Pre-Cal 40S (Fall '06)

The Adventure Continues ...

2007-02-05T14:21:00.001-06:00

Our adventures in blogging continue....

Watch for new blogs going live February 5, 2007 ...

Applied Math 40S (Winter '07) (Grade 12)

Pre-Cal 40S (Winter '07) (Grade 12)

AP Calculus AB continues ... (Grade 12)

So Long ...

2007-02-05T14:18:00.001-06:00

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! Watch this space for pointers to new blogs for each of my classes.

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

Podcapsule

2007-01-29T23:42:00.000-06:00

Last Thursday morning the students in this class wrote their final exam. That afternoon we wrapped up our class and recorded this podcast. It's called a podcapsule because, like a time capsule, it'll stay here as a permanent record of the students reflections and recommendations to themselves on how to guarantee their success in their next math class.

Although the class is over and the podcast is done you can keep it going as long as you like ...

First listen to the podcast (6.3 Mb, 13 minutes 11 seconds), then, if you like, you can keep it alive by adding an audio comment or text comment on this post.

Exam Rehearsal for Identities and Counting

2007-01-23T16:55:00.000-06:00

Hey people! It's JessiccaI_198 as scribe today. Today we had two exam rehearsals; one for identities in the morning and the other one was for counting in the afternoon. We talked a bit about the arc sine and about functions, but I didn't really get many notes on that. But here are the questions and solutions for the Exam Rehearsal Identities:

1) The expression (sinx + cosx)^2 - 1 is equivalent to:

a) sin2x

b) cos2x

c) tan^2x

d) 0

The answer is a) sin2x.

Solution:

sin^2x + 2sinxcosx + cos^2x - 1

2sinxcosx

sin2x

2) If sinA = 3/5 and sinB = 2/3, and A and B are acute angles, what is the value of cos(A - B)?

a) -2/3

b) (4√5 - 6)/15

c) (4√5 + 2)/5

d) (4√5 + 6)/15

The answer is B

Solution:

cos(A - B) = cosAcosB + sinAsinB

= (4/5)(√5/3) + (4/5)(√5/3)

= (4√5 +6)/15

3) Find all values of x in the interval 0 < x < cos2x =" -3sinx">

The answer is x = 7π/6, 11π/6

Solution:

cos2x = -3sinx - 1

1 - 2sin^2x = -3sinx - 1

0 = 2sin^2x - 3sinx - 1

= (2sinx + 1) (sinx - 2)

sinx = -1/2 sinx = 2

2 is rejected because in the unit circle, no value can be bigger than 1.

x = 7π/6, 11π/6

4) If sinΘ = √5/3, what is the value of cos2Θ?

The answer is -1/9

Solution:

cos2Θ = 1 - 2sin^2Θ

= 1 - 2(5/9)

= 1 - 10/9

= 9/9 - 10/9

= -1/9

5) Prove the identity: sinΘ/(1-cosΘ) + sinΘ/(1+cosΘ) = 2cscΘ

Here are the problems from this afternoons class on counting:

1) The number of different arrangements of 3 boys and 4 girls in a row, if the girls must stand together, is represented by:

a) 4!x4!

b) 3!x4!

c) 4!x4!x2!

d) 3!x4!x2!

The answer is a.

2) The students in a music department have practised 6 contemporary and 5 traditional choruses. For their concert, they will choose a program in which they present 4 of the contemporary and 3 of the traditional choruses. How many different programs can be presented, if the order of the choruses does not matter?

a) 25

b) 35

c) 150

d) 330

The answer is c.

Solution:

6C4 . 5C3

= 150

3) All telephone numbers are preceded by a 3-digit area code. In the original Bell Telephone System of assigning area codes, the first digit could be any number from 2 to 9, the second digit was either 0 or 1, and the third digit could be any number except 0. In this system, the number of different are codes possible was:

The answer is 144 different ways.

Solution:

8C1 . 2C1 . 9C1 = 144 different ways

8C1 - there are eight numbers in the first digit - 2, 3, 4, 5, 6, 7, 8, and 9.

2C1 - two number in the second digit - 0 and 1

9C1 - all the numbers, excluding 0

4) A paperboy who delivers papers on his bike can travel only on the trails represented in the diagram. The number of different trails that the paperboy can take to get from house A to house B without backtracking is:

Solution:
There are 60 different trails

5. a) How many groups of 3 chairs can be chosen from 7 chairs if the chairs are all different colours?

b) How many different ways can 7 chairs be arranges in a row if 2 of the chair are blue, 3 are yellow, 1 is red and 1 is green? (Assume that all of the chairs are identical except for colour.)

c) How many different ways can the chairs in (b) above be arranged in a circle?

5. Answers:
a) 35
b) 420
c) 60

Solution:
5a) 7C3 = 35
- since order doesn't matter. 7 is for the total number of chairs, and 3 is for the number of chairs to be chosen.

b) 7! / 3!2! = 420
- 7! represents the total number of arrangements, and 3! 2! represent the non-distinguishable objects.

c) 6! / 3!2! = 60
- one of the chairs is a reference point, and the rest are seated in a circle. 3!2! are the non-distinguishable objects.

Autograph that provincial exam with EXCELLENCE!!

2007-01-23T12:53:00.000-06:00

scribe post: Exam Rehearsal Probability

2007-01-22T13:51:00.000-06:00

Hello guys!!!!!
Our topics in mathematics just ended. For us it is a relief but we must always put in mind that learning in a never ending process so there are still maths stuffs out there waiting for us. Anyway, this morning we did a rehearsal examination in probability. Mr. K gave us 30 minutes to work on it then after that we had a group discussion where we talked about our answers and did some brainstorming. Afterwards, Mr. K discussed with to us the right answers with their respective solutions. The questions given are:

(1) A certain soccer player has scored on 82% of his penalty kicks throughout his career. Given this information, the probabality that he will score on exactly 4 of his next 5 penalty kicks, correct to the nearest hundredth, is

(a) 0.80
(b) 0.66
(c) 0.41
(d) 0.08

The answer is (c) 0.41 because the solution for this problem is

(2) If P(A) = 3/4 and P(A and B)= 1/2, where A and B are dependant events, then P(BA) equals:

(a) 1/4
(b) 3/8
(c) 2/3
(d) 5/4

The answer is (c) 2/3. The problem indicated that A and B occured in dependant events, which means that both have to happen.

Solution:

P(A) * P(B) = 1/2

3/4 * P(B) = 1/2

P(B) = 1/2 * 4/3

P(B) = 2/3

(3) Peter places 5 equal-sized tiles in a cloth bag. Each tile has a letter on it. The letters are P, E, T, E and R. The probability that Peter selects the 5 tiles, one at a time, in order such that they spell PETER, correct to the nearest hundredth, is

The answer is 0.02.
>

(4) A child 2 quarters, 2 dimes, and 3 nickels in his pocket, but he does not understand the value of any of the coins. He puts 35cents worth of candy on the counter at a store and randomly selects two coins from his pocket. The probability that the two coins he selects will have a total at least as high as the value of the candy is:

(5) It is known that 53% of graduating are boys. Three are chosen at random. Given that at least two of the three grads are boys, determine the probability that all three of the grads are boys. ( Answer accurate to at least 4 decimal places. )

Well that's all folks.....
The next scribe is JESSICA!!!!!!!!
I already told her that she is next scribe...
Anyway guys, good luck to all...
May the good news be ours after the provincial exam....
God bless us all!!!!!!!!!

Sequences & Series Notes

2007-01-22T10:56:00.001-06:00

Here are the notes to add to your math dictionaries for this last unit on Sequences and Series. The notes must be copied into your dictionaries by hand.

Here they are: Page 1 of 3, Page 2 of 3, Page 3 of 3.

Geometric Sequences

2007-01-20T14:56:00.000-06:00

sorry e, i didn't notice it was doing that..but here you go just click these links.

http://i57.photobucket.com/albums/g215/TRILLEST06/11807.jpg

http://i57.photobucket.com/albums/g215/TRILLEST06/11808.jpg

the next scribe is Daphne.

Jefferson's Scribe post

2007-01-18T21:54:00.000-06:00

the next scribe is JHO-ANN.

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

T₁ = 1
D = 1
N = 5000

‘the initial term is 1
‘the to from getting from 1 to 5000 is 1. on account that each term increases by one. And that it doesn’t say what the difference really is so its 1
‘n is the rank of the rank of the term we want to find. The rank is the sum in this case and we want to find what the sum is when the sequence reaches 5000.

S_n = n/2[2(t1) + (n – 1)d]
S₅₀₀₀ = (5000/2)[2(1) + (5000 – 1)1]
S₅₀₀₀ = 2500[2 + 4999]
S₅₀₀₀ = 2500 = [5001]
S₅₀₀₀ = 12 502 500

b) what is the sum of all multiple of 7 between 1 & 5000?

T₁ = 7
D = 7
N = 714

Since it says multiples of 7,the initial term is 7 because you start with 7 when it comes to “multiples of 7”. The difference is 7 because it increases by 7 each time.
Now we want to find the sum of sequence when it reaches to 5000. but we don’t really want to find 5000. since we’re looking for the sum of the multiples 7 between 7 & 5000 then there really aren’t 5000 terms if we’re increasing by 7 each time. To find what the rank is, we need to 5000/7. and you get 714. whatever. Don’t need to worry about the decimal answers if I were you.

S_n = n/2[2(t1) + (n – 1)d]
S₇₁₄ = (714/2)[2(7) + (714 – 1)7]
S₇₁₄ = 357[(14 + 7130)]
S₇₁₄ = 1 784 785

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

12 502 500 – 1 786 785
= 10 715 715

This is a no brainer but I’ll explain it to you anyways. To get the answer you take the sum of all integers from 1 to 5000 and subtracted by the sum of the multiples of 7.

A super ball is dropped from a height of 200 cm. It rebounds to ¾ of the distance it fell each time it hits the ground. What is the total vertical distance traveled by the ball when it hits the ground for the fourth time?

We know that each time it hits the ground,the ball comes back ¾ of the last distance it traveled. So it traveled down, up, down, up, down, up, and down. So in order to calculate the total distance it traveled. We need to brake this into 2 solutions first. We need to solve the “down” distance and then the “up” distance and then add the together.

Down:

So what are the sequences?

Well since the first term is 200 and it decreases every time by a ¼( bounces back to ¾ of the last term) and that it hits the ground for times…..

200 + 150 + 112.5 + 84.375

(but we really didn’t need to know that only the first term)

*DID YOU KNOW: that all the things we need to know to solve the question was in the question itself? Go check it out ; - )

T₁ = 200
R = ¾
N = 4

S₄ = 200(1 – (3/4)⁴)/(1 – ¾)
S₄ = 546.875

(thanks to the help of our righteous hero- the CALCULATOR!!)

The 3 term for the “down” distance is:
‘think everything opposite of what I said from the other part

150 + 112.5 + 84.375

T₁ = 150
R = ¾
N = 3

S₃ = 150( 1 – (3/4)³)/ (1 - ¾)
S₃ = 346.875

Now we add the two answer to get …….

546.875 + 346.875 = 893.75 (anybody else noticed that the is a difference of 200 for the 2 answers. Me thinks there is a shortcut for this problem….)

……………TAAADAAAAA!!

I've voice recorded mr k's lecture today. It covered about half the class. Now i have listened to it a bit and you have to crank up the volume super high to listen on account that the mic was far from mr k. I just wanted to test out if it worked or not. Feel free to add some comments about it. Apologies to the loud scratching sounds on account i was moving the mp3 play on the desk, i assumed that it picked up the sound.

test voice lecture

Sum of Arithmetic / Geometric Sequence

2007-01-17T20:03:00.000-06:00

This is RICHARD and I'm your scribe for today.

Before we started our lesson in Geometric sequences, Mr. K discussed the solutions on the on-line quiz on probability.

(1) A card is drawn randomly from a deck of 52 playing cards. What is the probability that it will be a spade or a face card?

(3) A certain partial deck of cards contains 6 red cards and 4 black cards. Two cards are chosen at random. What is the probability that both cards are black?

R = 6

B = 4

4C2
____ = 6/45

10C2

(6) The diagram shows a partial map of a certain city park, with walking paths located on the grid lines. A tourist starts at point A and randomly selects a path to point C, walking only to the south and east. What is the probability the tourist passes through point B?

The number of ways to get to point C is RRRDD which is 5!/3!2!. The number of ways to get to point B is RRD which is 3!/2!. The number of ways to get to point B over the number of ways to get to point C is 3/10

(7) It is known that about 5% of coffee cups used in a particular restaurant will be chipped. What is the probability that out of 10 cups selected at random, exactly 2 will be chipped?

CNCNNNNNNN

C - represents the chipped
N- represents not chipped

10!/8!2! (o.o5)² (0.95)^8
= 0.0746

(8) Three coins are tossed. What is the probability that at least two heads turn up?

These are the possibilities that you can get at least two heads

HHT = 1/8

HTH = 1/8

THH = 1/8

HHH = 1/8

The sum of these events is 1/2

(10) A jar contains 5 red and 7 blue marbles. What is the probability of pulling out 2 blue marbles in a row, without replacement?

(2) A certain kind of die has four sides, labelled 1 to 4. When the die is rolled, each side has a probability of 1/4 that it will appear on the bottom. If the die is rolled 5 times, what is the probability that the side labelled “3" appears on the bottom in exactly two of the five rolls?

3N3NN

= 5!/ 2!3! (1/4)² (3/4)³

(4) Over the course of his basketball career, Julius succeeded in making a free throw in 75% of his attempts. In a randomly selected game, he attempted to make a free throw 12 times. What is the probability that he succeeded in 8 of the 12 attempts? (You don't need to know anything about basketball to answer this problem.)

12C8 (0.75)^8 (0.25)^4

(5) Three different names are randomly selected from the following list of five names. Max, Kim, Codie, Lee and, Alex. Determine the probability that “Kim” is one of the three names selected.

(1· 4C2)/ 5C3

= 3/10

(9) Assuming equally likely probabilities for male and female births, what is the probability that a four-child family will have at least one boy?

P(all girls) = (1/2)^4 = 1/16

P( at least 1 boy) = 1 - 1/16 = 15/16

After we had discussed about the on-line quiz, we proceed to our topic in Geometric sequences.

A mathematician named Carl Friedrich Gauss is a brilliant student. One day his teacher got angry because the class is so noisy. She gave an activity to keep them busy. She asked them to add the numbers from 1 to 100. Gauss came out with a better solution. He instantly computed the answer 5050.

t1, t1 + d, t1 + 2d, t1 + 3d, . . . . . . t1 +97d , t1 + 98d, t1 + 99d

Find the sum of the arithmetic sequence 2, 7, 12 17, . . . . S 51

S51 = 51/2 [2(2) + (51-1) 5 ]

= 51/2 [ 4 + 250 ]

=51/2 (254)

= 6477

given the geometric sequence 16, 8, 4, 2, . . . .Find the sum of this sequence up to th 10th term.

r = 1/2

t1 = 16

that's it for today!! the next scribe will be Jefferson..

