JabbaMatheez 40S (Winter 2008)

So Long ...

2008-06-23T10:30:00.000-05:00

And so we begin where we left off ... don't let the sky be your limit. ;-)

I'm so glad we've had this time together,

Just to have a laugh or learn some math,

Seems we've just got started and before you know it,

Comes the time we have to say, "So Long!"

So long everybody, I'll see you in September! Watch this space for pointers to new blogs for each of my classes.

Farewell, Auf Wiedersehen, Adieu, and all those good bye things. ;-)

Class Survey

2008-06-16T08:54:00.005-05:00

The exam is over and we did a little survey in class. The results are below; 13 students participated. If you'd like to add another comment on what you see here email me or leave a comment below this post.

Without any further ado, here are the results of our class's survey. Please share your thoughts by commenting (anonymously if you wish) below .....

Classroom Environment
The questions in this section were ranked using this 5 point scale:

Strongly Disagree	Disagree	Neutral	Agree	Strongly Agree
1	2	3	4	5

The bold numbers after each item are the average ratings given by the entire class.

1. The teacher was enthusiastic about teaching the course. 4.92

2. The teacher made students feel welcome in seeking help in/outside of class. 4.69

3. My interest in math has increased because of this course. 4.23

4. Students were encouraged to ask questions and were given meaningful answers. 4.38

5. The teacher enhanced the class through the use of humour. 4.46

6. Course materials were well understood and explained clearly by the teacher. 4.31

7. Graded materials fairly represented student understanding and effort. 3.85

8. The teacher showed a genuine interest in individual students. 4.38

9. I have learned something that I consider valuable. 3.50

10. The teacher normally came to class well prepared. 4.54

Overall Impression of the Course
The questions in this section were ranked using this 5 point scale:

Very Poor	Poor	Average	Good	Very Good
1	2	3	4	5

1. Compared with other high school courses I have taken, I would say this course was: 4.77

2. Compared with other high school teachers I have had, I would say this teacher is: 4.77

3. As an overall rating, I would say this teacher is: 4.77

Course Characteristics

1. Course difficulty, compared to other high school courses:

Very Easy
0%

Easy
0%

Average
23.1%

Difficult
61.5%

Very Difficult
15.4%

2. Course workload, compared to other high school courses:

Very Easy
0%

Easy
7.7%

Average
61.5%

Difficult
30.8%

Very Difficult
0%

3. Hours per week required outside of class:

0 to 2
23.1%

2 to 3
23.1%

3 to 5
38.5%

5 to 7
15.4%

over 7
0%

4. Expected grade in the course:

F
7.7%

D
7.7%

C
38.5%

B
30.8%

A
15.4%

Specific Feedback
[Ed. Note: Numbers in parentheses indicate the number of students, over 1, that gave the same answer.]

What was your best learning experience in this course?

Workshop classes (4)
Explaining our work on the SMARTboard (2)
Confidence
Great teaching with humour
Understanding math concepts instead of just using formulae (2)
Being comfortable with everyone
Learning the language of mathematics
SMARTboard (2)
Interaction (2)
Humour
Learning not to be afraid of making mistakes
Developing Expert Voices project (2)
Getting feedback
This blog
Learning in other ways than with a textbook
Learning to solve problems (2)
Learning innovatively in many ways
Mathematics

What was your worst learning experience in this course?

None (5)
More 1 on 1 help in class
Class kept getting unfocused (3)
Graphing
Circular Functions Unit
Sometimes had to rush
Not asking questions
Logarithms

What changes would you suggest to improve the way this course is taught?

Fewer SMARTboard technical difficulties
Notes
Better scribe posts
Show formulae (how to "plug in values") first then explain or derive them
Mr. K. did an incredible job
I was motivated by Mr. K's enthusiasm
Laptops
Wanted to hear the "Mr. K. Quote" just before the exam
Study periods for tests
More workshop classes
More group work
More guidance for Developing Expert Voices project
None
Wanted to hear about The Golden Ratio
Give homework for marks
Too much humour
Make Developing Expert Voices projects simpler and more straight forward
We were occasionally distracted — could have got more done
Course was better than I expected
For once I actually enjoyed math
Stay on task
Don't get off topic (2)
Have an alternative to blogging

It's interesting to compare the items that were considered both the worst and best learning experiences. Also, take a look at the list of worst learning experiences compared to suggestions for next year. Help me do a better job next year by commenting on what you see here ....

Student Voices Episode 4: Justice, Lawrence, and Richard

2008-06-15T23:02:00.005-05:00

First an update on this podcast: While we have received few comments on this or any of our class blogs the number of times the audio files have been downloaded is remarkable ...

Episode 1: Jessie 2440 downloads

Episode 2: Tim_MATH_y 1766 downloads

Episode 3: Chris, Craig, Graeme 1367 downloads

Thanks to all our listeners. We might get one more published during this school year but this may be the last until September. In any case feel free to let us know your thoughts about what you heard; every comment is appreciated.

In this episode of Student Voices Justice, Lawrence, and Richard talk about how they put together their Developing Expert Voices project and what they learned in the process: how they they best learn math, how it can best be taught, and many other incidental things like team work and organizational skills.

They have titled their project with one of my favourite reminders to all my students: Mathematics is the Science of Patterns. If you watch any of the video content they created you'll hear several "in jokes", listen for them. Without any further ado, here is the podcast. A copy of the poster they made for their work is below.

(Download File 12.2Mb, 25 min. 30 sec.)

Photo Credit: Shadow singer by flickr user EugeniusD80

Judgement Day in Less Than Half a Day

2008-06-08T20:16:00.007-05:00

Hi friends,

Provincials are coming. It is TOMORROW!!! But don't fret, it is nothing bad. I believe that the reason why people do bad on the exams is because they are nervous or scared and that there isn't sufficient time. The exams are just longer tests. It isn't much more difficult than a test but a bit more tedious because of the lengthy and copious amounts of queBenofschool here. Seeing that my good ol' buddy m@rk didn't scribe I decided to do it for him. What we did TWO (YES TWO [2]) days ago was study for the provincial exams coming up. Thestions in under 3 hours. All that you have to do is just relax, feel confident in all of the studying you did (Don't be over confident because the will backfire. Just be enough that you are beginning to feel comfortable.) Studying would be the best way to cope with that pre-exam fear. Hopefully everybody asked others for help because listening to a peer is just as good if not better than asking a teacher. The voice of a peer is very valuable. There are many ways to be prepared for the exam and it varies among students. So there isn't a right way but only a wrong way to prepare. It all depends on you and what makes you feel comfortable in situations like tomorrow Provincial Math Exam. My method for studying might not be good for others but it makes me feel comfortable. Hopefully everybody has that comfort zone in their studying.

Remember that time is a major issue for many people. That is why studying helps. Studying helps us all see the path to solving that problem quicker, more elegantly, and most of all more efficiently. The exam may play around with words or problems. They might word it differently, so just take your time in reading the questions carefully. If it helps, read the questions out loud but quietly because when you read with just your eyes your brain doesn't register it like when you read it with sounds being heard.

To end this scribe as well as ending the final scribe I tell you this. Sorry if I can't be poetic like my friend Francis but I can be positive. Good Luck at the exam and remember not to be scared. Just feel comfortable and everything will be fine. Drink lots of water, bring water to the exam, bring pencils (note the plural), erasers (again the plural), and a lucky charm that are magically delicious. You can even bring a real rabbit's foot but not a real rabbit's foot because that is CRUEL!
Mr. K I have one favor to ask you. Could you use that before test line I just said above before the exam. I'm sure everybody will be happy when they hear that.

Tehran Sky by flickr user Hamed Saber

Don't let the sky be the limit. Surpass the sky. And like what my good friend Andrew says, "Carpe Diem!" Which means seize the day. Hope to see everyone again in Calculus!!!

Good Bye and Good Luck! =)

Today's Slides: June 6

2008-06-06T13:00:00.001-05:00

Here they are ...

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Last Class of the Jabbamatheez.

2008-06-05T19:27:00.007-05:00

Today was the end of of our classes together. It's been quite fun learning about everything from circular functions to Geometric sequences, especially because of all the skits Mr. K put in.

Well, we started the class by solving a geometric sequence, given the 2nd and 3rd term. We had to find the 8th term given this. Since it was a geometric sequence, we know that there's a common ratio. To find this common ratio, we found the product of the 2 terms. Which was 12/24 = 1/2. We then found where y = 0, if we were to graph this. We found the 1st term by going backwards, which means you would have to multiply the 2nd term (24) by 2. 24 x 2 = 48. Then to find the term of 0, we would double that, which was 96. We then multiplied 96 by (1/2)⁸ because we need to find the 8th term, and it had a common ratio of 1/2. The answer was 0.375

Last class we found out where the infinite geometric sequence came from, but today we learned that when the common ratio is greater than 1, then it is a diverging series, because it will keep going on and on, thus meaning it is infinity. So when |r| > 1 then r^∞ = ∞
When the common ratio is less than 1, it's called a converging series, because it hones in on one value, which is 0. So when |r|<>∞ = 0.

This was pretty much all we learned, so we did some questions after. There was a question about a super ball. At this point Paul pulled out a super ball for Mr. K to use, and it gave the class a little interaction. The ball would start at 200 cm and drop to the ground, it would rebound 3/4 of the distance it fell. We had to then find the total distance travelled by the 4th bounce. This question was easily accomplished using a diagram.
The initial distance being 200 cm, we can multiply by 3/4 to find the distance it bounces back up, then multiply that distance by 2, because it travels that distance twice, bouncing up and falling down, then we multiply the distance from the second bounce by 3/4 (150 x (3/4) = 112.5) to get the distance going up, then multiply by 2 for it going up and down,(150 x (3/4) = 112.5 x 2 = 225 cm) then multiply the distance from the 3rd bounce by 3/4 again (112.5 x (3/4) = 84.375) then multiply that by 2 because it goes up that distance and down hat distance, so it would be (84.375 x 2 = 168.75) add all the values together to get 893.75
You can also use the geometric sequence equation to find this also.
After this we had to find the distance travelled until the ball stopped, so we used the infinite geometric series which was 1200, given all the variables, and you would add the initial value which was 200, and that would be 1400 cm. This was the last question.

That was all we did for our last class. I know, we need something special, because I'm the last scribe and this whole course has been oh so special, because of this reason, I wrote a poem. It's a Kyrielle poem. With 8 syllables each line, and rhyming scheme being aabB ccbB ddbB. Enjoy.

One More Step

Unluckily today's the last,
Of this most enjoyable class.
Though it's not the end for us yet,
As we move on just one more step.

Mr. K, we give you homage,
For expanding on our knowledge.
Never dull, you were fun instead.
As we move on just one more step.

Leaving with a fantastic smile,
As every thing's been so worthwhile.
Now to continue, and to prep.
As we move on just one more step.

-Francis Bowers

er... Scribe List

2008-06-05T19:26:00.000-05:00

What Am I even doing this for? There's no more scribing! XD

Quote of the Cycle ;

"There's no such thing in the world as absolute reality. Most of what they call real is actually fiction. What you think you see is only as real as your brain tells you it is."

Stay Phi everyone,

Rence ~ Out

... and I wonder, if you know... what it means, to find your dreams come true...

Today's Slides: June 5

2008-06-05T12:42:00.002-05:00

Here they are ...

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Scribe List

2008-06-04T23:27:00.002-05:00

Cycle 6

Francis
Joyce
Eleven
benofschool
roxanne

JamieNeRd123C
~~zeph~~
Richard
Hi I'm Justus

~~nelsa~~
Rence
kristina
Paul

Quote of the Cycle ;

"There's no such thing in the world as absolute reality. Most of what they call real is actually fiction. What you think you see is only as real as your brain tells you it is."

I don't know about you guys, but honestly, I'm scared of the exam.

Stay Phi everyone,

Rence ~ Out

Infinite Geometric Series

2008-06-04T18:35:00.004-05:00

OVERVIEW:

Finding the Sum of Numbers in a Sequence
Sigma Notation
Infinite Geometric Series Formula

Finding the Sum of Numbers in a Sequence

(a) What is the sum of the integers from 1 to 5000?

Credit to Paul for being our Gauss today.

We imagine the sequence of all the numbers 1 to 5000 in the top row while the bottom row has all the numbers from 5000 to 1.

Example:
1, 2, 3, ... 5000
5000, 4999, 4998, ... 1

We find the sum of each column in our table and see that they're all 5001. Refer to slide 4 of June 3's slide for a better understanding of what I'm talking about.

We know that there are 5000 of these terms. We also know that if we want to find the sum of the integers between 1 to 5000, then we would have to multiply the number of terms by the total we have then divide by 2. We divided by 2 because there were two sequences we added.

Example: 5000 * 5001 / 2 = 12 502 500

Another way of solving this problem is to pull out the equation (found on June 3's slide #7) and plugging in the appropriate numbers, which Mr. K has a distaste for using this method because we're not really understanding the concept but just plugging numbers into an equation.

(b) What is the sum of all multiples of 7 between 1 & 5000?

The lowest multiple of 7 between 1 and 5000 is 7.
The highest multiple of 7 between 1 and 5000 is 4998.
There are 714 terms that are multiples of 7 between 1 and 5000.

We then imagine, again, that table, listing all the numbers in the sequence from 1 to 5000 (that are multiples of 7) in the top row and all the numbers in the sequence from 5000 to 1 (that are multiples of 7) in the bottom row, and find the sum of each column.

Example:
7, 14, 21, ..., 4998
4998, 4991, 4984, ..., 7

Add the top row and the bottom row, and you find that under each column in our imaginary table is 5005.

We multiplied the number of multiples of 7 between 1 and 5000 by the sum of the first and last terms of the sequence and divided that by 2 because we've counted the sequence twice and we get 1 786 785.

(c) What is the sum of all integers from 1 to 5000 inclusive that are not multiples of 7?

Well, we have the sum of all the integers from 1 to 5000. We have the sum of all the multiples of 7 between 1 and 5000. So if we find the difference of those two numbers, we find the sum of all the integers from 1 to 5000 inclusive that are not multiples of 7.

12 502 500 - 1 786 785 = 10 715 715

Sigma Notation

Left Question:

The k = 1 means that you start with 1 and evaluate 3k.
The number above the sigma symbol, 4, means to keep evaluating 3k until k = 4, then add all the terms.

Right Question:

The k = 0 means that you start with 0 and evaluate 2^k.
The number above the sigma symbol, 3, means to keep evaluating 2^k until k = 3, then add all the terms.

Then Mr. K says to find another sigma notation that is equivalent to the two questions on the slide. Of course, "mathematics is the science of patterns" is the common phrase that Mr. K always says, and yes, there're patterns here.

Also, a quick way of figuring out the answer to the sigma notation in question is to add the first and last term, then multiply by 2. Example: In the left question, [3(1) + 3(4)] * 2 = 30

Infinite Geometric Series Formula

The formula given in the above slide is something that we already derived from the previous class.

To answer the question "why is that the formula?" let's first use an example to help better our understanding of infinite geometric series. We make a square, cut it in half, and shade the half. We take the unshaded region of the square, cut that in half, and shade the half. Then we take another unshaded region of the square, cut that in half, and shade that half, and so on. So, firstly, we got 1/2 of the square shaded, then 1/2 + 1/4 shaded, then 1/2 + 1/4 + 1/8 shaded, then 1/2 + 1/4 + 1/8 + 1/16 shaded, and so on. If we continue this pattern and keep cutting and shading the square in half an infinite number of times, then we would've a shaded square.

1/2 + 1/4 + 1/8 + 1/16 + 1/32 ... = 1

The formula that we derived from the previous class was

Sn = t1*(1-r^infinity) / (1-r)

If we take r and raise it to the power of infinity, that would equal zero.
Example: r = 1/2

1/2 * 1/2 * 1/2 * 1/2 ... = 0

So, in the part of the formula where it says (1-r^infinity), that part of the formula is going to equal (1-0) = 1 anyways, and multiplying a number by 1 won't really have a visible effect. It's more efficient to just not include the (1-r^infinity) into the formula.

Plug in the numbers into the equation, BING! BANG! BOOM! you have the result. Have a nice day.

END NOTES

Start on Exercise 47: Infinite Geometric Series. You should have finished all the questions in the exercise book up to 47.
Tomorrow, we will finish off the lesson the infinite geometric series.
Hopefully, we will be on task and finish the course tomorrow, and do a review on Friday.
Next scribe is Francis.

Today's Slides: June 4

2008-06-04T15:56:00.002-05:00

Here they are ...

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Introducing Series

2008-06-03T17:42:00.007-05:00

The first thing we did in class today is answer some simple questions. Pretty straight forward, should be easy to do. Even if you weren't in class yesterday. Noo excuses, 'cuz I wasn't in class yesterday.

We also graphed the function 'y = [square root] x', which looks like half a sideways parabola.

Next, we learned about 'young Gauss', who was a pretty clever seven-year-old. His teacher told the class one day to add up all the numbers from 1 to 100. Instead of writing it all out and adding it, he found a pattern.

When the numbers are laid out like above, their sums equal 101. There should be a hundred 101's, hence the reason we multiply 100 and 101. We divide it because you added one hundred twice. That's what Karl Gauss figured out. If you want to know the details, you can go ahead and visit this link: http://www.sigmaxi.org/amscionline/gauss-snippets.html

Using the same idea as above, we found the formula for an arithmetic series.
Sn = (n/2)[2a+(n-1)d]

A series is the sum of numbers in a sequence to a particular term in a sequence (definition can be found on blog). You can't have a series if you don't have a sequence first, or else, what are you adding?

We also found the formula for a geometric series.
Sn = [a(1-r^n)]/(1-r)

Lastly, we touched a little bit on 'Sigma', which is located on the second-last slide (8).

Mmm, yeah, that's it. I'm tired now, I'm sorry. Hahaha, next scribe is Joseph, since he asked.. again.

Scribe List

2008-06-03T15:48:00.004-05:00

Cycle 6

Francis
Joyce
Eleven
benofschool
roxanne

JamieNeRd123C
zeph
Richard
Hi I'm Justus

~~nelsa~~
Rence
kristina
Paul

Quote of the Cycle ;

"There's no such thing in the world as absolute reality. Most of what they call real is actually fiction. What you think you see is only as real as your brain tells you it is."

And remember to watch...

... it's epic

Stay Phi everyone,

Rence ~ Out

Today's Slides: June 3

2008-06-03T13:09:00.002-05:00

Here they are ...

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Geometric Sequences

2008-06-02T19:12:00.006-05:00

WARNING: The geometric sequences unit does not have any new material; rather the unit gives a new perspective to what we already know from previous learning.

OVERVIEW:

What is a sequence?
Recursive definition vs. implicit definition
Common difference vs. common ratio
Determining any term (the nth term) in an arithmetic sequence vs. determining any term (the nth term) in a geometric sequence

SLIDE 2
An example of an arithmetic sequence is the first sequence of numbers found on SLIDE 2. Examples of geometric sequences is the second and the third sequence of numbers. Note that arithmetic sequences and geometric sequences aren't the only sequences that exists, such as the Fibonacci sequence, but the main scope of the unit focuses on geometric sequences.

We find the next three terms by determining a rule for each sequence of numbers.

For the first sequence of numbers (4, 7, 10, 13...), we see that 3 is added to any term to get the next term.
For the second sequence of numbers (3, 6, 12, 24...), we see that 2 is multiplied to any term to get the next term.
For the third sequence of numbers (32, 16, 8, 4...), we see that 1/2 is multiplied to any term to get the next term. (Remember, in grade 12, we multiply by 1/2 instead of dividing by 2; it's to make our calculations easier to do.)
For the fourth sequence of numbers (1, 1, 2, 3...), we see that we have to add the previous two terms to get the next term.

SLIDE 3
Let's take a closer look at the first sequence of numbers (4, 7, 10, 13...).

How did we determine that the rule is to add 3 to any term to get the next term? We find that out by determining that the common difference is 3 (green font). We then find out that if we're given any term, n, then 3n+1 is the equation we use to get what n equals. For example,

3n+1 = 3(1)+1 = 4.

So if the rank (n) is 1, then its term is 4. If the rank is 2, then its term is 7. If the rank is 7, then its term is 22. This is expressed with the equation 3n+1.

The rank represents the term. For example, in this sequence, the rank of the first term is 1, the rank of the second term is 2, the rank of the third term is 3, the rank of the nth term is 3n+1, and so on.

Graph 3n+1, and we find that it is the equation of a line. Also, we can determine the 0th term by plugging 0 into the equation.

