


y = -2x + 14
3x - 4z = 2
3x - y = 16

(1) 5
(2) 2
(3) 6
(4) 4

Answer: (4) 4
Substitute -2x + 14 for y in the 3rd equation
3x - (-2x + 14) = 16
3x + 2x - 14 = 16
5x = 30
x = 6
Substitute 6 for x in the 2nd equation
3(6) - 4z = 2
18 - 4z = 2
-4z = -16
z = 4



4. The hours of daylight, y, in Utica in days, x, from January 1, 2013 can be modeled by the equation y = 3.06 sin(0.017x - 1.40) + 12.23. How many hours of daylight, to the nearest tenth, does this model predict for February 14, 2013?
(1) 9.4
(2) 10.4
(3) 12.1
(4) 12.2

Answer: (2) 10.4
February 1 is 31 days after January 1. February 14 is 13 days after February 1. So 31 + 13 = 44.
3.06 sin(0017(44) - 1.40) + 12.23 = 10.3733...
Make sure the calculator is in radians mode, not degree mode (which would give you 12.1952 ... sneaky!)



Comments and questions welcome.

Tuesday, April 10, 2018

Algebra 2 Problems of the Day

Algebra 2 is not my usual subject, but I do get asked about the problems occasionally. So I've decided to run a couple of Regents problems daily for a while. If there's a positive reaction (or at least, a lack of negative reaction), I may continue it.

More Algebra 2 problems.

January 2018
1. The operator of the local mall wants to find out how many of the computations. mall’s employees make purchases in the food court when they are working. She hopes to use these data to increase the rent and attract new food vendors. In total, there are 1023 employees who work at the mall. The best method to obtain a random sample of the employees would be to survey
(1) all 170 employees at each of the larger stores
(2) 50% of the 90 employees of the food court
(3) every employee
(4) every 30th employee entering each mall entrance for one week

Answer: (4)
The only one of the choices that has a random element to it is the last one: every 30th employee who works at the mall. Notice that choices (1) and (2) only survey people working in certain location, leaving out the rest. "Every employee" is a terrible way to get a "sample".



2. What is the solution set for x in the equation below?


(1) {1}
(2) {0}
(3) {-1,0}
(4) {0,1}

Answer: (3) {-1, 0}
If you try each of the solution sets, you will quickly see that {1} is bad, so (1) and (4) are eliminated, and {0} is good. Next check {-1}, which is also good.

If you needed to actually solve it, as you might in part II, solve for x, as in the image below.
Move the -1 to the other side of the equation. Square both sides. Move the x + 1 to the other side of the equation, leaving 0. Factor the quadratic on the right side.





Comments and questions welcome.

Monday, April 09, 2018

Exterminate

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(C)Copyright 2018, C. Burke.

Yes, anyone who knows me, knows I telegraphed a Doctor Who pun just by the title.




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Friday, April 06, 2018

Sets: The More You Know

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(C)Copyright 2018, C. Burke.

~A: So much to learn




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Thursday, April 05, 2018

January 2018 Common Core Geometry Regents, Parts 3 and 4 (open-ended)

The following are some of the multiple questions from the recent January 2018 New York State Geometry Regents exam.
The questions and answers to Part I can be found here.
The questions and answers to Part II can be found here.

January 2018 Geometry, Part III

Each correct answer is worth up to 4 credits. Partial credit is available. Work must be shown. Correct answers without work receive only 1 point.


32. Triangle ABC and triangle ADE are graphed on the set of axes below.


Describe a transformation that maps triangle ABC onto triangle ADE.
Explain why this transformation makes triangle ADE similar to triangle ABC

Answer: Triangle ABC is dilated by a scale factor of 3 centered on point A.
You can see that to get from A to B, you go 4 units up and 1 to the right. To go from A to D, you go 12 units up and 3 to the right. That is three times the distance. (This explanation is not needed for credit.)
Dilations preserve the shape, so angle size is preserved. Therefore, angle ABC is congruent to angle ADE, and angle ACB is congruent to angle AED. Also, angle A is congruent to itself. By AA, triangle ABC is similar to triangle ADE.