Ashley's BoB

2007-01-17T09:54:00.000-06:00

I can't believe I totally forgot about this! Anyways Probability is alright. There are some questions that I find I'm able to do fast and other questions where I really have to think about it and I get stuck because I can't find the compliment. I actually think probablitity is a lot easier than combinations because probability uses combinations so I'm a lot more familiar with it. But overall I understand how to get some of the questions. But questions like the one on the pre-test I still need to work on. I'm just glad I made it on time to do the BoB. 2 minutes before math starts haha.... Do well guys (:

Richard's BOB on PROB

2007-01-16T23:46:00.000-06:00

One thing I can say about probability is that you need to understand the situation to get the right solution...
we follow some basic rules such as if it is "and" it means that we multiply probabilities and if it says "or", we add probabilities. But we also need to know if it is mutually exlcusive or not, independent or dependent...
...but there's only one thing that we shoulld always keep in mind regarding the probabilities, it is a number between 0 and 1 and the sum of the probabilities of the events will always equal to 1...

I hope everyone will do good on the test..Good luck and God Bless!!

Jefferson's blogging on blogging

2007-01-16T23:04:00.000-06:00

WEEEEEEELLLLLLLLLLLLL, This sort of hard to explain on my part. In probability it's easy to understand after the teacher explains it to you. But when you apply it, It's all a puzzle, and i hate puzzles! I know how to do probability questions that involve the word "given that", but I'm just a little confused with the independent and dependent and mutually exclusive and not mutually exclusive. I mean what do you do after you find that out? what sort of math is involved if these events are mutually exclusive. I have the hardest time understanding it. They rest of probability is A-OK for me! the tree diagram always helps and it gives you more than you're looking for if you know what I mean.

BOB.... how about probability

2007-01-16T22:57:00.000-06:00

What was the probability that after taking the day off school to sleep so that maybe i'd get my voice back, that i would remember to do my blogging on blogging?? all i can say is Thanks to the kids that posted them before now, cause when i went to check the scribe post to see what i missed, BOBs were posted!! Anywho enough about that, how about that probability unit?? I don't know about any of you, but I thought it was a good unit... I think there were only a couple times that i didn't know what i was doing, and those times were cleared up rather quickly. The other day when we did the pretest my group did so well.... minus a slip up where we were soo close. My favourite thing about the unit was doing the "what is the probability of A given that you already B?" those ones were no sweat... but the combinatoric ones still trip me up every once in a while... especially when i get distracted :D ... so i hope that i can keep focused and do well on that test tomorrow... and of course that everyone else does well too.... ok that's all for me tonight cause i'm getting some more sleep... who knows i might even get my whole voice back tomorrow.... till then, like mr. k says "Learn hard!!"

jho's blog

2007-01-16T22:02:00.000-06:00

This unit had it's pros and cons. In the beggining i was getting confused, but since we were exposed to this unit for quite some time i was able to understand the concept, and therefore i am confident i will do well in this test. Hmmm maybe..haahha can't be too confident..but that pre-test was good, i was able to answer most of the questions without a problem. Hopefully i will do well tomorrow. Man i have been missing a lot of work lately, because of my loss of internet connection, got to earn some marks. GOOD LUCK everybody =D

Tenny's BOB

2007-01-16T21:43:00.000-06:00

wow this unit has it's up and down. You would have to read twice ,to be able to write down your answer. sometime the answer you think its right might be even wrong.The first couple of part of unit was easy but it gets harder ,and harder. These questions you would have to do alot to be able to be familar with it. So practice,practice. Well thats all i have to say. bye

Bloggin on Blogging

2007-01-16T21:26:00.000-06:00

In probability well I'm not so good at it and let's just leave it at that. On plenty of questions I end up making plenty of mistakes, whether it's to using the incorrect method or leaving something out when solving the problem. While practice does help improve something, I've had close to a decade of problem solving questions and I doubt another few days is going to get me ready for those questions.

Scribe post

2007-01-16T20:24:00.000-06:00

Hey everyone its Tennyson ,well today we just did some questions but i was unable to finish. i will just put down the one that we did only.The next scriber can carry out the other questions.

1). List the 1st four terms in each sequence

a).tn= 3(n-1)
b).tn=(n-1)2
c).tn=2(power of)n-1

a).tn= 3(n-1)

-tn=3(1-1)
-tn=3(0)
-tn=0

-tn=3(2-1)
-tn=3(1)
-tn=3

-tn=3(3-1)
-tn=3(2)
-tn=6

-tn=3(4-1)
-tn=3(3)
-tn=9

Answer:0,3,6,9

b).tn=(n-1)2

-tn=(1-1)2
-tn=(0)2
-tn=0

-tn=(2-1)2
-tn=(1)2
-tn=1

-tn=(3-1)2
-tn=(2)2
-tn=4

-tn=(4-1)2
-tn=(3)2
-tn=9

Answer: 0,1,4,9

c).tn=2(power of)n-1

-tn=2(power of)1-1
-tn=2(power of)0
-tn=1

-tn=2(power of)2-1
-tn=2(power of)1
-tn=2

-tn=2(power of)3-1
-tn=2(power of)2
-tn=4

-tn=2(power of)4-1
-tn=2(power of)3
-tn=8

Answer: 1,2,4,8

Later on , we talked if the question was a Geometric sequences ,Arithmetic sequences, or neither.

a).This question is an arithmetic sequence because when the number is the same ,you add or subtract from each of the numbers to get the next number in the sequence.

b).This question is neither because They dont involve in multiplying ,divid,adding,or subtract , it just does have the patterns like the other sequences.

c). This question is an geometric sequences, because the common ratio is a number that you can multiply term in order to get the other next term like,Also can be divided by the previous term.

Those question are all related to a graph,tables, and many other things

next person is ricardo , but the list was confusing so yeah , hope u didnt do 4 scribs

by3by3by3

ruschev Blogging on Blogging

2007-01-16T20:11:00.000-06:00

I don't really know where I am right now in the probability unit....T_T.. I get some of the answers sometimes correct and sometimes I get some of the answers wrong and sometimes I know I have the correct answer and got it wrong.....sometimes I have the right answer but get the part where you solve for the correct answer wrong............. I guess I just need a Little more practice!

good luck tomorrow~

Oliver B's blogging on blogging.

2007-01-16T20:04:00.000-06:00

Well now, I sorta like probability, but the part I didn't or well, I wasn't fond of much was, the part where combinatorix was combiend with probability, only because I don't think I did that good during combinatorix, which I need to review on. I guess it is kinda neat though how they're together like that. Overall I did like this probability unit, I hope I have as much luck tomorrow during the test though! I wish everyone luck on that, okay bye bye.

daph's seventh BOB

2007-01-16T19:05:00.000-06:00

Probability for me is one of the scariest topics in mathematics. I found it hard to solve questions about cards and its decks. Moreover, thinking for the starting solution such as the tree diagram is another burden for me because I cannot really imagine things to be fitted in the diagram. On the contrary, I found it a little challenging. Some of the questions are really tough but sometimes I managed to solve it. Another is I want to acknowledge the effort that Mr. K gave to us in order to understand the probability well because he explained everything in details and still willing to explain it more if some of us was not able to understand it.
Honestly saying I'm not doing my assignments in probability. It is the reason why I cannot push myself in to a higher level when it comes to probability.
GOOD LUCK!!!!!!!!!!!!!
God bless us all in the upcoming test!!!!!!!!

JessiccaI_198's BOB for Probability

2007-01-16T19:00:00.000-06:00

I think that this unit was interesting, and I enjoyed doing it. I like making the trees for the problems which make everything easier and more organized. I just have to remember the methods to be used for certain problems. Using some of the Counting Principles helps a lot in some of the problems, and makes it easier to calculate too, so I'm glad we get to use some of that. Hopefully I won't forget to use some of the rules in the problems, but I'll probably just look at some of the problems we did in class so I would probably know which one I would need to use for the ones that will be on the test. The pre-test was a little hard, but I think I'm prepared for the unit test. Everyone Good Luck!!

BOB

2007-01-16T17:34:00.000-06:00

This unit is very interesting and I get most of it. It's just that sometimes when it seems like you have the right answer, you don't..does that make any sense? What I mean by that is that a wrong answer can seem right as to the point where it looks possible so you end up getting the wrong answer. So multiple choice questions for this unit are pretty difficult. I really like the long answers where something is assumed given that an event happens, because they're a breeze. What's really hard to remember is that you always have to have the probability over the sample space and I always forget that. The pre-test was okay so I'm sure the test will be okay as well. Good luck everyone.

Probability Quiz

2007-01-15T16:17:00.000-06:00

Here is it! You have until midnight tomorrow to complete it. We'll talk about in class on Wednesday morning. The test is Wednesday afternoon.

Learn Hard!

Wouldn't it be nice if the world really worked like this ... ;-)

scribe post; geometric sequences.

2007-01-15T14:40:00.000-06:00

today we started off with mr. k giving us questions on the board.
find the next three numbers..
1, 2, 4, 8, 16, 32
how you would get the next set of numbers was multiplying the number by two
OR 1, 2, 4, 7, 11, 16
how you would get this one was to find the pattern in the sequence. so you would have to find the difference between the numbers. for example: 2 and 1, the difference is 1. and the difference between 4 and 2 is 2. so you would get the pattern. you know that the number is increasing sequentially. so the next number must be a 7 - since the next number is a 7, the difference between 7 and 4 is three. and so on..

6, 11, 16, 21, 26, 31
for this one, the difference between the numbers is 5 and it's always constant.

3, 6, 12, 24, 48, 96
this one is basically like the first one, you multiply the numbers by two.

1, 1, 2, 3, 5, 8, 13, 21
for this one.. you find a pattern by adding the numbers.. for instance: 1 plus 1 is 2. then 2 plus 3 is 5. and 3 plus 5 is 8. so you get the point..

1, 3, 6, 10, 15, 21, 28
this one is basically like the first one.. but the second answer.. but this one doesn't start at 1. it starts at two - meaning.. the difference between 3 and 1 is 2. 6 and 3 is 3. and it's going up sequentially, and so on..

x, -x², x³, -x⁴, x⁵, -x⁶, x⁷
for this one, you would multiply by -(-x)

3, 4, 7, 11, 18, 29, 47
this one, you add the numbers together. 3 and 4 is 7. 4 and 7 is 11. 7 and 11 is 18. etc etc..

77, 49, 36, 18, 8, 8, 8
this one was kinda tricky. you would split the numbers - so you would get 7 and 7. you would then multiply them together and you would get 49. 4 times 9 is 36. 3 times 6 is 18, and soo on..

in the afternoon class, we had a pretest on probability. and here are the questions and solutions to them.
1. twelve people, including you, are members of a choir. the choir director is going to choose three members to attend a workshop. the probability you and two other members will be chosen is:
a) 1/4 b) 3/10 c) 1/12 d) 1/10

first of all, YOU have to go. so you would get 1 x 11C2 because you're part of that twelve so there are eleven people remaining to choose from - and you're going to choose two. and the denominator would be 12C3 since there are twelve people and you have to choose three. so that would look like..
1 x 11C2 ALL OVER 12C3
and that would all work out to 1/4, which would be a.

2. rex is playing a guessing game. the probability he will guess each question correct is 0.3. what is the probability he will guess exactly 5 out of 10 questions correctly?
a) 0.15 b) 0.10 c) 0.29 d) 0.80

so the P(correct)= 0.3 and the P(wrong)= 0.7
since there can be different ways of getting the questions wrong or right, but you still have to get five correct. so you would get..
10!/5!5!
the 5! represents the correct ones and the other 5! represents the wrong ones. you would then get..(C)⁵(W)⁵
(0.3)⁵(0.7)⁵
and when you work that all out, you would get 0.10 and that's the answer b.

3. you choose four digits from the numbers 0-9 (ten digits in total), with no repeated digits. to win a prize, all four of your numbers need to math the four selected by a computer. the probability that you will win a prize is:
P(all four digits)= 4C4 ALL OVER 10C4
and when you work that all out, you would get 1/210 (and that's the answer)

4. the serial nuber of a $10 bill contains 8 digits. if your $10 bill contains the digit 7 at least once, you will win a prize. what is the probability that your $10 bill will win?
first find the probability that you wouldn't get any 7's. so, the probability that you would get a 7 is 1/10 and the probability that you wouldn't get a 7 is 9/10. soo..
P(not seven)= (0.9)⁸
since you now have the probability of not getting any 7's, you can now get the probability of getting at least one 7.
P(7 at least once)= 1-(0.9)⁸
so, you would get 0.57 (and that's your answer)

5. jeff takes a lunch to school on two days, and on the other three days he buys it. if he takes lunch - he is late for his period 4 class 15% of the time, but if he buys lunch - he is late 35% of the time.
a) what is the probability that jeff arrives on time?

b( if jeff was late, what is the probability that he took his lunch to school?
first make a tree diagram

P(arrives ontime)= P(takes&ontime) + P(buys&ontime)
= (0.4)(0.85) + (0.6)(0.65)
= 0.34 + 0.39
=0.73
= 73%

P(late|took)= P(takes&late)/P(takes&late) + P(buys&late)
= 0.06/0.06 + 0.21
= 0.06/0.27
= 0.2222
= 22%

oh yeah, the test is now on wednesday.
umm.. the next scriber is tennyson.
gooday&goodnight (:

Challenge

2007-01-15T07:53:00.000-06:00

It's a new year with new units, new tests, and new learning. And how will you meet those challenges?!

Scribe Post

2007-01-13T12:43:00.000-06:00

Okay, sorry this scribe is so late but I got it done heh. Leave some comments if you think there could have been some changes to make it more funny or just funny. Click on the pictures to view. And the next scribe is on my other scribe post which is up to date.

Scribe Post

2007-01-13T11:21:00.000-06:00

Okay, today Mr.K started and finished off, with the following questions:

1)Tickets numbered 3,6,9,12,15 and 18 are placed in Box A. Tickets numbered 6,12,18,24 and 30 are placed in Box B. A ticked is chosen at random from each box. Find the probability:

a)That both tickets have the same number?

First find the numbers that are in both Box A and Box B. You will see that only 6,12, and 18 are in both Box A and Box B. Then calculate the probability for each, as shown:

P(A6B6)=(1/6)(1/5)= (1/30)
P(A12B12)= (1/30)
P(A18B18)= (1/30)

Now add them up:

(1/30)+(1/30)+(1/30)= (3/30)= (1/10)

Therefore the probability that both numbers are the same is (1/10).

b) That there are different numbers on the two tickets?

This question is simple to answer, because we have just figured out how many times the numbers will be the same and it was (1/10), so now all we have to do is subtract this fraction from 1, resulting in the probability of the numbers not being the same.

1-(1/10)= (9/10)

2)John uses Google 50% of the time, Lycos 30% of the time and Altavista 20%.If he is using Google, there is a 40% chance he is searching for info about cell phones. If he's using Lycos or Altavista the probability he is researching cell phones is 30%.

First we should create a tree to organise our information, and making the question more easier to calculate. SORRY BUT MY PICTURE WILL NOT LOAD AT THE MOMENT, I will quickly look into it and post it as soon as possible.

a)What is the Probability that John decides for info about cell phones?

This will be simple we just need to look at our tree (which i currently don't have up so the change to this question will be made once i clear up my problem.)

b)You see John looking at a website that is not about cell phones. What is the probability he used Google to find it?

