Friday, May 25, 2018

Peace-wise

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(C)Copyright 2018, C. Burke.

Before you ask, it also can't be represented as piecewise. It's a system of equations: an absolute value and a vertical line. It's less clear if the circle is graphed or added for emphasis. Who can figure out what Ken is thinking?

Then again, who can figure out what Michele is thinking? She's the one marrying him!




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Thursday, May 24, 2018

(x, why?) Mini: The Cheese Stands Alone

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(C)Copyright 2018, C. Burke.

To isolate the cheese, you need to hold the pickles, hold the lettuce, hold everything else and maybe switch franchises.
At least you don't have to hold the chicken.

I debated whether or not to use an actual trademarked property, but I didn't want to use the word cheese on the right side of the equation.




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Algebra 2 Problems of the Day (open ended)

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

August 2017, Part IV

The Question in Part IV is worth 6 credits. Work need be shown (or explained or justified) for full credit. Correct numerical answers with no work receive one credit.


37. The value of a certain small passenger car based on its use in years is modeled by V(t) = 28482.698(0.684)t, where V(t) is the value in dollars and t is the time in years. Zach had to take out a loan to purchase the small passenger car. The function Z(t) = 22151.327(0.778)t, where Z(t) is measured in dollars, and t is the time in years, models the unpaid amount of Zach’s loan over time.

Graph V(t) and Z(t) over the interval 0 < t < 5, on the set of axes below.

State when V(t) = Z(t), to the nearest hundredth, and interpret its meaning in the context of the problem.

Zach takes out an insurance policy that requires him to pay a $3000 deductible in case of a collision. Zach will cancel the collision policy when the value of his car equals his deductible. To the nearest year, how long will it take Zach to cancel this policy? Justify your answer.

Answer:
Create Tables of Values for both V(t) and Z(t). You will see that an interval of 0 to 5 will be sufficient. Next, determine the scale. There is enough room on the y-axis for the scale to be 1 box to equal $2,000. The x-axis will measure the time, and you can use every 2, 3 or 4 boxes for each year (be consistent!) to spread the graph out, for readability.
The plot the points and label the graphs and the axes.
See the images below.

The point where V(t) = Z(t) can be found be finding the point of intersection in your calculator. This happens at about 1.95233, or 1.95, to the nearest hundredth. In the context of the problem, this is the point where the value of the car is equal to the remaining balance of the loan.

For the final portion of the question, since they want the answer rounded the nearest year, you don't need to solve algebraically. You only need to continue the table and find when the car will be worth less than $3000. At year 6, the car's value is V(6) = 28482.698(0.684)6 = 2916.87.
It is reasonable after six years, because that is the first time that the car will be worth less than $3000.



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Wednesday, May 23, 2018

Radio

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(C)Copyright 2018, C. Burke.

BOOM!




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Tuesday, May 22, 2018

The Glass

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(C)Copyright 2018, C. Burke.

It's just another problem to solve.




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Monday, May 21, 2018

Unit Analysis

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(C)Copyright 2018, C. Burke.

'Wizard of floz' or 'floz after every meal'? Which should I use?




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Algebra 2 Problems of the Day (open ended)

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

August 2017, Part III

All Questions in Part III are worth 4 credits. Work need be shown (or explained or justified) for full credit. Correct numerical answers with no work receive one credit.


35.a) On the axes below, sketch at least one cycle of a sine curve with an amplitude of 2, a midline at y = -3/2 and a period of 2π.

b) Explain any differences between a sketch of y = 2 sin (x - π/3) - 3/2 and the sketch from part a.

Answer:
The sine curve starts at the midline value. In y = sin x, the midline would be zero, and it would start at the origin. This curve will start at (0, -3/2). The amplitude is 2, so it has a maximum at -3/2 + 2 = 0.5 (or 1/2), and a minimum at -3/2 - 2 = -3.5 (or -7/2). Label the graph!

They didn't ask for it, but you can create the equation from the description above. (This may also be useful if you wish to use your calculator.)
With an amplitude of 2, period of 2π, and midline of y = -3/2, the sine curve would be

y = 2 sin x - 3/2

See the image below.

In part b, the only difference between the original curve and the new curve is the "- π/3" in the parentheses. This represents a translation of the original curve.
The original sketch will shift π/3 units to the right.





36. Using a microscope, a researcher observed and recorded the number of bacteria spores on a large sample of uniformly sized pieces of meat kept at room temperature. A summary of the data she recorded is shown in the table below.


Using these data, write an exponential regression equation, rounding all values to the nearest thousandth.

The researcher knows that people are likely to suffer from food-borne illness if the number of spores exceeds 100. Using the exponential regression equation, determine the maximum amount of time, to the nearest quarter hour, that the meat can be kept at room temperature safely

Answer:
Put the data into lists L1 and L2 in your calculator and do an exponential regression. You will get, to the nearest thousandth:
y = 4.168(3.981)x

In part b, substitute 100 for y and solve for x.

100 = 4.168(3.981)x
100/4.168 = (3.981)x
23.99232 = (3.981)x
log (23.99232) = log (3.981)x
log (23.99232) = x log (3.981)
log (23.99232) / log (3.981) = x
x = 2.300...

The nearest quarter hour is 2 1/4 hours or 2 hours and 15 minutes.



