## Monday, December 31, 2018

### New Years Eve Plans!

(Click on the comic if you can't see the full image.) Now I know why I like math puns so much. When people talk, it gets so wordy!

And there's a lot of standing around so that they get a chance to say this stuff.

And I deleted and rewrote a lot of dialogue.

Which is why this didn't actually appear on the last day before Christmas break!

I considered labeling this one "School Life", but I wanted to keep that for the students. And I've toyed with "After Hours" for the lives of the characters, but this is obviously school time. When else would Mike and Judy be around each other? Such is the life of a comic writer.

Oh, and Happy New Year! May all your plans be successful!

Come back often for more funny math and geeky comics. ## Friday, December 28, 2018

### (x, why?) Mini: Shapes Are Cool!

(Click on the comic if you can't see the full image.) Their shapes are compatible though trapezoids are less elegant. Trapezoids have no concept of elegance.

Come back often for more funny math and geeky comics. ## Thursday, December 27, 2018

### Algebra 2 Problems of the Day

(not quite) Daily Algebra 2 questions and answers.

More Algebra 2 problems.

June 2017, Part IV

The Question in Part IV is worth up to 6 credits. Partial credit is possible.

37. A radioactive substance has a mass of 140 g at 3 p.m. and 100 g at 8 p.m. Write an equation in the form A = A0(1/2)(t/h) that models this situation, where h is the constant representing the number of hours in the half-life, A0 is the initial mass, and A is the mass t hours after 3 p.m.

Using this equation, solve for h, to the nearest ten thousandth.

Determine when the mass of the radioactive substance will be 40 g. Round your answer to the nearest tenth of an hour.

We are given A = 100, A0 = 140, and t = (8 - 3) = 5. We want h.
A = A0(1/2)(t/h)
100 = 140 (1/2)(5/h)
100/140 = (1/2)(5/h)
5 / 7 = (1/2)(5/h)
(log 5/7) / (log 1/2) = (5/h) log (1/2) / log (1/2)
(log 5/7) / (log 1/2) = 5 / h
(log 5/7) * h = (log 1/2) * 5
h = (log 1/2) / (log 5/7) * 5
h = 10.3002, to the nearest ten-thousandth.

Second part:
We are given A = 40, A0 = 140. We now have h = 10.3002. We want t.
A = A0(1/2)(t/h)
40 = 140 (1/2)(t/10.3002)
40 / 140 = (1/2)(t/10.3002)
2 / 7 = (1/2)(t/10.3002)
(log 2 / 7) / (log 1/2) = t / 10.3002
t = 10.3002 * (log 2/7) / (log 1/2)
t = 18.61... = 18.6 hours to the nearest tenth of an hour.

Logs were never my favorite subject in school, even if the rules are pretty straightforward. It's just one more inverse operation.

Next up: January 2017.

More Algebra 2 problems.

## Wednesday, December 26, 2018

### Can Hardly Wait!

(Click on the comic if you can't see the full image.) Plot twist: Mom works outside of the house, too, but she took off this week to spend time with the kids!

Come back often for more funny math and geeky comics. ## Tuesday, December 25, 2018

### Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

June 2017, Part III

All Questions in Part III are worth up to 4 credits. Partial credit is possible.

35. Graph y = log2(x + 3) - 5 on the set of axes below. Use an appropriate scale to include both intercepts.

Describe the behavior of the given function as x approaches -3 and as x approaches positive infinity.

If you graph y = log2(x + 3) - 5, you will see that it is not defined for x %lt 3.
The y-intercept is at y = log2(0 + 3) - 5 = -3.42.
Solving 0 = log2(x + 3) - 5 gives you the x-intercept at 29.
You can use the calculator, or work it out:

0 = log2(x + 3) - 5
5 = log2(x + 3)
x + 3 = 25
x + 3 = 32
x = 29

The best scale to use for the x-axis is 3. Use a scale of 1 for the y-axis.
Check the image below. The table of values is provided for you. It wasn't necessary for the exam, but it would be a good idea to label the intercepts.

Second part: As x approaches -3, the function goes to negative infinity. As x approaches infinity, the function approaches infinity.

36. Charlie’s Automotive Dealership is considering implementing a new check-in procedure for customers who are bringing their vehicles for routine maintenance. The dealership will launch the procedure if 50% or more of the customers give the new procedure a favorable rating when compared to the current procedure. The dealership devises a simulation based on the minimal requirement that 50% of the customers prefer the new procedure. Each dot on the graph below represents the proportion of the customers who preferred the new check-in procedure, each of sample size 40, simulated 100 times.