P(G/NCP)= (P(GNCP))/(P(GNCP)+ P(LNCP)+P(ANCP)
=(30/100)/((30/100)+(21/100)+(14/100))
=(30/100)/(65/100)
=(30/100)X(100/65)
=(3000/6500)
=(3/65)
=4.6%

Therefore the probability he used Google is 4.6%

*P(G/NCP) meaning: prob. used Google but not for cell phones.
*P(GNCP) meaning: prob. used Google not for cell phones.
*P(LNCP) meaning: prob. used Lycos not for cell phones.
*P(ANCP) meaning: prob. not used Altavista.

3)The letters AADKRRV are written on separate pieces of paper and placed in a bag. If they are drawn from the bag at random, what is the probability that the order of the letters spell AARDVARK?

There where many ways to calculate this problem. I will show you how I have completed the question.

First I reflected on the unit on Counting, and the question became simple. A method we where tought involved lines and multiplying the factorials that go in them. Then I applied it to my knowledge of going about solving probability questions. I came up with this:

3!/8 chances of A 1st.
2!/7 chances of A 2nd.
1!/6 chances of R 3rd.
1!/5 chances of D 4th.
1!/4 chances of V 5th.
1!/3 chances of A 6th.
1!/2 chances of R 7th.
1!/2 chances of K 8th.

Then I simplified the answer and this is what I got:

(3!2!)/(8!)

4)Two traffic lights operate independently. The probability that the first is red is 0.4. The prob. that the second is red is 0.7. What's the probability that neither are red?

P(NR1)= 0.6
P(NR2)= 0.3

(0.6)(0.3)=.18

Unfortunately the Big Space has come back,heh, I was able to kinda fix but its still there.

THE NEXT SCRIBE IS THE JESSICA THAT HAS'NT DONE 3 SCRIBES YET.

Scribe Post

2007-01-08T17:02:00.000-06:00

First day of class after the winter break and of course not much happened today, as we went through a few questions which are listed below. Other then that we discussed more about probability.

Suppose a test for cancer is 98% accurate. This means that the result of the test is correct 98% of the time. Suppose that 0.5% of the population has cancer. What is the probability that a person who tests positive has cancer? Suppose 1, 000, 000 randomly selected people are tested. There are 4 possibilities.

A person with cancer tests positive
A person with cancer tests negative
A healthy person tests positive
A healthy person tests negative

1) a) How many of the people tested have cancer? (1, 000, 000)(.005) = 5000
b) How many do not have cancer? 1, 000, 000 - 5000 = 995, 000

2) Assume the test is 98% accurate for people with cancer
a) How many people with cancer test positive? (5000)(.98) = 4900
b) How many people with cancer test negative? 5000 - 4900 = 100

3) Assume the test is 98% accurate for people without cancer
a) How many people without cancer test positive? 995, 000 - 975, 100 = 19, 900
b) How many people without cancer test negative? (995, 000)(.98) = 975, 100

4) a) How many people tested positive for cancer? 4900 + 19, 900 = 24, 800
b) How many of these people do have cancer? 4900
c) What is the probability that a person who tested positive for cancer actually does have cancer?

P(C|P) = 4900/(4900 + 19900)

= 19.76%

P(P) = 0.0049 + 0.0199

= 0.0248

P(C|P) = P(CP)/(P(CP) + P(HP))

= 0.0049/0.0248

= 0.1976

P(H|N) = P(HN)/(P(CN) + (PHN))

= 0.9751/(0.975 + 0.0001)

= 0.9999

I'm not too sure about what's going to happen tomorrow since I believe that a lot of people from class will be taking the English Provincial Exams so unless the scribe will be skipped to Wednesday the next person is Gerald.

Peter Donnelly

2007-01-08T16:58:00.001-06:00

Oxford statistician Peter Donnelly explores the common mistakes humans make in interpreting statistics, and the devastating impact these errors can have on the outcome of criminal trials.

You learned the math he's talking about here today.

Click on the picture. (22 min. 6 sec.)

Scribe post

2007-01-05T16:15:00.000-06:00

Probability

1st period class

it was the day before the Holiday break so we didn't do much but as usually we started the day by doing question......^_^

Blood types in North America are distributed as follows:

Type O: 44% Type A: 42% Type B: 10% Type AB: 4%

a) are the blood types mutually exclusive?

b)what is the probability that a couple both have type O blood?

answers

a) no (because the blood type can be both be "A" "B" or "AB" and "O" is the universal blood Donner meaning that it can fit into any blood type category)

b) "tree diagram"!! HO = "has blood type "O" NO = No blood type "O"

the first tree branch representing one of the couple has 0.44 chance of getting blood type "O" and the second branch representing the second couple has 0.44 chance getting type "O" also. we use "Multiplying Probability" since we want to find out what is the probability that both couples have the same type "O". (HO)HO) = Santa clause? HO NO HO NO I'm just kidding the answer is (0.44)(0.44) = .1936 or 19% that both of them have the same blood type "O".

2nd period class

In a group of 30 students what is is the probability that at least two have the same birthday?

*find the probability that nobody else have the same birthday

* use Pick formula since the order matters

*(# of days in a year)P(# of students) divided by (# of days in a year)^(# of students)

365P30/365^30 = 0.26 then subtract it from 1 ( 1- 0.26 =74%)

therefore the probability of at least two students have the same birthday is 74% (3o students)

and then one of my classmate asked him if our answer for this question was really true....in a 30 people group is there really a 74% chance of picking at least two people that have the same birthday? but most of our classmates are "missing"*wink**wink* so he started getting students on the hall way to make a group of 30 students and then he asked us one by one our birthdays but in our 30 group of students no one has the same birthday so he went back on the hallway to get some more people to raise the chances............were like HO HO HO HO laughing but we forgot to make a list and everyone was like HO NO HO NO no one made a list oi that was fun..............

sorry I couldn't get all the notes for this question :)

THE SCRIBE IS TENNYSON .....................FOR MONDAY

Ben Saunders

2006-12-29T09:00:00.000-06:00

Arctic explorer Ben Saunders recounts his harrowing solo journey to the North Pole, complete with gorgeous images, amusing anecdotes and previously unseen video footage from the Pole. At 26, Saunders became the youngest person ever to ski solo to the North Pole, updating his blog daily during the trek. He's now planning the next journey, SOUTH, an unprecedented, roundtrip expedition across Antarctica and back.

His story is spellbinding. There's a really important message for you in the last two minutes ... but you won't really understand it unless you watch the whole thing.

Click on the picture. (18 min. 48 sec.)

Bono - TED Prize Winner

2006-12-28T09:00:00.000-06:00

For the past 20 years, members of the TED community have gathered together to share ideas and passions that are big enough to change the world. Each year they will honor a maximum of three individuals who have shown that they can, in some way, positively impact life on this planet.

Rather than simply receiving financial support, winners of the TED Prize are granted something extraordinary: something which children dream about, but which adults assume is merely the stuff of fairy-tales.

They are granted three WISHES to change the world.

They may wish for anything. And TED will seek to make their wishes come true.

Bono won the TED prize in 2005.

Rock star and activist Bono accepts the 2005 TEDPrize with a riveting talk about our moral obligation (and economic incentive) to help lift Africa out of poverty. He unveils his TEDPrize wishes by challenging the TED community to help build a social movement of more than one million American activists for Africa; to tell people one billion times about the ONE campaign; and to connect every hospital, health clinic, and school in one African country, Ethiopia, to the Internet.

Click on the picture. (28 min. 37 sec.)

For more: Read the update on Bono's wishes.

Richard St. John

2006-12-27T09:00:00.000-06:00

Why do people succeed? Richard St. John compacts seven years of research into an unmissable 3-minute slideshow on the real secrets of success (Hint: Passion, persistence, and pushy mothers help).

Click on the picture. (3 min. 40 sec.)

Gregory Colbert

2006-12-26T09:00:00.000-06:00

Photographer Gregory Colbert shares the remarkable images and film footage from his exhibit, "Ashes and Snow," and announces his founding of the Animal Copyright Foundation, which will require that royalties be paid when images of nature are used for commercial purposes.

The pictures are just stunning.

Click on the picture. (18 min. 42 sec.)

Hans Rowling

2006-12-25T09:00:00.000-06:00

Hans Rowling is a professor of of international health at Sweden's world-renowned Karolinska Institute.

Watch how he displays and talks about statistics. He'll make you laugh and he'll make you think.

Click on the picture. (20 min. 35 sec.)

Pi and e ... their first date

2006-12-22T15:30:00.000-06:00

You know enough math to get all the jokes in this now ... ;-)

Have an excellent holiday!

Cheers,
Mr. K.

Flickr Assignment Rubric v1.0 - We're Out of Beta!

2006-12-22T15:00:00.000-06:00

Here is the rubric we've settled on together.

Thanks to everyone who helped put this together. I found this to be a great experience for me as a teacher I hope it was also valuable to you as a student.

Anyway, here it is, version 1.0 fresh out of beta. ;-)

Flickr Assignment Rubric
It is paramount that the picture be in tune with the purpose of the assignment. It should show, first of all, the student's understanding of how the photo is related to mathematics. The hot spots are important too, because that's essentially your way of teaching other people. Creativity is a factor, because keeping one's interest in the photo contributes to the learning process. Finally, the picture quality should be kept in mind too. If we can't see the picture, it's going to be hard achieving all the other requirements.

Scribe post

2006-12-21T19:27:00.000-06:00

the 3rd scribe post of ruschev

PROBABILITY

note:

- Mutually Exclusive : unable to be both true at the same time ; one event taking place prevents the other event from taking place ex) flipping a coin is a mutually exclusive since it can not be heads at the same time.

- drawing a tree diagram will help allot when solving probability questions

- drawing a Venn diagram will also help

- adding probability. P(a)+(b) when the question is an "or" question. ex) 2 boys are running what is the probability of one of these boys win if one boy has 1/4 of winning and one has 2/3 of winning.

-multiplying probability. P(a)*(b) when the event is independent. ex) a boy flips a coin and gets a heads, he flip it again and get a tails. independent does not change the outcome of the next occurring event.

- (~) is the symbol for "not"

- "1" is the highest possible value you can get ; a 100% chance ; more than 1 will result in an incorrect answer ; when you add up all of your probability the numbers you should only get is equal to or less than 1 or zero.

- changing percent into a decimal will make calculation easier

- fraction is our Friend!

Questions:

1. man of war and secretarial are in a horse race, man of war has a 2/5 chance of winning and secretarial has a 1/3 chance of winning..

a) are these events mutually exclusive?

b) what is the probability that one of these horses wins the race?

c) what is the probability that both win?

answers:

a) no

b) probability of man of war winning is 2/5 ; probability of secretarial winning is 1/3 so...

(2/5) + (1/3) = (11/15)

c) (2/5) x (1/3) = (2/15)

Questions:

10 grade 12....40% of the students take math, 35% take history and 15% take both. If a student is randomly chosen. what is the probability that the student takes..

a) math but not history?

b) neither math or history?

Answers:

a) change percent into decimal. P(math) = .40 P(history) = .35 P(~m)= .60 P(~h) = .65

P(M x ~H)= P(.40 x .65) = .26

"green" = students taking math ( 40%-15% = 25% or .25) "brown" = students taking history (35%-15% = 20% or .20) "blue" = students taking both subjects ( 15% or .15) *add up all of our values ( .20+.15+.25 = .60) meaning that there are .40 students that are not taking neither subjects (1-.60 = .40)

Question:

you are dealt a 5 card poker hand, what is the probability that you have "at least one" heart?

Answer:

52 cards in a standard deck of cards and there are 13 hearts * 1 - none leaves all the possibility of "at least one"* *1- none leaves all the possibility of "at least one"* * 1 subtract none leaves all the possibility of "at least one"* * 1 take away all the cards that are not hearts will leave all the cards that are hearts*

39C5 = 39 ( the number of cards that are not hearts) 5 ( 5 card poker hand)

52C5 = 52 (total number of cards) 5 ( 5 card poker hand)

1- 39C5/52C5 = .77

numerator: 39C5 <- 39 cards that are not hearts choosing 5 random cards

denominator: 52C5 <- total cards choosing 5 random cards

"1" <- the highest possible value that you can get ; 100%

First things first? One day left to go--

2006-12-21T11:19:00.001-06:00

Wondering if you're having trouble putting first things first in these last days before winter break?

And

Wanting to wish each of you happiness, peace, and joy during this season! Be safe and enjoy!

Best,
Lani

BOBing the conic way :D

2006-12-20T16:58:00.000-06:00

Looks like its that time again folks... that's right BOB time, which
means a test is really close... and this was a short unit, but that
didn't mean we didn't learn a lot :D ... conics was not my favourite
unit, but it was far from the worst, so overall its not too bad. I
thought i was doing ok...  until i hit the online quiz... but i figure
my barely passing mark was because of some poor
concentration.. i like doing the quizzes at home, but this
time there were just too many distractions, but if i had paid
attention (i just went and reviewed my mistakes... jeeze i made
a lot of dumb ones) i could have done really well... so this means
that during the test i really need to just buckle
down, stay calm and believe that i know what I'm doing... then i will
do just fine... other than that the hardest part for me is just trying 
to make everything work out correctly without missing steps because
I'm rushing.. again if i stay calm and take my time I'll end up with
a much better result.. well that's just about it from me, and soon
for conics too (until the going for gold assignment and exam anyways :D )... 
So good luck everyone, and I'll see you on test day :D *learn hard!!* :P

2006-12-20T16:41:00.001-06:00

Don't forget to check your spelling. You can delete this now.

Blogging on Blogging

2006-12-19T22:42:00.000-06:00

Alright well, I'd like to say, conics was okay, it didn't last very long, but I like this unit, But I hope I do good tomorrow. I usually don't do as good at online quizes, But I believe that's a ood way of reviewing, or testing what you may know already, and see where you need improvements. Anways I wish everyone good luck tomorrow, and myself as well! Bye.

Conic's Blogging on Blogging

2006-12-19T22:18:00.000-06:00

It's the end of the conics unit, and I think this one was one of the hardest units yet! I think I'm okay with it though, I just hope I don't blank out or anything. Hopefully I'll be able to remember how to apply the rules for the test tomorrow.

Hopefully everyone does well on the test tomorrow!! ^_^

bob

2006-12-19T22:14:00.000-06:00

well here's my last minute blog...almost forgot. For this unit I thought it was going fine and then we had the group work for parabolas which kind of scared me because I couldn't get the answer since I forgot what to do...but then afterwards i remembered which was good...we should have handed in the second parabola question..my group actually got that one correct..yayy..well anyway back to the test, the quiz for some odd reason i didn't do too well in which really sucks because I thought I knew everything so now I think I don't know anything.I think it'll be okay I just have to try my best..I think I got all of my thoughts for this test out so see everyone tomorrow and good luck on the test.

Probability

2006-12-19T22:06:00.000-06:00

Sorry guys this came late, and I gotta sleep soon too, My net's very .. bad .. and I haven't gotten on till now, but yes .. In the morning we were given 3 questions to do on the board.

