## Friday, November 30, 2007

## Thursday, November 29, 2007

## Wednesday, November 28, 2007

## Tuesday, November 27, 2007

### Permutations

Next scribe is... Precious.

## Monday, November 26, 2007

### Combinatorics

Okay, so that was my looong post on combinatories. Sorry for the late post, I was having trouble in how I would publish it. Anyways, I thought I would be able to get away with that really short scribe, but I guess not .. ooh well, it's all good. Oh yeah, before we started class Mr. K showed us this really cool video of some guy figuring out the a rubiks cube; it was craaazy.

and the next scribe is still .. shane : )

### Today's Slides and Homework: November 26

Answers are at the bottom of this post.

Here are today's slides ...

Answers to homework questions:

1. 120

2. 20 160

3. (a) 840 (b) 240 (c) 144

4. (a) 2160 (b) 1260 (c) 660

5. 144

6. 4320

7. (a) 4536 (b) 2240 (c) 952

8. (a) 5040 (b) 720

9. (a) 576 (b) 72

10. (a) 891 (b) 445 (c) 179

11. 240

12. (a) 10 416 (b) 5082

## Sunday, November 25, 2007

### TEST DAY

the next scribe is .. Sharmaine

## Thursday, November 22, 2007

### BOB

### Blogging on Blogging

Well then, I'll see you next semester Meesah K. If you feel what I'm saying.

### combinatories and functional notation

Domain:(2,infinity)

Range:(-infinity, infinity)

y-int: none

x-int: 7/3

aymp:2

domain:(-infinity,infinity)

range:(-infinity,1)

aymp:1

y-int:3/4

x-int:2

*ln is like a log. This is change of base law.

* We can't expand ln(x-1) and ln(x+3)any further since ln are not distributive!

*First I expand the lnx and lny using the product law and by quotient law, I divide lnz.

The class start with Mr. k's social commentaries. He talks about smoking and numbers from the lottery, I don't really get it cause I'm not listening... well anyways we are face with a new lesson and this is what we talked about for the mornin' class...

***There are 2 ways for n to be tossed. For every n1 and n2, there are 2 ways that q1 and q2 to be tossed and so that would be 4.**

And for every q, there are 2 ways that d1 and d2 to be tossed and so that would be 8.

Now multiply all the numbers :2x2x2=8

And for every q, there are 2 ways that d1 and d2 to be tossed and so that would be 8.

Now multiply all the numbers :2x2x2=8

***The difference between the Combinatories and Modeling is that combinatories shows how many ways that can be done and arrange items, the following examples show how to use it.**

*There are 3 ways to wear the P1, P2, P3 then there are P1 x P2 x P3 ways to do three things.

*There are 5 people that can be seated in this straight line.

*we used up one number, so we have 4 people to be seated.

*3 people are remaining

*2people remaining

*1person remaining

~if ever you want to see the original and clear images just go to this site: www.http://slideshare.net/nocture/inverse-3/

Sorry for the inconvenience in reading my scribe...i know it's blurry....now i can sleep...next scribe is michelle.....thank u thank u...ajah ajah...

### "can't stop B0BB'n ittt" cont`d

## Wednesday, November 21, 2007

### BOB

as for the formulas

i find them to be very easy to use all i need to improve on though is when given little information such as the original amount, half life and the period instead of wasting time on a test trying to solve for rate then using the Pe^rt formula i can just use the model(A2=A1(m)^(t/p)) formula and save some time.

i really think i should start doing far more amount of homework and study for test because i may know this stuff now but say when the exam comes along.. well ill try not to let that happen

well i guess thats it good luck everyone on the test.

### BOB

### BOB

### BOB

Good luck to everyone on the test tomorrow! :)

### B.O.B.

Hmm, at the beginning of this unit I thought that it was easy. Especially when Mr. K said “A Logarithm is an exponent”. First few questions in this unit was easy but when Mr. K introduce to us the natural logarithm, oh god, it start me to confused but when I remember say the word “A Logarithm is an exponent” over and over again I understand how to solved those problems. But when Mr. K discuss on how to solve the word problems with the three formulas that Mr. K gave to us and how to graph the exponential functions it starts me to confused more. But with the help of my seatmates and the help of the pre-test and the review I barely understand now this unit....

Well, good luck on our exam tomorrow!

### Today's Slides: November 21

and from this afternoon ...

### World Population Clock

### can't stop BOB'n itt

## Tuesday, November 20, 2007

### BOB

### BOB

### BOB

### BOB

### BOB

### Scribe Post Logs and Exponents

Tomorrow's class we will begin our new unit. Also a reminder, Thursday is our Logs and Exponent test.