3n+1 = 3(0)+1 = 1.

For the zeroth term, the output 1. It is also the y-intercept of the graph.

Term 1 (t1), or in some textbooks it's called a, equals 4. The common difference, d, is 3.

The recursive definition, in this sequence of numbers, is to take a term and add 3 continuously.

The implicit definition, in this sequence of numbers, is the equation of the sequence, tn = 3n+1.

SLIDE 4

Let's take a closer look at the second sequence of numbers (3, 6, 12, 24...).

How did we determine that the rule is to multiply 2 to any term to get the next term? Firstly, there isn't a common difference in this case but a common ratio. The common ratio is 2. We then find out that if we're given any term, n, then (3/2)(2^n) is the equation we use to get what n equals.

tn = (3/2)(2^n) = [3 * (1/2) * 2^n] = [3 * 2^(n-1)]

(Refer to the bottom-right corner of SLIDE 4.)

So if the rank (n) is 1, then its term is 3. Similarly, if the rank is 2, then its term is 6. If the rank is 7, then its term is 192. Etcetera.

Graph [3 * 2^(n-1)], and we find that it is the equation of an exponential function. Also, we can determine the 0th term by plugging 0 into the equation.

tn = [3 * 2^(n-1)]
t(0) = [3 * 2^(0-1] = 3/2

For the zeroth term, the output 3/2. It is also the y-intercept of the graph.

The recursive definition, in this sequence of numbers, is to take a term and multiply it continuously by 2.

The implicit definition, in this sequence of numbers, is the equation of the sequence, tn = 3*2^(n-1).

SLIDE 5
Let's take a closer look at the third sequence of numbers (32, 16, 8, 4...).

The recursive definition, in this sequence of numbers, is to take a term and multiply it continuously by 1/2.

The implicit definition, in this sequence of numbers, is the equation of the sequence.

VOCABULARY/SUMMARY is found on SLIDES 6 to 9.

HOMEWORK

Exercise 45: Geometric Sequences

Next Scribe is Nelsa.

Today's Slides: June 2

2008-06-02T15:53:00.002-05:00

Here they are ...

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Scribe List

2008-06-02T00:42:00.000-05:00

Cycle 5

~~Francis~~
~~Joyce~~
~~Eleven~~
~~benofschool~~
~~roxanne~~

~~JamieNeRd123C~~
zeph
AnhThi
~~Richard~~
~~Hi I'm Justus~~

~~nelsa~~
~~Rence~~
~~kristina~~
~~Paul~~

Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

Stay Phi everyone,

Rence ~ Out

Probability: More Practice

2008-05-31T21:39:00.021-05:00

Instead of starting a new unit, we worked on Probability. Mostly we went over the practice sheet we were given on Thursday. The questions we reviewed were one, two, six and nine.

Question One:

The probability of a randomly chosen car being defective is 1/3. Four cars are chosen randomly in order. Given that at least two cars are defective, what is the probability that the first car is defective?

The first thing we did, was figure out what the question was asking.
P(1st is defective I at least two is defective)

Next we found the sample space. It's always easier if you think of it as a word, and then asking yourself how many times the letters in that word can be rearranged, like so:
DDGG -- meaning two of the cars are defective
DDDG -- three cars are defective
DDDD -- all four cars are defective
Then you can find the probability that the cars will be chosen that way, and that becomes your sample space, or denominator.

Next we looked at all the possibilities that the first car will be defective using slots, which, if you remember, was from combinatorics.

Put that numerator on top of the denominator that we found earlier, and you have your answer.

Question Two:

Two jars contain red and green marbles. Jar I contains 3 red and 2 green marbles. Jar II contains 4 red and 3 green marbles. A jar is picked at random and two marbles are picked out of that jar in order. If it is known that the first marble is red, what is the probability of the second marble being red?

You can start this problem by drawing a tree, and writing down the probability that each possibility will occur - if that makes sense. For example, if you look at the picture above, you'll see that the first branch is 'Jar I' followed by 'Jar II', the probability of choosing either jar is one half. The next branches are 'red' and 'green', and the probability of choosing red in 'Jar I' is 3/5, while the probability of choosing green is 2/5. So on and so forth. Keep in mind that the marbles are not replaced after you take it out, so the sample space is different everytime you draw.

In this question, what we're looking for is the probability of getting a red after drawing a red marble first. That becomes the numerator, and the denominator consists of all the possibilities, as shown above. So you plug in the proper numbers in their proper places, and you have that question solved.

Question Six:

Susan sees her friend, Tim, at his locker with a worried look on his face. She asks, "What's wrong?" Tim has to open his locker and change clothes within the next five minutes. However, he has forgotten the combination to his new lock. He knows that the lock requires three different numbers. He also remembers that all of the numbers are odd, and all of the numbers are divisible by seven. It takes 10 seconds to dial a locker combination and 1.5 minutes to change clothes. Is Tim likely to be ready for gym class on time?

I think you'll understand how we solved this problem just by looking at the above picture. The '1 min 30 sec', written in pink, is how long it takes Tim to change.

Question Nine:

Three identical boxes each contain two drawers. In one box, each drawer contains a gold coin. In another box, each drawer contains a silver coin. The remaining box has a silver coin in one drawer and a gold coin in the other. One drawer is opened and a gold coin is found. What is the probability that the other drawer in that box also contains a gold coin.
Michael claims that the probability is 1/3. Jessica claims it is 1/2. Raymond says the probability is 2/3. Explain how each person may have arrived at their answer. Who is correct? Justify your answer.

The probability of choosing the right box is 1/3, and the probability of choosing either a gold or silver coin is 1/2, since there's two drawers. Obviously if you get the box that only have two gold coins, you're not going to get a silver coin. But I'm sure we're all smart enough to realize that.

The probability of getting a second gold coin is shown by:
[P(IG)/(P(IIIG) + P(IG) + P(IG))] + [P(IG)/(P(IIIG) + P(IG) + P(IG))]

If we plug in the correct numbers, we have:
[(1/6)/((1/6) + (1/6) + (1/6))] + [(1/6)/((1/6) + (1/6) + (1/6))]
[(1/6)/(1/2)] + [(1/6)/(1/2)]
(1/3) + (1/3)
2/3

Therefore, Raymond is correct. Michael could've gotten his answer by thinking that there's only one box that contains two gold coins, and the probability of choosing that box is 1/3. Jessica could've simply thought, well there's two drawers, so there's a 1/2 chance of getting a gold coin.

So that's the end of that. There's also questions in the slides (9-18) that we could do for practice. In the afternoon class we had a pre-test that we're going to go over tomorrow. There isn't going to be a test for this unit because we don't have time. Soo.. we're starting a new unit next week.

And that's that. The next scribe will be Joseph because he asked me yesterday. Whoot. Good luck to everyone doing the English Provincial Exams tomorrow! Remember to pace yourself and.. uhh, I forgot what else Mr. V and Mrs. Strecker said.. but.. yes!

Today's Slides: May 30

2008-05-30T12:34:00.002-05:00

Here they are ...

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Scribe List

2008-05-29T23:45:00.004-05:00

Cycle 5

~~Francis~~
~~Joyce~~
~~Eleven~~
~~benofschool~~
~~roxanne~~

~~JamieNeRd123C~~
zeph
AnhThi
~~Richard~~
~~Hi I'm Justus~~

nelsa
~~Rence~~
~~kristina~~
~~Paul~~

Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

Reminder: Roxanne, Francis and Elven - Your D.E.V. Project is due FRIDAY, yeah that's right, FRIDAY by midnight.

AND!

If you haven't already, just give Mr.K a quick email to have him add you to the D.E.V. Blog. Do it now! 'Cause I know you want to.

On a side note: That dot on the Visitor map is getting awfully large. Makes us seem like a big target =( Lol, just kidding.

Stay Phi everyone,

Rence ~ Out

Workshop Period

2008-05-29T22:17:00.005-05:00

Ohhh my goodness. I nearly forgot I was scribe today because of the basketball game XD. (Go lakers!)

So the title above describes todays class. Mr. K was away again but no worries, he'll be back tomorrow...i think. Anyways, we were given a worksheet that we worked on in our little groups for the whole class. I thought the sheet had some challenging question like 1 and 2. I hope we go over those tomorrow. Yeeaah it was a pretty uneventful period other than that.

So I'll get to the point. Next scriibe is NELSA. "Beautiful" right Nelsa?

Scribe List

2008-05-29T00:30:00.001-05:00

Cycle 5

~~Francis~~
Joyce
~~Eleven~~
~~benofschool~~
~~roxanne~~

~~JamieNeRd123C~~
zeph
AnhThi
~~Richard~~
~~Hi I'm Justus~~

nelsa
~~Rence~~
~~kristina~~
~~Paul~~

Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

AND NO, Unicorns aren't real. :) Sorry to burst your bubble guys. Lol.

Stay Phi everyone,

Rence ~ Out

DEV Work Period!

2008-05-28T17:58:00.002-05:00

Well, today we had Ms. Gonzaga come back as our substitute since Mr. K was away today. Surprisingly, and thankfully, there was no quiz or anything (which some of us may have been expecting since we usually get one when there's a sub). But anywho, who cares about that right?! So then, basically, this class was dedicated to working on our DEV projects. The due dates for them are really close that's why. We were also told to work on Exercise 44 as well, and that's it! Same deal for the afternoon class! Okay, that's it!

Aaaannnnd!!! Here's something Jamie is probably expecting (inside-outside joke?!)...she's not scribe by the way, Joyce is, and no I did not mean harm to her by putting her name in red!

Well then, that's it for today!!!! Did I mention the next scribe? Well for those who don't care about reading scribe posts and just like scrolling RIGHT TO THE BOTTOM immediately, the scribe is Joyce.

Scribe List

2008-05-27T23:42:00.003-05:00

Cycle 5

~~Francis~~
Joyce
~~Eleven~~
~~benofschool~~
~~roxanne~~

~~JamieNeRd123C~~
zeph
AnhThi
~~Richard~~
~~Hi I'm Justus~~

nelsa
~~Rence~~
kristina
~~Paul~~

Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

Stay Phi everyone,

Rence ~ Out

Today's Slides: May 27

2008-05-27T22:51:00.001-05:00

Here they are ...

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Combinations of Permutations and Probability and Combinations O_o

2008-05-27T16:31:00.005-05:00

Why hello there!
Benofschool here. I never scribed for a while, I was beginning to think that you guys forgot about me =(. Well anyways time for a scribe. Mr.K was later than usual today but he wasn't late for school which was good. Today's class was a workshop. We were broke up into groups like every other workshop class. Unfortunately the slides aren't up. Questions were put up and we were off...

Question 1

We had to find the probability of getting an ace or a diamond from a deck of cards. Fairly simple. Francis went up to answer it. To get the answer we just add the probabilities of getting an Ace and a Diamond and subtract the probability of getting the Ace of Diamonds. We add because it is an "or" question and we subtract that single card because it was counted twice when we calculated the probability of getting an Ace or Diamond. So the answer was 4/13.

Question 2

The second question involved a venn-diagram and venn skills from grade 11 logic in pre-calculus. The question was what is the probability of picking a person who doesn't like Dr. Pepper or the diet version if 7 people liked Dr.Pepper, 11 people liked the diet version, and 3 people liked both. So first we have to create a two circle venn-diagram. Rule is start with the inside and work outwards. So we put in the middle which means that 3 people prefer both drinks. Then lets work with on side. But remember from grade 11 we have to subtract the number of objects in the center from the separate values. If we look at the first question, we didn't count the Ace of Diamonds because the probability of getting an Ace or a Diamond is not mutually exclusive as well as the Dr. Pepper situation. If one thing occurs, the other probability can still happen. So that means we have to subtract the number of people that liked both drinks from the separate preferences. We get the sum of those numbers (4 +3+8=15) and subtract from the total number of students in the class (25-15=10). That will be the numerator in the probability and the sample space would be the total number of students. So the answer is 2/5.

Question 3

The 3rd question is about picking dresses. There are 15 dresses: 6 green, 5 blue and 4 yellow dresses. We wanted to know what is the probability of getting exactly 2 green dresses from picking 6. This question involves both combinations and probability. So 6 choose 2 because they are indistinguishable. multiplied by 9 choose 4 for the remainder of the dresses divided by 15 choose 6 which is the ways that we can choose 6 dresses out of 15. So we get an answer of 37.8%.

Question 4

Another combination question. A couple has 4 children that are about to be born. What is the probability of getting at least 2 girls. So what we do is find the probability of getting 2 girls first then 3 girls and finally all girls. Then we just find the sum of those values and we get the answer. The answer is:

That were all of the questions we did in today's class. Remember DEV due dates are arriving in a future now. Check the calendar on the right side bar to check for your due dates. Tomorrow's double will be used as a DEV work day. So bring the stuff you want to bring to work on your DEVs. That is all I have to say about today's class. The next scribe will be kristina. Good Night and see you all tomorrow!!!

Costly Test

2008-05-27T01:38:00.002-05:00

So yeah, the reason why I'm up so late was because I just got back from the Glow in the Dark concert... and I've completely lost my voice. Anyways, I found this really cool story so I thought I'd share it and hope that you get a laugh out of it too.

A professor was giving a big test one day to his students.

He handed out all of the tests and went back to his desk to wait. Once the test was over the students all handed the tests back in. The professor noticed that one of the students had attached a$100 bill to his test with a note saying, "A dollar per point."

The next class the professor handed the tests back out.

This student got his test back and $64 change.

why DOCTORS are so DARN CAGEY!

2008-05-26T21:44:00.004-05:00

and no ...
- being cagey does NOT mean you're inside a cage alot ...

So I was chosen for being the scribe for today. Let me tell you guys something. REVENGE GETS YOU NOWHERE. ~~That's why I'm always going to steal Paul's seat from now on~~. Thanks for choosing me as scribe Paul.

[ SECTION A ]
- SLIDE 2, 3
- Identifying events / dice diagram

[ SECTION B ]
- SLIDE 5, 4
- Probabilities involving "AND" and "OR"
- Testing for independence

[ SECTION C ]
- SLIDE 6, 7, 8, 9
- A test for cancer

[ SECTION D ]
- SLIDE 10
- A test for industrial disease

[ SECTION A ]
- identifying events / dice diagram

SLIDE 2 - IDENTIFYING EVENTS

- and so we started the class finishing off what we had last done in class on Friday.
DRAG N' DROP BABY!
- a list of events which the class had to determine whether it was dependent or independent. Mutually exclusive or not mutually exclusive. I'll take two examples.

b) one card - a red card or a king - is randomly drawn from a deck of cards.
- because it is only ONE event we know that it must be independent
- a red card can also be a king. Therefore it is not mutually exclusive

c) A class president and a class treasurer are randomly selected from a group of 16 students.
- because we can only have ONE president and ONE class treasurer, the event must be dependent. One student cannot be the president and class treasurer. Well in this case it can't. Therefore it is mutually exclusive

SLIDE 3 - THE DICE DIAGRAM

FORGIVE ME IF I AM DOING AN ILLEGAL ACT ON THE INTERNET.

I had actually found this diagram on another blog of Mr. K's class.

This was done to answer the question:
e) Rolling two dice and getting an even sum or a double
- It is independent because the first roll does not affect the chances of the probability for the next roll.
- As you can see on the diagram, you are able to roll a double AND get an even sum. There fore it is not mutually exclusive.

[ SECTION B ]
- Probabilities involving "AND" or "OR"
- Testing for independence

SLIDE 5 - PROBABILITIES INVOLVING "AND" OR "OR"
AKA "THE ADDITION RULE"

here is the slide ...

SLIDE 4 - TESTING FOR INDEPENDENCE

We compared the result of the probability of getting the flu shot and getting the flu. (0.10)
To the result of the probability of the seniors getting the flu (0.15)

There fore the event is dependent if you get the flu shot or not.

[ SECTION C ]
- A test for cancer

SLIDE 6, 7, 8, 9 - A TEST FOR CANCER

- The given question, information and possibilities.

0.5% of 1,000,000 = 1,000,000 x 0.005 = 5000
To find who does not have cancer you would subtract 5000 from 1,000,000
-
98% of the time, the test will be positive.
The amount of people who have cancer and will test positive would be ..
5000 x .98 = 4900
The amount of people who have cancer and will test negative would be ..
5000 x .02 = 100
982% of the time, the test will be negative.
The amount of people who don't have cancer and will test positive would be ..
995,000 x .02 = 19,900
The amount of people who don't have cancer and will test negative would be ..
995,000 x .98 = 975100
-
Adding it up the amount of people who tested positive for cancer
4,900 + 19,900 = 24,800
When only 4,900 actually do have cancer.

The probability that a person who actually does have cancer and tests positive is the amount of people who do have cancer divided by the total amount.

4,900 / 24,800 = 19.75%

Here are the tree diagrams to see how this all works out.

[ SECTION D ]
- A test for industrial disease

SLIDE 610 - A TEST FOR INDUSTRIAL DISEASE

The probability of a person who tested positive and actually does have the industrial disease is much similar to what was done in SECTION C.

The probability of people with industrial disease and tested positive.
DIVIDED BY:
The probability of people with industrial disease and tested positive,
ADDED WITH:
The probability of people without industrial disease and tested positive.

^ does that make sense?!

In other words, and in this case ..

0.0099 / 0.0099 + 0.0099 = 0.5%

NO WONDER DOCTORS ARE SO DAMN CAGEY! ...
... ~~BECAUSE THEY'RE ALWAYS INSIDE CAGES! WOW!~~
because there is still a great chance that even if you do NOT have cancer, the result may still test POSITIVE

SO YOU MIGHT HAVE CANCER, MUAHAHAHAHA
- not intentionally trying to scare any of you, or myself.
- not intentionally trying to offend anyone who is reading this.
- It's just a joke! Humor in OUR math class is NO laugh matter.

That's it, That's all ladies and 'gents! Have yourself a great day. Go out there and commit random acts of kindness! Cheers!

The next scribe will be ...benofschool

Today's Slides: May 26

2008-05-26T15:02:00.001-05:00

Here they are ...

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Types of Probability and Exclusivity

2008-05-25T19:40:00.005-05:00

So, this is a double scribe post, which covers the lessons we did on May 22 and May 23 (Thursday and Friday respectively).

So lets get started, shall we? First off is our lesson from May 22...

TYPES OF PROBABILITY: MAY 22

So last Thursday, we talked about the two types of probability. There are dependent and independent probabilities, and thats what we'll explore here.

Slide Two:
The first thing we did was make a tree, which we've all done before. This tree displays all the possible outcomes of flipping two coins. Easy right? We know probability of each outcome is 1/4 because there are four possible outcomes, and each one is one of those four.

Slide Three:
Moving forward, we have "an entirely different question that is not at all similar to the one we just did." Please note the quotation marks. Because, this question is very similar to the one we just did. Instead of heads and tails, there are reds and blues. But this is similar but not the same. Why? Because unlike our previous probability question, this question deals with a dependent probability, whereas our previous one dealt with an independent probability. Remember, this lesson was supposed to be about dependent and independent probabilities?

But Paul, you say, what the heck are dependent and independent probabilities anyway? Im confused!

Well, to put it simply, independent probabilities are probabilities that are affected by the steps before them. Why don't you think about it in context, like this:

If you flip a coin, and then you flip another entirely different coin, does the chance of you getting tails on the second coin change because you flipped the previous coin?

Answer? Of course not, excluding far-out possibilities such as your hand got so tired flipping the previous coin that you put less energy into flipping your second coin and gave it 0.001% better chance of landing on tails. This is an example of an independent probability, where both probabilities are, well, they're independent of each other. Makes sense, right?

However, with our second question, our probabilities are dependent. Why? Because when you choose a marble (wearing a blindfold, earmuffs, and nose plugs so you cannot possibly see, hear or smell which marble you're picking), you also remove that marble, thus changing the probability of you getting the same colour marble when you draw a second one. This is all displayed in the slide two tree, where if you drew a red marble and removed that, your chance of drawing a red marble again the second time is 2/5, whereas if you drew a blue marble the first time, you have a 3/5 chance of getting a red marble when you draw a second time. See? They're different, which is exactly why this is a dependent variable.

And that is basically the gist of this entire lesson, so I'll just summarize the rest of the slides...

Slide Four:
Okay, something changed in this question, and I'll give you a cookie if you can spot it.