1. Find the sample space for the genders of 3 children?

2. What is the probability of 2 children in a family both being boys?

3. What is the probability of drawing a red card and then a black card out of a deck of 52 playing cards if:

a) The first card is replaced before the second is drawn?
b) The first card is not replaced before the second is drawn?

The answer to this question will have to be drawn, but it would have to be listed and these are it:
B= Boys
G= Girls

BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG

2. The answer to this question is well 1/4 because when you want something to happen you have to multiply. It would help making a diagram =).

3.a) The answer for the first one would be 1/4.
b) The answer for this would be
P(RB)= (26/52)(26/52)
which would equal 0.2549

For the last part of class we wrote a bit of definitions in out math dictionary:

Dependant events: 2 ( or more ) events are dependant if the outcome of the second event depends on what occurred in the first event.

Example: A bag contains 3 red and blue marbles. A marble is drawn without replacement. The probability that a second marble will be blue depends on what the first marble was.

Independent events: 2 ( or more ) events are independent if the outcome of the first event has no effect on the outcome of the second event.

Example: Given the same bag of marbles. A marble is drawn with replacement. The probability that the second marble is blue is unaffected by which marble was drawn first.

Alright well, that's all everybody, sorry there are no diagrams for the moment, I'm trying to fix my computer, it's hard enough trying to fully fix our connections. Anyways, it's that time again, the NEXT SCRIBE WILL BE PATRICK. GOOD LUCK ! =).

jessica.'s bob.

2006-12-19T21:17:00.000-06:00

hmm, conics. i think this unit was pretty straight forward. i pretty much know what i'm doing. i've actually learned a different way of doing the equations. it's all good (: goodluck on the test tomorrow.

gooday&goodnight.

Tennyson's BOB

2006-12-19T17:45:00.000-06:00

man , i am trying my best on this one , everytime we do a test , i always mess up. I know how to do the questions , but somehow, the question always different , very little is the same that we do on our assignments. Like i do well in homework , but when the test comes , its like everything is different then what i have done before. Well like Mr k said , " your mark determines , how much time you put into it". This unit is ok , its not bad, very short.well i hope i pass this test good luck everyone

JessicaJill's BOB

2006-12-19T08:42:00.000-06:00

Whoa, I almost forgot to do this. Well, the unit has come to an end, and the question is, have I learned anything? Indeed I have. A quick and easy unit I thought, but there were new things for me to pick up and learn. All in all, a good unit. Now I'll just head to class. =)

Ashley's BOB on Conics

2006-12-18T22:06:00.000-06:00

So this unit Conics is my absolute favourite! I actually understand what I'm doing and I finally know how to find the foci because for the longest time now I had no idea what the focus was:P I think I'm doing in this unit but I don't want to get too happy because the test could be hard. Some questions still confuse me, I mean I tend to make little mistakes and forget that y^2 = x that means that the parabola is opening sideways. Stuff like that I always forget so I need to learn how to read the question properly but other than that I'm doing fine. Good luck everyone!

Probability and math dictionary notes on it

2006-12-18T17:12:00.001-06:00

Hey it's Ashley and I'm the scribe for today!
In the morning we started our new unit Probability. Mr.K said that Pascal invented Probability and its the same Pascal that did the Pascal's Triangle.
Mr.K asked us "What can happen if you roll six-sided die?"
Well you could get 6 possible answers. (i forgot what he said oops)

Soon we were asked "What can happen if you flip a coin twice?"
What you could do is draw a tree diagram

Next question was a family of two children - "What is the probability they have one of each, a boy and a girl?"
Draw the tree diagram.

That question and the flipping of the coin question are the same.

Next was "What if you roll a die twice?"
You could for the question draw a tree diagram but that would be too complicated so instead you could draw a table.

Last question was "What if you flip a coin and roll a dice?"
Mr. K made us draw a table and a tree diagram.

Towards the end of the period Mr.K handed us 2 booklets. One was yellow and one was green. Their both from the June 2005 exam. The title for this assignment is drum rolls please..... Go for Gold!!!

For this assignment you must:
- get 100% on it.
- complete the two booklets with no mistakes or errors.
- show all work including for the multiple choice questions either by writing a sentence explaining yourself but don't write a book. If there is no work shown or a sentence you get 0.
- must write in pencil! no pen or else you get 0.

In Go for Gold there are only two possible marks available. 0 and 100. It's worth 5% on our final class mark. Due on the Friday we get from winter break and you may get help from others and Mr.K.

In the afternoon Mr. K went over the questions on the quiz. If you want to see which ones you got right or wrong you could check now because it will show the answers. Mr.K got a question wrong on "Determine the foci of the hyperbola (y2/16) - (x2/25) = 1" by placing the answer on the wrong letter so that question won't count.

Next we took some notes down into our Math Dictionary

Probability

Probability: the branch of mathematics that deals with chance.
Sample Space (Ω): the set of all possible outcomes for a given "Experiment" represented by capital omega Ω.
Event (E): A subject of the sample space. A particular occurence in a given experiment.
Simple Event: The result of an experiment that is carried out in a single step.
Example: Flip a coin. The result is heads
(a simple event)
Compound Event: The result of an experiment carried out in two (or more) steps.
Example: 1. Flip a coin and roll a die - 6
The result is {H,6}
2. Flip a coin twice.
The result is {H,T}
Probability of an Event: (suppose to be sides to the box below but i don't know how to draw a rectangle so just put a box around it, i only managed the top and bottom borders)

_________________________________________________________

P(E) = number of favourable outcomes

Total number of possible outcomes (sample space) ________________________________________________________

Probability can be expressed as:

- a Ratio

- a Fraction

- a Decimal

- a Percent

(rectangle box below but i only got the top and bottom border, not the sides)

IMPORTANT: Probability of any event is always a number between 0 and _______________________________________________________________

Certain Event: an event whose probability is 1

example: roll a die - 6 and get a result less than 10
Impossible Event: an event whose probability is 0.

example: roll a die - 6 and get a 7

Complimentary Event: the compliment of E is E',

If P(E) = a then P(E') = 1-a

example: Given a standard deck of cards, a card is drawn at random.

P(spade) = (13/52) = (1/4) P (not a spade) = 1 - (1/4) = (3/4)

And that was the end of the math dictionary notes

Mr. K recommended we visit these sites:

math40s.com because it has the entire curriculum on it and some other stuff on it which I forgot but apparently it's a very good site.

webct.merlin.mb.ca because it too has the entire curriculum on it. you need to log on to this site.

username: demo

password: demo

it asks if you trust this site I think and you say yes.

both sites are good for reviewing.

** Don't forget to check the rubric for the Flickr assignment and leave your comments and what you like and what you want to change because the Rubric will be finalized on Friday. To leave a comment just press the Insert Comment button. The next Flickr assignment is on Exponential function which is due the Tuesday after winter break. The tag will be expfunc.

So that's it for my scribe.

Homework assignment is Ex.39 by the way.

Next scribe is........ well you know who you are *cough*Oliver*cough*

scribe post

2006-12-15T22:07:00.000-06:00

Hey it's Anh for your last scribe of the cycle. Class started with a discussion of the flickr assignments. We talked about how the assignment would be marked. The smaller hotspots should be on top of bigger hotspots so that Mr.K is able to access it. Also the photos should be clear. Yesterday Mr.K sent everyone an email invitation to a document where we can have a input to the marking rubric for the flickr assignments. So please participate.

*** TEST IS MOVED TO WEDNESDAY!

*** QUIZ IS ACTIVE TILL MIDNIGHT SUNDAY!

We pretty much did our pre test all class.

CONIC PRE-TEST

The pictures are still being stupid.

I guess this concludes my scribe post...umm good luck on the quiz.

Next scribe: Ashley :)

Online Conics Quiz

2006-12-15T14:59:00.000-06:00

Your online conics quiz ~~will be~~ is live at this link ~~very soon~~ now. ~~Working on it now. I will update this post as soon as it's ready.~~

Remember the deadline to write is Sunday night at midnight; which means you need to start it no later than 11:15 Sunday night. There are 10 multiple choice questions and 5 true or false questions. The quiz is timed. You will have 45 minutes to complete it once you begin.

BTW, the light in the picture forms a hyperbola; two of them actually. The light itself is a circle. ;-)

daphne's SAIS blogging on blogging

2006-12-15T14:16:00.000-06:00

The CONICS was a very interesting, mind-squizzing, and mental blocking unit (...for me!!!). It was a lot of "It seems like a complicated problem" remark but in reality, most of Mr. K's problem are the basics. Unfornately in my part, I found it hard to solve word problems especially when we are looking for the equation such as the ellipses and hyperbola. Parabola equations are way easier to solve than ellipses and hyperbola because I sometimes mixed up the information needed to find the equation. Futhermore, complicated word problems sort of gave me a headache specifically when I'm in the part of solving the equation then suddenly I'm stock because I do not know what will be my next step but in all fairness, Mr. K did a great job in teaching us this unit. He explained every unsolved problems in details and he never gave up on us until we are confident enough to solve CONICS problems with a guaranteed high mark. In addition, I would to thank those who helped me in this unit because without I may not be able to understand some of the topics.

Go pre-cal 40S!!!!!!!!!!!!....We can all do it on top!!!!!!!!!!!!!!

azn_chilly's style of bloggin on itself

2006-12-14T23:12:00.000-06:00

this unit about conic is somewhat easy but i still need to understand how to get those equations from those problem questions but with the full equation out and i have to make the graph its easy to understand it but it all the problem solving, its giving me a hard time and its giving me a negative feeling that i did all i can on the question and trying to figure it out soo.. i think i have to say for this unit, and one thing good luck to everyone on the big test

Word Problems with Conics

2006-12-14T19:56:00.000-06:00

Hey kids, I was yesterday's scribe post. This is going up a day late because I had to work late last night and I went straight to bed as soon as I got home. Anyways yesterday was a one period class, and we worked on some conic word problems from previous exams. Here are the five questions given to us on the board:

1. Find the equation of an ellipse with major axis AB and minor axis CD if the following coordinates are the vertices: A(2,7) B(2,-1) C(0,3) D(4,3).

2. Identify each conic:

3. Find the equation of the circle with the centre (4,0) which passes through the orgin.

4. y²-2y-3=0

a) Find all intercepts.
b) Sketch a clearly labeled graph.

5. A balloon arch in the shape of one branch of a hyperbola is made for a wedding party. The arch is to span 4m and have a max clearance of 2.2m.

a) Find an equation for the hyperbola.
b) What would be the height of the arch 0.8m from one end?

ANSWERS:

1. This is what I know. AB have the same x-coordinates, so that means that the line of AB runs vertically. So I just added the absolute values of the y-coordinates to get the length of AB: -1 + 7 = 8. It's an absolute value because distances don't have negative values. So AB is 8 units long, and we divide it by 2 to get the value of a, which is 4. So, we can say that line of CD runs horizontally and we add together their x-coordinates: 0 + 4 = 4. We divide 4 by 2 to get the value of b and we get 2.

To get the centre, we take the y-coordinate of A and subtract the value of a from the y-coordinates, just as we take the x-coordinate of D and subtract the value of b from the x-coordinates.

If you wanted to add the value of a to find the y-coordinate of the centre, you'd have to add from B's y-coordinate. To add the value of b to find the x-coordinate of the centre you'd have to add it from C's x-coordinate.

Subtracting

AB: 7 - 4 = 3
CD: 4 - 2 = 2

Adding:

AB: -1 + 4 = 3
CD: 0 + 2 = 2

So the center is (2,3).

Now we can start plugging in our values to an equation.

2. a) We can say that it is a hyperbola, because we have the subtraction sign between a fraction of x² and a fraction of y².

b) In the expansion we have positive squared values for x and y, which tells us it can't be a parabola or a hyperbola, but they have different coefficients so that tells us it is has to be an ellipse.

3. The equation of a circle is: (x-h)² + (y-k)² = r²

(h,k): (4,0)
point: (0,0)

From this we can tell that both points rest on the x-axis, and have are 4 units away from each other. Hence, we can say that the radius of this circle is 4 units.

(x-4)² + y² = 4²
(x-4)² + y² = 16

4. We take our eqation and change it to standard form.

y² - x - 2y - 3 = 0
(y² - 2y + 1 ) = x + 3 + 1
(y-1)² = x + 4

a) To find intercepts we just plug in zero for our variable.

x--intercept; y = 0

(0 - 1)² = x + 4
(-1)² = x + 4
1 = x + 4
-3 = x

y-intercepts; x = 0

(y - 1)² = 0 + 4
(y - 1)² = 4
y - 1 = ±2 [square root both sides]

y -1 = 2
y = 3

y - 1 = -2
y = -1

Therfore, our x-intercept is at -3, and our y-intercepts are at 3 and -1.

b)

5.

After realizing how lengthy question 5 was, we got to doing some group work. I kind of lost the sheet with the question on it, so I don't have much else to post up. I'll fix this up when I do find it, I just kind of forgot I never finished this. Sorry for the huge delay!

Analytic Geometry ( Conics )

2006-12-14T19:41:00.000-06:00

Hello everybody!, I hope everyone is doing fine tonight. I am today's scribe and well here's a short summary of today's class. In the morning we came in class, like any other day and then Mr. K put up 3 kinds of problem solving questions on the board and we had to answer them. He gave us sometime then we took up the questions. Here are some of my answers. The first question was:

An earthquake observation centre is located at coordinates (-2,5). An earthquake epicentre is located at a distance 120k, from the observation centre. What is the equation of the circle that defines where the epicentre ca be located?

** I would just like to say, I couldn't get most of the explanations on how we got the answers for these questions, because in the morning there was 2 fire drills. Very sorry about that.

Well the answer for this first quesion was (x+2)^2 + (y-5)^2 = 120^2.

This is the second quetion:

A reflector dish has a diamtere of 1.8 and is 24cm deep. Model the dish as aparabola opening up with its vertex at the origin. Find its equation and the coordinates of the focus.

Well around this time, we got the fire drill, and then we stood outside for a bit, so I didn't have too much time to copy the answers. We gt back in and then we somehow got into the convo about the sun, and I learned that in the galaxy our solar system is caled the "orion's arm" or something like that.

In the afternoon, we were supposed to have a pre-test but we finished up the last question from the morning, and this was when tennyson came up to the bored, and it was quite funny actually. Then a little after that we got into groups and we did some problem solving questions together. I personally found the questions, not hard, but confusing, because for some reason we kept mixing up numbers and what not. Anyways I hope everyone is ready fo the test on monday, and tomrrow is our pre-test! Good luck Everyone! That is all. Till next time, BYE !

THE NEXT SRIBE POSTER IS, **ANH !!=)**

Blaging on blagging-Jefferson style

2006-12-14T14:44:00.000-06:00

CONICS IS SO FUN AND EASY TO UNDERSTAND!. Unfortunately I'm having a lot of problems on the word problem questions. i just don't seem to know where to start. after one particular word problem is solved, i begin to think why i wasn't able to solve it with ease. There is nothing different about the idea. But if it's rewritten in a way that seems like it's from another language then it's really really hard.