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Sunday, May 20, 2018

Algebra 2 Problems of the Day (open ended)

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

August 2017, Part III

All Questions in Part III are worth 4 credits. Work need be shown (or explained or justified) for full credit. Correct numerical answers with no work receive one credit.


33. Solve for all values of p:

3p / (p - 5) - 2 / (p + 3) = p / (p + 3)

Answer:
Move the fraction with the (p + 3) denominator to the right side and combine the fractions. Next cross-multiply to get a quadratic equation. Set the equation equal to 0 by moving everything to the left side. Divide the equation by 2 to get rid of the leading coefficient. The remaining trinomial is ridiculously easy to factor into (p + 5) and (p - 1). That means that the solutions are p = -5 and p = -1.
According to original equation, +5 and -3 would have to be rejected as solutions. However, -5 is fine, so do not reject it (or -1, for that matter).

Refer to the image below:





34. Simon lost his library card and has an overdue library book. When the book was 5 days late, he owed $2.25 to replace his library card and pay the fine for the overdue book. When the book was 21 days late, he owed $6.25 to replace his library card and pay the fine for the overdue book.
Suppose the total amount Simon owes when the book is n days late can be determined by an arithmetic sequence. Determine a formula for an, the nth term of this sequence.

Answer:
The difference between the 5th term and the 21st term is $4.00 over 21-5 = 16 days. That's a constant different of $.25 per day.
Using the formula an = a1 + (n-1)d, we can find the fine for the first day
2.25 = a1 + 4(.25)
2.25 = a1 + 1.00
1.25 = a1 is the first day.
an = 1.25 + (n - 1)(.25)

To find the cost on the 60th day, substitute 60 for n:
an = 1.25 + (60 - 1)(.25) = 16.00
The fine will be $16.00



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Using Sine!

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(C)Copyright 2018, C. Burke.

Shall we bow our heads and pray for an answer?




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Friday, May 18, 2018

Algebra 2 Problems of the Day (open ended)

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

August 2017, Part II

All Questions in Part II are worth 2 credits. Work need be shown (or explained or justified) for full credit. Correct numerical answers with no work receive one credit.


31. Algebraically determine whether the function j(x)= x4 - 3x2 - 4 is odd, even, or neither.

Answer:
A function is even if it is symmetrical about the y-axis. If reflected across the y-axis, it will be mapped onto itself. Because of this symmetry, for any value of x, f(x) = f(-x).
A function is odd if it is rotational symmetrical about the origin. If rotated 180 degrees about the origin, it will be mapped onto itself. Because of this symmetry, for any value of x, f(-x) = -f(x).

All the exponents (including the constant) have even exponents, so we know that the function is even. We need to show it algebraically, but knowing this tells you which situation you should start with.

f(-x) = (-x)4 - 3(-x)2 - 4
f(-x) = x4 - 3x2 - 4
f(-x) = f(x), therefore f(x) is even.





32. On the axes below, sketch a possible function p(x) = (x - a)(x - b)(x + c), where a, b, and c are positive, a > b, and p(x) has a positive y-intercept of d. Label all intercepts.


Answer:
In the given function, a and b are roots on the right side of the y-axis (positive), and -c will be on left side (negative). Also, b comes before a because a > b. Finally, d is on the y-axis somewhere above the x-axis. This is just a sketch, it doesn't have to be perfect. At a minimum, please make sure your sketch passes the vertical line test -- don't be sloppy.

The line has to go from -c to d, so it starts in Quadrant III, through -c to d, then down to b and back up through a and beyond.
See the image below.
Note: d does not have to be a local maximum, but it could be. And c is a positive number, so the axis must be labeled -c.





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Thursday, May 17, 2018

Algebra 2 Problems of the Day (open ended)

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

August 2017, Part II

All Questions in Part II are worth 2 credits. Work need be shown (or explained or justified) for full credit. Correct numerical answers with no work receive one credit.


29. While experimenting with her calculator, Candy creates the sequence 4, 9, 19, 39, 79, … .
Write a recursive formula for Candy’s sequence.
Determine the eighth term in Candy’s sequence.

Answer:
If you look at the differences, you see that the sequence increases by 5, then 10, then 20, then 40, etc. It does not have a common difference, but it does have a common ratio of 2, so it's a geometric sequence.
The initial value is 4, and 4 times 2 is 8, so you have to add another 1 to get 9, the second number in the sequence.

The recursive formula for this sequence will be:

a1 = 4
an = an-1 + 1

To get the 8th term in the sequence, just continue where the question left off. You were given the first 5 terms.
a6 = a5 + 1 = 2(79) + 1 = 159
a7 = a6 + 1 = 2(159) + 1 = 319
a8 = a7 + 1 = 2(319) + 1 = 639





30. In New York State, the minimum wage has grown exponentially. In 1966, the minimum wage was $1.25 an hour and in 2015, it was $8.75. Algebraically determine the rate of growth to the nearest percent.

Answer:
2015 - 1966 = 49 years, which will be the exponent.
8.75 = 1.25(x)49
7 = x49
(7)(1/49) = x --> Yes, you are going to take the 49th root of 7.
x = 1.04051.., which is approximately 104%
This is about 4% growth to the nearest percent.

(Don't stop at 1.04, or round that to 1)



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Octillion

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(C)Copyright 2018, C. Burke.

I skipped the 'what to wear' joke -- you figure she'll round up something.




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