Alright, found it? Here, have a cookie. You clearly noticed that our question changed in that the marble is no longer simply discarded (or "thrown out the window" as Mr. K would say [hey, I rhymed]). Instead of removing a marble when we draw it, we just put it back. Because we put the marble back, our probability doesnt change, and it becomes independent. Now that its independent, all you'd have to do is change those R's to H's and B's to T's, and you've got your original heads and tails question! Magic right?

Which brings up another point Mr. K mentioned. A lot of the time, your question boils down to stuff much like the heads and tails question, just wrapped up in a pretty little. Remember that (say it to yourself five times, or something), and you'll find things will be a lot easier.

Slides Five & Six:
Just a rehash of what I already wrote, basically definitions of what dependent and independent variables are (in simpler, less rambling and rant-y [is that a word?] terms).

Slide Seven:
A few questions, dealing with figuring out whether or not the question is describing an independent or dependent probability. Simple, right? I'll explain why quickly.

a) Independent because the outcome of the coin toss doesn't affect the outcome of the dice roll (again, this is factoring out such insane possibilities and how your possibility might be changed if you did both at once, or something along those lines).

b) Dependent because when you draw the first card, you remove it from the deck and change the amount of cards in the deck. So when you draw your second card, your deck is one card smaller than it would be had you not removed the first card, and you can no longer draw that particular card. Thus, the first outcome affects the second outcome.

c) Independent because now your deck is static and the probabilities always remain the same (excluding time travel and such nonsense).

Slide Eight:
Slide eight takes our second slide's question one step further by adding another coin flip. Nothing terribly new here, it just makes the chances of getting a particular combination smaller. I won't bother explaining this in great detail, but I'd like to point out that the final probability of an outcome (say, HHH, or flipping three heads in a row) is equal to the product of its previous steps probabilities (in other words, 1/2 * 1/2 * 1/2 = 1/8, not a coincidence), where each coin flip had a 1/2 chance of landing Heads.

Slide Nine:
Now to follow our trend of making small changes to our previous questions, we take our old question and put a fresh splash of paint on it and volia, we get something entirely new. Fortunately for me, all I have to do is take our old solution, paste it on to our new slide and say Boy = H, Girl = T. But wait, our question says, whats the probability of mom and pop getting exactly two girls and one boy. So what do we do? We find all the answers that have two girls (T's) and one boy (H's). Since there are three possible outcomes out of 8 that have two T's and one H, we determine that the chance of getting two girls and one boy exactly is 3/8.

Slide Ten:
Ignore this, this was my poor and not at all though out attempt at solving the question, and it is completely wrong.

Slide Eleven:
Our final slide, with the correct solution, courtesy of Kristina. Another variation on our girl/boy question, except with a different amount of steps and different conditions. So for the first draw, if we draw a blue marble first, our bag with 3 red and 3 blue becomes a bag with 4 red and 2 blue, and vice versa. So when you draw a second marble, you either have a 1/6 or 2/6 chance of drawing a blue marble the second time depending on the colour of the first marble you drew.

And that concludes our lesson on the Types of Probabilities.

EXCLUSIVITY IN PROBABILITIES: MAY 23

And without delay, I give your my scribe post for our May 23rd (Friday) class on Mutual Exclusive Events.

Slide One:
Contains a fox that is entirely unrelated to probability. However, I would like to dub him (or her) the Probability Fox, just for fun.

Slide Two:
A simple slide with a simple question. If you have 56 listed and 144 unlisted phone numbers, you have a total of 200 phone numbers. So if you want the probability of choosing a listed phone number from those 200 total numbers, you have a 56/200 chance, or a 28/100 or 7/25, or .28, or 28% chance.

Slide Three:
Okay, so we have a horse named Gallant Fox (makes total sense, right?) and another horse named Nashau. Gallant Fox runs a race and has a 2/5 chance of winning. Nashau runs an entirely separate race and has a 1/3 chance of winning. What is the probability that:

a) Both Nashau and Gallant Fox win their respective races. That's easy, we simply multiply their probabilities together and get 2/15. Remember earlier how our final probability was the product of all the probabilities of the steps before it? This is just like that, where our first step is Gallant Fox wins (2/5) or he loses (3/5), and then our second step is Nashau wins (1/3) or he loses (2/3). Since we want the path that has both Nashau and Gallant Fox winning, we want to multiply 1/3 by 2/5, which gives us our final probability for that outcome.

b) So this is just a in reverse, and instead of taking the winning path, we take the losing path. So 2/3 * 3/5 = 6/15.

c) Well, there are two ways we could solve this question. We could find the probabilities for all the routes where one of the horses wins their race and then add those together. But thats just long and tedious, and considering what we already know, we have a much simpler solution at our doorstep. The question asks us what the probability is that one of the horses wins, so as long as they both don't lose, we're good right? Wait, don't we already know the probability of them both losing? Well... if we know that, cant we just take all the possibilities and remove the ones we don't want to get the ones we do want? Something like this... 15/15 is our total possibilities, but 6/15 of them are ones that have both our horses losing their races. So we just subtract 6/15 from 15/15 and... our answer is 9/15.

Slide Four:
This slide was probably the most confusing question we had encountered in a while, and it took a while to figure out. But basically, it goes like this:

Chad wants to meet his girlfriend in either the Library or the Lounge. If he goes to the Lounge, he has a 1/3 chance of meeting her (apparently Chad has poor arranging skills as he still has no idea where they're meeting despite the fact that this has all been pre-arranged, but I digress), but if he goes to the Library, he has a 2/9 chance of meeting her.

a) What is the total probability he'll actually meet up with her, either in the Library or the Lounge. Because its an "or" probability, we add the probabilities together (the Library OR the Lounge), and get 5/9 chance they actually meet.

b) Now this is where it kinda gets confusing. The probability of them not meeting is 2/3 * 7/9? Wait, just were adding just a moment ago, why did we suddenly switch to multiplication? Now we're dealing with an "and" probability, where she is not in the Library AND not in the Lounge. Thus, we get our 14/27 chance they don't meet.

Slide Five:
This slide is a little demonstration on how Chad cannot go to the Lounge AND the Library. Thus, going to the Library or going to the Lounge are mutually exclusive events, which is what this lesson is all about.

Slides Six & Seven:
These slides give you the basic definition of Mututally Exclusive Events, and some examples. These are pretty self-explanatory.

Basically, if you have one event that makes another impossible, then they are mutually exclusive. You cannot turn left and right at the same time, you cannot be facing North and South at the same time. Facing North is mutually exclusive to facing South.

Slide Eight:
So we want the probability of drawing a King or a Spade in a single draw from a pack of 52 playing cards.

Our first event is to draw a King. Our second event is to draw a Spade. Drawing a king has 4/52 chances, and drawing a Spade has 13/52 chances. However, because of the one card that is both a King and a Spade, we must remove that card (1/52). So our formula:

Probability(Event A or Event B) = Probability(Event A) + Probability(Event B) - Probability(Event A and Event B)

Means this:

The probability of Event A or Event B is equal to Event A's probability plus Event B's probability subtract any Probabilities that fulfill both Event A and Event B.

And inputting our data, we get the Probability of drawing either a king or a spade as 16/52.

Slide Nine:
And finally, a couple practice questions. Drag'n Drop Baby!

a) Independent because the marble is returned, mutually exclusive because you cannot draw a red and blue marble at the same time.

b) Independent because there is only one step, not mutually exclusive because it is possible to draw a red king.

c) Dependent because once a president or treasurer is selected, the number of people in the selection pool changes and thus so do the probabilities for the next selection. Mutually exclusive because the president cannot be the treasurer at the same time (they are removed from the selection pool once they have been chosen).

d) Independent because there is only one step, mutually exclusive because there is no card that is a red king and black queen.

e) Independent because the second step does not change the probability of the second, mutually exclusive because its not possible to get an even and odd number at the same time.

And that concludes this super long double buttery extra flavoured scribe post, again posted in the middle of the night and written in the dark. I'll see you all in a few hours. Please feel free to point out any mistakes, I'm fairly sure I made a few.

And because he asked so nicely, my next scribe will not be Thi.

No, the next scribe will be SOMEONE WHO HASNT BLOGGED YET, WHY ISNT THE SCRIBE LIST UP TO DATE WHAAAAAARGHADSADJKSAHDJKAHSJDL.

>:(

I'll just tell you guys in the morning since nobody will read this until then anyway. Good night, and have a pleasant tomorrow.

P.S. That Google translate thing works, atleast, I think so (I cannot read Japanese and verify this). However, from this I have learned that "Eleven" is "11" in Japanese.

Today's Slides: May 23

2008-05-23T13:31:00.000-05:00

Here they are ...

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Today's Slides: May 22

2008-05-22T14:45:00.001-05:00

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FOCUS ON MY BOB FOR [ice cream] CONES

2008-05-22T08:20:00.001-05:00

Once again, it’s been days since I’ve gotten access to the blog. But I still manage to get things done. Conics—the in-depth look at the aspects of a cone. It sounds like I’m writing a feature article on it or something. I sort of enjoyed this unit, it had its moments. It was pretty reminiscent of the days when I was master paper folder in origami club. By the end of emphasizing each shape that evolved from a cone, I simply wondered, “what happened to just making cranes?” I also commend Mr. K for his innovative [yet deadly] ways of disarticulating a cone with a shward.

This may have been a short unit but I really have to admit that I learned a lot. I didn’t know that so many shapes could come from a cone, or two cones to be precise. A “hypercone”, just kidding… The proper term of course would be the double-napped cone that gives way to a hyperbola, a parabola on steroids. Sarcasm, my native tongue.

The class learned each of the formulas and also learned how to graph the formulas. Call me juvenile, but I still have trouble with the grade 12 version of the parabola. I think the 4p always throws me off. I also feel like using the other formula from grade 11 though. But using that idea again should be avoided since thinking would not be lateral. This is why practice makes perfect. I just need to do a little adapting. Otherwise, I’d think I’m ready for the test today. Just keep my hocus pocus my focus, look it’s a locus.

BOB: Conics

2008-05-22T07:58:00.002-05:00

WOAAAHH!! I almost forgot to do my BOB again! Sorry this is really really late XD

Okay, this unit was one of my favorite units. I actually enjoyed graphing for this unit and I thought that the concepts were very simple. It was a good break from the Counting unit, which I didn't really like much. I also enjoyed doing all that folding during class, such easy homework! This unit was also full of laughs, we really couldn't seem to stay serious most of the time. As for the parts that I found the hardest for this unit, I would probably have to say that the hyperbola and ellipse were definitely the most difficult, although not by much. I got confused at first with how to find the foci and the asymptotes (hyperbola) but then it all made sense after some explaining. I also remember messing up graphing my vertical hyperbolas since I switched around the transverse and conjugate axis. All in all, tis was a good unit, now I got to get ready for school! Good luck on the test guys :P

BOB on Conics

2008-05-21T23:43:00.003-05:00

Hi guys this is Richard and here is my bob on the unit of conics.

This unit was not as hard as the other units there were some parts that i got really easy like the ellipse. I probably found that this topic was easiest because i had to scribe for it or it is just the simpler out of the conic shapes. Another thing that was cool was that we got to do a lot of fun folding. That led to the geometry of the conic shapes.

There were also topics that i found difficult for example the parabola. I think that the main reason i don't understand it is because i was not present for that Pre cal class. I almost forgot that the graphing of the parabola is also hard. i also felt the same was that Francis felt on that Pre test. I was also puzzled by the transforming parabola question . Well that's just me ah ha

Richard signing off.

Conics Bob

2008-05-21T23:22:00.002-05:00

Ahhh Conics. Too bad Mr.K didn't show us any more of his samurai skills with this metre-stick-katana.

To the point. I will admit, like every section in this course, I had troubles in one way or another. The origami was cool, as it went with the whole Japanese theme Mr.K was trying to capture (TOYOTA-HONDA-MITSUBISHI-MAZDA-YOKOHAMA-SONY-etc...) Like every other type of graphing, I don't like to graph. Granted it does give me a better look at whats happening and I can tell what's happening, but like everything else in life, just because you know how to do it doesn't mean you're supposed to like it. Like graphing.

I liked how parabolas, circles, ellipses and hyperbolas had their distinct way to tell each other apart without having to think hard but rather just look at what's switching to make it graph that way. But, that's probably about it. The math was pretty simple once I took the time to sit down and study it all, like I am tonight.

I didn't like word problems however, how they make it seem like it's supposed to be something it's not, but yeah, I know, I'm just over complicating it.

I will be deep in study for a while because well, exams are closing in surprisingly fast. OH and don't forget about your DEV projects.

Rence ~ Out

BOB on Conics

2008-05-21T22:18:00.003-05:00

This unit was quite short, but I'm not celebrating. Personally, I need a bit more time with this unit, only because I don't think we went far enough with transformations that can happen with each conic shape. On the pre-test I saw that one question where the graph of a hyperbola was stretched 4 units, and moved down 1 unit, I was sitting there staring at the page for like 5 minutes, forgot all about those transformations. Plus, I don't think I can remember all those equations such as the equation of a circle, ellipse, hyperbola and the new parabola equation. Hopefully we'll get a formula sheet, or some other type of aid. On top of that, I don't fully understand the total anatomy of all these conic shapes, such as the major, minor, semi-major, semi-minor, conjugate, and transverse axis', and some, I suppose. I still don't exactly know about graphing these guys either. Other than that, I really enjoyed folding the paper, being interactive in a class is always a plus.

Good luck on the test everyone.

Until next time,
-Francis

Bobbing on Shwards and Things

2008-05-21T21:36:00.004-05:00

So, the infamous conic sections unit was completed late last week, and this would be a bob for said unit.

Where to begin? the good? the bad? the ugly? Lets start with the ugly

THE UGLY *dun dun dun*

-GRAPHS! Yes I'm terrible with graphs but I am getting better at them, thanks to a certain "perspective." More on that later though.

The Bad

Not to much here this time, thankfully.

-One of the most difficult parts of this unit is how most of the equations, have another equation which is nearly identical. This made memorizing/remembering which was which difficult at times.
-Going from standard to general form, and remembering which is which was often a pain.

The Good

-Short unit
-Easily understood by used of "visual" tricks (what I meant in the ugly section. By thinking of what the graph of some equation looks like, the question is often made much easier.
-Overall the concepts in this unit, and the difficultly level of the questions was not very demanding.

So yeah, there is it, my small and not very impressive bob post. :P Now I'm back to studying haha.

Ciao!

An Introduction to Probability

2008-05-21T18:13:00.006-05:00

Alright young fellers, american idol is almost on so I'll try to finish this before it starts but if I don't..I'm sure I can do both! (:

Slide One
The first thing Mr. K asked us today was, "What is the probability of the sky falling?" (chicken little yay!) Each of us were given about a few minutes to write down our answer and we all shared our answers to each other. *refer to the first slide* Most of us came up with different numbers but some of us had the same answer as others. Most of us just chose a random number, like me, but some of us also took some time to why they choose that number. After this dicussion, he then asked us what the numerator and the denominator represented. The numerator is the number of chances of the sky actually falling while the denominator is the set of all possible outcomes.

Slide Two
The next slide explains the defintion of an inderterminate form. So if we take any number such as 6 and multiply it by one then divide by one your answer would be 6. And if you take the 6 again and multiply it by 1/10 then divide by 1/10, you still get the answer 6. Continuing the pattern, you'd eventually reach zero. So you would multiply 6 by zero then divide by zero. Therfore it is in an inderterminate form. If you recall on slide one, 0/0 and 0/∞ are examples of indeterminates as well as ∞/∞ and ∞/0. 1/0 is not in an indeterminate form, it is however, an undefined value.

Slide Three
The next slide includes various vocabulary words which we should know for this unit as well as different examples to explain the meaning of each words.

Slide Four
Calculating the Probability of Event A
This slide includes the formula P(A) = n(A)/n
P(A) - Denotes the probablity of A
n(A) - number of favourable outcomes (number of accourences of A)
n - sample space (the total number of possible outcomes
Probability can not only be expressed as a fraction but also as a ratio, decimal and a percent. However we will not be looking at it as a ratio in this unit.
IMPORTANT TO REMEMBER: Probability of any event is always a number between 0 or 1. This means that if your answer is bigger than one or it is a negative, then your answer must be wrong.

Slide Five
The fifth slide is also a continuation of the different vocabulary. In this specific slide we dicussed mainly about the complimentary event. The example he asked was, if there was a sudden death single tennis tournament and there were 54 players, how many games would be played if there could only be one winner? The easiest way of coming up with the answer is asking yourself, "how many losers would there be" The answer is 53 because only one player could win. Another example is using a coin or a die. Basically, you're looking for the opposite. So if P(E) = a then P(E¹) = 1-a.

Slide Six
In the next slide, we had to determine the sample space when a fair die is rolled once. Since it is possible to roll either 1, 2, 3, 4, 5 or 6, the answer to the question is {1, 2, 3, 4, 5, 6} *note: if you are asked to determine the sample space you must use brace brackets { }*
The next question was to determine the sample space for rolling a six sided die and flipping a coin. You could either use a tree diagram or a chart to find the answer. In this case we first used the tree diagram to find the answer. The first two branches consited of Heads and Tails and each had 6 branches to indicate the numbers on the die. Each number could either get H1 to H6 or T1 to T6 so the answer was 12.

Slide 7
We then had to determine the probability of rolling a 2 when rolling a fair die. The answer is 1/6 because there's only once chance of getting a 2 out of 6 different outcomes. We write this as P(2) = 1/6
In the next question it asked to determine the probability of getting a head and an odd number when rolling a die and flipping a coin. In this case we used the previous chart we made on slide 6 to find the answer. We then found that there are 3 possible chances out of 12 possible outcomes which also redues to 1/4. *note: remember to always reduce your fractions.

I believe I covered most of what we learned and talked about in class, so that means..I'm finished! (: This scribe actually took longer than I clearly expected considering I was running back and forth to my room and my sister's to watch American Idol. But enough of my life..toodles everyone, American Idol isn't over yet!

BTW..Homework is on slides 8-10 and don't forget to study for tomorrow's test! Goodluck and Ciao! OH and.. Next scribe is... "what it do PAUL wall" (:

BOB For Conics

2008-05-21T17:53:00.003-05:00

This was a short unit that I was hardly there for because of the music trip to Moosejaw, which was fun, buut now it's time to get back to school and all that piled up work that I have to catch up on. Whooo.

Missing classes for a long period of time has never been a good thing, especially if it's your weakest, and most especially if it's pre-calculus. But some things can't be helped, I think. Well, either way I get in trouble soo..

The concepts were all there for me. Probably the hardest part of this unit was breaking down the question and figuring out which numbers I could use to make an equation. Of course graphing was hard (when is it not hard), but after you have that image, it makes everything so much easier. Everything could've probably been more straight forward if I was actually there (I'm not much of a self-learner when it comes to math) and that's just my fault I guess. Honestly, I'm not looking forward to this test, but I don't know, I think I've done what I can. Hope for the best I guess.

Hahah, as if I'm the very last one to BOB (I think so anyways), but better late than never. Good luck everyone.

Today's Slides: May 21

2008-05-21T11:09:00.001-05:00

Here they are ...

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BOB: Conics

2008-05-20T22:19:00.002-05:00

Time to Bob! (:

I'm quite happy that this unit was pretty short but I'm not going lie because truth be told, I did have some trouble in a few areas. The most trouble I had in this unit was graphing the parabola and I have no idea why I find that difficult compared to the others. The circle, ellipse and the hyperbola were probably the easiest because it was sort of similar to each other. I suppose I had trouble with the parabola because the formula seemed much different compared to others because it included 4p. Hopefully, with a bit more practice I'll feel much more comfortable graphing them. (:

The best part of the unit was folding the paper and visualizing the geometry in each conic section. I would probably have a harder time understanding the whole unit if Mr. K just explained each conic section without folding the paper. Though it took a lot of time and made me quite frustrated beacause the paper wouldn't fold the way I wanted it to, it helped me a lot to understand the anatomy of each conic.

Overall, this unit was pretty straight forward and hopefully I'll do better on the test on Thursday than the pretest we had today. *cross fingers and toes* Goodluck to everyone! Ciao! (:
PS. school's almost over! yay!!

Conic Workshops and Pre-Tests

2008-05-20T19:58:00.002-05:00

This will be basic and short as I don't really have any time to spare.

Our first class consisted of

A workshop on Conic word problems

Our Second class consisted of a Pre-Test.