Richard's 6th BOB

2006-12-14T14:39:00.000-06:00

In this unit, CONICS, ahhh! I find it difficult.
(How many times did I say the word difficult in my BOB?)
I somehow understand the lesson but still I don't have enough background in this unit.
Maybe because I didn't take grade 11 here. It's really hard for me to adjust.
There are lots of formula to memorize and lots of critical cases.
This unit is so painful..but I still believe in "No pain, No gain"
I should really work harder for this unit. It's so sad that I don't have the time to ask Mr. K (or rather shy).
Anyway...good luck everyone to the test! We still had a long time to prepare.

Blogging on Blogging

2006-12-13T23:12:00.000-06:00

this unit about conics is very easy.................but.......I seem to be having a trouble on vertical parabola and vertical Ellipse type of questions. I keep on forgetting that the formula for vertical is different from horizontal parabola and Ellipse, like the one we did today in class and I don't really get that word problem question about the balloon with the arc circle and angel...arg..ugh..ah..(brain shuts down....reboting...back online). other then that, I think I understand most of the horizontal type of questions. good luck on the test on friday

ruschev~~out

Blogging on Blogging

2006-12-13T18:10:00.000-06:00

Might as well get this done before I forget, conics I'm having an easy time with if you can believe it. The only problems that I seem to be having are just a few of the problems similar to the ones involving a semi-ellipse or half of a hyperbola like the questions in class today. Well that and the random bits of mistake that I make on occasions when I forget that an equation like "y^2 = x-4" is suppose to open vertically. Other then that's there's nothing too important to mention about how I'm doing for this unit.

Next Scriber?????????????

2006-12-12T12:10:00.000-06:00

i dont know if this is a new cycle of scribers or not but the person who will be the next scribe for tomorrow and i choice, choice..... plus i will be not finished by 12am i came home at 10 pm and am like 3/4 done typing it all out and starting my graphs soo.... the next person will be..... jessicajill who was to be today's scriber (mad mood) and now she own me 20 dollars in cash tomorrow hahahha j/k

azn_chilly's scribe post (problems + learning)

2006-12-12T11:35:00.000-06:00

This morning we started off with three questions off the bored and i was late for that class and was who willing to the scribe today hahaha!!!!!!. so we stared off by this .....

graph this conic section and label all information points
4x(square)+9x(square)-16x+18y-11=0

4x(square)-16x + 9x(square)+18y =11
(4x(square)-16x+4-4) + (9x(square)+18y+1-1)=11+16+9
4(x-2)(square)/36 + 9(y+1)(square)/36 = 36/36
(x-2)(square)/9 + (y+1)(square)/4 =1
(x-2)(square)/3 + (y+1)(square)/2 =1

A(square) - B(square)=C(square)
3(square) - 2(square)=C(square)
9-4=C(square)
5=C(square)
(square root)5=C

find the equation of an ellipse with center at (2,-5), major axis at length 6 units parallel to the x-axis and major axis 2 units long

(x-2)(square)/9 - (y+5)(square)/1=1

a tunnel has a semi-elliptical cross-section it is 20m wide and 5m at its highest point, find the equation of the ellipse and use it to find the height of the tunnel 2m away from one end

(x-h)(square)/a(square) + (y-k)(square)/b(square)=1
x(square)/100 + y(square)/25=1

8(square)/100 + y(square)/25=1
y(square)/25=100/100-64/100
y(square)/25=36/100
(25)y(square)/25=36/100(25)
y(square)=9
y(square)-9=0
(y+3)(y-3)=0
y=-3 y=3
y=-3 is rejected cause it will be in the negative side

in the afternoon we talked about of the construction of the hyperbola and its properties...

(x-h)(square)/a(square) - (y-k)(square)/b(square)=1 is the hyperbola
(x-h)(square)/a(square) + (y-k)(square)/b(square)=1 is the ellipse

hyperbola: a locus of points that moves in such a way so that the absolute value of the distance(focal radii) from two fixed points(foci) is constant
|PF1-pf2|= 2a[a constant]

the anatomy of a hyperbola
0 is the centre
A1 + A2 are the verticies
A1A2 is the transverse axis its length is 2a
B1 +B2 are the end point of the conjugate axis its length is 2b
F1 + F2 are the foci they are for the centre
PF1 and PF2 are the focal radii
|PF1 - PF2|= 2a

the picture of the hyperbola is horizontal and centred at the the origin, the equation of the of the asymptotes are: y= b/ax and y=-b/ax
the transvers and conjugate axis are lines of symmetry for the hyperbola

Scribe Post

2006-12-11T17:42:00.000-06:00

The morning started out with the usual problems nothing new about that.

1) Find the center and radius of the circle: 4x²-12x+4y²-30 = 0

4[(x²-3x+(3/2)²-(3/2)²)]+4y² = 30

4[(x-3/2)²-9/4]+4y² = 30

4(x-3/2)²-9+4y² = 30

4(x-3/2)²+4y² = 39

4[(x-3/2)²+y² = 39

(x-3/2)²+y² = 39/4

The center is at (3/2,0) while the radius is the sqr. root of 39 over 2

For what value of k will the circle x²+y²-2x+4y+k = 0 have a radius of 4?

(x²-2x)+(y²+4y) = -k

(x²-2x+1)+(y²+4y+4) = -k+1+4

(x-1)²+(y+2)² = -k+5

Since we want a radius of 4 we add it in a new equation

4² = -k+5

16 = -k+5

k = -11

Find the equations of the ellipse with center (3,-1), major axis of length 10 and parallel to the y-axis, minor axis is 4 units long.

(x-h)² + (y-k)²= 1

b² a²

(x-3)² + (y+1)² = 1

4 25

Foci:

C² = A²-B²

=5²-2²

=25-4

=21

C =sqr(21)

The foci is measured from the center of the ellipse moving up and down hanging in between sqr(16) and sqr(25) since sqr(21) is the next perfect square above sqr(16) and below sqr(25).

The second half of the class was us working on the hyperbolas that we drew last week, the changes I'm afraid you'll have to ask people in class about since the new additions wouldbe a little confusing for me to show on computer.

Next scribe is JessicaJill

Conic Scribe Post

2006-12-10T19:56:00.000-06:00

Hey everyone it's tennyson in the house!!!!

During the morning we started to work out some problems that we had from the last homework that everybody had troubles with.

From the homework Exercise sheet

#2.

Determine the equation of the circle which passes throguht the points
J(-3,2) k(4,1) L(6,5).

Points:
J(-3,2)
k(4,1)
L(6,5)

this is an example of how we know where the centre is
( spelling error know = knowing )

we have to find the midpoint.

So we used Rise/Run

Rise/run
m(KL)= 4/2 = 2
m(AL)= -1/2

midpoint
midKe=(5,3)

y-3=-1/2(x-5)

After that the whole class has to do the paper folding again, this time for the "Hyperbola"

Here was a sample of what it turn out to be:

After that we had to calculate what line measures up to.

Pf1 - Pf2 = 56mm-135mm = 79

Qf1-Qf2 =146mm-69mm = 77

Rf1-Rf2 =143mm - 69mm = 76

A1A2= 81

( note: my calculations are different from other students)

During the afternoon mr.K was away , so we just did a quiz & homework on exercise 38

well by3 everyone , next person would be : JOHN!!!!!

Did you know?

2006-12-10T09:50:00.000-06:00

Did you know I can see your classroom from two windows?!

My first window is your blog. I am excited by what I see and hear! I never cease to be amazed by the quality and sophistication of your scribes; you constantly achieve new heights in illustrating and annotating your scribes. More than that I am so impressed when you celebrate each others’ learning, when you are creative, and when you critically reflect upon your learning in your BOBs.

Did you know Mr. K’s blog is my second window? I admire and respect what I see and hear here too! Did you know that Mr. K celebrates your learning on his blog? that he reflects upon what best helped you to learn and why? that he unselfishly shares all he knows with those who read his blog? that he learns from the conversations on his blog? that he writes with passion and is creative? and that he commits many random acts of kindness by honoring other teachers’ accomplishments in his posts? Did you know all he expects of you, he shares those same expectations for himself?

Did you know that because of all that and more, Mr. K.’s blog has been nominated for “Best Teacher Blog 2006” on the Edublog Awards website?

I just happen to think that no one deserves this honor more than Mr. K.

What about you?

Jefferson's scribe post

2006-12-07T16:37:00.000-06:00

Today Mr.K was away so we had a substitute. All we did today was work on a stencil for the rest of the class and if you didn't finish it in class, it's for homework.

The next scribe will be Tennyson

Conics: The Parabola, Circle and Ellipse (Cheek's 3rd Scribe post)

2006-12-06T16:30:00.000-06:00

Hello its cheeks again working on today's scribe post on Conics: The Parabola, Circle and Ellipse. We started off by talking about our website reaching its max on 300 mb due the cause of all the images we posted up. So for now on we will be using photobucket to upload our images into our blog. We then watched some videos that were quite hilarious based and mathematics in relation with the real world. Our final video was about a tire then went loose as the driver was driving and how it rolled, flyed, bumped into other cars and reaching its final destination back into the drivers car again and the probability of that was 0.0000001%. Think its possible? Comment it. Our teacher then wanted all of us to comment about how mathematics is used in our other subjects. Think it is? Comment that too.

We continued our lesson where we left off from the anatomy of the parabola...

Our lesson began and ended with our dictionaries:

Deriving the standard form for the equation of a parabola

__ __
PF=PD (by def'n)
√[(x-h)²+[y-(k+p)]]² = √[(x-x)²+[y-(k-p)]]² (Distance formula)
(x-h)²+[y-k-p][y-k-p] = [y-k+p][y-k+p] (Square both and simplify)
(x-h)²+y²-ky-py-ky+k²+kp-py+kp+p² = y²-ky+py-ky+k²-kp+py-kp+p² (expan)
if we were to bring the terms that cancel from the left to the right side we simplify to:
(x-h)² = 4py - 4pk (balance)
(x-h)² = 4p(y-k) (factored)
***Standard form for equation of a vertical formula

__ __
PF=PD (by def'n)
√[[x-(h+p)]²+(y-k)²] = √[[x-(h-p)]²+(y-k)²] (Distance formula)
[x-h-p][x-h-p] +(y-k)² = [x-h+p][x-h+p] (Square both and simplify)
(y-k)²+x²-hx-px-hx+h²+hp-px+hp+p² = x²-hx+xp-xh+h²-hp+px-ph+p² (expan)
if we were to bring the terms that cancel from the left to the right side we simplify to:
(y-k)² = 4px - 4ph (balance)
(y-k)² = 4p(x-h) (factored)
***Standard form for equation of a horizontal formula
__________________________________________________________

Circle: A locus of points that move such that they are a fixed distance (radius) from a fixed point (centre).

Deriving the standard form for the equation of a circle

__
OP = r (by def'n)
√[(x-h)²+(y-k)²] = r (Distance formula)
(x-h)²+(y-k)² = r² (standard form for a circle)

__________________________________________

Ellipse: A locus of points that move such that the sum of the distances (focal radii) from fixed points (foci) is constant.

The anatomy of the ellipse

A₁ and A₂ are the vertices of the ellipse
A₁A₂ is the major axis of the ellipse. It is a line of symmetry and its length is 2a
B₁B_{2 are the endpoints of the minor axis.}B₁B_{2 is the minor axis of ellipse. It is a line of symmetry and its length is 2b.}
_{F₁ and F₂ are teh foci of the ellipse each focus is C units from the centre.
PF₁ and PF₂ are ca}_{lled the focus radii. Their sum is always 2a, the length of the major axis.}
The standard form for the equation of an ellipse

Standard form for a horizontal equation

Standard form for a vertical equation

The focal radii property

Suppose P is at A
PF₁ + PF₂ = A₁F₁ + A₁F₂ (by def'n)
PF₁ + PF₂ = A₂F₂ + A₁F₂ (by symmetry
PF₁ + PF₂ = A₁ A₂
PF₁ + PF₂ = 2a

The Pythagorean property

Suppose P is at B
PF₁ + PF₂ = B₁F₁ + B₁F₂
By symmetry we know that
B₁F₂ = B₁F₁
there fore...
PF₁ + PF₂ = B₁F₁ + B₁F₁
PF₁ + PF₂ = 2B₁F₁
By the focal radii property
PF₁ + PF₂ = 2a
therefore..
2a = 2B₁F₁
a = B₁F₁
and by Pythagorean theorm
b² + c² = a² or c² = a² - b²

Thats all for tonight the next scribe is Jefferson

Conics: Parabola Exercises

2006-12-05T15:58:00.000-06:00

Here are the homework exercises and answers for tonight. ;-)

Download the pdf file here.
~~That link is broken ... working on fixing it asap. I might be late getting it up. Sorry about that. :-(~~ fixed!

ScubaSteve's Third Scribe Post

2006-12-04T23:32:00.000-06:00

This just in! Mr.K recently informed that the Flickr assignment has been extended! Memo hurrahs around. So for those who are thinking about flying under the radar, get camera happy! Plus, the favorite number picture will still be allowed for uploading. Aside from the pc40sf06 tag, the favorite number should also be tagged with a 'interesting' and the recent trig assignment be tagged with a 'trigonometry' tag =). Enjoy.

This is ScubaSteve reporting for the third time. I know so far i've topped my post with the previous one, and hopefully you guys are not expecting me to top this one ;_;. The last one was a big one, so many pictures almost jepordized my scribe post (it was spacing out around the end, good thing i finished before it dragged on). Now, today was a double class. We were introduced to Conic Sections, a new unit that i seemed to be already confused about. Its not that i dont understand it, its just tough to re-write in my own words or explain it, so bare with me ^.^; I promise if i scribe again, i'll make up for this post if this is unsatisfactory ;]. Without further adieu, read on my cyber-math classmates.

The first class was an introduction. We learned how to find the eq'n by looking at the graph. Lets look at a basic one first.

Given what is drawn here the formula to find an eq'n of a graph is: (PF=PD) (root->),(x-h)^2+[y-(k-p)]^2 = ,(x-x)^2+[y-(k-p)]^2. First we properly expand the monster that is [y-(k-p)]^2 which ends up being:

(y-k-p)(y-k-p) => y^2-yp-yk+k^2+kp-yp+kp+p2 => y2-2yk-2yp+2kp+k^2+p^2 (its ugly i know but we'll have to bare with it) So putting it back into the original eq'n we'll have...

(x-h)^2+y^2+k^2+p^2-2yk-2yp+2kp=y^2+k^2+p^2-2yk-2yp+2kp => (x-h)^2=4yp-4kp => (x-h)^2=4p(y-k)

This graphs eq'n is (x+2)^2 = 8(y-3)

Could you recognize simularities in the eq'n that is familiarly applied to the graph? Vertex?