Next scribe for tomorrow is .... Luis.

## Monday, November 19, 2007

### BOB Log and Exponents

### B-o-B

Bobbin' again... In this unit, Logarithms and exponents, I find it confusing, especially the world problems and graphing. I believe that is my muddiest point. The laws are pretty straight forward. This unit is like trig identities because there's so many ways you can solve a problem. Whatever base you choose. Oh well, that's it for this unit. Good Luck to everyone! Test is on THURSDAY!!!

### Scribe Post: Logs and Exponents

Today in our class we did a somewhat review the whole morning. Actually first Mr. K. hand out a new exponetial modeling sheet and most of us, Im just guessing, did not get a good chance to finish the whole sheet that period because majority of the class want to discuss the sheet he hand out last week. Well most of the sheet were explained and I guess everybody kind of understands it. Im not really sure how to explain everything because its mostly straight to the point. Just remember which one is which in using the formula.

The formula we need to memorize are:

P= A (1+ (r/n) ) ^t(n)

A = Pe^rt

A = Ao (model) ^t

Well that's basically what we did in the mornign class. Ooh and we also solve more logarithmic equations.

e^(2t-1) =5

lne^(2t-1) = ln 5

2t-1 = 5

t = (ln 5 + 1) / 2

The other one is this one:

ln (x+1) = 1 + lnx

ln(x+1) - lnx = 1

ln ( (x+1) / x) = 1

e = (x+1) / x

ex = x+1

ex - x = 1

x (e - 1) =1

x = 1 / (e-1)

That's everything we did this morning.

PM class

All we did that afternoon was more review but its all about the things most of us had troubles with.

The next were graphing, I cant really explain it because its hard to explain. So far I guess I explained everything. Since we did not really learn something new, we mostly did review today and I hope it somewhat helps to the upsocming test about this unit. Well I guess I covered everything. I have to do other homeworks now.

### B.O.B.

### BOBBING FOR MATH

That's about all the problems I have for the unit Logarithms. Everything else I understood and hope that I remember everything for the test. Wish everyone luck and practice hard.

-SAMUS

### BOB

Ant.. I just had to know that your gonna work out your shoulders tomorrow.

### BOB'in once again

No double tommorow!

So Im working Shoulders 4th and Back 5th.

## Sunday, November 18, 2007

### Bobbing on Logarithms

I was away for one class where Mr. K taught the change of base law. A fellow classmate explained it to me, but I still didn't get a good idea of it.

What I found extremely useful was when we got into groups again and answered questions on the review sheet. I guess being with the people I was with was a benefit for me since I understood things more clearly and I was actually able to do some of the questions on my own. I was able to do the word problems which I usually have a lot of hard time doing.

A logarithm is an exponent! It's funny how a lot of people, including myself, are still forgetting this very important rule. It's so easily said, but really difficult to remember.

### Bob ing

Well I hope that my Bob on this unit. If i guesses are correct on how much effects each part is worth than I am in some hot water...

### Log-a-rithms BoB

### Logarithms BOB

### Log and Exponents BoB

Well this unit wasn't that difficult in my opinion, but maybe that's just because I'm trying to concentrate more. I think the most difficult concept to grasp at first was that a logarithm is an exponent. Just now I was doing a problem: 27^log[base3]9 (How am I suppose to type that?). I remember we had a similar question on a quiz, which I got wrong. I get it now though because a logarithm is an exponent therefore log[base3]9 is equal to 2. So solving the question is easy because you just take 27^2. It's such a simple concept, but so difficult to grasp! I suppose that's my "breakthrough" moment. Hopefully I can always keep this is mind. "A logarithm is an exponent."

I've been studying diligently this weekend (for the most part) and doing the problems from the textbook. It isn't that difficult especially when they show you how to do it at the back when you get stumped. I've attempted these exercises before, but now I'm trying them all over again so hopefully this proves to be beneficial.

I understand how to do the questions when I have my notes in front of me for quick reference, but I am ultimately worried that I will forget when and where to apply the laws and rules. There's that math anxiety.

Well, the one thing I am confused about at the moment is when to use the "A=Pe^rt" method or the "A=P(1+r/n)^tn" method when problem solving. I think I understand the model stuff though. I'm going to have to work on those problem solving questions.

I've noticed from that my percentage mark on tests is slightly higher on each passing test than the one previous to it. So hopefully by studying diligently, I might pass this test. A little sad, but I'm hoping I do well.

Well, this has been my, b-b-bob. Good luck on the test everyone.

Good day.