Our first class was pretty straight forward and Mr. K just put us in groups to solve a number of word problems. When we started on the first one. Some of us had an answer, but we didn't put it up, whether it was right or wrong. Mr. K pointed out that we were afraid to get out our Ideas because most of us, if not, all, are afraid of getting the wrong answer, and we'll be "bad people". He also pointed out that we shouldn't be scared in getting our idea's out, because it wouldn't make us "bad people" and if everyone was always right, we wouldn't need to attend the class. Because how else do you learn? By making mistakes. So take that into consideration.

Anyways, we worked on a few questions but we sometimes made it a little complicated by looking at it wrong. Remember Mr.K's block of wood. Look at it in different perspectives. I won't really go into detail, and you can more or less pick your parts and pieces from the slides.

In the afternoon class, we had a Pre-Test, and again, the answers are posted up on the slides. Our test is on THURSDAY. Do recall that Mark has kindly posted Links for us to brush up on our Conics. I know I'm gonna hit those links up later, so you should too :) Mr. K also has practice exam's and exam information in his LINKS section (because you know, the exam is in a bout two weeks roughly) Sorry this is short and not into detail but I have to go. Kthxbye!

Oh right.. Scribe.. The next scribe shall be... ROXANNE :)

Today's Slides: May 20

2008-05-20T14:46:00.003-05:00

Here they are ...

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BOB for CONICS

2008-05-20T00:54:00.004-05:00

Here is my BOB for the unit Conic Sections.

What more can I say than .. "I've missed too many classes" ..
Although I've missed probably more than half the classes for this unit, it really seems like a fun unit. It seems like a unit that involves lots of algebra and equation handling. Anyways I'm hoping to participate in this last couple of days in this math class. Good luck to everyone! Especially me. I've read the blog, but it might not be enough. Hope I can keep up.

- This is Elven, cheers.

Conics quizzes

2008-05-19T23:17:00.003-05:00

Well, since i heard that the test for this unit is coming very soon I decided to compile a bunch of links that will help all of us to study, so without further ado here it is:

Link 1
Link 2
Link 3
Link 4
Link 5
Link 6
Link 7
Link 8

I hope that those links will be very useful to all of us on the upcoming test and especially the provincial exams which is only 20 days away. Leave a feedback if any of those links doesn't work.

-m@rk

Scribe List

2008-05-19T22:09:00.008-05:00

Cycle 5

~~Francis~~
Joyce
Eleven
benofschool
roxanne

~~JamieNeRd123C~~
zeph
AnhThi
~~Richard~~
~~Hi I'm Justus~~

nelsa
~~Rence~~
kristina
Paul

* Note; Rence is crossed off as he is scribe today [May 20th].

Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

"'Cause it hurts me soul" by hanimeh

Stay Phi everyone,

Rence ~ Out

Bob for Conics

2008-05-19T12:34:00.003-05:00

*Sigh* What a nice time to have a long weekend. Now I can bob early.

Anyways, this unit was really fun! Not to mention it was also short and simple. What really helped make me understand this unit more was the folding exercises with the ellipse, parabola, and hyperbola. Yes it really did help I'm not just making my bob extra fluffy this time XD. Folding was great for homework assignments too! I usually have trouble with anything that deals with graphing but not in this case. I found it really simple and *coughs* I enjoyed doing them. But what I'm worried about is mixing up all the standard formulas for the ellipse and hyperbolas because they're so similar. Other than that there wasn't anything that troubled me surprisingly and I'm really confident with this unit! Hopefully I'll still feel the same way about this unit after we get through word problems.

BOB Version 6: Conics

2008-05-18T19:08:00.003-05:00

Haha I like Ben's creativity... Unfortunately, time isn't a luxury for me.

Anyways, this unit was nice and short, even though the unit had an emphasis on graphing. Once I was able to visualize the graph in my mind's eye--the expression that Mr.K likes using--graphing the graph made things a lot easier, including the origami portions of the unit as well.

Along with remembering the terms, remembering the equations of the circle, ellipse, parabola, and hyperbola is a bit of a doozy, but mathematics is the science of patterns, and contrasting and comparing the formulas between the horizontal and vertical orientations really helped.

I really liked the shward fiasco, the horrible dubbing by DJ K, and his whole samurai act. We may have been fooling around excessively during that class--like what Benz says--but that's what gives this unit its uniqueness. (And maybe there's going to be more unique classes like that when we enter AP Calculus.)

There were some questions left unanswered though, like how golf is evidence that aliens visited our planet Earth and how the focus of the circle is infinity, but if time really isn't a luxury for us, then I guess it's understandable, seeing that THE PROVINCIAL EXAM IS ONLY 14 SCHOOL DAYS AWAY.

If I were to rate this unit, I would give it 5 stars, for being nice, short, unique, and fun.

P.S. Vote ZEPH for V.P.

Conically BOBING

2008-05-18T11:59:00.004-05:00

The Conics Unit was pretty simple. The thing that really helped me in this unit was actually visualizing the geometry of each conic section. I started to see how each formula was related to the sections. Overall I'm doing fine so far in the unit. Just like what happened during the quiz. The circles and ellipses were easier to work with than the parabolas. I guess it was because that the
circle formula didn't change at all and that the ellipse formula
was very similar to the circle formula. The parabola
formula was a bit of a change. The a became 4p
and there are 2 binomials now. But it is like
what they all say, "Practice makes
perfect!" The Hyperbola
is still pretty new to
me but what I
noticed
was
that it
used a lot
of our older
analytic geometry
skills to graph the section.
It used the linear graphing skills
as well as many others. Now for my
thoughts about the class. I'm sorry if I am
being a bit mean but I would like it if we stayed
on task a bit more. It just seems like we were fooling
around quite a bit excessively, especially during the second
period. We would get a lot more done if we stayed on task. I don't
mind if we do have a little fun. But it would be great if we toned it down
just a tad. That's all I really have to say. I'm looking forward to the DEVs and
the exam. I'm not really nervous about it at all. As you can see I'm having some fun with my BOB. I am also looking forward to the word problems with conic sections as well as the test that is coming up. So until next time good luck on the test and YAY Calculus is going to work out. Calculus class of the future. Bye and TABBERNACK

WORKING WITH A(N) HYPERBOLA - CONCLUSION [beware shameless advertising.]

2008-05-17T14:26:00.004-05:00

Sorry the post is late, but I guess this is why I needed the weekend. Thank you Justus. Nice doodle there, by the way. I don’t think I can incorporate images of the slides in my post, but I’ll still try to use some form of an outline and I’ll also parody a certain [we]blogging method. Yes, I said PARODY. I thank Richard for this.

On Friday’s class, our goal was to conclude our lesson about the hyperbola. We leaed to: rn

Write the standard forms of a hyperbola [horizontal and vertical]
Find the similarities and differences between vertical and horizontal hyperbolas
Find the Pythagorean theorem property in hyperbolas
Finally, apply everything we’ve learned by graphing and finding the coordinates required in every hyperbola.
As well as that, we also had a sort of English epiphany, but we’ll discuss that later.

STANDARD FORMS

On Thursday, the standard formulas of a hyperbola were:
HORIZONTAL: [[(x – h)² / a²] - [(y - h)² / b²]]
VERTICAL: [[(y – h)² / a²] - [(x - h)² / b²]]
As soon as the class first caught a glimpse of the equations, we wondered why it was the same as the ellipse, but as we examined closely, there were differences as well as there were similarities between them.

SIMILARITIES & DIFFERENCES BETWEEN HYPERBOLAS

SIMILARITIES:

they have the same terms
they both equal 1
both have the same denominators
both have square binomials
(h, k) indicates centre

DIFFERENCES:

numerators have different orientation
numerators have opposite signs

We also emphasized the differences between the ellipse and the hyperbola by cloning the slides from the ellipse and altered them. We rewrote the formulas and redrew the diagrams, interactively of course. There was even a point when Mr. K was going to redraw a diagram that was on the other page. For the sake of elegancy that rubbed off on the class, they argued for him to clone it, cut it, paste it and shrink it…[that’s sounded like I was singing a new version of “Technologic” or something haha DAFT PUNK man. All the way.] The point is “Just Shrink It”, man. That should be the new Nike slogan. Jeebers, I shouldn’t call out on Mr. K, it’s not even June yet. By the way, it was deemed necessary to rotate the “O” in the graph.

Anyhow, we then realized that a² [pronunciation: “ay squared”] could be smaller than b² [bee squared] after seeing the animations that proved so. These diagrams revealed what would happen if there were variations in the lengths of a [semi-transverse] and b [semi-conjugate]. There were also additional animations for the variation of points h [ey-ch] and k [kay]. “h” moves center and hyperbolas horizontally because it is an “x” coordinate. “k” moves center and hyperbolas vertically since it is a “y” coordinate.

The class then asked themselves this question: if a² can be greater than b², how will a vertical graph be distinguished from a horizontal graph? The answer lies in this statement:

If the “x” coordinate is positive and the “y” coordinate is negative [y being subtracted from x], then the hyperbolas are HORIZONTAL. If the “y” coordinate happens to be positive and the “y” coordinate negative [x subtracted from y], then graphically, the hyperbola is VERTICAL.

PYTHAGOREAN PROPERTY OF A(N) HYPERBOLA

This is where the English lesson came in. Mr. K insisted that it was “an” hyperbola since using an in a consonant was true in “an hour”. We realized that the English language is at times illogical and contradictory. If all of the rules were followed, we would either choose to have the silent letters or disregard them in the words: “talk”, “milk” and “walk”. Imagine what the Hulk would sound like. OMG Edward Norton as the new hulk. Unexpected, but I'm okay with it. By the way, you don't want to make Mr. K mad. He'll turn green [environmentally friendly?]

I'm such a subliminal message. But let’s not derail the train of thought here. The formation of a(n) hyperbola on a graph depends on a rectangular box, which forms a right angle triangle inside. The legs of the right triangle are composed of:

semi-conjugate + semi-transverse = c [the hypotenuse also equal to the length of the center (O) to a foci point.]

simply, this is where the Pythagorean [pie-tha-gore-eeiy-an] theorem is applied; a² + b² = c2

APPLICATIONS OF SKILLS ACQUIRED ON FRIDAY BY SCHWINGING THE SCHWARD.

Then we spent the remainder of the class solving a series of hyperbolas in two different perspectives: graphically and symbolically.

FOR EXAMPLE ON SLIDE 7:
To start we find the standard form by multiplying by (1/9) and (1/25) on each side
leaving us with [(x² / 9) - (y² / 25)] = 1

From this equation, we can derive the rest:

Center (0, 0) because nothing is being added to either of the x or y coordinates.

TRANSVERSE AXIS is found by finding the square root of 9 (a²) and multiplying by 2 (2a), in this case 2a = 6

CONJUGATE AXIS is found by finding the square root of 25 (b²) which is 5 (b) and then multiplying by 2. 2b = 10

VERTICES are the distances from the center to the vertex of both parabolas in the hyperbola. It helps if you know the length of the semi-transverse axis the coordinates of the vertices are (3, 0) and (-3, 0)

To find the FOCI, we start by recalling that they are the same distance away from the center as the hypotenuse of the right angle triangle formed by the semi-conjugate and semi-transverse axis serving as the legs of the triangle. Therefore we use a² + b² = c².
In this case, 9 + 25 = c²
c² = 34

Find the root of that and add/subtract from the x coordinate of the centre because this is a horizontal hyperbola, resulting in (root 34, 0) and (-root 34, 0)

Finally the asymptotes that help graph the two parabolas in a(n) hyperbola, it is important to recall grade 10 pre-cal where we spent time learning about oblique lines and how we find the slope. y = mx + b or simply "rise over run" and use a corner of the "box" made, meaning that we use the two legs of the triangle which happen to be the length of the semi-transverse axis and semi-conjugate axis. Rise 5, run 3 in both directions.
ASYMPTOTE = (5/3) and -(5/3)

Sorry if the post is messed up, but the library computers are timed... As for the next scribe.. I can't doodle, but nevertheless I dub with my very own schward the next scribe Rence.

PS. [post scribe, hahah] I know I said I was a hopeless apolitic when it comes to school elections and that I'd rather not vote. But I still encourage all of you to vote JOSEPH vice-prez.

Today's Slides: May 16

2008-05-16T11:17:00.002-05:00

Here they are ...

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Hyperbolae Unveiled...Sorta

2008-05-15T23:24:00.010-05:00

Alright guys, no shenanigans this time, I'm tired, just got off work, and would like to sleep. So I apologize if I'm a little bit frank, and a little less funny an animated in my scribe post compared to my usual ones. Hopefully the quality is still there though, if not, I'll fix it up tomorrow or w/e. Alrighty, here we go.

*note* links added under appropriate images for full sized goodness. Like I mean, it wont matter on the smartboard, since that things huge, but at home I doubt everyone has like 70 inch moniters, so ja. Thats why there there. Why arent they hotlinked or w/e its called? Cause its 1:24am haha.

Morning Class

So this mornings class wasn't much of anything special, which is kinda different from the norm (kinda paradoxical I know.) Anyways we started going through Mr. K's slides, and we solved the little equation for one of the friendly Ellipses which can be seen on the following slide.

As always, fully more large sized versions of these slides can be found in the slides themselves. Which Mr.K posts quite consistently :P

Moving on. After that we filled out a health survey. That basically took up the rest of the morning class. However, we did have a moment to do a quiz, which was marked the following afternoon class. If you were not there *cough cough* then talk to Mr.K, see if you can write it. The solutions are on the slides. If you were here, then you don't really need an explanation right? Cool.

Moving on.

Afternoon Class

The afternoon class was slightly more work oriented, and was not interrupted by some unnecessary forms/paperwork.

We began by correcting the quiz thinger from the first period.

Some usefull things to take out of that include the following.

How to distinguish Various conics from their equations.

-If the Equation has an x² term but no y² term it is a Horizontal Parabola
-If the Equation has an y² term but no x² term it is a Vertical Parabola
-If the Equation has an y² AND an x² term, with coefficients that are THE SAME, it is a Circle.
-If the Equation has an y² AND an x² term, with coefficients that are DIFFERENT, it is an Ellipse.

Remember those :)

Next we took out our previous nights homework (that being measuring, and finding the difference, of the lengths from one point on a branch of the hyperbola, to both foci. That will make more sense when the image comes up in a moment.)

The results of taking up that homework are shown as follows.

Okay, I'm going to try to explain this as best as I can, but like I said before, I just got off work, I'm crazy tired, not running on anywhere near enough sleep and a host of other not so great things. So if I phail (yes thats a "ph" fail. That doesn't mean an acidic fail but rather, means a really bad fail for all you en zero zero be's out there.)

So, the homework was to pick a point on one of the branches of the hyperbola, and measure that distance from the point, to both foci (that is to say, measure from say, point P, to F₁, and then from point P to F₂). We then too those values, and found the difference, (PF₁-PF₂). Voila! homework done. By now it should have come as no surprise that the result were values which should have been basically the same.

Once we did that, things started to get tricky. The next thing we did, was find the Vertices (sp?) of the hyperbola. To do this, you take your ruler, and line it up so it goes through both focus points. The POINTS at which the line made by the ruler intersects with the curve of the hyperbola make up it's vertices.

Okay, so if you followed my last paragraph's directions correctly, you should end up with two points at what are the vertices of the hyperbola. The next step is to connect those two points with your handy dandy ultra straight ruler. This creates what we called the TRANSVERSE AXIS. This axis can be described as the line which connects the two vertices of a hyperbola, and is seen AND labeled in green on the slide crop up there. Got it? I hope so. Moving along.

Next, we bisected this transverse axis.

"How do we do that Mr.k?"
"Well I'll show you."

Thats a rough translation of what probably everyone was thinking. Except the Ogre 1 and 2 of our class. They knew the answer already, (Outside joke.)

So to bisect the transverse axis, we folded our papers for one more time. This time, we folded one focus point, directly on top of the other. Thus we obtained a crease which defined not only the bisector, but the perpendicular Bisector of the Transverse Axis. We see this as a vertical red line on the image above. Lets step things up a notch (I know Mr. K did.)

So, do you remember the Transverse axis? and how we just cut it in two? Were going to take this line, and modify it a little bit. Using our rulers again, we measured the line from one of the FOCI (in this case F₂) to that vertical line (which we now call the CONJUGATE AXIS.) This gave us some length.

Now this next bit is tricky, we took the length we measured from the PREVIOUS STEP (that being from the conjugate axis, to Focal point F₂) and moved it so that one end of it was on the (whatever the singular of vertices is. Verticii?) of the nearest branch, and so that the other end of the length lies on the Conjugate axis. Since I bet your saying "what the frig" I made a quick little paint mockup to get the point across, or rather, to help you "see" what I mean.

http://i24.photobucket.com/albums/c21/skyL193/Slide3.jpg

Okay, so as I wrote on the picture, the light green is the length Conjugate Axis to F₂. Then by moving it like I mentioned before, we get the makings of a triangle.

Step 3. Measure the distance from the point where the light green line meets the conjugate axis, to where the dark green axis meets the conjugate axis. This we can call length b for explanations sake. The next step is to take this length, and go from the intersection of the Transverse axis and the conjugate axis, (aka the cross in the centre) and "b" units down. I'll show you what I mean again.

http://i24.photobucket.com/albums/c21/skyL193/Slide4.jpg

So, as I wrote again, the light blue, and dark blue lines are the ones I was talking about before, and they are of equal length. Thus we get those points G1 and G2. This brings us to our next step.

By drawing perfectly 90 degree lines from the vertices (going through them, vertically) and from the end points of the light and dark blue lines (going through points G1 and G2, horizontally) we can construct a box of sorts. This is seen once again on the slides.

http://i24.photobucket.com/albums/c21/skyL193/Slide5.jpg

So you see the box right? The black dotted lines there are actually it :P So yeah after that we took our ruler again, and drew that orange cross that goes far off the page kinda. We drew that cross by going from one corner of the box to the other (and beyond.) After we had done this Mr.K revealed to us that those orange lines, are actually asymptotes for the graph. In other words, neither branch of the hyperbola will actually ever touch those orange crossey lines.

The final slide here, is the equation of the hyperbola, along with some similarities, and differences, which we came up with together, to help us remember the equation, and keep it distinct from the others.

http://i24.photobucket.com/albums/c21/skyL193/FinalSlide.jpg

To Conclude: The basic parts within the anatomy of a hyperbola
- Transverse Axis.
- Conjugate Axis
- Focus Points
- The Asymptotes
- And finally, you should know the equation, which is shown in the above slide image.

Alrighty, I think that about wraps this blog post up. I'm not entirely sure, because at 1:00am I kinda start to get a bit loopy and my thoughts aren't so coherent (thats probably evident based on how my writing progressed.) If you guys have any questions or whatever lemme know. I'll try to answer them, and post em up. Or if you find any errors, or know something I dont/forgot to add, lemme know so I can add it.

Well I'm off, but of course, not without letting you know who the next scribe poster is :)

http://i24.photobucket.com/albums/c21/skyL193/JamieSlide.jpg

Remember, vote for Zeph for VP.

K? Kthxbai

Justus- FINALLY out.

Today's Slides: May 15

2008-05-15T15:50:00.001-05:00

Here they are ...

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Hyperbolas

2008-05-14T23:17:00.003-05:00

Today's class we didn't do a whole lot of anything other then hyperbolas. We recapped on the equation of an ellipse, refer to last class: Richard's scribe post.

We learned about how to put subscripts and super scripts, that are used to right those tiny numbers, or letters, that are used to show various things, such as an exponent on a power
Example: 2¹ - The 1 in this equation would be the super scripted number.
Use: This is given with <.sup> and end with <./sup> Get rid of the periods.I put them in so it wouldn't actually work.
Example2: log₂ - The 2 in this log would be the subscripted number.
Use: This is given with <.sub> and end with <./sub> Get rid of the periods. I put them in so it wouldn't actually work.
This is a pretty useful html code to use. Remember you can only use these codes in the blog if you are in the "Edit Html" tab, shown beside the "Compose" tab at the top right, just above the various posting tools (spell check, insert picture, etc.).

We were given 1 choice of 3 different questions that we were asked to solve. We started by evaluating this equation: 16x² + 9y² = 144

The first part of the question was to change it to standard form, which was quite easy. To do this, you would have to reduce the coefficients so there wouldn't be any on the x or y values. This could be done by multiplying each side by (1/(16)9). By doing this, it would reduce the coefficients, giving us the standard form of the equation: (x²)/9 + (y²)/16 = 1.
For the 2nd part of the question, we were asked to find the centre, major axis, minor axis, vertices and the focii points.

Centre:
The centre being (h,k) but because there is no h and k values in this equation, the centre is at (0,0).