The p is the distance between the Focus (F) to the vertex. It is equal to the distance between the vertex and the Directrix (D)

P can be found by 4p=(coefficient), in this case it's 8, therefore 2

Back in grade 11 we learned the eq'n of a parabola was y=a(x-h)^2+b. Although this gets the job done back then, it does not provide us sufficient information when it comes to Conic Sections. y=a(x-h)^2+b can be expressed as the formula we use now, just to prove it is legitament.

y=a(x-h)^2+k

a(x-h)^2+k=y

a(x-h)^2=y-k

[(x-h)^2=(1/a)(y-k)]

That ended the first period. When we came back, we had a couple of questions on the board. These questions were incomplete squares with eq'ns that were un-usuable. We had to retrace our steps back a year to balance out the eq'n before we could find any information on them.

y^2+8x-6y+1=0 find... (work in red text to find the solutions to answers)

(y is the one ^2'd so it opens sideways) (if x was ^2'd it would open up or down)

Vertex;(1,3)

Focus;(-1,3) (vertex's x value -2) <-2=p (4p is negitive, therefore it opens left, parabola is around focus)

Eq'n Of Directrix;x=3, (vertex's x +2) <-2=p (4p is negitive, therefore it opens left, parabola opens away from directrix)

Eq'n of Axis of Sym;y=2 (axis of sym = vertex's y value if open sideways, x value if up or down)

Domain;(-oo, 1] (4p is negitive, therefore it opens left)

Range;(-oo, oo)

y^2+8x-6y+1=0

y^2-6y=-8x-1

y^2-6y+'9'=-8x-1+'9'

(y-3)^2=-8(x-1)

4p=-8, p=-2

The rest of the class consisted of a math dictionary insert, enjoy!

MATH DICTIONARY IS THE FOLLOWING:

Conic Sections

The word 'conic' means 'cone-like'. Four different types of cruves are generated by taking cross sections of a double napped cone. They are: i)Circles, ii)Ellipses, iii)Parabolas, iv)Hyperbolas

Locus: Set of points that follow a certain rules

Focus: A fixed point (or points) that determines a locus

Directrix: A fixed line that determines a Locus

Parabola: A locus of points that are equidistant from a fixed point (focus) and a fixed line (directrix)

The Anatomy of a Parabola

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

vertex(h-k)

focus(h,k+p)

eq'n of Directrix y=k-p

Eq'n of axis of sym: x=h

note: p

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

vertex (h,k)

focus (h,k+p)

eq'n of Directrix: x=k-p

eq'n of axis of sym: y=k

note: p

That ends it for today's scribe post~. It's really late now, i should wrap it up. Homework is Excercise the next =o, i believe it's called 'Parabola'. Serious of not, Mr.k said test was on friday, so learn hard this week! ScubaSteve out.

Sorry For the inconvenience

2006-12-03T19:36:00.000-06:00

Sorry but im having technical difficulties at the moment, but i should be able to post up the scribe post by tomorrow night, and the next scribe is ScubaSteve, have fun!

daphne's BOB number five

2006-12-01T12:14:00.000-06:00

The lesson that we had last unit was really tough. When I first looked at it, it seems like everything is complicated because it twisted my mind by just thinking of the right solutions for it. The good thing about it is that Mr. K never stop helping us to improve our skills in solving problems like this. Unfortunately, in my case, I'm still at the bottom of understanding the proper technique in solving cases like it. In fact, I'm not really confident that I can make it in this unit but I hope that from all of the efforts my classmates, Mr. K and I did, I passed.

SCRIBE POST NOVEMBER 17

2006-11-30T20:37:00.001-06:00

Hey everyone its Gerald. I understand my scribe post is pretty late and i'm very sorry.
To get things started Mr. K put some questions on the board for us to do.

HOW MANY WAYS CAN 5 PEOPLE (A,B,C,D,E) SIT IN A ROW IF A MUST:

A) SIT NEXT TO B
B) NOT SIT NEXT TO B
C) SIT TO THE LEFT OF B BUT NOT NECESSARILY NEXT TO B

SOLUTION:

A) 4!2! = 48 ways to be seated in a row if A must be seated next to B

2! = AB seated next to eachother
4! = C,D,E,AB = since AB must be seated together

B) 5! = 120

120 - 48 = 72 ways to be seated not next to B

5! = all ways they can be seated regardless who they sit with
48 = the ways AB are seated together

C) 5! divided by 2 = 60 ways A sit next to be but not necessarily next to B

5! = all ways they can be seated regardless who they sit with
2 = A sitted to the left of B

IN HOW MANY WAYS CAN 5 PEOPLE SIT AT A ROUND TABLE

the formula for this is (n-1)!

(5-1)! = 4!
4! = 24 ways 5 people can sit around a table

MATH DICTIONARY NOTES:

CIRCULAR PERMUTATIONS

the number of ordered arrangements that can be made of "n" objects in a circle is given by:

(n-1)!

example: how many ways can 6 people be arranged around a circular table?
(6-1)! = 5!
5! = 120 ways 6 people can be arranged around a circular table

SPECIAL CASE: a bracelet is a circle that can be flipped over. The number of arrangements that can be made of object on a bracelet is:

(n-1)! divided by 2

example: how many different bracelets can be made using 5 different coloured beads?
(5-1)! divided by 2 = 4! divded 2
24 divided by 2 = 12 ways a bracelet is made using 5 beads

thanks and sorry once again. I hope everyone did good on the recent test we just had. See everyone next class. Good night and goodbye =)

scribe post for today

2006-11-30T16:01:00.000-06:00

Another winter day had past and we had an uneasy day due to the test we had this afternoon. In our first period pre-calculus, we had a pretest in relation to our actual examination where we did individual worksheets then group work and then a two and a half minute spy time. After that Mr. K revealed the answers for the following questions:

1) 7 teams compete in a men's hockey league. If each team plays each other twice, how many games are necessary to complete the league's schedule?

(a) 42
(b) 21
(c) 84
(d) 52

The answer is (a) 42 because the solution is "2*(7C2)" where 7 is the number of teams and there must be at least first two teams to compete then multiplied by 2 because they must compete with each other twice.

2) The sum of the seventh row of Pascal's triangle is the same solution to:

(a) The number of solutions a student could get if they guessed on six questions of a multiple choice exam, each question having four answers.
(b) The number of solutions a student could get if they guessed on six questions of a true false exam.
(c) The number of solutions a student could get if they guessed on seven questions of a multiple choice exam, each question having four answers.
(d)The number of solutions a student could get if they guessed on seven questions of a true false exam.

The answer is (b) because in six questions there are two choices for each, which is either true or false. Therefore, 2*2*2*2*2*2 or 2^6 equals 264.

Solution:
_ _ _ _ _ _
2 2 2 2 2 2 = 264

3) Suppose the last four digits of a telephone number must include at least one repeated digit. How many such numbers are there?

The answer is 4,960 because there are 10^4 or 10,000 ways of having last four digit numbers if there is no restriction and there are 10*9*8*7 or 5040 ways of having last four digit numbers with restriction. In order to solve for the number of ways to have last four numbers with at least one repeated number, we have to subtract 5,040 from 10,000. As a result, we got 4,960.

Solution: 10,000-5,040= 4,960

4)A multiple choice exam has 20 questions, each with four possible answers, and 10 additional questions, each with five possible answers. How many different answer sheets are possible?

There are (4^20)*(5^10) or 1.073741824 x 10^19 possible answer sheets because in 20 questions there are four possible answers, 4^20, for each and in 10 questions there are five possible answers, 5^10, for each. In order to get the answer we have to multiply both (4^20) and (5^10) with each other because we are looking for the differenr number of answer sheet possible.

Solution:
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4= 4^20

_ _ _ _ _ _ _ _ _ _
5 5 5 5 5 5 5 5 5 5= 5^10

(4^20)*(5^10)= 1.073741824 x 10^19

5) In a 52 card deck, how many 5 card poker hands that have exactly two pair? one pair?

in two pairs: There are 123,552 possible cards with exactly two pairs.

Solution: 13C2* 4C2* 4C2* 11C1* 4C1= 123,552

where:
13 is the number of face values from ace to king
4 is the number of suits per face value (hearts, clubs, diamonds, and spades)

13C2 is the equation for a person to choose 2 face values out of 13. Two 4C2 are given because when a person have already chosen two face values, he will then choose one pair of number from four suits, which are the hearts, diamonds, clubs, and spades and another pair from the other suit. 11C1 is the equation for the remaining face values where the person has to choose only one from it. 4C1 is given because after the selection from the remaining 11 face values, the person has to choose one number from any suits of the same number.

in one pair: There are 1,098,240 possible cards with exactly one pair.

Solution: 13C1* 4C2* 12C3* 4C1* 4C1* 4C1 or 13C1* 4C2* 12C3* (4C1)^3= 1,098,240

13C1 is the equation for the 13 face values where a person has to choose only one from it. 4C2 is given because after he has chosen the face value, he has to choose one pair from four different suits of the same number. 12C3 is the equation for the remaining face values where a person has to choose three from it. Three 4C1 are given because in different face values there are four suits each and he has to choose only one number from the four suits of the same number resulting into having three different numbers.

In our second period pre-calculus, we started answering our Combinatorics test where all of us are hoping for the best test results.

That's all for now guys.
The next scribe is Allen.
Goodluck to all of us.....
God bless you all..........

------(",) daphne.......

Jho's Bob =)

2006-11-30T12:41:00.000-06:00

Well this unit was an okay unit for me, it was kind of a refreshment of what i left of last year. I remember last year this was one of my best units, but for this year, i've learned some new stuff, like the pascals triangle and the bionomial expansion, unless i just forgot. But yeah i think i've learned alot more this year about this unit when i was teached about it last year. I hope i pass the test, i'm not oh to convinced that ill be superior for this quiz, because of my absenses and my failure to keep up with what we learnt. But i will try my best.

Oh yeah sorry about not posting up a scribe post yesterday, i couldn't get my internet going, after work, but i have informed who will be the scribe for today. I'll try to have my scribe up ASAP. GOOD luck everybody

JessicaJill's Blogging on Blogging

2006-11-30T00:23:00.000-06:00

So, the test is tomorrow, and the question is, am I prepared? Well, despite my being absent from school for over a week, I feel comfortable with this unit. It does get a little tricky every now and then, but I think that just a little more effort will help me through. I'm just hoping I do as well tomorrow as I did on my quiz tonight. Last year I think I had some trouble with this unit, probably because of my excessive absenteeism, and I'm not being cocky, but I feel like I'll do exceptionally well with our test.

Good luck everyone.

Ashley's BOB#4

2006-11-29T23:19:00.000-06:00

Okay so this unit is alright. I mean I liked it at first when we started learning it because it seemed simple but now i'm not so sure. after doing the quiz i'm not sure how well i'll do on the test. i guess i still get confused on the pick and choose stuff. like i'm not sure which one to choose even though i know that you choose, choose when order doesn't matter and pick for when order does matter but then i start double guessing myself on deciding whether the question says that order matters or not. i don't know anymore but i still like this unit for some reason even though my brain hurts everytime i'm in class=/ oh well....just means i have to practice more on the questions. Good luck guys!

azn_chilly's blog'n on blog'n

2006-11-29T23:03:00.000-06:00

this unit was somewhat difficult and confusing on some parts, like on choosing on which formulas to use for which questions. but i get most of it but then suddenly i think its a different answer and all of them are tricky at the start but i read more carefully the understanding for it gets to me then i can be able to answer it, but practice and practice make better then say and its working for me for some questions......

thats all i think i have to say at this point so...

goodbye to all & good luck to all on the pre-test and real test hahhah uhhhhh so...
latez to all

Blogging on Bloggnig

2006-11-29T22:56:00.000-06:00

I did about average for this particular work, but then again I've never been very good at making estimates about myself. Homework I went through to try and understand as much as possible and at the least the only thing I think I have to worry about is where to use the pick formula and where to use the choose formula. Truthfully I kinda wish we had some more time before the test but that's how I feel about all my tests.

jessica.'s blogging on blogging.

2006-11-29T22:45:00.000-06:00

perms and combs.. what a great unit. i actually understand this more than last year. this year we went through this unit more throughly. there are some problems that i would get.. like when to use the "pick" or "choose", sometimes you just have to re-read the question a couple of times to get it. i think that's all that's bothering me at the moment =/ goodluck on the test tomorrow (:

gooday&goodnight.

blogging on blogging

2006-11-29T21:48:00.000-06:00

at first, I'm having fun in the lesson. but when the time past, i getting confused in what we are discussing. it really gave me a hard to understand it because i can say im a slow learner. when i i'm doing my homework. its a great help for me. before the unit end. I understand everything then. Its just a matter of practice.
good luck 2 everyone. enjoy your day.
Hope I'll pass this test.

Richard's BOB on combinatorics

2006-11-29T20:42:00.000-06:00

This unit, Combinatorics, is an interesting unit. It doesn't give me stress at all. But be careful...because it is tricky. You don't need to follow the formula. Just carefully analyze the problem and you will surely get the correct answer. I'm having trouble about arranging things with different kind of restrictions. I'm always got stuck on how to deal with those problems. Pick and choose is easy as long as you know which one is to use.
In this unit, I also learned about patterns especially the Pascal's triangle. Binomial Expansion theorem is also a great lesson. In card problems, I'm familiar with the terms that is being used. But it seems like it is still complicated.

Good luck everyone for the test! I hope we all did not just good but the BEST!

BLOGGING ON BLOGGING

2006-11-29T20:02:00.000-06:00

Holy moly ! it's not even funny, this unit went by fast, but I mean, I didn't get much, the beggining made a it of sense, then I lsot myself somewhere, tried to do some homework and I got what was happening, but then once we got in class, I totally dunno what was going on, wowsers, I need to study hardcore tonight! but good luck to everyone else ! =).

bob

2006-11-29T19:48:00.000-06:00

This unit was pretty straight and went by really fast...the only problem I have with this unit is when to use "pick" and "choose" because for the bus problem we used "pick" so does that mean we use "pick" for the desk questions too? because what if the student wanted to sit in a particular desk, how are we to know, and for the bus question how do we know if they want a specific seat like one near the window?...and for the quiz I thought I seriously knew all the answer but it didn't turn out that way, which surprised me. As for when we draw the "slots", how do we know when to do that or factorial...because on the quiz I did a question in factorial and the answer didn't turn out right. Anyway I'm sure it'll be fine because I understand all of the group work and in-class board questions..well good luck to everyone and as Mr.K says it "may the best news be yours" =)

BOB time again!!!

2006-11-29T18:52:00.000-06:00

I would just like to say that this has definitely been my favorite unit!! I have also done permutations and combinations before, but alas my teacher didn't exactly explain it.. at all... which meant it was really hard for me to understand... luckily mr. k is sooo much better at making things easier to understand!!! this unit actually made total sense to me and instead of feeling like i was completely lost all the time, i actually answered some questions!! woo hoo... and i've already done much better on the online quiz then i did during logs so i feel much better... with a little more practice on my part, and some discussion in class first class tomorrow, i think i might actually do well on the unit test... which would be nice for once... i really need to get my mark up, and i'm going to have to work on it, but i think i can do it if i really try!! ok well i'm done, yay for permutations and combinations... i hope i didn't just jinx myself for the test :P ... ah ha.. well good luck everyone and i'll see you in the morning!!!

BoB- counting(combinatorics)

2006-11-29T18:07:00.000-06:00

well I can definitely say that i understand counting a whole lot more than last year. but I'm still having trouble on problems that involving playing cards. like what you need to you need to think about first IE, suits, face values etc. the binomial theorem is easy up until you have to find the term containing something like x^20 given (x-y^4)^6. i know how to find the middle term now which is uber easy. I think that's pretty much it i can't really grasp the concept of how to deal with playing card problems.