33. A storage tank is in the shape of a cylinder with a hemisphere on the top. The highest point on the inside of the storage tank is 13 meters above the floor of the storage tank, and the diameter inside the cylinder is 8 meters. Determine and state, to the nearest cubic meter, the total volume inside the storage tank.

Answer: The radius of the cylinder is 1/2 of 8m, which is 4m, so the radius of the hemisphere is 4m, too.
This means that the height of the cylinder portion of the tank is 13 - 4 = 9m.
Volume of the cylinder = Area of Base X Height = (3.141592...)(4)2(9) (Note: do NOT use 3.14.)
Volume of the cylinder = 452.389342116...
Volume of the hemisphere = (1/2)(4/3)(3.141592...)(4)3 = 134.041286553
Total Volume = 452.39 + 134.04 = 586.43 = 586 m3. (Units are not necessary, but if included, MUST be correct.)


34. As shown in the diagram below, an island (I) is due north of a marina (M). A boat house (H) is 4.5 miles due west of the marina. From the boat house, the island is located at an angle of 54° from the marina.


Determine and state, to the nearest tenth of a mile, the distance from the boat house (H) to the island (I).
Determine and state, to the nearest tenth of a mile, the distance from the island (I) to the marina (M).

Answer: In the first part, you need to find the hypotenuse of the triangle. You are given the side adjacent to the given angle. Adjacent and hypotenuse mean that you need to use cosine.
cos 54 = 4.5/x
Then x = 4.5/(cos 54) = 7.65585727517 = 7.7.
In the second part, you can use either the tangent ratio or you can use Pythagorean Theorem. The problem with Pythagorean Theorem is that if you use 7.7 then you will not get the correct answer because you have rounded the data in the middle of the problem. (Using, say, 7.6559, for example, should be enough.)
tan 54 = y/4.5
Then y = (4.5)(tan 54) = 6.19371864212 = 6.2


Part IV

A correct answer is worth up to 6 credits. Partial credit is available.

35. In the coordinate plane, the vertices of triangle PAT are P(-1,-6), A(-4,5), and T(5,-2). Prove that triangle PAT is an isosceles triangle. [The use of the set of axes is optional.]
State the coordinates of R so that quadrilateral PART is a parallelogram.

Answer: To show that a triangle is isosceles, you must show that the length of two of its sides are the same. Use the distance formula three times to find the lengths of PA, AT and TP.

d(PA) = ( (-4 - -1)2 + (5 - -6)2 )(1/2)) = ( (-3)2 + (11)2 )(1/2)) = (9 + 121)(1/2)) = (130)(1/2))

d(AT) = ( (5 - -4)2 + (-2 - 5)2 )(1/2)) = ( (9)2 + (-7)2 )(1/2)) = (81 + 49)(1/2)) = (130)(1/2))

d(TP) = ( (-1 - 5)2 + (-6 - -2)2 )(1/2)) = ( (-6)2 + (-4)2 )(1/2)) = (36 + 16)(1/2)) = (52)(1/2))
Since PA and AT are the same length, triangle PAT is isosceles

Finding the coordinates of R is fairly simple using the grid. PA must to parallel and congruent to RT.
To go from P to A, move 3 units to the left and 11 units up.
From point T, go 3 units left and 11 units up and you get R(5-3, -2+11) = R(2, 9).

End of Exam

How did you do?
Comments and corrections welcome. (I get many of the latter!)

Wednesday, April 04, 2018

No Buts About It

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(C)Copyright 2018, C. Burke.

Or just a simple set up.

Mathematically, AND and BUT are the same thing.

I meant to do my homework BUT I forgot.

I meant to do my homework AND I forgot.




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Sunday, April 01, 2018

Happy Easter 2018

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(C)Copyright 2018, C. Burke.

Like mother, like son, like licorice.




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Thursday, March 29, 2018

Mercury in Retrograde

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(C)Copyright 2018, C. Burke.

Go retro or go home




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Monday, March 26, 2018

(x, why?) Mini: Inclusive

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(C)Copyright 2018, C. Burke.

Can it really be inclusive if it excludes potential members?




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Wednesday, March 21, 2018

Sample Space

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(C)Copyright 2018, C. Burke.

Knowing me, this was the only possible outcome.

The photo is public domain, created by NASA.