Major Axis:
the major axis being 8, because 16 = b², and b is the length of the semi-major axis from the centre to one vertices, so square root 16, to get a semi-major length, which is 4, then multiply that by 2, to get the other semi-major axis, which gives us the major axis(two semi-majors = 1 major), which is 8.

MinorAxis:
Same idea with the minor axis, but using the a² value, which is 9, square rooted giving us 3, which is the semi-minor axis, then multiplying by 2, to get the minor axis, which is 6. the vertices would be the endpoint of the semi-minor axes, and semi-major axes, starting at the centre (0,0).

Vertices:
The vertices of the minor-axis spanning a value of 6, from the centre and outwards horizontally, because it's a vertical ellipse, due to the fact that b² is below the y-value (need more info?, refer to Richard's post). So if a value of 6 was spread outward horizontally, from the centre (0,0), it would give the vertices a value of (3,0) and (-3,0). The other vertices would come from the value, of 8, spanning outward vertically from the centre (0,0) giving the other vertices points at (0,4) and (0,-4).

The 3rd part of the question was to graph it: (I'll edit this in later, when the slides come out, so you can see the actual graph)

Hyperbola:
Finally, is the hyperbola, the last section of todays class, we made these hyperbola by folding a piece of paper. It started off by having a circle that was off centre on the paper, and the circle had a given centre point. We were asked to make a point about 2 cm off, out of the circle. If you were to draw a line perpendicular to the edge of the paper in landscape view that bisected this centre point, the point out of the circle should be close to touching this line. We were then asked to draw atleast 25 dots with 5 of the dots bunched in the area on the edge of the circle closest to the outside point. These points should be touching the edge of the circle. We then folded all these points, onto the points outside of the circle. This gave us an outline of a hyperbola, with the centre point of the circle, and the outside point the focus points of the hyperbola. Voila a hyperbola.

For homework, we were asked to pick a point on the left branch of the hyperbola and 2 points on the right branch of the hyperbola, after that, we measure the distance from this point to both of the focal points, then find the difference. What do you get? Find out.

That was pretty much everything we did for today's pre-cal class, and now I feel like a world-class math shaped origami teacher. I''m sure everyone feels this way, after we folded 3 different different shapes. I can't complain though, I found it quite fun. I propose the next scribe will be: eeeny meeny miineyy, Justus. Surprised?.. I thought so.

Until next time,
Francis

Today's Slides: May 14

2008-05-14T15:35:00.003-05:00

Here they are ...

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The Whisper Room

2008-05-13T17:44:00.014-05:00

The Whisper Room and What Does it Have to do with Ellipses?
The Anatomy of the ellipse.
The Vertical Ellipse
Horizontal and Vertical Ellipse Equations
The Pythagorean Property
Focal Radii Property
Sketching an Ellipse when given the equation of an ellipse and finding the length of the Major and Minor axes, The coordinates of the center, Foci's and vertices's.
From Standard to General Form and Back Again ...

The Whisper Room and What Does it Have to do with Ellipses?

Now to start of with the question What Does The Whisper Room Have to do with Ellipses? Well the whisper room is a room found in the white house where the president of the united states lives. This room also has two marked point on the floor of the room. If you were to stand on one of those points and a friend, aunt or uncle, or anyone were to stand on the other point and if you were to say something, the other person on the other point would hear it too. Even if you whispered and talked towards any where (ex the wall) he would still here it and anyone else in between wouldn't hear it.

On the picture above or found on slide 2 of today's slides is a picture of how the whisper room in the white house works. And if you are wondering these points are called Foci points.

The Anatomy of the ellipse.

On The picture above you are probably wondering what are those numbers for. Well the numbers are our classmates measurements of a point that we picked on the edge of the ellipse, to each of the foci's, and then we added them together. We then came to a conclusion no matter where the point is the measurement of the point to each Foci and added them together the sum would be the same.

On slide four is just the The Pythagorean Property. But I will discuss that later in my blog.

On slide Five on just below the writing there is a beautiful diagram of the Anatomy of The Ellipse.

The Vertical Ellipse

Which is pretty self explanatory. If you need a visual just talk a circular device and put it in between your hands with your fingers pointing up and try squishing it together. And that is how you get an ellipse. If you need it more descriptive its where the minor axis is horizontal to the floor and where the major axis is vertical to the floor. If you do not understand me there is a vertical ellipse found on slide 6 of today's slides or you can just look at the picture below.

Horizontal and Vertical Ellipse Equations

The Horizontal and Vertical Ellipse Equations are almost identical but there is one thing different. On the horizontal ellipse equation it is read as: open bracket x - h closed bracket squared all over a(ey) squared plus open bracket y - k closed bracket squared all over b(bee) squared equals one . And on the vertical ellipse the a(ey) squared and b(bee) squared are switched. So it is read as: open bracket x - h closed bracket squared all over b(bee) squared plus open bracket y - k closed bracket squared all over a(ey) squared equals one.

If you need too see it written in numbers it is found on slide number seven of today's blog or just below.

The Pythagorean Property

On Slides 4 and 8 we talked a little about the Pythagorean property Regularly the Pythagorean Theory is a^ 2 + b^ 2 = c^2 but in this case of an ellipse it is c^2 + b^ 2 = a^ 2. If you don't believe me you can look below or check on slides 4 or 8 of today's slides. You need to use this property so that you can solve for the Foci because the foci is equal to c units from the center of the ellipse. Well to solve for c you will of course need to rearrange the equation which is c^2 = a^2 - b^2.

Focal Radii Property

On slide nine of today's slides and just above us there is the Focal Radii Property in action. On the first line it says the distance from any point of the circumference of the ellipse to the first foci plus the distance from the same point to the second foci is to be constantly the same. Now on the second line it says that the distance from P(pee) to foci1 plus the distance from P(pee) to foci2 is equal to the distance of A(ey)1 to foci1 plus A(ey)1 to foci2. So if that is true then the distance from A(ey)1 toFoci1 is equal to the distance of A(ey)2 to Foci2 by symmetry of course. So if that is true then the distance from P(pee) to foci1 plus the distance from P(pee) to foci2 is equal to the distance of A(ey)2 to foci2 plus A(ey)1 to foci2. And now if that is true the distance from P(pee) to foci1 plus the distance from P(pee) to foci2 is equal to 2a.

Sketching an Ellipse when given the equation of an ellipse and finding the length of the Major and Minor axes, The coordinates of the center, Foci's and vertices's.

There were some examples of these type's of questions found on slides 10 and 11 I am only going to go over one because they are practically identical on how to solve it.

If you are given an equation the first thing that i do jot down the center on the graph and on the side of the graph , Where on an ellipse the center is (h, k) in this case the center is (2, -1). The second thing i would do is check which denominator is bigger so you know which one is a(ey) and which one is b(bee) you do this to solve for the major and minor axis. The major axis is 2a (two of ey) The bigger one in this case is 16 SO you square root it and then double it which you will get 8. For the minor axis you use the smaller number b(bee) but this time it is 2b (two of bee) so you do the same process and this time the number is 9 which become 6. To solve for the Vertices you find the end points of the major axis. in this case the Vertices are (2, 3) and (2, -5). Now the hard part The foci is tuff because you need to a little more work. This is also one of the various reasons for the Pythagorean Property. To solve for the foci point you will need to know c because the two foci points are c units away from the center. first you write down the Pythagorean Property which is c^2 = a^2 - b^2. You then plug in your given values and solve for c. In this special case the value for c is The square root of 7. The answer is found in green and red they are both correct but we would perfer to use exact values. so the one in red is the best fit answer.

Oh i almost forgot sketching the graph first you jot down the center. After you have done that you then find out if it is a vertical ellipse or a horizontal ellipse. now that you know that you can then jot down four important points. You jot down the two vertices and The two endpoints of the minor axis. For the two vertices you take half of the major axis which is the semi major-axis and then you add it to the x or y coordinates of the center depending on if it is a vertical or horizontal ellipse and you then subtract it to the x or y coordinates of the center depending if it is a vertical or horizontal ellipse. Now for the two endpoints of the minor axis you take half of the minor axis which is the semi minor-axis and then you add it to the x or y coordinates of the center depending if it is a vertical or horizontal ellipse and you then subtract it to the x or y coordinates of the center depending if it is a vertical or horizontal ellipse. After you have those points you then can draw your ellipse.

From Standard to General Form and Back Again ...

on slide 12 or just below this text is a perfect example on how to go from standard to general form and back again.

If You have an equation in standard form and you want it to make it into general form. The first thing that you would do is get rid of the denominators by multiplying it by the products of the denominators. You then expand the equation. After that you then distribute the term outside of the bracket. Then after that you rearrange the equation with the Squared variables(x^2) first then the none squared variables(x) and the the number(1)s to equal 0(zero). And there you go you get General form

Now if you have an equation in general form but you want to make it into standard form you need to first collect the like terms like (x^2 and x ) you then factor out the most common factor. After you have that you then a have some missing numbers so you then have to complete the square by taking the middle term and dividing it by 2 and then squaring it. After you have done that remember to add it on to the other side. also keep in mind to distribute the numbers. Example 16(x^2 +6x + 9) = 0 You don't just add nine to the other side you add 9 times 16 which is 144 big huge difference. after you have that you then simplify your equation and then divide it by the product of two factors on the outside of the brackets. And there you go you get standard form.

The next scriber is going to be Francis.

Today's Slides: May 13

2008-05-13T15:33:00.004-05:00

Here they are ...

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Scribe List - Cycle 5 -

2008-05-13T00:57:00.003-05:00

Cycle 5

Francis
Joyce
Eleven
benofschool
roxanne

JamieNeRd123C
zeph
AnhThi
~~Richard~~
Hi I'm Justus

nelsa
Rence
kristina
Paul

* Note; Richard's crossed off as he scribe today [May 13th].

Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

Stay Phi everyone,

Rence ~ Out

Student Voices Episode 3: Chris, Craig, and Graeme

2008-05-12T23:22:00.000-05:00

In this episode of Student Voices three Advanced Placement Calculus students, Chris, Craig, and Graeme, talk about a wiki assignment they did to prepare for the exam. Then the conversation transitions to a discussion of the many things they learned while doing their Developing Expert Voices project. It ends with a challenge, the result of which will be featured in a future podcast.

Let Chris, Craig, and Graeme know what you thought about the podcast by leaving a comment here on this post or on the mirror of this post on their class blog.

(Download File 31.8Mb, 26 min. 30 sec.)

The video mentioned near the end of the podcast is called Daft Hands. Here it is:

Photo Credit: Shadow singer by flickr user EugeniusD80

Bit of Review and...Origami?

2008-05-12T21:19:00.008-05:00

Highlights for today's class are:

-Reviewing how to write equations of a parabola when given the vertex, focus, and directrix.
-Reviewing what we learned about circles from last class
-Origami-ish fun with circles!

This slide was basically a review from last class to see if we really understood the concepts.

When given the vertex and focus, it is easier to draw a sketch of the parabola first to see which way it opens. Once you know whether it opens in a vertical or horizontal way, you will be able to see which equation you should use (either the horizontal or vertical equation).

P equals the coordinate value that represents whether the equation has the y side squared or the x side squared. In the first question where the focus is (4,0), the P value would equal 4 since the x side is squared and the parabola also opens horizontally. To find what 4P is in the equation, just multiply the P value by 4. Also, if the parabola opens in a negative way, 4P will also be negative.

More review from last class. Since the distance between P and F is equal to the radius, we can use the distance formula to find the standard form for the equation of a circle.

We already did the question on the left side last class, so we worked on the one on the right. We were asked to put it in the standard form (circle). After solving it, questions arose about what to do with -2 since the squareroot of -2 is basically an imaginary number. We came to the conclusion that it wasn't a circle since you can't have a negative value for the radius, thus the not a circle.

*Gasp* A circle with two dots on it...and a doodle I made because I was bored. We were all given a sheet of blank paper and were told to draw a perfectly round circle on it. Then we added the center dot and another dot out on the side somewhere near the circumference. After that we made many ticks on the circumference (minimum 25) and made sure that we clumped the ticks together around the outer dot. We then folded our ticks onto the outer dot. The end result was an elipse on the back side of the paper.

Homework is the above slide. Place points P, Q, and R anywhere on the outside of the elipse. Draw a line connecting them to F1 and F2. Add them together as shown and see what happens ;D

Yep, I joined the advertising bandwagon! Watch Batman: The Dark Knight on July 18 (R.I.P. Heath Ledger)! XD......Okay..time for next scribe. Looks like a new cycle! I choose........Richard. Ahah, well then I'm off! Oh and vote for Joseph for vice president! K, have fun ;D

Today's Slides: May 12

2008-05-12T15:49:00.001-05:00

Here they are ...

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Parabolas and More...

2008-05-10T20:55:00.010-05:00

What we did in todays class:

1) learned formula of a horizontal and vertical parabola

2) graphing the focus, directrix, vertex and I almost forgot the parabola

3) completing the square

How we get (x-h)^2=4p(y-k)

From that piece of paper where we did all those folds, we realized that the distance of PF was equal to the distance of PD. So by putting the coordinates of PF and PD into the distance formula we can get the formula for vertical parabolas. As you can see from the slide we first expanded (y-(k+p))^2 and (y-(k-p))^2. Also (x-x)^2 = 0 if you were wondering where that went. After expanding, we see that most of the like terms cancel out except the py's and the kp's. Then after isolating (x-h)^2 and factoring out the 4p from the right side, we're left with (x-h)^2 = 4p(y-k). The equation for the vertical parabola. The same process is done with finding the horizontal parabola's equation except the coordinates are different.

y=a(x-h)^2+k vs. (x-h)^2=4p(y-k)

-the binomial (x-h) is squared on both

-by moving k to the left side of first equation we'll see that 1/a is equal to 4p

-if a>1 we use to say that the parabola would be narrow, however in our new equation when 4p >1 then it will be wide

From looking at the equation we know:

-vertical parabola since equation starts of as (x-2)^2

-vertex coordinates are (2,-1)

-direction of parabolas opening, which is up

-wider than a standard parabola

To find the focus and directrix, we must find the value of p. p is the distance of the focus to the vertex and the distance between the vertex and the directrix. From the equation we can see that 4p=4 and we get p =1. Now we know the focus is 1 unit above the vertex and the directrix is down 1 unit of the vertex. Then you can find the roots by making y = 0 and voila there's our forgotten parabola.

(y+1)^2 = -8(x-3)...equation tells us:

-horizontal parabola since it's (y+1)^2

-4p is negative so opens to the left-parabola going to be very wide because 4p > 1

-v=(3,-1)

*note* kinda tricky because x and h aren't in the same position as last question

Just like last question, we should find p first. p = -2 since we know the -8 = 4p. We'll ignore the negative sign since distance is always positive. So now we know the the directrix is 2 units to the right of the vertex and the focus is 2 units to the left of the vertex.

Completing the Square!

Feels like grade 11 all over again. Anyways, Mr. K said that most of the time equations will be given to us in general form. General form is when all the terms are on one side of the equation. So to get it back into standard form we complete the square. First we get all the like terms to one side. We usually put the variable that is squared along with its like terms to the left so it'll match the equations for horizontal/vertical parabolas. So to complete the square we take the second terms coefficient and divide it by 2 then square it. Now what we add to the left side of the equation we must do to the right. Then we factor the left side and factor out the -8 from the right and we're back to standard form.

Well there is my short and early scribe post. You better be reading this Nelsa *squints*. Anyways, next scribe is.....who to chooose?! Wait there's only Kristina left psha i want more choices jk. Yeah so kristina your next.

Today's Slides: May 9

2008-05-09T15:53:00.003-05:00

Here they are ...

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MY PREMEDITATED COMBINATORIC BOB

2008-05-08T08:27:00.002-05:00

Cominatoric BOB— sounds like a transformer or a robot or something. IRON MAN! eheheh Look at us and our shameless plugs, advertising movies. I am a sublinminal message. Anyhow, I know this is almost really late, but I couldn’t BOB previously because I don’t have internet access anymore. It’s a long story. So I wrote this on Microsoft Word and I’m going to paste it on. I have to admit, this was a fun unit. We learned about “counting” how could that not be fun? But straight to the point. These are only some of the things I learned this unit:

- There IS a difference between PICK and CHOOSE
- A combination lock is not really supposed to be a combination lock, but rather a permutation lock because the order of which the numbers are inputted matter.
- 49 nCr 6 is the actual number of different combinations that can be entered in 6/49 lottery which totals up to 13 983 816. What an investment that is…especially for those depending on it. Sorry for the sarcasm, but I just had to say.
- A circle was once again to be proven as such a special case as we subtracted 1 so that it is the REFERENCE point and then chose a number.
- Pascal’s Triangle [STOLEN] contained many different patterns and sequences, especially sequences significant to quadratic functions and whatever they’re called if there are more than just four…They deal with binomials so to say.
- Poker is a game of chance! [not factorial, but I mean it. Don’t gamble. Count cards and play Blackjack. It works…Winner Winner Chicken Dinner]

On the other hand, we all have our troubles, and these were some of mine:
- I still sometimes have a hard time knowing whether or not the problem deals with factorials or not. Because it could be using exponents like the one question in the pre-test with the number of questions vs. the number of answers. (4^20)(5^10) or something… with four being the possible number of answers to the twentieth because there are 20 questions classified this way and five to the tenth.
- Remembering how to calculate the chances of getting each type of poker hand when you’re considering what suit it is, how many there are, etc…
- When to normally multiply and not use factorials, I think it was also something Mr. K mentioned also.

I know this BOB was a little paint-by-numbers but I think I reached the objective of why we do these BOB’s. I didn’t actually entertain much, but that doesn’t matter because I did actually REFLECT today and I hope I’m ready for the test today. I’m going to slash it with my conic shward. You have to love subtitles and dubbing.

BOB Version 5: Combinatorics

2008-05-08T06:03:00.003-05:00

I didn't like this unit. This unit wasn't very orderly like logarithms where there was sort of like a "process" to answer a question. Studying this unit felt studying like something that's abstract, but nonetheless, this is math. Although there wasn't any graphing to do...

I was a bit slow in this unit at first (and yeah I'll admit I struggled a bit), but I managed to learn the basics of permutations and combinations. The poker combinations took a while to wrap my head around, so I may still not be comfortable doing those types of questions, although being scribe for the class on the poker combination class really helped.

On the other hand, learning about phi, the Fibonacci sequence, and Pascal's triangle was... is eye-opening a good word?

Well, I hope I do at least satisfactory on the test today. Good luck.

Bob(ing for Combinatorics...instead of apples)

2008-05-08T00:25:00.003-05:00

Haha I just realized I make up my bob titles as I decide to write the bob itself. Thus the title is basically just a direct brain to keyboard link of the first thing that pops into my head :]

Anyways to the point, the unit combinatorics (also known as counting, for those who missed that little piece of unit defining info.)

For starters I'd like to say that I thoroughly enjoyed this unit, a first for any of our units to date. Now that isn't to say that Mr.K didn't teach the other units adequately, I just had to really work at those ones to even get semi decent at them. This unit on the other hand, came kinda easy to me (some of it anyways) and thus, I enjoyed it!

Some of the things I liked about the unit were...
- The practicality of it. This unit gave me alot of knowledge that I can *gasp* actually use in REAL LIFE!!!!11!111!!!11!1!
- The fact that it wasn't really concept, or practice, or anything extensive, but seemed to have just the right amount of everything in it (unlike some other units *cough*logarithmsandUNFUNexponents*cough*)
- All the clever little things we learned along the way, that were totally unrelated to the unit itself. (We should have more of these)
- All the clever little things we learned along the way, that were actually related to the unit itself (and these also)
- The whole unit? Almost

On the other hand (or the second option here, which would be 1 choose 1, because I already went over the first topic.)

I didn't so much enjoy
- The formulas. Overall I found them bulky, and annoying, and hindering to my understanding of the unit as a whole. They just made everything more complicated I think.
- The algebra involving the whatever choose something else, and or the whatever pick something else. The whole factorial deal kinda gets me sometimes (errors with my algebra cause of the ! in the middle of it. Makes no sense I know.)
-The fact I left this bob till so late (I was horseback riding for my sisters birthday party. There goes half my day D:)

So ja, overall this unit gets 8 out of 10 justus league points, not 10, because I didn't much enjoy some of the algebra involved in it but meh, its pre-cal, OBVIOUSLY theres going to be algebra, so I'll just get over it then.