Counting 1..2..3!

2006-11-29T15:56:00.000-06:00

At the beginning of today's class, the class was divided into 5 groups of 4. Each group were given the same 5 questions on counting. As seen below. I will now explain briefly how to get these answers and why.Solutions and Explanations

1.a) As seen in question one the answer is 21 terms. Determining how many terms there are is defined by (n +1) , n is the exponent after the binomial expansion. So ( 9+1) =10 and (10 +1)=11. Therefore 10+11 = 21 terms.

1.b) The answer is (x +y) ^10. This is because in order for you to have a middle term, you need to have the same amount of numbers before and after that middle number, and this can only be done with an odd number of terms, but since our formula is (n+1), in finding how many terms, in a binomial expansion it would have to be an even number. (x +y) ^10, has 11 terms therefore having 5 before and 5 after. The middle term would be 6 in this expansion.

c) As said above, If "n" is an even number.

2. The answer is there is no middle term because the "n" was an odd number.

3. a) self explanatory. (18)(10)(20)(10)(20)(10) <given to us,and 10 stands for numbers you can choose between 0-9.

b) KEY WORD IS NEW. So without restrictions it would look like this (26)(10)(26)(10)(26)10) but since how many new ones are formed you would subtract the answer from, part a, from the "no restrictions" to get all the new codes.

4. It is 7P4. It is pick, because order matters, enemies do not want to sit beside enemies, or window sitters. 7 stands for the empty seats the can pick from, and 4 stands for the people who will be sitting in their picked spots.

5. a) 18C6 because order does not matter. 18 stands for number of people. 6 stands for people choosen to be on committee.

b) 8C4(10C2), exactly 4 boys, so since there are 8 boys they have to choose exactly four, and remaining 2 people will have to be chosen from the 10 girls, to make the committee of 6 people.

c) 10C4(8C2)Same explanation as above, but exactly 4 girls have to be chosen from the 10 girls , then the other 2 people will have to be chosen from the 8 boys, to form the committee.

Counting Quiz

2006-11-29T12:42:00.000-06:00

As we discussed in class today you will be writing an online quiz in preparation for the test tomorrow. Here's what you have to do to get started:

Write the quiz online here. It is live now and will no longer be available as of midnight on tonight. When you get to the site follow these instructions:

If you hit any snags email me and we'll sort it out together.

This quiz is timed. You'll only have 45 minutes to complete it once you've begun. It consists of 13 multiple choice questions. DON'T PANIC and I know you'll all do well. We'll talk more about this in class tomorrow. ;-)

Remember ...

There are three kinds of people ... those who can count and those who can't. ;-)

Counting Problems

2006-11-28T16:43:00.000-06:00

Okay, today Mr.K explained more about the online assignment we had to do on the Flickr site. For anyone who wasn't in class, you have to get one picture that represents something of Trigonometry. It has to be in the real world, not something drawn up and then photographed, but you can also set it up how you want it to be. Just remember it has to be a trigonometry picture; it could represent a triangle, a sine graph, angles, area, degrees, etc. When you get your picture try to incorporate as many different concepts of trigonometry as you can. After showing us he went to putting up the questions on the board.

From a 52 card deck, how many 5 card hands have:

a) 3 hearts and 2 clubs?

b) 4 of a kind (4 cards, same face value, and 1 card of a different face value)

c) Royal flush (A, K, Q, J, 10, with the same suit)

d) straight flush (5 cards, in sequence, of the same suit)

2) Find the 6th term of (x - 2y)^9

3) Find the middle term of (3a + 2b^2)^6

4) Which term of (x^2 - 1/x)^6 is constant?

The Answers:

1. a) 13C2 . 13C3 = 22 308

b) First we got the total number of hands dealt
52C5 = 2 598 960
13C1
We get this value because of the 13 different face values
4C4
We get this value because there are four suits in each face value, and since we need four ofa kind it's 4C4.
12C1
Since this choice is supposed to be a different suit then one of the face values will be eliminated
4C1
We only need one of the face values and only one suit.

Afterwards we just multiply it all together:

13C1 . 4C4 . 12C1 . 4C1 = 624

c) The answer would be:
1 . 4c1
There reason we are multiplying 4c1 by 1, is because there is only one way to get the sequence
4C1 just means that it could be any one of the 4 suits.

d) This one is different because you don't count the royal flush.
This problem is also related to a problem we did in previous classes, where there were 4 students and 9 seats to fill.
the answer would be:
9 . 4C1
Because there are nine different ways to get the sequence, and 4C1 means that the whole had has to be one suit.

2) To find the sixth term of (x - 2y)^9, you have to remember that 9C0 also counts as a term.
Solution:
9C5 (x)^4 (-2y)^5 = 126 x^4 (-32y^5)
= -4032 x^4 y^5

3) Let x = 3a y = 2b^2
6C3 x^3 y^3 = 20 (3a)^3 (2b^2)^3
= 20 (27a^3) (8b^6)

4) 1st term:
6C0 (x^2)^6 = x^12
2nd term:
6C1 (x^2)^5 (-x^-1)^1 = x^9
3rd term:
6C2 (x^2)^4 (-x^-1)^2 = x^6
...
The 5th term is the constant term.
6C4 (x^2)^2 (-x^-1)^4
= 15 <--- This is the value for the constant term
(a + b)^p
pCd (a^c) (b^d)
c + d = 6
2c - d = 0

SECOND CLASS...

(2x^3 - 1/x^2)^10 Find the constant term

The 4th term of (x - 1/2)^n is -15x^7, find n.

How many ways can a person get from point A to B?

4. In a 52 card deck, find the number of ways to get a 5 card hand that is:

a) a full house (3 cards of one face value, 2 of another)

b) Flush (5 cards, same suit, not in sequence)

c) Straight (5 cards in sequence, not all the same suit. Ace high or low)

d) 3 of a kind

e) Two pair

f) One pair

g) No pairs

10C6 a^4 b^6

10!/(6!4!) (2x^3)^4 (-x^-2)^6

210 (16x^12) (-x^-12)

3360

An extra question...Find the term that contains x^15

10C7 (2x^3)^7 (-x^-2)^3

-120 (128x^21) (-x^-6)

-15 360 x^15

2. nC3 a^(n-3) b^3 - - 15x^7

n!/((n-3)!3!) (x)^(n-3) (-1/2)^3

x^(n-3) = x^7

n-3 = 7

n= 10

3a) The answer is

7!/(4!3!) = 35

Going right, then down.

We get 4! because you're going RRRR. You're making four rights on the squares.

3! because you're going DDD. You're down 3 squares.

The combination would be RRRRDDD, it's how you're getting from point A to B.

Another way was using Pascal's Triangle:

As you can see, when you look at it at an angle, you can see pascal's triangle.

B) RD = 2

c) RRRRD----5!/4 = 5

d) RRRRDD . RRD . RRDD

6!/(4!2!) . 3!/2! . 4!/(2!2!)

15 . 3 . 6 = 270

4a) 13P2 . 4C3 . 4C2

We used 13P2 because the order does matter.

b) 4C1 . 13C5 - 10.4C2

the 10.4C1 is the total number of flushes in sequence.

Well that's the end of my scribe. I hope it was informative to all of you. The next scribe shall be......Jhoann.

Can you say..... Patterns are IMPORTANT??

2006-11-27T16:42:00.000-06:00

Hi guys its me again!! I hope everyone is enjoying their evening... so
this is what we did today in class.....

We started class off by going over some of the things we did last
Thursday, so Mr. K wrote these on the board.

(a+b)^0 = 1
(a+b)^1 = 1a+1b
(a+b)^2 = 1a^2+ 2ab+1b^2
(a+b)^3 = 1a^3+ 3a^2b+ 3ab^2+ 1b^3
(a+b)^4 = 1a^4+ 4a^3b+6a^2b^2+4ab^3+1b^4

1
11 1
11211
1 13 131 1
11 4 16 14 11
1115110 10 15 11

oCo
Co 1C1
2Co 2C1 2C2
3Co 3C1 3C2 3C3
4Co 4C1 4C2 4C3 4C4
5Co 5C1 5C2 5C3 5C4 5C5

Do you remember those... if you don't you might want to
check Ashley's lovely scribe post from yesterday for reference

After he finished putting all of these back up on the board, we went
over patterns that we fount yesterday like:

Powers of 2

1
1 1
1 2 1

1 3 3 1

1 4 6 4 1
1 5 10 10 5 1
2^3 = 8 111111111111+3+3+1= 8
2^4 = 16111111111+4+6+4+1= 16

And powers of 11

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

11^2 = 121
11^3 = 1331
11^5 = 161051

Whoa that doesn't look
the same... well it still works, you just
have to make sure to do the carrying.
The 5 turns into 6 and the tens turn
into 0 and 1 because 10 carry the 1
leave the 0, and 11 carry the 1 and
leave 1 behind and that 5+ the carried
1 = 6

Ok now here's the new pattern that Mr. K showed us... it shows the
Fibonicci numbers in Pascal's triangle (even though it wasn't pascal
who made it :P )

Pay attention cause its hard to see at first!!

The fibonicci numbers can be found on the diagonal edges... the numbers that are cut by this line are added together to show these fibonicci numbers.
Can you see the diagonal edges... they are the pink lines.... the first line crosses through only 1... the next crosses only 1... the next passes through 1 and 1... add these together to get 2... the next row passes through 1 and 2.. add these together to get the 4th fibonicci number 3... and on and on

Can you see it?? if not feel free to come and ask me in class, i'll do
my best to show you and make sure you understand.

After everyone found this pattern Mr. K wanted to point out other interesting
facts... so he told us to take out our calculators...

if you take one fibonicci number, and divide it by the fibonicci number before
it, you will find that as you travel down the triangle, the values get closer and
closer to the "GOLDEN RATIO" or PHI !!! ∅(phi) = 1.61803399

1/1 = 1
2/1= 2 Can you see how the numbers seem to be
3/2= 1.5 converging (narrowing down) to a single value?
5/3= 1.6667 All getting closer and closer to ∅!!
8/5= 1.6
13/8=1.625

As I've already mentioned, ∅ is the GOLDEN RATIO... but why is
this so you ask... well because if an item has a ratio that is equal to, or
close to ∅ than it just looks better... unexplainably but it really does..

For example have you ever wondered why 4 by 6 pictures look better
than 5 by 7s... they're smaller, yet they look better... this is because

6/4 = 1.5 * because this has a ratio closer to ∅ it is more visually appealing*
7/5 = 1.4

Another example of this is the Parthenon. This is a temple in Greece dedicated
to the goddess Athena (who is the goddess of wisdom). This temple is also
known as 1 of the 7 wonders of the world as it is one of the most visually
appealing structures in the world. This is because if you look at the Parthenon
it is constructed of rectangles. If you take the larger length of each rectangle and
divide it by the shorter length you will find that the ratio is 1.618... or really
close to the GOLDEN RATIO !

Everything is connected!!! The Fibonicci ratio is not only encoded Pascal's
Triangle, but it is also in various aspects of life, including the human body.
This is the Vitruvian Man done by Leonardo Davinci. (Found on Flickr,
posted by absolutesteven) It is known as the most accurate drawing of
a human being ever done. Why is this... i bet you can guess... that's right
Davinci made sure to measure every and make sure that every
part came out close or equal to the GOLDEN RATIO ....

ex.
Man's height / distance from ground to navel
Man's arm / distance from elbow to tip of middle finger

Don't you just love it when everything fits together like that?? :D

After his speech on the GOLDEN RATIO, Mr. K brought us back to
the (a+b)^x expansions....

What are the patterns???

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

oCo
1Co 1C1
2Co 2C1 2C2
3Co 3C1 3C2 3C3
4Co 4C1 4C2 4C3 4C4
5Co 5C1 5C2 5C3 5C4 5C5

Firstly the co-efficents for the expansion can be found within
the triangle (look at above triangle with matching colour)

(a+b)^4 = 1a^4+ 4a^3b+6a^2b^2+4ab^3+1b^4

Can you see the co-efficents??
1, 4, 6, 4, and 1
according to the 5th row

and 4Co row...
tells us the exponents for the expansion

a^(e-x) b^x
e = what it is being chosen out of.. in this case 4

4Co 4C1 4C2 4C3 4C4

so this expansion is

a^(4-0) + a^(4-1) b^1 + a^(4-2) b^ 2 + a^(4-3) b^3 + a^(4-4) b^4

a^4 + a^3b + a^2 b^2 + a b^3 + b^4

If you add these two patterns together you will get...

1a^4 + 4 a^3b + 6a^2 b^2 +4 a b^3 + 1b^4

the expansion of (a+b)^4

Now, if asked to find the 3rd term of (a+b)^4 do you have to
figure, and write out the whole expansion. Sure why not,
(a+b)^4 is easy... but what if i asked for the 7th term of
(a+b)^11?? do you really want to write that out??

I can guess that your answer, like mine, is NO WAY!!!

Ok now its time to show you how to do this much QUICKER
and much EASIER....

First of all i must make sure that it is clear to you that as we go along
the triangle from the left to right the terms are numbered from 1
increasing... 1, 2, 3, 4, 5 etc.

oCo
1Co 1C1
2Co 2C1 2C2
3Co 3C1 3C2 3C3
4Co 4C1 4C2 4C3 4C4
5Co 5C1 5C2 5C3 5C4 5C5
11112222333444455556666

So,
5Co is term 1
5C1 is term 2
5C2 is term 3
5C5 is term 6

Do you see the pattern, the term number is always 1 larger
than the amount being chosen!!

Therefore, if the number being chosen is always 1 larger,
when we ask for the Xth term, we must subtract 1 so it will
look like this nC(x-1) where n equals the amount being
chosen from.

Ok given this question, 7th term of (a+b)^11
lets use what we know, and the new equation

11 is the number being chosen out of and takes the place of
the n. looking like this 11C(x-1)

7 is the term we want, and takes the place of x.
looking like this 11C (7-1)

Do the math... 11C6 is what you are left with. Now use
rules from before that state that give exponents to a and b

a^(e-x) b^x
e = what it is being chosen out of.. in this case 11
b = the number chosen, in this case 6
a^(11-6) b^6

a^5 b^6

Now put both portions together!!

11C6 a^5 b^6
This will be the 7th term in the expansion of (a+b)^11

Lets try this again to make sure you've got
it....

find the 5th term of (a+b)^13

13 is amount being chosen from
5-1 = 4 this becomes the number being chosen!!
Therefore this becomes 13C4

next comes the exponents
13-4 = 9 which becomes the exponent on a
4 is the exponent on b

put them together

13C4 a^9 b^4

is the 5th term of (a+b)^13

Mr. K tested the class a few more times, and soon a couple
people could find the Xth term of an expansion in less than
10 seconds, and the rest of the class could do it in under
30 seconds.... can you?? Time yourself... If you work at
it you'll get there... if you have any questions just ask me
i'm always willing to help if i can!!

Ok guys time to turn up the heat!!

How many terms are in each expansion?? How do we know this??

Well the number of terms in an expansion can be found by taking the
power of the expansion...

ex. (a+b) ^11
the power is 11

take this power and add 1
11+1 = 12

12 is the number of terms in the expansion!!