Assume the set of data is approximately normal and the dealership wants to be 95% confident of its results. Determine an interval containing the plausible sample values for which the dealership will launch the new procedure. Round your answer to the nearest hundredth.

Forty customers are selected randomly to undergo the new check-in procedure and the proportion of customers who prefer the new procedure is 32.5%. The dealership decides not to implement the new check-in procedure based on the results of the study. Use statistical evidence to explain this decision.

The mean is given as 0.506, and one standard deviation is 0.078. To be 95% confident in the result requires two standard deviations, or 2 * 0.078 above or below the mean. So the interval would be:

0.506 - 2 * 0.078 < 0.506 < 0.506 + 2 * 0.078
0.506 - 0.156 < 0.506 < 0.506 + 2 * 0.156
0.354 < 0.506 < 0.656
0.35 < 0.506 < 0.66
Be sure to round to the nearest hundredth.

In the second part, since 32.5% is below 35.4%, the amount is outside of the 95% confidence interval.

More Algebra 2 problems.

## Monday, December 24, 2018

### Christmas Once a Venn

(Click on the comic if you can't see the full image.) Guide us to Thy perfect light!

Merry Christmas, from (x, why?)

Come back often for more funny math and geeky comics. ## Friday, December 21, 2018

### No L

(Click on the comic if you can't see the full image.) I guess he needs to pay Caesar what's Caesar's. Everyone else is doing a "No L" comic, so why not, right?

Come back often for more funny math and geeky comics. ## Wednesday, December 19, 2018

### Greetings

(Click on the comic if you can't see the full image.) I actually considered putting images of the 12 Days of Christmas on the dodecahedron. Then I got sane.

Seriously, after 5 Gold Rings, I wasn't getting anything onto those faces.

However, there are three words on the five dice.

Come back often for more funny math and geeky comics. ## Monday, December 17, 2018

### Wassailing

(Click on the comic if you can't see the full image.) Waaaaaaaaa-saaaaaaaaaaaaaaaail!

Of the two definitions, it's easy to see how one could lead to the other.

Come back often for more funny math and geeky comics. ## Friday, December 14, 2018

### Dodec the Halls

(Click on the comic if you can't see the full image.) And I wish I could take credit for that, but someone actually made these!
I am not affiliated with these, but I've seen them on Twitter, and they're cute.
https://mathsgear.co.uk/products/dodecorations

Come back often for more funny math and geeky comics. ## Thursday, December 13, 2018

### Skate Fall

(Click on the comic if you can't see the full image.) Can a figure 8 make a figure 8? I'm lucky if both feet are headed in the same direction!

So this is another back burner comic. I think originally, it was to feature the Keegan children and I wasn't ready to draw them like that yet. One thing I do know is that I wrote down this comment in 2010! Probably after a school outing one Saturday to the rink at Bryant Park.

Come back often for more funny math and geeky comics. ## Sunday, December 09, 2018

### Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

June 2017, Part III

All Questions in Part III are worth up to 4 credits. Partial credit is possible.

33. Solve the following system of equations algebraically for all values of x, y, and z:

x + y + z = 1
2x + 4y + 6z = 2
-x + 3y - 5z = 11

Any system of three equations with three variables can be solved by eliminating one of the variables and then solving the remaining system of equations with two variables.

I was away from the PC for a while the other day, so I worked the solution out long hand, two different ways. First by using elimination, then by using substitution, to remove the x. After that, the work would be the same.

Note that the 2x + 2y + 2z = 2 came from multiplying the first equation by 2.
At this point, by either method, you should be left with the two equations:

2y + 4z = 0 and
4y - 4z = 12
Add these together, and the z term is eliminated, allowing you to solve for y.
Once you have y, you can find z.
Once you have y and z, you can find x.

34. Jim is looking to buy a vacation home for \$172,600 near his favorite southern beach. The formula to compute a mortgage payment, M, is M = P • ( r (1+r)N) / ( (1 + r)N - 1) where P is the principal amount of the loan, r is the monthly interest rate, and N is the number of monthly payments. Jim’s bank offers a monthly interest rate of 0.305% for a 15-year mortgage.

With no down payment, determine Jim’s mortgage payment, rounded to the nearest dollar.

Algebraically determine and state the down payment, rounded to the nearest dollar, that Jim needs to make in order for his mortgage payment to be \$1100.