I think thats everything, so I'm going to study for awhile, then sleep, then hopefully mega super power up overshield bxrrxyy final boss strongside ogre 2 wavedash fox shine infinite aleph not pwnzor the test tomorrow. I wish you all the same amount of goodness on your tests :]

Goodnight! and Ciao!

Justus - out, and studying now.

Edit: ohmg! I just remembered I'm missing like a bajillion delicious box links. I'm super sorry guys I forgot all about that thing until I scrolled down and saw it D: I promise to go and find a bunch of super kewl links tomorrow, like lots of them, cause I owe you guys -_-; Hopefully you can find it in your textbooks (or TI-83's) to forgive me. :]

Okay, now I'm really gone :p

BOB FOR COMBINATORICS

2008-05-08T00:21:00.002-05:00

This unit was very, very confusing. From beginning to end, it didn't seem to let up. Well of course that depends on me as well. I basically missed a quarter of the lessons which really hurt what I could be capable of learning. My brain literally feels empty. I mean we might have not gone through a lot of workshops or work thoroughly with this unit. Or perhaps I wasn't just here, and haven't been doing my homework.

I wouldn't say that it is hard to catch on. For me though, having to miss many classes and was not catching up, had even more trouble. It was slow for me to come up with solutions and ideas to solve problems.

I found the poker combinations fairly easy though.

Overall this unit for me was a failure. I shouldn't let this slip away again. I've got homework to do!

Cheers!

BOB For Combinatorics

2008-05-07T23:32:00.003-05:00

well this is my bob for Combinatorics. When we started this unit i thought that counting should be a piece of pi ah ah get it (pi 3.14 ) anyways this unit was a lot harder then i thought it would be. well it wasn't harder then logs but it was still a little confusing.

One of the things that confused me was the circle table questions. another was that zero is definitely a problem. And the one that really caught me of guard was that if it is bracelet it is a whole new ball game.

and now that my xbox is broken because it has got the red rings of death. i guess i will study now

Richard signing off.

THE MiTSUBISHSONYTOSHIBaHONDATOYOTASAmSUNG PLAYSTATION PANASONIcCASIoJVCCLARIonYAMAHAAAAFUJIKODAK SANYOHITACHI.. BlOG

2008-05-07T22:34:00.009-05:00

Well first off.. I'd just like to say that this blog isn't entirely all about japanese brands, or brands that sound japanese. Here's a little guide line of what I will be discussing in this blog. (Again sorry if it is late for you early birds.)

#1) THE NEXT SCRIBE WILL BE .. Zeph! .. cool hey?! Just returning the favor.

#2) Poker Combinations

#3) Pre-Test Thoughts

#4) Conic Sections

.. Because #1) is said, #2) is automatically first.

Instead of starting the class with the pre-test, which was how it was supposed to be scheduled. We discussed last night's homework instead.
Poker hands consist of 5 cards.

We started with the FLUSH.
A flush is getting any card in any order, but all 5 cards having the same suit.
So we use 13C5 to pick 5 cards in 1 suit.
4C1 to pick all cards in 1 suit
and because we are only looking for flush NOT INCLUDING straight flushes and royal flush.
We would subtract 4x10, because that's all the ways for straight flushes and royal flush.
4C1*13C5-40 = 5108

Next we have the STRAIGHT.
A straight is an order of 5 face cards. A, 2, 3, 4, 5 is an example. A can be low and high.
10 is the number of ways we can have the cards in sequence.
[A,2,3,4,5](1),6(2),7(3),8(4),9(5),10(6),J(7),Q(8),K(9),A(10)
we would want to avoid having the same suit, because then it would be a STRAIGHT FLUSH.
So we have 10, the number of ways we can have the cards in sequence.
4C1 to pick each card for it's suit.
(4C1)^5 because there are 5 cards.
40 is the number of straight flushes.

(4C1)^5 * 10 - 40 = 10200

Next we have the 3 of a kind!
As the name says, 3 face cards of the same kind. 2 other random face cards.
So we have 13C1 to pick 1 face card for the 3 first cards.
4C3 to pick the suits (Can't have 3 of the same suit and face card.)
12C2 to pick 2 more face cards.
(4C1)^2 for the suits of the other 2 cards. (Making sure they are all different suits.)

13C1 * 4C3 * 12C2 * (4C1)^2 = 54912

After 3 of a kind, we have 2 pairs!
A pair means 2. And 2 pair means another 2. So we need 2 cards of the same face card, and another 2 cards with the same face card. And a random 5th.
So we have 13C2 for picking 2 face cards for 2 pairs.
4C2 for picking 2 different suits for the 2 pairs.
11C1 is for the 5th card.
4C1 is for the 5th card's suit.

13C2 * 4C2 * 4C2 * 11C1 * 4C1 = 123552

If we can have 2 pairs, then there must be 1 pair!
Meaning 2 cards of the same face card. Then 3 random 3. Heh that's catchy.
So we have 13C1 to pick 1 card for the pair.
4C2 for the two cards.
12C3 for the other 3 cards.
(4C1)^3 for the suits of the other 3 cards.

13C1 * 4C2 * 12C3 * (4C1)^3 = 1098240

If we do not have any of those .. No pairs.
Although this may seem to be the easiest one to figure out, it was actually the most confusing for the class.
First, we know we can have any card that is not the same, in no order and are all different.
13C5 * (4C1)^5 is the ways to have all different face values.
13C1 * 4C1 is the ways to have all the cards same suit (FLUSH).
10 * (4C1)^5 - 10*4 is the ways to have all cards in sequence.

13C5 * (4C1)^5 - 13C1 * 4C1 - (10 * (4C1)^5 - 40) = 1307428

That is the end of the poker combinations! Now to move on to ..

#3) Pre test ..

Mr K. said there was only one problem that everyone had a problem on. That question was number 4!
"A multiple choice exam has 20 questions each with four possible answers, and 10 additional question, each with five possible answers. How many different answer sheets are possible?"

Each of the 20 questions has 4 possible answers. 1 - 4, 2 - 4, 3 - 4, etc ...
Each of the 10 questions has 5 possible answers. 1 - 5, 2 -5 etc ...
4^20 + 5^10
= 1099511627776 + 9765625
= 1099521393401

Now moving on to the .. very wild afternoon ..
So starting off .. It kinda took a while for everyone to settle down. When we finally did. Mr K. just couldn't help it. He was turning into a tomato! With all our japanese sounding words .. and .. His very bad dubbing .. And the cow bell was it ? eh anyways ..

#4) The Conic Section
How do I explain this? Mr. K used his lightning fast skills and chopped off a cone. Yes A CONE! To show us where the circle, parabola, ellipse and hyperbola came from. As also shown in SLIDE 2.
ANNND THENNNNN .. because my lack of sleep I began to fall asleep.

Next Mr. K kindly handed out some nice white paper for an experiment on our own. Of course skipping all the laughter and stupid jokes popping out of our mouths. Especially something about hamburgers and hotdogs. Going on ..

REFERRING TO SLIDE 3 ..

We found where the parabola comes from. It is curved by the FOCUS POINT ..
Then, on our parabola, we were asked to place another point on the parabola and call it p, and then draw a vertical line from P to the edge of our paper, which is the DIRECTRIX. Then connect P to F. We found out that PD is the same length as PF .. Yay! We know where the vertex is.. It has the same X coordinate of the focus point. Also the lengths in between F and V and V and D are the same. We call this line lower case p.
The Vertex having the coordinates (H,K)
Then the focus point must have the coordinates (H, K + P)
Ya'll Dig? ..
Then that must mean the Directrix must have the coordinates .. (H, K - P)
We don't know P's coordinates, which is why it is (X, Y)

So we know PD is equal to PF .. The unit will continue.

And that's all! One again the Scribe is ZEPH. Good luck on the test everyone. I know I'LL need it.

Cheers!

BOB For Combinatorics

2008-05-07T22:26:00.002-05:00

When I heard counting I thought, "Weeell, how hard could that be?" It turns out, pretty hard. It didn't take me long to realize that counting, was like probability, and I don't like probability. =/ Just like I didn't enjoy this unit.

It was probably the wording or word problems that got me. I also didn't like how sometimes you had to experiment until you got it right, kind of like Trig Identities.. except less fun. It took me a while to understand 'pick' and 'choose'. Well, not really understand, understand, but it took me a while to see how to apply it, I guess. There were a lot of stuff in this unit that made me go, 'what?' at first. It's just, a whole bunch of numbers everywhere. Especially with the binomial theorem idea. But after I noticed the pattern, it wasn't too bad, though I still had to concentrate so as not to lose track of what I'm doing. There was that one question we did in class where I wasn't really paying attention and I ended up missing b^1 so the whole thing ended up being wrong. So yes, I have to pay attention.

I didn't like this unit, honestly. I'm happy and sad at the same time that the test is tomorrow. It's like, "Yes! The test is tomorrow! IT'S OVER!" and then at the same time it's like, "OH NO! The test is tomorrow!" Youu know? Hahah, I agree with Kristina, I miss logs too.

ComBOBinatorics

2008-05-07T22:06:00.003-05:00

Combinatorics. I found this unit quite easy at the beginning. It was one of the units I felt very comfortable doing, 'cause more or less, it's just combinations and permutations. After that, it got a bit harder, with the whole poker hands and what not.

Basically you just had to focus on what was being said.

The Pick and Choose was easy enough to grasp, but then after that you throw in restrictions and all that which created a bit of a turmoil for some people, including me at times. At first, the table question stumped us all, and then we understood it after we got the whole reference part down because you know, a circle doesn't end =P

Then after that, bracelets threw us off because we thought "oh, it's just like the table." But it wasn't. Bummer huh.

I mostly had trouble with the poker hands, with all the choosing and restrictions =/

Overall, this was a good unit for me. Nothing tricky, I just have to observe it closely. Look for the clues, and you'll know what to do more or less.

BOB on Combinatorics

2008-05-07T21:46:00.002-05:00

During the first class, I was quite excited because I was pretty anxious to learn how to use simple equations and numbers to figure out the chances of a certain subject, such as how many ways can 3 people sit in 4 ways. I thought that was pretty cool. At first it was really simple with the factorials, I got my head around that quite quick, my head lassoed those factorial no problem. Then we fondled with permutations and permutations that include 0's. Those were slightly more challenging but nevertheless, I eventually came to understand those as well.

Now circular permutations were just plain frightening. Then Mr. K explained it, and I was still confused which made me even more scared. Then he gave us an equation: (n-1)!, that was easy enough to memorize, and I wasn't so worried anymore. Same with bracelets, the only difference being that you can flip them, so just divide the equations by 2: (n-1)!/2.

Pascal's triangle, seemed to be difficult at first, but I started to notice simple little patterns, then learn more challenging patterns, like the Fibonacci sequence, and the hockey stick sequence, that I probably would have never discovered. Phi was pretty cool, learned a bit of photography that class. It was interesting to know that "Phi" was attractive, and how the ancient Greek's knew about Phi and how they built an entire temple with objects that were related to Phi. The binomial theorem was challenging and I don't understand everything about it, but I should do just fine.

The last thing were the Poker hands. That class, I really felt like playing poker, but then my head started to spin, with all these different equations put together like a puzzle with so many rules put into play, I didn't really get it. I hope I won't flunk this test, like I flunk in actual poker. Good Luck on the test everyone.

-Francis

Bob for Combinatorics

2008-05-07T21:44:00.001-05:00

This is the most difficult unit out of them all for me. I have a long list of problems in this unit. What I am having trouble with are the word problems. I have trouble figuring out where to start first of all like since there's so many ways to solve these problems. But todays class really helped me. I have a better understanding of how to tackle these word problems after seeing how Mr. K solves them by breaking down the question into parts. I also realize I am making a lot of mistakes with questions because I use pick when I should really be using choose. There was also that binomial theorem stuff that was really a pain but I get that now fortunately. It was just so confusing to stare at all those numbers and letters for the first time. Oh and how could I forget those circle seating questions?! I get half the work right for these questions and mess it up at the end by like multiplying numbers I shouldn't be but I think I've learned my lesson...

There wasn't much I was good at in this unit but I liked doing the easy things of course like the simplifying of those factorial questions. I'm relieved this unit is almost over...I can't wait actually. Knowing those numbers in Pascal's triangle are everywhere will haunt me forever though XD.

BOB: Combinatorics

2008-05-07T19:48:00.002-05:00

Its time for another BOB! Man, to sum up my feelings about this unit, I'd have to say that it was definitely my least favorite out of the units we've done so far. It was really difficult to wrap my head around the various concepts and it took me a while to understand how questions were done. The easiest parts for me were definitely the circle questions. For some reason, those questions really clicked with me, unlike the other ones. There were just so many different things to do and solve while figuring out how to solve those other types of problems that it just made my head hurt. By the end of most of the classes, I would always feel a bit dizzy after intaking all the information and trying to sort it all out in my head. I also remember thinking to myself when we were told to solve a question, "Man..I miss logarithms.."

Yep, that's how I felt about this unit. As much as I may dislike it, I still have to get through the test tomorrow. I hope I, and all of you, do well on the test! Oh yeah, one more thing....JOIN AP CALCULUS! one of us..one of us....

Bob: Combinatorics

2008-05-07T18:35:00.003-05:00

Test tomorrow! =( So here I am once again for my Bob ..

First thing's first, I didn't really enjoy this unit. I just had to get that out of the way. LoL I had the most trouble with the word problems and I knew from the start that I would have the most trouble in that area. Particulary the poker problems and the book questions where you had to arrange a certain amount of books but 3 must stay together. As well as the questions where there's a certain amount of chairs and it's asking how many ways can lets say, 4 people be seated consecutively. I know its not THAT hard of a question because it seems to me that most of the class understands it but I guess I confuse myself at times. On another note, the binomial theorem was pretty easy to understand and I think I'd do better on that area than the word problems. Well at least I hope so! But I'm not going to lie, some of the problems on Excercise 34 were somewhat difficult. =(

All in all, this unit wasn't my favorite -__- but I hope I'll do really well on the test tomorrow! And I won't tottally bomb it haha. Goodluck to everyone and especially to me!! lol *cross fingers* Lataa!

Today's Slides: May 7

2008-05-07T14:22:00.003-05:00

Here are the slides from this morning ...

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and from the afternoon ...

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Poker Combinations

2008-05-06T17:46:00.009-05:00

OVERVIEW:

SLIDES 2 to 6 are a review the basics of combinations and permutations
SLIDES 7 to 16 deal with poker combinations

SLIDE 2

In how many ways can 8 books be arranged on a shelf if 3 particular books must be together?

There are 8 books. We grouped 3 of the books together. (If you imagine that the 3 books are together in a bag, this might be helpful to you.) So now we have 6 objects. We can shuffle the 6 objects 6! ways for a different arrangement. We can also shuffle the 3 books in the bag 3! ways for a different arrangement.

Answer: 6!3! = 4320 ways

SLIDE 3-4

There are 10 football teams in a certain conference. How many games must be played if each team is to play every other team just once?

SLIDE 3 is wrong. SLIDE 4 is correct.

To understand this problem, let's simplify the problem to 3 football teams. Let's imagine Mr.K's team, Rence's team, and AnhThi's team are together in a conference, and they're to play each other in a game just once. Mr.K versus Rence in one game, and Mr.K versus AnhThi in another game. So far, there are 2 games played. Rence versus AnhThi in a game, so the maximum number of games that can be played between the 3 teams is 2+1 = 3. Here we see a pattern that we're adding. WE'RE NOT USING FACTORIAL. The student who did SLIDE 3 made that mistake and got the wrong answer.

So if there are 10 teams that are to play against each other and they're to play each other in a game just once, then the first team can play any of the 9 other teams, the second team can play any of the 8 other teams, the third team can play any of the 7 other teams, the fourth team can play any of the 6 other team, etc. And so, the maximum possible number of total games played in a game is:

Answer: 9+8+7+6+5+4+3+2+1 = 45.
---

Another way to solve the problem is to see that there are 10 games and 2 teams are chosen to play against each other. This can be expressed as:

Answer: 10C2 = 10!/(8!2!) = 45

SLIDE 5

There are 9 chairs in a row. In how many ways can 4 students be seated in consecutive chairs? (Hint: First find the number of ways of choosing 4 consecutive chairs.)

First, we counted the number of ways of choosing 4 consecutive chairs. By moving the red "container" (that holds only 4 chairs) at a time to the right, we find that there are 6 ways to get 4 consecutive chairs.

Now, we ask ourselves, how many ways can 4 students sit in a row of 4 chairs? Answer is 4!.

6*4! = 144

SLIDE 6

Seven people reach a fork in a road. In how many ways can they continue their walk so that 4 go one way and 3 the other?

There are 7 people. These are your slots.
_ _ _ _ _ _ _

Each person (represented by a slot) either goes left or right because there's a fork in the road, but there must be 4 that go in one direction, while the other 3 go the other direction.
_ _ _ _ _ _ _
L L L L R R R

Remember, the formula for permutations of non-distinguishable objects says:

n! / (k1! k2! k3!)

n is the number of objects that contain k1, k2, k3..., which are non-distinguishable objects.

Answer: 7!/(4!3!) = 35
---

Another way of solving the problem is to choose 4 people from the 7 to go in one direction, while you choose 3 people to go in the other direction, which is expressed like so:

7C4 * 3C3 = 35
---

The rest of today's lesson is the "juice" of today's lesson.

POKER COMBINATIONS

Given a standard deck of 52 cards, how many ways are there to draw 5 cards to obtain each hand? (SLIDE 7)

In other words, "how many different 5 card poker hands are possible?"

Well, there are 52 cards and we're choosing 5 from them. This is expressed as:
52C5 = 2 598 960 ways.

(a) Royal Flush [ace, king, queen, jack, ten in the same suit]

There is only one way to get the sequence: ace, king, queen, jack, and ten if they're all in the same suit. This is expressed as:
4C1 = 4.

(b) Straight Flush [five cards in sequence and of the same suit, but not ace, king, queen, jack, ten] (SLIDE 8)

* This problem is similar to the problem with the red container and chairs.
* The black work on SLIDE 8 is wrong, by the way.

Here are your "slots."

A 2 3 4 5 6 7 8 9 10 J Q K

Because of my limitations of only describing things to you and not by means of animation, please bear with me.

Imagine the container again and this time the container can hold up to five slots. Starting from [A, 2, 3, 4, 5], we move our container to the right and count how many ways the container can hold consecutive slots. We count that there are 9 ways to have five different slots consecutively in the container.

By definition, a 'straight flush' must be of the same suit, so we can only use one out of the four suits. This can be expressed as 4C1, "4 choose 1". (From the four suits, only one of them is being chosen.)

Answer: 9 * 4C1 = 36

(c) Four of a kind [four cards of one face value and one other card] (SLIDE 9)

The first four cards...
There are 13 different face values per suit, and 1 face value is being chosen of that thirteen; this is expressed as 13C1. In total, there are 4 face values of each suit, and any of those 4 face values of each suit can be chosen; this is expressed as 4C4.

The fifth card...
The fifth card can't have the same face value of the first four cards, which leaves us with 12 different face values remaining. Of those 12 face values, we just want 1 of them; this is expressed as 12C1. There are 4 suits, only 1 will be chosen; this is expressed as 4C1.

Answer: 13C1 * 4C4 * 12C1*4C1 = 48

(d) Full house [3 cards of one face value and 2 cards of another face value] (SLIDE 10)

"3 cards of one face value"
There are 13 different face values, and only 1 of the face values is being chosen; this is expressed as 13C1. There are 4 different suites, and 3 of one face value is being chosen; this is expressed as 4C3. Now we have our three cards of one face value.

"2 cards of another face value"
Since, by definition, a full house includes 3 cards of one face value plus 2 cards of ANOTHER face value, we can only choose from the remaining 12 face values that hasn't yet been chosen by the first 3 cards. Of the remaining 12 face values, only 1 is being chosen. There are 4 different suits, and only 2 of one face value are being chosen; this is expressed as 4C2. Now we have our 2 cards of another face value.

Answer: 13P2 * 4C3 * 4C2 = 3744.

HOMEWORK

You should be at Exercise 35.
Slides 11 - 16.

NEXT SCRIBE
By the process of elimination, Eleven is scribe.

Today's Slides: May 6

2008-05-06T12:28:00.001-05:00

Here they are ...