What would happen if we were asked to find the middle term of the
expansion??

for odd numbers this is easy, all you need to do is add 1 to the power,
then divide by 2

13 +1 = 14 14/2 = 7

Therefore the middle term in this expansion is the 7th term!!

* this formula will only work if the power you start off with is odd*

In order to find the middle term for an even power, we take the
power, divide by 2 and add 1

12 / 2 = 6 6+ 1 = 7

Therefore the middle term in this expansion is the 7th term!!

* this formula will only work if the power you start off with is even*

Now that we can find the middle term of the expansions we've practiced,
can we find the middle terms of other binomial expansions??? YES!!!

how about [x-(2/x)]^7 ?? what is the middle term of this??

First of all lets make this into something we can understand shall we??

Let a = x
Let b = - 2/x to make this even simpler we're going to change
this into - 2x^(-1)

so now the binomial looks like this

(a+b)^7 and that is much better....

now find the middle term... 7 is odd therefore we must add 1 to make 8 and divide by 2.. the middle term is 4!!

now show the term 4 of this expansion.

7C3 a^4 b^3 *reference above if you don't remember how
1111111111111to do this step*

now rewrite 7C3 in terms of factorials....

7! / (4!3!) a^4 b^3
4! reduces leaving--> 7x6x5 / 3! a^4 b^3
6 and 3! reduce as 6 = 3! leaving--> 7x5 a^4 b^3
35 a^4 b^3

Now substitute the real values of a and b back in.

35(x)^4 (-2x^-1)^3
35 x^4 (-8x ^-3)
-280x^1

therefore -280x^1 is the middle term in the binomial expansion
of [x-(2/x)]^7

Lets try this again to make sure you've got it!

Find the middle term of (x^2 - 2/x)^8

now make this simpler for yourself
Let a = x^2
Let b = -2/x even simpler -2x^-1

now the binomial looks like

(a+b)^8

middle term for the power of 8 is the 5th term
therefore 8C4 a^4 b^4

write in terms of factorials

8! / (4!4!) a^4 b^4
4! reduces leaving--> 8x7x6x5 / 4! a^4 b^4

Now substitute the real values back in

70 (x^2)^4 (-2x^-1)^4

70 x^8 (16x ^-4) 111

1120x^4 1111111111
1111111111111111
A common mistake made is when the exponent
is an even number, the student forgets to make
the coefficient positive. Watch for this!!

Therefore 1120x^4 is the middle term in the binomial
expansion of (x^2 - 2/x)^8

CONNECTIONS ARE EVERYWHERE!!!!!!

As you have seen through class and scribe posts for this
unit, everything is connected and therefore

Combinations and binomials go hand in hand!!

The last of the class was spent reviewing most of the above one more time, in order to really make sure that we understood!!

I hope everything in this scribe post is useful to you. If you have any questions or concerns don't hesitate to ask me, I'm here to make this scribe post make perfect sense to you!! Enjoy the rest of your evening, and i'll see you all tomorrow!! Good Night!!

p.s please make sure to have at least 5 pictures on your flickr account if you are a new user. If you do not, your account will not be reviewed, and therefore not deemed safe to make your pictures public for the assignment!!

Oh ya... and just to finish this off, jessiccal_198 your scribe tomorrow :)

Ok i believe that the spacing is fixed now... see you later!!

All about Patterns

2006-11-26T15:48:00.000-06:00

Hey it's Ashley and I'm the scribe for today and yes I know i didn't do it right away but for those who missed you should have really been there. It was a pretty interesting class because the class was all about PATTERNS and mr.k telling us about how God built the universe and blowing our minds (not literally of course...)

To start off Mr.k posted up the questions that he previously posted the day before on the board.

Expand and Simplify

(a+b)^0
(a+b)^1
(a+b)^2
(a+b)^3
(a+b)^4

To answer this question it's simple just expand and simplify

(a+b)^0 = 1 (because its exponent is 0)
(a+b)^1 = a+b (because its exponent is 1)
(a+b)^2 = (a+b)(a+b) => a^2 + 2ab + b^2

now be smart about the next question
instead of doing this (a+b)(a+b)(a+b) just multiply a^2 + 2ab + b^2 by (a+b) since it will be easier since a^2 + 2ab + b^2 is the same as (a+b)(a+b) and you are just multiplying one more (a+b) since its to the power of 3.
(a+b)^3 = a^2 + 2ab + b^2 * (a+b)
now distribute the a

= a^3 + 2a^2b + ab^2

then distribute the b

= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3

then gather the like terms
= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3
= a^3 + 3a^2b + 3ab^2 + b^3

Now do the same for (a+b)^4. Distribute and gather the like terms.
*Do not foil because foiling only works with binomials*

(a+b)^4 = (a^3 + 3a^2b + 3ab^2 + b^3) * (a+b)

Distribute a
= a^4 + 3a^3b + 3a^2b^2 + ab^3

Then Distribute b
= a^4 + 3a^3b + 3a^2b^2 + ab^3 +a^3b + 3a^2b^2 + 3ab^3 + b^4

Gather the like terms
= a^4 + 3a^3b + 3a^2b^2 + ab^3 +a^3b + 3a^2b^2 + 3ab^3 + b^4
= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

The other question posted on the board
Look for a pattern

Write the next two rows

1
1 1
1 2 1
1 3 3 1

1 4 6 4 1

Now looking at this question having down the previous question a few of us realized the pattern

The first pattern:

Adding the numbers like this (sorry it's hard for me to explain so a diagram will hopefully help):

The Second Pattern:

The numbers in the triangle are the coefficients to the expand and simplify question.

The Third Pattern:

The exponents on the variable are: a descending and b ascending.
example:

The last question posted on the board:

Evaluate each term (you can use your calculator):

OCO

1CO 1C1

2CO 2C1 2C2

3CO 3C1 3C2 3C3

4CO 4C1 4C2 4C3 4C4

Now if you did this on your calculator you would get

OCO = 1
1 = 1CO 1C1 = 1
1 = 2CO 2C1 = 2 2C2 = 1
1 = 3CO 3C1 = 3 3C2 = 3 3C3 = 1
1 = 4CO 4C1 = 4 4C2 = 6 4C3 = 4 4C4 = 4

In other words.....

1 1

1 2 1

1 3 3 1

1 4 6 4 1

It's the same as the second question. Neat huh?!

It's interesting as to how all of the three questions written on the board are all related to each other somehow. So how is the last question connected to the first question? Well.... it's better if i show an example:

Then just when I thought there couldn't be more to talk about Mr. K handed us a worksheet which I'll post tomorrow. But it basically showed Pascal's Triangle which is this and I forgot what Mr.k said about pascal so ask him if you want to know who he is.

Now at the bottom of the worksheet it asked if you can find powers of 2, powers of 11 etc....

To find the powers of 2 in the triangle you just add the numbers in each row.

To find the powers of 11 in the triangle just look at it.

Sierpinski triangle is a fractal and well...... a guy named Sierpinski decided to punch out triangles in an equilateral triangle where the area is fixed but the perimeter is infinite. I think that's what Mr.k said but here's what Sierpinski did. He used an equilateral triangle and punched out other equilateral triangles inside it like this: (sorry bad explanation)

Then Mr.K told us to shade in the odd numbers on the triangle which felt like I was back in elementary. We should color more often!

Looks like the Sierpinski triangle doesn't it? If you continued on shading the odd numbers it would be even more clearer. Cool right?!

Next was on the talk about the Fibonacci numbers 1, 1, 2, 3, 5, 8,....

Fibonacci numbers work by adding the last two previous numbers.

1+ 0 = 1, 1+1 = 2, 1+2 = 3, 2+3 = 5, 3+5 = 8

Apparently bees, yes bees follow the fibinacci sequence because a male has one parent which is it's mother and a female has two parents, both a mother and a father.

What else was mentioned in class? Ah yes flowers also respect the fibonacci sequence. You won't find a flower with 6 petals but you will find a flower petal with 5 petals but you won't find a flower petal with 7 petals but you will find a flower with 8 petals, so that's why it's rare to find a four leaf clover.

And Lastly Mr.K mentioned a number which i forgot what the number is called but it has to do with something about dividing your body into parts and it always equals to that number. Sorry but i forgot what he said but it sounded cool. However I'm still not sure as to what part of the lesson was on God building the universe? Oh well....

The next scribe will be.........LISA!!!

and sorry about the picture not showing. i wonder why that is?

Countings and Combinations

2006-11-22T17:26:00.000-06:00

Hi im John,the scribe for today.
In today's class, we started our lesson about a chance of winning 6/49 lotteries.
Mr. K explain to us our chance to win in the lottery. i'll give you some of his explanation.
In 6/49, we have to choose 6 numbers in order for us to get the ticket.
Here is the explanation in the problem.
Assuming there's 49 letters word, 6 of that is the one we like and the rest are those we doesn't chose..

49 C 6 = 49! / (49-6)! 6!
= 49! / 43!*6!
= 13,983,816 no. of ways tickets can be sold in lotto 6/49.
So, if you buy 1 ticket: 1 / 13,983,816 is your chance.
: 7.15 x 10^-8
: 0.0000000715 % a chance in choosing a number in lottery.

He ask the class, do you still buy a lottery? For me, Yes. I will. because still i have a single chance to win in the lottery and maybe that's the time what I'm waiting for.
Anyway, back into class..
If we have 20 people, can we still do to become a lottery winner assuming the pot money is 14 million or above.?
let say, every 15 sec. is the time of processing the ticket.
so 15*4 is 60.this is minutes.
60mins = 1 hour therefore 24x60 = 1440 tickets can be print in an hour.
1440 x 24 = 34560 because we're looking for day.
the answer multiply by 20 people equals 691200.
therefore:
13983816 / 691200 = 20.23121528
We say that there are 20 days will take to get the lottery card to finished.

Again, the question is, are u still going to buy a ticket because there's just 4 days to wait before the 6/49 drawn..

EXPECTED VALUE:
P(WIN)*GAIN + P(LOSE) *LOSS
G + L = 0.9284886
Everytime we play lotto we loss 0.9284886 cents.

JOKE TIME:
There is a mathematician guy, i forgot his name, always thinking "there is a bomb in a plane." but when he go to plane, he says "there is 2 bomb in a plane." so Every time he flies he brings a bomb.

MATH DICTIONARY

COMBINATIONS : An arrangement of objects where order does not matter.

NOTATIONS: n C r or (r/n)
> means from a set of n objects, how many different groups of r objects can be chosen where order doesn't matter.

FORMULA: n C r = n! / (n-r)! r!

EXAMPLE:
How many tickets can be sold in the 6/49 lottery?

49 C 6 = 49! / (49-6)! 6! OR
= 49 / 43!6!
= 13,983,816 ways ticket can be sold.

we ask to solve the problem:

How many different ways can a group of 3 students be selected from 7 students?
Solutions:
7 C 3 = 7! / 4!3!
= 7*6*5 / 3*2*1
= 35 ways

A sample of 5 people are selected from 3 smokers and 12 non-smokers. In how many ways can the 5 people be selected?
Solutions:
15 C 5 = 15! / 10!5!
= 15*14*13*12*11 / 5*4*3*2*1
= 3003

You have a penny, nickel, dime, quarter, loonie and toonie in your pocket. If you pull out a 3 coins, how many different sums are possible?
Solutions:
6 C 3 = 6! / 3!3!
= 6*5*4 / 3*2*1
= 20

Ten people, including one married couple, are eligible to attend a conference, four people can go and the married couple only go as a pair. How many different groups of 4 can go?
Solutions:
> 10 C 4 if the couple split them up
>8 C 4 if the couple is the problem
there is only 1 way to send the couple
8 C 2 is the other people
therefore:
8 C 4 + 1 * 8 C 2 = 98 ways

GROUP PROBLEM

FOOTBALL
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 one?
Solution:
9+8+7+6+5+4+3+2+1 = 45 or

10 C 2 = 45

PRESENTS
a.) In how many ways can 9 presents be given to two children?
9 C 2 = 9! / (9-2)! 2!
= 36

b.) In how many of these ways will younger child recieve just 5 presents?
9 C 5 * 4 C 4 = 126*1
=126 ways

THE ROAD LESS TRAVEL
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?
solution:
7 C 4 * 3 C 3 = 35
7! / 4!3! 3! / (3-3)! 3!
= 35 ways

POINTS
there are 9 points marked in a plane.No three of which lie in a straight line.
a.) How many straight lines can be drawn,each containing 2 of the points?
solution:
9 C 2 = 36

8+7+6+5+4+3+2+1 = 36 straight lines

b.) How many of these pass through one or more of 3 specified points in the set?
solution:
8+7+6 = 21 ways

classmate, i forgot to ask the 2nd group problem to my groupmate this afternoon because we're running out of time. Because you know i'm having a hard time to understand the lesson thats why i have to listen to the explanation of Mr. K. Im going to post it tomorrow.

If you dont understand my post, you can talk to me anytime.
THANKS.. I'll choose ASHLEY to be the next scribe.

Counting Review

2006-11-22T15:51:00.000-06:00

Mrs. Ingram pointed me to a great site for our course. You can review and get practice tests and old exams ... all free, all online, 24/7/365. The site is called Math40s.com.

Spend some time in the Perms & Combs section tonight. ;-)

Don't forget about your flickr assignment.

Cheers!
Mr. K.

BTW, the picture shows that 7 x 7 = 49 ... click on it to find out how. ;-)

Classification	Mathematical Content (50%)	Hot Spots (35%)	Photograph (15%)
Level 4	Packed with mathematical concepts/facts. (Minimum 7 concepts/facts.)	All hot spots accessible; i.e. "smaller" hot spots are "on top" of larger ones, they do not obscure each other. All hot spots are actually labels and relate to parts of the photo (not on blank space with filled in notes). One or more hot spots include a link to a relevant supporting resource on the internet. Minimum 7 hot spots.	In focus or appropriately focused for effect. The subject of the picture occurs "naturally," it is not a contrived shot. Really makes the viewer "see" math in a place they hadn't realized it existed. (Example: trigonometry)
Level 3	Significant number of concepts/facts included. (Minimum 5 concepts/facts.)	All hot spots accessible. Most hot spots are actually labels and relate to parts of the photo. Not more than one hot spot on blank space. One or more hot spots may include a link to a relevant supporting resource on the internet. Minimum 5 hot spots.	In focus or appropriately focused for effect. The subject of the photo has been "set up" or contrived yet still illustrates math found in "the real world." (Example: derivative)
Level 2	Some effort to include content evident. (Minimum 3 concepts/facts.)	Most hot spots accessible. Most hot spots are actually labels and relate to parts of the photo. More than one hot spot is on "blank" space. May or may not include links to relevant supporting resource on the internet. Minimum 3 hot spots.	In focus or appropriately focused for effect. Although it is a "real world" picture, objects have been used to "draw" the math. An obviously contrived shot. (Example: trigonometry)
Level 1	Very scarce content related to assignment.	Less than three hot spots are visible or have information related to the theme of the assignment.	It is evident that little effort went into finding and shooting a picture that reflects the theme of the assignment.
Level 0	Content unrelated to theme of assignment.	No hot spots or mostly unrelated to the theme of the assignment.	Out of focus and/or otherwise difficult to look at.