### Logs and Exponents BOB

### November 16, 2007 : Scribe

## Saturday, November 17, 2007

### Logs and Exponents BOB

## Thursday, November 15, 2007

### Exponential Modeling

f(x)= ab^x is the basic function

How we model real life situations depends on what kind, or how much information we are given.

When you have:

Limited Information

A= A_0 (model)^t

Lots of Information

A= A_0 〖(m)〗^(t/p)

A = amount of the “substance” at the end of the time period.

A_0 = original amount of the “substance” at the beginning of the time period

Model = is our model for growth or decay of the “substance”, it is usually an exponential expression in base 10 or e although any base can be use.

t = is the amount of time that has passed.

m = is the multiplication factor

p = is the period; the amount of time required to multiply by “m” once

Okay, so having that information we can now solve the following problems.

The first problem has Limited Information.

The population of the earth was 5.3 billion in 1990. In 2000 it was 6.1 billion.

Model the population growth using an exponential function.

What was the population in 2008?

Before solving this problem, Mr. K showed us a really cool stuff in the internet. It is a world population clock. Here’s the link to the site World Population Growth .

Anyways here’s how you solve it:

There are two ways we solved this one with base 10 and the other with base e although any base can be use.

A= A_0 (model)^t

6.1=5.3(model)^10

6.1/5.3= 〖model〗^10

Base 10

log(6.1/5.3)〖= log〖(〖model)〗^10 〗 〗

(1/10) log(6.1/5.3)=log〖(model)〗

0.0061= log(model)

〖10〗^0.0061=model

P=5.3(〖10〗^(0.0061t) )

Base e

ln(6.1/5.3)=ln〖(model)^10 〗

(1/10) ln〖(6.1/5.3)=ln(model) 〗

0.0141= ln(model)

e^0.00141=model

P=5.3(e^(0.0141t) )

∴ 〖10〗^(0.0061)= e^(0.00141)=1.0142

e^x

*Solving this kind of problem with base e is preferable because the exponent in base e, when you rewrite it in percent, tells you the percent rate of growth or decay per time.

t=0 in 1990

t=18 in 2008

P=5.3(〖10〗^(0.0061t) )

P=5.3(〖10〗^((0.0061)(18) )

P=6.8261

P=5.3(e^(0.0141t) )

P=5.3(e^((0.0141)(18)) )

P=6.8261

Okay, so moving on to the next problem which we are given Lots of Information.

A colony of bacteria doubles every 6 days. If there were 3000 bacteria to begin with how many bacteria will there be in 15 days?

A= A_0 〖(m)〗^(t/p)

A=〖3000(2)〗^((15/6) )

A=16970.5628

A ≈16971

Okay so that it is for our morning class.

Now for our afternoon class, we watched a short movie clip about star trek and tribbles. Here’s the link for the clip mailto:http://youtube.com/watch?v=lZvmxmVVdk8.

Here is another example of exponential modeling given with Lots of Information.

The mass in (grams) of radioactive material in a sample is given by:

M(t)= 100e^(-0.0017t)

where t is measured in years.

Find the half-life of this radioactive substance.

Create a model using the half-life found in (a). How much of a 10 gram sample of the material will remain after 40 years?

M(t)= 100e^(-0.0017t)

50= 100e^(-0.0017t)

1/2= e^(-0.0017t)

ln〖(1/2)=ln〖e^(-0.0017t) 〗 〗

ln(1/2)= -0.0017t

(1/(-0.0017)) ln(1/2)=t

t=407.7336 years

The half-life of this radioactive substance is 407.7336 years approximately.

ALWAYS REMEMBER THAT A LOGARITHM IS AN EXPONENT!

A= A_0 〖(m)〗^(t/p)

A=10(1/2)^((t/407.7336) )

A=10(1/2)^((40/407.7336) )

A=9.3426 grams

OR

A= e^(-0.0017t)

A= e^((-0.0017)(40))

A=9.3426 grams

*Most of the class didn`t know the story about the life of ants and wasps. So Mr. K had to tell how they grow exponentially.*

So that’s it for Exponential Modeling. Let’s now go back to the Consumer Math stuff.

We’re given this problem to solve:

A $5000 investment earns interest at the annual rate of 8.4% compounded monthly.

What is the investment worth after one year?

What is it worth after 10 years?

How much interest is earned in 10 years?

We use this formula in compound interest.

A = The new Amount

P = Principal/original amount

R = The rate of which it will accumulate

N = Number of times per time amount given (rate will be divided by this value)

T = Amount of Time that it has been accumulating for (yearly, seconds, months, whatever it may be)

What is the investment worth after one year?

A=5000(1+0.084/12)^12

A=$5436.55

What is it worth after 10 years?