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Counting my BOBs

2008-05-06T06:04:00.003-05:00

Hi this is Benofschool and this is my BOB for the Combinatorics Unit. The easiest part of this subject is the simplifying factorials and the Binomial Theorem. I understood those parts of the unit pretty quickly but the part that I am having the most difficulty are the Word Problems like how many ways can these books be sorted. Other examples would be the worksheet that was given to us when Mr.K had to leave because he was late for a meeting. Those question I am having difficulty on. I can't seem to find out how to solve them. I don't understand which formula to use nor do I know what the questions are asking. I might seem angry but I'm not see? :) I'm just a bit stressed out. I'll pull through...

The Binomial Theorum, Fibonacci, Vitruvian Dan

2008-05-05T23:00:00.000-05:00

Today, Mr.K continued on his journey to show us how The Good Lord made the world and his attempt to blow our minds... I'll tell you how that went in a bit BUT, strap yourselves in because I tried to take down notes on basically everything he went through and said so, needless to say, this one's gonna be long.

To easily locate what you are looking for, either press Ctrl + F or Apple + F, and insert one of the subjects to quickly jump to what you'd like to review.

First Class consists of ;

Vitruvian Man
Pascal / Shijie Triangle Review
The Hockey Stick Pattern
Revealing The Fibonacci Pattern
Phi - The Golden Ratio
Binomial Solving

Second Class consists of;

Workshop Groups
Binomial Solving - Continued
Using a System of Equations

The First Class

The Vitruvian Man
First he presented to us 'The Vitruvian Man', which was originally a portrait, drawing, painting... uh.. artistic depiction of man, drawn by none other than Leonardo Da Vinci...

ANYWAYS, The Vitruvian man is the most accurate depiction of man, with everything being of most accurate measurements. Fingers, arm length, leg length etc. I stated that if no mutations had occurred in a human body, A person's Wing Span (Length from arm to the other) is equal to that person's height. And so, Mr. K measured me. Turns out, with my hair, and shoes, I seemed to be longer than my wing span. Subtract my hair and shoes, it would be just about right. He then talked about the symmetry of people and that if someones face or body was more symmetrical, the closer they would be to being "beautiful".

Pascal / Shijie Triangle Review
Concluding that, we reviewed 'Pascals' Triangle. Notice the quotations there. We first went over the patterns. Such as...

- when the binomial is expanded, A's exponent is decreasing, and B's exponent is increasing.
- the degree of any term, (a, ab, b) is equal to the term of the binomial.
ex.) (a + b)² = a²+ 2ab + b² <-- the exponent on the A and B terms of AB have a sum of two.

The yellow line is simply the counting numbers.
The green line in pascal's triangle are the triangular numbers.

- Triangular numbers are the sum of the counting numbers. 1 = 1, 2 = 3, 3 = 6 ...
- Justus, [in the previous class] found the tetrahedral numbers. (Follow the link to get a better definition etc)

- The powers of Two are embedded in each row. Example; 2° = 1 (Zeroth Row) 2¹ = 2 (First row) 1 + 1 = 2 ... and so on.
- The powers of Eleven [Word up to Elven] are also embedded in each row.

Now getting on with the Regularly Scheduled programming...

The Hockey Stick Pattern
He showed us new patterns within Zhu Shijie's Triangle. Such as the 'Hockey Stick' Pattern... How Canadian :) Basically, the hockey stick pattern is starting from any side (starting with one), go down diagonally, then cut back in the direction you came in and voila, the sum of those numbers are equal to the last number you cut into. Coincidence? I don't think so.

What the H, E, quadruple hockey sticks!?!?

Revealing The Fibonacci Pattern

Next, he showed us how to find the Fibonacci numbers within the triangle. In this slide, it is found by adding the numbers in a diagonal orientation.

He showed us that the 'Nacci sequence is found all around us, like in tree's and well, plants. For example. They're found in Pine Cones, Flowers, and their leaves. Bee's also respect the 'Nacci numbers when reproducing, as the male can only have one parent, and the female two, and from there you can pretty much depict how that goes down.

Phi - The Golden Ratio
He then started to divide each Fibonacci number by the number before that one. We found that the resulting number seemed to be reaching a certain number. He then introduced to us... Phi! "PHI is One H of a lot COOLER than Pi!!!" Get it? GET IT?! *crickets* Ok cool

So we have this here. If you want to look into it more, go to http://goldennumber.net/. I believe it's the 'Phinest' Source on Phi, or the Golden Sequence, or Golden Ratio, or ... You get where I'm goin'. So he did some stuff, like explaining that a 4 x 6 picture is better because it approaches Phi more than the 5 x 7 photographs. Apparently if you take pictures like the way he has set up in the slide above (at the black spots), you'll get a better looking, more attractive picture, if your object is concentrated at one of those points. This was where he started to tell us how to get 10 % more on the things we hand in... by making it approach Phi. So then he did all this stuff to 'Build' a golden ratio. So you...

Take a square, cut it in half
Make a semi-diagonal, in which you get a Pythagorean triangle...
and then I got lost after that.

He then went to the point where the golden ratio is 1 + √5 / 2 = 0. He then went on to explain Phi is everywhere, in the proportions of your body, (elbow to finger, wingspan, feet, height etc.) Everything is divided Long by Small to get Phi. So basically to get 10% of everything you hand in, hand it in so that it respect's Phi, and it'll look that more attractive... I think.

Basically, the more symmetrical you are, the closer you are to 'beautiful'. This is why a famous building in Athens (My bad, I couldn't recall what the building was called), is crumbling, but still looks appealing to people.

INTERMISSION

This Scribe post is sponsored by AVP

[No not Alien vs. Predator]

Binomial Solving
He went on an then introduced this...

So then we all screamed in terror (At least I did) as if Mr.K adopted the Dark Side due to this very long expression. But really it's the same thing we've been doing all along. nC2 = N! / (N-2)! 2! = N(N-1)/2! Which we find is one of the expressions within the algebraic expansion. So don't run away.

Remember, The number of terms is one more than the degree of the binomial. Mr. K then showed us the patterns.

The coefficient of the 'ith' term is nC(i-1) ex.) for the 5th term, you'd choose 4
The exponent on the "a-term" is given by: [n - (i - 1)]
The exponent on the "b-term" is given by: (i - 1)
The relation holds for each term in the expansion [exp a] + [exp b] = n
The number of terms in any binomial expansion is: ( N + 1 )

Remember, There's always one more term than the degree of the binomial.

THE SECOND CLASS

Then we started to do some Binomial Solving. Mr.K asked us to find the 4th term in the expansion of (2 + x) ^7

To make this easier, we changed it into the (a + b) format we were using earlier. So we let a = 2 and b = x. We then have it as (a + b) ^7

So to find the 4th term we set it up as so;

t4 = 7C3 a^4 b^3

B's exponent is 3 because of (i - 1), 4 -1 in this case.

A's exponent is 4 because of [n - (i - 1) ], [7 - 3] in this case.

So then we came down with 7! / 4!3! (2)^4(x) ³ <-- We put back in the variables.

Then we came down with 35 (16) (x³) = 560x³

We then did about two more of them, but what I found really cool about how he did one of them is how he multiplied 56 an -8.

Basically he broke up 56 into 50 and 6. He then showed the concept of basically doubling those three times (2 cubed).

So first he doubled 50 and 6 --> 50 * 2 = 100 6 * 2 = 12
... then he doubled it again --> 100 * 2 = 200 12 * 2 = 24
... then he doubled it again --> 200 * 2 = 400 24 * 2 = 48
then he added them together --> 400 + 48 = 448.
And because the 8 was negative... --> -448.

So try it on your calculators... 56 * -8 = ... - 448!

Yeah I know, crazy.

Ok, lets try to wrap this up quick. I'm just showing this slide, as it's many student's mistakes to multiply the -1 by 3, but really, you're just cubing it... which still makes it -1 :)

Here, we discussed finding a certain term within an expansion.

First we did it a long and tedious way ( I didn't put up the slide for it), and once we found the pattern, which is the exponent changing by the same difference, we can then find the term we are looking for. But Mr. K showed us a more elegant way.

First, we wrote out the way we'd choose a number from an amount. And we'd use dummy variables so that we can just get the structure up.

So we set it up like ; qCp a^p b^q

This is where our *Remember* segment comes in.

REMEMBER, the exponent's sum must match the degree of the Binomial! So take a quick look above in the slide and we notice the exponent of the Binomial is 9, therefore, p + q MUST, and I mean MUST equal 9. Now let's put back in the variables so we can solve this little sucker.

(x^3) ^p (-x^-1) ^q So since we put back in the variables, the exponents now are 3p and -q. So then because we're looking for x^7, we set it up like;

3p - q = 7. well at least it should equal 7, because that's how we built it.

Then create a system of equations. p+q + 3p - q = 9+7 --> 4p = 16. Therefore, p = 4 and since p = 4, and the exponents MUST equal 9...

9 - 4 = 5. Therefore q = 5.

Now we can find that term.

9C5 (x^3)(-x^-1)^5 = x^7

= 126x^7

Class, this concludes today's lesson and work.

REMEMBER!

The sum of the exponents on any term MUST equal the degree of the Binomial!

Don't mistake exponents and multiply the coefficient. It must be raised to that exponent.

Don't forget about the Zeroth term!

And finally... A LOGARITHM IS AN EXPONENT! ... Just in case you forgot.

A side note;

Watch Iron Man. It's Dope. ... or should I say Phi? Oh yeah before I forget. The Next Scribe is ZEPH.

Now then, I gotta get some sleep and stay PHI! 8)

Rence ~ Out

Scribe List

2008-05-05T22:59:00.002-05:00

It's been a LONG WHILE since Mr.K updated the 'Hit-List' so I thought I'd be a good samaritan and just post it up :)

Cycle 4

~~Francis~~
Joyce
Eleven
~~benofschool~~
~~roxanne~~

~~JamieNeRd123C~~
~~zeph~~
~~AnhThi~~
~~Richard~~
~~Hi I'm Justus~~

~~nelsa~~
~~Rence~~
kristina
~~Paul~~

* Note; Zeph's crossed off as he is today's scribe.

Stay Phi everyone,

JabbaRence ~ Out

Today's Slides: May 5

2008-05-05T13:42:00.003-05:00

Here they are ...

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Pascal's Triangle That Wasn't Really Pascal's Triangle...

2008-05-02T17:03:00.012-05:00

Hi everybody...Benofschool here to scribe what we did today in class. Today Mr. K promised us that today's class will show a way that G-D built the universe and that it will blow our minds. But we didn't have enough time so we didn't get to G-D building the universe but the lesson did blow our minds.

Lets start with the first slide. The first slide had a picture of an ancient Chinese diagram. Hey doesn't it look a bit like Pascal's Triangle? Well it was Zhu Shijie's triangle created 400 years before Pascal.
When I asked my grandfather what the characters meant but unfortunately he said he couldn't understand it. The characters were of the ancient Chinese characters. The Chinese language has evolved a lot. But my grandpa and I found a connection of why the number 7 (fourth character from the left) was there. It probably had to do with how many rows the triangle had but there seems to be something different about the first 4 rows because the last row on the triangle says something about the number 7.

After a little conversation about that triangle and teaching Mr.K some Chinese pronunciation lessons we continued onto the next slide. He asked us to expand and simplify the binomials. It was a bit tedious but we did it. Since some of them were a bit long he enlarged the hidden answers like he always does (that sneaky guy...). It looks like any other expansions we've done but there was something special about them. So let's look at the next slide. It may give us some insight into what was so special about those binomials.

On the next slide...
we were asked to find the next two rows from what was shown. It was pretty easy to see. To find the next row just add the numbers on the previous row to obtain the next row for example: since there are no numbers before the 1st 1 and after the last 1 we can assume there is a zero there so 0 + 1= 1 so the next row starts with a 1. The next number can be obtained by adding 1 and 4 to get 5 on the next row. 4 + 6 to get 10 to get the next number, and so on until we get to
1+ 0=1 that ends the row.

So if we look at the numbers in each row and compare them to the coefficients of the simplified binomials on the last slide, they are the same. Lets look at (a+b)³: we get that the coefficients are 1,3,3,1 just like on the 4th row of the triangle. But wait that's not all of the patterns found on the binomial work. If we look at the exponents of the variables a and b. The coefficients of a decreases counting from the exponent on the original binomial until the exponents is zero. The coefficients of b are increasing until we get to the exponent on the original binomial. We were all commenting on how much more easier that would be to expand those binomials if we knew that patter so we went on to finding (a+b)^6.

Then on the 4th slide...

we were asked to evaluate the following. So to do that all we need to do is to plug it into our
TI-Calcumulators or just plug it into the Choose or Combination formula seen on the picture to evaluate it and it seems the numbers were the same on the triangle. Cool huh? Pascal originally "made" the triangle to help a friend gamble. Gambling is bad...

The next few slides we were just looking for more patterns on "Pascals" Triangle. We found the counting numbers (in yellow). On the other slide we found the powers of 2 on the triangle. If we find the sum of the numbers on each row we get a power of two. Cool. And the digits of the power of 11 is found on it too but a few rows down there is a problem... or is there? Since the 10 is on the 6th row we carry that one to the 10 on the left which makes it 11 and then we carry that 1 to the 5 to the left and we get six so 11^5 is 161051. Those slides totally blew our minds...

Okay for some Pascallian/ Zhu Shijie vocabulary...
Rows- The row of numbers: First Row, Second Row
Terms- The terms are the numbers on each row: First Term, Second Term
That vocabulary only works in high school. In University the first row is referred to as the zeroth row and the first term is referred to as the zeroth term. So if we combine the high school vocabulary the 5th term on the 3rd row is the number 6.

A little challenge for tonight is to find the Fibonacci Sequence (1,1,2,3,5,8...) on the triangle. And homework I thought will be posted on the blog but there doesn't seem to be any homework. I guess homework should be exercise the next like Mr.K always says and get practice on the unit because the pretest is on Wednesday and the test is on Thursday. So get those BOB's posted. Monday will be the day that Mr.K will show a way that G-D built the universe.

I just proved to my parents that the male seahorses give birth. The next scribe will be Rence. Until next class good bye and good night....

Today's Slides: May 2

2008-05-02T14:15:00.002-05:00

Here they are ...

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Prologue to the Secret of the Universe

2008-05-01T17:08:00.011-05:00

Unfortunately today, Mr. K had to leave for a meeting about curriculum changes so we were left to fend for ourselves. Luckily, we packed some survival tools: pencil, eraser and a calculator! We would need them to fight off the Fundamental Principle of Counting Sheet.

In the afternoon, with the return of our teacher, we wrote a quiz. Apparently, it consisted of the stuff we had been learning so far.Then, we went over the results of our efforts and assembled into 4 groups and continued to advance down the slides one by one solving each problem that came with each slide. Though as we may, we only managed to complete three but a surprise was in store for us next class. As Mr. K described the sneak preview of what would happen tomorrow. The next lesson will blow our minds out, teach us about the universe and poker...? Sadly that is all we did in two periods. Hopefully we can pick up the slack in the future.

NO HOMEWORK! well except the for the Fundamental Principle of Counting sheet

Tomorrow's scribe: Benofschool (upon request)
Next time in class: One of the ways god intended the universe and the game of poker??? Good luck Ben! you're gonna need it...

Today's Slides: May 1

2008-05-01T15:44:00.001-05:00

Here they are ...

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Combinations and Permutations: The Difference

2008-04-30T17:55:00.014-05:00

The Mastermind

Today's class started with Mr. K explaining the game shown above, "Mastermind". The concept is pretty simple, the mastermind, chooses any colours they want and arrange them in any order they want. Then the other player tries to guess the order each colour occurs. If you haven't already caught on, this is a permutation, because the order matters. If the other player guesses a colour correctly, the mastermind puts down a white peg, and if the other player guesses the right colour and places it in the right spot, then the mastermind puts down a black peg. But those pegs aren't placed in any particular order, so they're known as a combination, because order doesn't matter.

Did you get that? A permutation is an ordered arrangement of objects without repetition and a combination is an arrangement of objects where order doesn't matter. Using Mr. K's definitions of course. =)

After that, we watched a video that reviewed the past couple of days. It's called "Probability and Statistics" and is narrated by Ms. Jenkins. Watching this video felt like I was thrown back in grade seven, and sitting in my social studies class watching another educational video and absently taking notes. Hahahah. But it was helpful, so you should go watch it. Makes you want a tablet, hahaha.

ANYWAYS. The next few slides are just some problems that we worked on today. (I just realized that I'm not using the whole 'outline' format. Meh, too lazy to start over and.. meh, don't have time)

1. Use the 'bracelet' formula because a necklace is a circle. Circles have no beginning and no end and can be flipped over. Fill in your values.
(n-1)!/2
(12-1)!/2

2. Now solve!
11!/2
In your calculator, you would enter [11][MATH][<][4]
19958400

With this question we had slightly more trouble. I can't speak for everyone right now, but for me, what made this question confusing was the whole deal with alternating. I also didn't think of using 'slots' so I really made everything more difficult. But basically, mom and dad always has to be together. In this diagram, Paul (who came up to solve it) seated mom first and placed dad to her right. On her left side, one of the three boys have to sit there, and beside him, one of the three girls, and so on and so forth, filling in the slots as we went. Our original answer was 36. What we didn't realize was that dad could've have also sat on the left side of mom, which opens up a whole new set of numbers. So we changed the 1 in the second slot to a 2, therefore changing our answer to 72.

This question seemed easy enough, there are seven people and only three slots. So we thought, "well why not just fill in the blanks or use the 'pick' formula?"
7 x 6 x 5 = 210
7 P 3 = 210
We were all thinking, "that was too easy", and then Mr. K comes in and he's like, "are you sure?" Well, no he didn't really say that.. but he might as well have. THE POINT IS, it was wrong.

You see, when choosing who was going to be part of the committee, the order didn't matter. The 'pick' formula wouldn't apply. So he showed us another way of looking at it, basically, you're saying 'yes' to three people and 'no' to the rest (in this case, four). If you think about this as a word, you can just apply what was taught yesterday. There are two non-distinguishable objects, 'y' and 'n'.
7!/(3!4!) = 35

Lastly, we learned the 'choose' formula.

This is also in the slide if it's too small to see.

Aaannd.. that's the end of that. Thanks for waiting so long. I had to stop near the end to go to bible study at my church, and I almost lost everything earlier. SO YES, lots of problems today. But it's all good. xD

Oh, and the scribe shall be.. shall be.. (uhh, Mr. K, the scribe list is still not correct) Thi. Goodbye everyone!

Today's Slides: April 30

2008-04-30T13:47:00.002-05:00

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Permutations of Non-Distinguishable Objects and Circular Permutations

2008-04-30T00:14:00.005-05:00

Okay, so its that time again, time for a class blog. I doubt this one will be as long or epic as say, a blog by Justus or Francis (seriously, just make a book or something), but hopefully it will be just as informative.

Today's topic was "Permutations of Non-Distinguishable Objects and Circular Permutations."

Whoa, that's pretty long. Someone, create Tinytopic.com, quick!

So does everyone remember what a permutation is?

(As defined by Mr. K)
Permutation - An ordered arrangement of objects without repetition.

Or (as defined by me)
Permutation - A set of objects where the order matters.

A permutation is not to be confused with a combination as we discussed in the last class. In short, a combination is a set of objects where the order does not matter. A permutation is a set of objects where the order does matter. That's why a "combination lock", should technically be a "permutation lock."

Remember that? Yeah, you better, because it's important for our formula, the aptly named "Pick" formula, which goes like this:

nPr = n!/(n-r)

(n and r are subscript, they are not being multiplied)

Where:

n is the number of objects to "pick" from
r is the number of objects to arrange

And when we use this formula, nPr is read as "n 'pick' r."

3P₄ is read as "three pick four."

What's totally awesome though is that this function can be accessed through your calculator!
So it goes something like this...

Enter your "n" value first
Press the [MATH] button
Press the [<] button
Press [2]
Enter your "r" value
Press [ENTER]

Isn't technology wonderful?

So after that, we talked a little bit more, but since there's not much to the "Pick" formula, we went on to talk about really huge permutations.

TinyURL.com is a site that takes immensely long website URLs and turns it into a short and sweet (although probably not so easy to remember) URL. So http://www.internetisseriousbusiness.com/index.html would turn into something like http://www.tinyurl.com/9aidso. Thus, it generates a unique (this means it's a permutation and not a combination) url every time someone wants to make a TinyURL.

So, Mr. K brought up a good point. Just how many URLs can TinyURL generate, and how long would it take to generate all those URLs?