For the first part of the question, use the given formula with the following substitutions:
P = 172,600, r = 0.00305, N = 12 * 15 = 180 payments
M = P • ( r (1+r)N) / ( (1 + r)N - 1)
M = 172600 • ( 0.00305 (1+0.00305)180) / ( (1 + 0.00305)180 - 1)
M = 1247.4933... = 1247 to the nearest dollar. (Round 49 cents down.)

The down payment reduces the Principal that needs to be borrowed. To answer the second part, we need to solve for P when M = 1100. Once we know that, we need to subtract 172600 - P to find the amount of the down payment.
M = P • ( r (1+r)N) / ( (1 + r)N - 1)
1100 = P • ( 0.00305 (1+0.00305)180) / ( (1 + 0.00305)180 - 1)
P = 1100 / • ( 0.00305 (1+0.00305)180) / ( (1 + 0.00305)180 - 1)
P = 152193.19...
The down payment is 172600 - 152193.19 = 20406.81 = 20407.

More Algebra 2 problems.

## Friday, December 07, 2018

### (x, why?) Mini: Revolution

(Click on the comic if you can't see the full image.) Why, yes, I have been covering science classes again. Why do you ask?

Come back often for more funny math and geeky comics. ## Thursday, December 06, 2018

### Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

June 2017, Part II

All Questions in Part II are worth up tp 2 credits. Partial credit is possible.

31. Write (x)(1/3) * (x)(1/2) as a single term with a rational exponent.

First of all, please look at the image. This is how the question was asked. I typed it the best way that I could for this medium.
That said, you're first step is to rewrite the equation as:

(x)(1/3) * (x)(1/2)

Now use the rules of exponents:
(x)(1/3) * (x)(1/2) = (x)(5/6)

Leave your answer in this form with the rational (fractional) exponent. Do NOT rewrite it with a radical symbol.

Note: (1/3) + (1/2) = (2/6) + (3/6) = (5/6)

32. Data collected about jogging from students with two older siblings in the table below.

 Neither Sibling Jogs One Sibling Jogs Both Sibling Jog Student Does Not Jog 1168 1823 1380 Student Jogs 188 416 400

Using these data, determine whether a student with two older siblings is more likely to jog if one sibling jogs or if both siblings job. Justify your answer.

Sum up the second and the third columns. The number of students with one sibling who jogs is 1823 + 416 = 2239. The number of students with two siblings who jog is 1380 + 400 = 1780.

The probability that a student jogs if one sibling jogs is
P(student | one sibling) = 416/2239 = .186

The probability that a student jogs if both siblings jog is
P(student | both sibling) = 400/1780 = .225

It is more likely that a student with two older siblings will jog if both of their older siblings jog.

More Algebra 2 problems.

## Tuesday, December 04, 2018

### Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

June 2017, Part II

All Questions in Part II are worth up tp 2 credits. Partial credit is possible.

29. Graph y = 400(.85)2x - 6 on the set of axes below. (image omitted, see below)

See the graph below.
The x-axis has a scale of 0.5, but using integer values for x is sufficient for full credit. The y-axis has a scale of 20.
I included a table of values for reference. It isn't needed, but it's not a bad idea to include it, or to label the points.
This was a very straightforward probably, using the graphing calculator. A little time-consuming, but simple.

30.Solve algebraically for all values of x:

sqrt(x - 4) + x = 6

Follow along on the image below. Subtract x from both sides of the equation to isolate the radical. Next, square both sides of the equation. Subtract (x - 4) from both sides to set the quadratic equation equal to 0. Factor the trinomial.

More Algebra 2 problems.

## Monday, December 03, 2018

### Elemeno P

(Click on the comic if you can't see the full image.) ''Elemental P'' might've been a rockier pun.

This is the dude from the Alphabet Song.

Come back often for more funny math and geeky comics. ## Saturday, December 01, 2018

### Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

June 2017, Part II

All Questions in Part II are worth up tp 2 credits. Partial credit is possible.

27.Over the set of integers, factor the expression 4x3 - x2 + 16x - 4 completely.

Group the terms into the pairs. You will notice that in the first two terms, you can factor out x2, as follows: x2(4x - 1).
In the last two terms, you can factor out a 4, as follows: (4)(4x - 1).
This gives you x2(4x - 1) + (4)(4x - 1).
Factor out the (4x - 1) from the two terms, and you get (x2 + 4)(4x - 1).
(Distributive Property in reverse.)

28. The graph below represents the height above the ground, h, in inches, of a point on a triathlete's bike wheel during a training ride in terms of time, t, in seconds.

Identify the period of the graph and describe what the period represents in this context.