A=5000(1+0.084/12)^(12*10)

A=5000(1+0.084/12)^120

A=$11547.99

How much interest is earned in 10 years?

I_T=A-P

I_T= $11547.99-$5436.55

I_T=$6111.44

Okay so that’s pretty much all we did this day. I hope my scribe post helped you!

Ciao

NEXT SCRIBE is

CHRYCEL

the better version ( i think) of my post is

here....

## Wednesday, November 14, 2007

### Logarithms and Exponents

We should be able to picture what y= e^x looks like.. and WE can easily figured out two points (0,1) and (1,e), e which is a little bit of less than three. In y=e^x-2, we just have to shift our graph two units down. our asymptote is y=-2, our graph should not cross over that line.

Before our period ends, Mr. K. had introduced us to our new topic, Exponential Modeling. Bare with me guys, but I didn't get most of it...but I'm pretty sure he will talk about this again tomorrow. Ask him lots of questions...OK? And don't forget guys, Logarithms are Exponents.

That's all about we learn this morning...The next Scribe is Jordan.

### A little help... would help.

## Tuesday, November 13, 2007

### Scribe Post - Logs and Exponents

looking at

**19^(x-5)=3^(x+2)**we know that to get an exact answer we must get the log of both sides..

which then becomes:

**Log19^(x-5)=Log3^(x+2)**

__That Log is an exponent__

**REMEMBER**:so when an exponent is increasing to another exponent we can multiply them:

which we can then expand and move x to one side which becomes:

**now something a little new....**

compound interest is growth from the original amount by a certain rate that will continue to grow exponentially from the last amount

we were given a formula that can help us calculate this value:

**A**= The new Amount

**P**= Principal/original amount

**R**= The rate of which it will accumulate

**N**= Number of times per time amount given (rate will be divided by this value)

**T**= Amount of Time that it has been accumulating for (yearly, seconds, months, whatever it may be)

knowing that we can plug in what ever information is given to find what we need to know:

we see that that question sayes "How much money will you have after 5 years if you invested $300" "5 years" becomes our **T** value and "$300" becomes our **P** value

then its goes to say"at 6% interest compounded annually? monthly?"

**T**=5

**P=300**

well **r**=0.06 (6%)

**n**=1 annually means yearly

**n** also will = 12 (monthly)

which means you have use this formula twice to get a monthly and yearly amount

now when this interest is compounded continuously we would use a different formula but will look very much the same:

and so in this next question we can see how we would use it.first question is pretty straight forward, much like the last one we did only difference is that it is compounded continuously. Sooo **p=300 r=0.06 t=5**

the second question ask if you made if you have invested $300(**P=300**) at 6%(**r=0.06**) how long will it take to get double your money (**A=600**). so it wants to no T time it will take to get $600

600 and 300 will reduce to 2 and 1:**2 = e^0.06t****if we take the ln (log e log base e)** **of both sides then divide by 0.06 we can isolate t****ln2=lne^0.06t**__ln2 __=__0.06t__**0.06 0.06**__ln2 __= t **0.06 **

that's about all we learn in the morning in the afternoon we had a short log quiz and then we did exercise the next for the rest of the class time**The Next Scribe Isssssssssssss..................... not ****lina **as i said before, but instead will

be **luis** voluntarily.

sorry for the mix up i was in a bit of a rush

## Monday, November 12, 2007

### Logarithms and Exponents

The first question:

Using the Product law, you can multiply the powers of the logs together.

In the second line, you notice that the equation is in exponential form. You put it into exponential form since you have the base and the exponent already. Once both sides have been multiplied out, you make everything equal to zero. Since it doesn't factor well, you can pull out the Quadratic Formula to solve what x is:

You can always check if your answers are correct if you plug them correctly into you calculator or if you approximate the values of x. Note: if you plug it into you calculator make sure the you've made the whole equation into base 10 (the common log).

The second question we got was:

This almost the exact same thing as the one above. The only exception being that both sides of the equations have logs. The answer for this question:

The "antilog" on the third line undoes the log on both sides of the equation, so you're left with a quadratic. The -1 was rejected because when plugged in to the original equation Log(base)2 (of) (x-2), it becomes a negative. No number can equal a negative no matter how many times the number gets multiplied by itself.

After finishing these questions we then looked at the Properties of Exponential Functions and the Properties of Logarithmic Functions. Refer to November 12, 2007 slides to see the graphs and properties of the graphs. Remember if you want to see a log graph on the calculator or calculate values, make everything to base 10 or, in other words, use the Change of Base Law we learned to do on Friday.

The next scribe is JoeS.