Well, lets think about this for a second, and look at the example url TinyURL generated for us.

http://www.tinyurl.com/4kaqlv

Well, the fact that there is a number and letters there tells us, the value of a slot can be either a number (0 to 9) or a letter (a to z). Lets assume for now that:

a) The number can be in any slot, but there can only be 1 number
b) The letters are not case sensitive, so we can assume they'll always be upper or lower case (if it was a combination, we'd have to count upper and lower case a's as two seperate objects, but more on that later).
c) Letters can repeat (so your string could look like "1aaaaa"). This means you cannot use the "Pick" formula!

Why cant you use the pick formula?
Because the pick formula calculates the number of permutations where once an object or character is placed in the set, it cannot be used again. This is obviously not the case here.

So, we've already determined that a slot can have one of 36 values. 10 of those values are for each of the 10 numbers, and 26 of those values are for each of the 26 letters.

And we have 6 slots. So for every slot, we have 36 possible values. Thus our number of possible permutations is 36*36*36*36*36*36 or36^6 .

Which equals:

36^6 = 2 176 782 336

So thats two billion one hundred seventy six million seven hundred eighty-two thousand three hundred thirty six. (Say that 10 times fast!) And remember, a billion is big. Godzilla big.

Infact, its so big, we determined that it would take ~32 (approximately thirty-two) years to give Lawrence a billion dollars if we gave him 1 loonie (1 dollar) per second consistently. Times that by 2.176782336 and you have approximately how long it would take to generate every possible URL if one URL were generated every second consistently:

2.176782336 * 32 = ~69 years. The one on the slides says 64 years, because there we simply multiplied 32 by 2 (which is how many whole billions we have). I counted the fraction that would have been generated all the way up till the 2 176 782 336th day. However, since the 32 itself is an approximation, this is not completely accurate (hence the tilde [~], which means approxmiately).

And thats just if we're not counting capital letters!

If we did count capital letters, our number would be even huger! Because now our number of possible values in each slot is now larger. Instead of just "t", we now have "T" and "t". Thus, we don't just have 36^6 possible values, no. Now we have 62 possible values, because we have 52 possible letters and 10 possible numbers.

So how many permutations is that? Thats 62^6. Which equals...

5.680023558 x 10^10Which equals to 56 800 235 580, or fifty-six billion, eight hundred million, two hundred thirty-five thousand five hundred eighty. Permutations.

So, 56 * 32? Thats ~1792. That's how many years it would take to generate all those permutations if one were generated every second!

But enough about that. Lets look at distinguishable objects versus non-distinguishable objects.

(Im going to speed up here because it's now 2am and I'm getting sleepy)

So on slide 7, we have a little example with the word book. How many ways can you rearrange the word book? How many unique permutations are there.

Well, we figure that out easy: 4! (the number of slots) divided by the number or repeating letters in this case, that letter is "o" and it repeats twice). Thus, our solution is 4!/2!, which is 12.

But then there a plot twist. What happens if one of the "o"'s is red, and is counted as a unique letter? Then we have even more permutations! Because now, instead of having only 3 possible values for a slot (where one slot has the same value as another slot due to one of the values repeating), we have 4 possible values. Thus, we have 4x3x2x1 = 24 possible permutations. But if you turn that red "o" back into a regular "o", some of your permutations become repeats. Thus, the number of permutations is halved, and you get your original answer of 12 permutations. Wow!

After being introduced that concept, we do a little practice with some other words. Then its on the the main course. Circular permutations.

Circular permutations looks wacky and complicated, and it takes a while to wrap your head around it, but really its just an expansion of the concept of distinguishable objects. The basic idea here is that when you have a circle, the first point you put on it serves as your reference point, and you work from that point on. Because your reference point is wherever you want it to be, your perspective as to the rotation of the table can be anything you want. Thus, with circular permutations, your duplicates stem from the fact that some permutations are in fact existing permutations that have just been rotated.

Alright, I give up now. This scribe post is incomplete, and I apologise to everyone that I'm posting so late. I'll expand and finish this post properly later today and make sure it's done as it should be, but for now I cant think when I'm half asleep.

Tomorrow's scribe (or should I say, today's scribe) is now Nelsa.

Also, a million points and a PHD to someone who can tell me the adjective of "bracelet." Seriously, a braceular table? I don't think so. I wonder if that's even an actual legitimate shape, or is it just a concept...

Today's Slides: April 29

2008-04-29T15:37:00.001-05:00

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Permutations

2008-04-28T14:43:00.005-05:00

We had a fairly long chat about Justus' awesome scribe post, and his BOB before that and how awesome that was, and Mr. K introduced for the first time to us, a "Scribing Hall of Fame" and how to get inducted into this Hall of Fame. For you to be accepted, you must have a vote with the class, and if enough students vote to induct your special scribe post then congratulations you're scribe post is now a Hall o' Famer. Justus, you have my vote, thumbs way up. If you want to see what it takes to be just like Justus, then just read all the past scribe posts that are in the Hall of Fame. Found just a click away at this link: http://thescribepost.pbwiki.com/HallOfFame

At the start of today's class Mr. K had us confused on why combination locks shouldn't be called "combination" and how they should be called permutation locks. This had me quite confused. I started to wonder, that if three different numbers in different orders isn't a combination, then what is? If you want to find out, you must read on. At the start of class, we were given questions to do, these are found on slide 2 of today's slide show. These questions were similar to last class' questions, refer Justus' scribe post at: http://pc40sw08.blogspot.com/2008/04/intro-to-combinatorics.html

There was one difference though, these new questions had the number zero included as an option. With the question: "How many numbers of 5 different digits each can be formed from the digits 0, 1, 2, 3, 4, 5, 6?" Using a "slot system" (as found in the 2nd slide) is easiest to use. The first slot can usually be any number, but not if that number is zero, and this is where the confusion and difficulty kicks in. It's because if we are referring to a 3 digit number, we can't say that number is 012, because this is just a 2 digit number with a zero placed at the front, so the zero has no real value. If you included zero as a first digit option, then it would be a 4 digit number, not 5 which was asked for in the question. The first digit therefore only has 6 options. Now for the 2nd slot, you can use zero, because it will now have value (ex. 01 = one, but 10 = ten), so you have 6 options, instead of 7 because one option was used for the first slot. The 3rd slot would have one less options, which would be 5 options, and so on until all slots have been taken up. Multiply all these number and you should get 2160, which is the amount of options you have of 5 digit numbers created from the number 0, 1, 2, 3, 4, 5, and 6.

For part b of the question we had to find how many of the 2160 options are even numbers, which means the number would have to end in 0, 2, 4, or 6. This question has 2 parts to it, which means 2 different slot systems, as found in slide 2. One slot system would be numbers that don't end in zero, if these numbers don't end in zero, you would have 2 less options for the first digit. (options: 1, 2, 3, 4, 5, 6. (total 6 options)) and it wouldn't start with the number used in the last digit (total: 5 options), it would be easier to start with the last digit on this question. The slots would then be 5, 5, 4, 3, 3. The last digit would be 3 instead of 4 because you're not ending it in zero this time (leave the 2nd slot system for numbers ending in zero). When the numbers in this slot system are multiplied this would give you 900 options. For the 2nd option you would use numbers ending in only zero, so the last slot would be a 1 (the only option being zero). Since the last slot is zero, the first slot can have 6 options (1, 2, 3, 4, 5, 6) because the last slot doesn't take up any of these numbers. The next slot would be 5, then the 3rd slot would be 4, and so on, with the last slot being 1. When all slots are multiplied, you're left with 360 options. These options are even numbers ending with zero. Now add the total number of options from 1st slot system, and 2nd slot system, and you would have 1260 options which would be even numbers.

Part C is "How many of these numbers are divisible by 5?" For a number to be divisible by 5 it would have to end in either 0, or 5. We now know that if it has to end in zero, we need to do 2 different slot systems, a system ending in 0, and another system, not ending in 0, for this particular question, the only other number is 5. The first system will end in zero: so our first slot would be 6 different options, then 2nd slot would be 4 options because zero is used and 5 is used for the different slot system, and another options is used for the 1st slot of this system. The 2rd slot would be one less, and so on, with the last slot being 1 option. When all slot are multiplied together, you get 360 options. Now for the next slot system, which ends in 1 options, which is 5. This would give us 5 options for slot 1, because it can't include 0 or 5. Then this would give us 5 options for slot 2, because it cant include, 5, or the digit used for slot 1, then 3rd slot would be 4, and so on, with the last slot having 1 options. When multiplied all together we have 300 different options. Now add the results from both slot systems and we have 660 options that are divisible by 5.

In this way, these questions are a problem, because they include the number zero. If these systems end in 0, there's not much trouble, but if they don't be positive that the leading number won't be zero.

We then worked on factorials. On slide 3. We would simplify these by expanding them, then reducing. (Ex. 8!/7! = 8 because 8!/7! = (8 x 7!)/7! and the 7!'s reduce to get 8) This is quite simple, and if you don't understand factorials, refer to Justus' scribe post, the link is mentioned in the 2nd paragraph.

In the factorial notation: n! The variable n, can't be negative or a decimal, because of a domain error. Remember that the factorial notation is a definition.

We did another problem similar to the first on slide 4. All you have to do it pay attention to what the question is asking you, and it's quite simple.

Permutation is an ordered arrangement of objects without repetition, as found on slide 5. Although permutations is in combination so permutation is an ordered arrangement of objects without repetition, in combination.

We also learned an equation entitled the "Pick" formula, which is like the slot system but simplified, found on slide 5. Mr. K says we won't be using this equation a lot. I take this as to not sweat over this equation.

In a combination of numbers, the order doesn't matter and it can repeat numbers, but in a permutation order does matter, and there is no repetition. This is pretty much why combination locks should be called permutation locks instead.

This is everything that has to do about anything in our class today. Enjoy and don't stress. Sorry if this blog seemed a bit rushed, I just had to rush it so I can get to work on time. Apologies.

One last thing, the next scribe will be Paul.

-Francis

Today's Slides: April 28

2008-04-28T11:41:00.001-05:00

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Student Voices Episode 2: Tim_MATH_y

2008-04-27T19:22:00.000-05:00

In this episode Timothy came back to school on Friday afternoon to talk about his week attending the miniUniversity program at the University of Winnipeg. He talks about the differences he finds between teaching and learning at high school and university and describes learning in the university classroom using a thought provoking metaphor, listen for it. Also, we have a cameo appearance by two very special people at the very end.

Please feel free to leave Tim_MATH_y your comments here on this post.

(Download File 7.2Mb, 15 min. 3 sec.)

Photo Credit: Shadow singer by flickr user EugeniusD80

Intro to Combinatorics

2008-04-27T16:32:00.006-05:00

So, the long awaited blog post on combinatorics from last Friday. At least its not 12 at night right? Haha. Anyways, here we gooo!

During the first period class on Friday we had our Logarithms and Exponents test. It wasn't fun. If you werent there for the test It would probably be a good idea to talk to Mr. K about a time in which you can re-write the test. They are worth marks you know :P

The afternoon class introduced us to this wonderful thing called combinatorics. Now Mr.K told us that this is basically a branch of math that involves counting. When prompted as to why it was called combinatorics rather then something like, countinatorics or something he gave us a sample problem.

Given 5 students and 5 chairs, how many different ways can those students be seated in those chairs? Now it is important to note that the question is HOW MANY different ways, and now WHAT ARE the ways. When asked what are the ways, we are prompted to list all possible combinations, which is a long and tedious (although not necessarily difficult) task. In the following image, we used a tree diagram, to go through all the possible combinations or students (named a,b,c,d and e).

(*note* this image can also be found in the slides, in case this is too small to read.)

Although we found the answer to our question of how many but listing out what all the combinations were, there is an easier way to do this, thanks to that handy dandy thing called a calculator. By hitting 5x4x3x2x1 (five times four times three times two times one) on our calculators we can find the total number of calculations without drawing out every single one. We did this by looking at the possibilities we had left. So if theres 5 open seats, and 5 students, we have a 5 choices. When we pick one, we're left with 4 students to choose for the 4 seats. When we pick another student theres 3 spots left, and so on. Thus you multiply 5 x 4 x 3 x 2 x 1 (1 because once you've used 4 of the 5, there aren't really any choices left.)

So Lets recap.

So Far...
-Unit is Called Combinatorics
-Tree Diagrams help
-Looking at your options and choices allows you to multiply to find HOW MANY different ways.
-HOW MANY, and WHAT ARE are two totally different things.
-Theres a difference between Long and Tedious, and Hard/Difficult.

Now Moving onward!

the next problem presented a twist, what happens when there are more then 1 option for each choice? An example of this was in the next problem.

HOW MANY different ways can a nickel, dime, and quarter land on a table?

So to solve this we have to look at the options available to us (or use a tree diagram again :] ).

To solve this we looked at our options. For any 1 coin there are 2 possible outcomes, either heads or tails. now because order doesn't matter in this case, we set up our tree like this. you read the tree, by following its branches. For every flip of the nickel that lands heads or tails, the dime can land either heads or tails, and for every flip of the dime, the quarter can land either heads or tails. This the total number of combinations is, 2 x 2 x 2 = 8 (as shown in the red there.)

Now the next problem changed one of the coins to a die, thus adding a bit of a wrench to the system. BUT since we're smarter then the average bears, we figured out quite quickly that the total number would be 2 x 2 x 6 = 24 (because the die has 6 sides...normally.)

After having gallivanted through a few problems like this Mr. K unveiled the pattern behind this specific branch of math.

Fundamental Principal of Counting
-If you have "M" number of ways to do one thing, and "N" number of things to do another thing, then there are M times N number of ways to do both things.

Above is the simplified version of the Principal of Counting. Basically if you can do one thing x number of ways and another y number of ways, then you can do both of them at the same time, x times y number of ways. (haha I basically just repeated it :P)

Shortly after we were introduced to this we were introduced to Factorial Notation

Factorial Notation
n! = n · (n-1)(n-2)(n-3)...3·2·1

Basically what that means is that (correct me if I'm wrong here ^_^;) is n factorial equals n times, n minus 1, times n minus 2, times n minus 3, etc etc until you get down to, 3 times 2 times 1. Now because not everyone wants to input that into their calculator, they have an EVEN EASIER WAY!

By hitting, [n][math][<][4] you can get the factorial of whatever you want. This input reads in normal terms, means, hit your n value (whatever it is, 4, 7, 234872398, 0.0000000000002) followed by math, the arrow left key, and then 4. You should end up with your n value followed by a ! symbol (which means factorial, not I really mean the letter n)

Now we come to the final little bit of this blog, using the slot method, and irregular combinations.

In the first problem illustrated on this slide, the question reads, "How Many "words" of 4 different letters can be made from the letters A,E,I,O,R,S,T?" On the slide you can see that we set up our slots. Because there is a 4 letter limit, we only used 4. With 7 letters, we simply filled in the slots as we went down the line, multiplying as we went. The result was 7 x 6 x 5 x 4 = 840 words.

The next question gives even more restrictions, asking how many of the words begin with a vowel, and end with a consonant. Yet again we set up our slots, with 4 places. This time though, we set up our restrictions first. Because there are only 4 vowels, and 3 consonants, we fill those in first (as shown in the diagram.) Then we take the remaining letters and fill in the two slots in the middles (5 and 4 respectively, because we already used two with the first and last letter).

Finally in the last question, all the stops are pulled, and we are required to use two separate "slot" mechanisms, to solve it. Using all the same methods as before we came up with 144 as our answer. Thus we completed the class and our introduction to Combinatorics and I've just about finished my scribe post.

To conclude
-Using Tree diagrams helps, but is sometimes really long
-Factorial Notation saves the day (on your calculator [n][math][<][4])
-Fundamental Principal of Counting is if there are M ways for one thing, and N ways for another, there are MN ways for both.
-Using a slot system like those seen in the slides can greatly help
-When there are restrictions on a problem, solve the restrictions FIRST, then everything else.

Alrighty, I think that about sums it up. You know the deal if somethings wrong, tell me, or edit it or whatever, or if you don't get something ask me and I'll try to put it a different way, and edit the post on here.

So the next scribe post is....

*checks scribe list*

Francis....

If francis already did it

Lawrence. Okay? so lemme make sure yous all gots it.

I chose francis, but cause the scribe list seems a bit behind maybe, IN THE EVENT THAT FRANCIS ALREADY WENT. Lawrence. So we dont have to do the whole shenanigans in class thing. k? Alrighty, I'm going to skate :]

Ciao!

Justus out.

Today's Slides: April 25

2008-04-25T15:46:00.001-05:00

Here they are ...

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Dr. L. Og, Robert

2008-04-25T07:18:00.003-05:00

I know, confusing name... Dr. L. Og Robert. [hey its like that Beatles song w. Dr. Robert] Logarithm BOB... I can't believe I didn't BOB yet it's pathetic. I was so preoccupied with bio....but I'm here now.

Well... I have to admit that I liked this unit. But there will always be some part of it you don't like. For example, me and graphs, we're not exactly in a good understanding with each other, but we've decided to keep our relationship civil. I'm not exactly comfortable with them yet, but I'm trying. I just have to look back on the other units like transformations to help me solve that.

I seem to have no problem with using the laws, but I also have to confess that I've occasionally forgotten that a LOGARITHM IS AN EXPONENT. Otherwise, I think I'm pretty confident about today, but who knows what'll happen.

BOB on Logarithms

2008-04-24T23:36:00.004-05:00

I almost forgot to BOB, it's a good thing I looked at this blog...before bed. Logarithms are exponents. Yes I remember. Logarithms are also pretty tough. At first I was quite confused about the logarithms are exponents, and the whole anatomy of it, but a little ways into the unit I grasped that concept. I found it quite easy to solve for a value using logarithms, and I thought it was going to be easy street. It's good to remember then, when you see the letter e, use ln, but when it's 10 use log, and if you see neither, then use ln, because its one less letter, good stuff. As I strolled along into the pre-test I unfortunately forgot all about graphing logarithms and I froze thinking about it, but I still don't fully understand it, but I can always study! Compounding interest was pretty interesting, and I never really thought about how interest worked, I just watched my savings grow, but it was nice to learn about how those bankers do it. I found half-life's also pretty interesting and quite straight forward and I loved the story behind it, and how scientists use it to date dinosaur bones. I always wondered why people were so sure about the age of dinosaurs, and now I know, so that was good stuff. It was all good fun, and I learned tons of stuff that weren't directly about pre-cal and it was really interesting. That's pretty much it, good luck everyone!

-Francis

BOB version whatever we're at -_-;

2008-04-24T23:35:00.003-05:00

Okay well I almost forgot about this. I had to work tonight and was studying and doing crazy amounts of english and blah blah and when I finished I was just ready for bed. Then I remembered the bob for this unit of ours, logarithms and exponents. So because I'm tired, lets get this show on the road.

Pros (or other assorted things I found somewhat easy):

-I found this unit was helped along quite well with the phrase, "A Logarithm is an exponent." Many times when I found myself stuck on a question, I'd turn to this, and usually a eureka moment was soon to follow.

-Much of this unit was algebra, so as long as you had that locked down, almost half the work was done for you.

Cons (or other assorted things I found somewhat difficult):

-For me, one of the hardest parts of this unit was grasping the concepts in it. Usually when I learn something new, I take what I'm taught, and translate it in my head, and change it into a form I can remember. Now for the most part this process goes off without a hitch, but with this unit I found parts of it...trying. For example, I knew the change of base law worked, and I kinda knew where and when to use it, but I had nooo idea HOW it worked, until recently, when I got that cleared up with Mr. K

-Alot of the time I found it downright confusing. Like much of the unit made sense, but seeing the logs and ln's and such all over the place was sometimes disorienting, and a bit smothering. I had to learn to break it down, and take those problems one step at a time, or else I'd get overwhelmed.

Overall:

Not my favourite unit. Not by a long shot. It was interesting yes, and as always Mr. K's teaching methods were inventive and got the point across, and when I needed help he clarified things quite well. However I am almost looking forward to this test tomorrow, because that means they're finished, done, finito; until exam time. I struggled the most with this unit out of all of them by far.

Anyways I'm going to sleep haha. z_z

Justus out,

Ciao.

Logarithms and Exponents Practice

2008-04-24T22:12:00.003-05:00

I can see that most of you guys are loving this unit, but I still made a collection of online quizzes that can be found around the net.There are many good sites out there that you can practice on, so I decided to make a list of few sites that will definitely help all of you. Here it goes:

Link 1
Link 2
Link 3
Link 4
Link 5
Link 6
Link 7
Link 8
Link 9
Link 10

I planned to post this earlier but my internet connection is acting weird because of the weather outside.

-m@rk