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Monday, March 19, 2018

Negative Vibes

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(C)Copyright 2018, C. Burke.

He doesn't really use that window pole for much else. Too bulky for drawing straight lines.




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Friday, March 16, 2018

Lucky Clover

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(C)Copyright 2018, C. Burke.

Getting this comic put together today took a little luck of its own!

Happy St. Patrick's Day!

Funny thing, when I graphed this, I thought I could get it shaded by using < instead of "=", and then I couldn't understand when two of the petals went away ... Sigh.

UPDATE: Pretty much as soon as this hit Twitter, there was a comment by William Ricker (@n1vux), mentioning STEM. D'oh! I hate missing the obvious.




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Wednesday, March 14, 2018

Happy Pi Day 2018!

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(C)Copyright 2018, C. Burke.

They're going to occu-pie the Teacher Center.




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Tuesday, March 13, 2018

Putting the Science (& History!) in Science Fiction Convention: Heliosphere NY

As longtime readers of this blog may know, I generally go away for one science fiction convention weekend per year. For the past two years (last one in this one), that convention is Heliosphere, a new convention held in Tarrytown, NY, right off the Hudson River and the Tappan Zee Bridge, and just a stone's throw from Sleepy Hallow.

Heliosphere is a small con, but growing. It's not one of the big flashy events with all the media guests. It has writers and editors in attendance, and they'll happily offer advice along with telling you of their latest projects.

Unlike the inaugural outing last year, I was not a panelist or program participant this time around. I was a plain old fan, with no commitments, free to go where I wanted. (That also make this review a tad more independent, I guess.) And there was pretty to do.

For the science fan, there were panels devoted to the mechanics of sci-fi, including a Sunday morning discussion on Quantum Mechanics, and their applications in real life.

But the big draw would be for the History buffs (and the Alternate History buffs), because the con hosted a 1632 Mini-con, based on the works of, and the world created, by Guest of Honor. In this alternate timeline, a piece of land that included the fictional town of Grantville, West Virginia, was transported in time and space to Germany in 1632, in the middle of the Thirty Years War. The residents had to adapt to their new home and survive hostile encounters. Their "future" tech is helpful to a point, but they have to start an industrial revolution of their own even as they form their own United States over a century early.

A fun panel on Friday consisted of the "Weird Tech" that they could create based on the knowledge they brought with them and the raw materials on hand.

The Gaming room was a good place to pass some time, although I didn't play too much. Personally, I don't want to start a game that'll pull me in for a couple of hours when there are other things going on. Card games and word games usually work best for me -- but those can fool you, too, so be wary!

Another highlight is the popular Books & Brews panels, where the "brew" is coffee. I had signed up in advance to sit in with a group with another Guest of Honor, Dr. Charles E. Gannon, author of the Caine Riordan series of novels, as well as some entries in the 1632 series. Rather than took about his own work, Gannon quite eagerly chose to speak to the attendees about their writing, as nearly everyone at the table had done some kind of writing, or was at least trying. He sympathized with my comment that most of my writing credits happened in a different century.

What made this a highlight was running into Dr. Gannon again, later in the evening, at one of the parties. He came up to me, and asked me about my writing, and where I wanted it to go. If he hadn't had a fan before, well, he sealed the deal here. The guy's for real. (And now I have to make sure I have something written and submitted -- and accepted?? -- if I encounter him again next year.)

I've already registered for next year, April 5-7, 2019. Guests to be announced. More information can be found on their website: http://www.heliosphereny.org/

Friday, March 09, 2018

The Fact of FOIL

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(C)Copyright 2018, C. Burke.

Okay, so maybe I'm Disturbed.




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Tuesday, March 06, 2018

Building a Butter Algorithm

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(C)Copyright 2018, C. Burke.

There's another routine somewhere in there about washing and drying the butter dish, too.

In my house, there's actually an extra step because we keep a spare, unfrozen stick of butter in the refrigerator, which goes in the dish, and then a frozen stick replaces the unfrozen stick.




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Friday, March 02, 2018

Twins

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(C)Copyright 2018, C. Burke.

Just friends, Five.

11 is very popular. It can be a twin prime, a cousin prime, a sexy prime, and even an octupus prime if such a thing existed.




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