ScubaSteves Scribe Post #2-Review of assignments?!

Well hello there from Cyber-Math World, fellow classmates! This is ScubaSteve reporting for the second time. First off, today was a double class, and we also had a test. But thanks to somebody (not pointing any fingers >_>) we had it on the second class, but I for one, like many knowing alot of us are just dying to get it over with.

Well well, we had a hand out, and some online questions. A handful of people were bravely enough to put up their hand and ask. (I say bravely, I know there would be more, hoping their questions would be answered by a fellow sucke-- err im mean classmate. =]) Well, lets get on it shall we? Of course!. (talking to himself)

This Applies to the Hand-Out we Had. *Questions done below are the ones ask'd*

Our first question that we worked with was #2b

2) In 1995, the population of Calgary was 828,500 and was increasing at the rate of 2.2% per year.

b)Calculate how many years it would take for the population to double.

In terms of real numbers
To begin with, we give ourselves the oh so sexy formula that Mr.K made up to help solve this kind of problem, simple put as:

A=Ao(model)^t

Yes, quite 'hawt' and easy to use. Now lets look at the question(s) and assign the variables with real numbers for us to crunch.
t= 0 in 1995
Ao= 1 (we put 1 because it can be used to describe the "first" amount)

A= 2 (2 is so convenient because it works well with 1, simple because it's twice the amount which we want.)

r= 0.022

Now lets crunch them down in terms of t

A=Ao(model)^t

2=1(1.022)^t (we put in it as 1.022 because we'd want it to increase every year by 102.2%)

ln2=(t)ln1.022 (put them in ln form to bring the t down! >=o)

ln2/ln(1.022)=t

t=31.8520
It will take 31.8520 years for Calgary's population to double.

In terms of e

2=e^0.022(t)

ln2=0.022(t)

ln2/0.022=t

t= 31.5067 (as you can see the two answers are a biiiit off from eachother, but for the record, in terms of e is more favoured than the first one.)

It will take 31.5067 years for Calgary's population to double.

c)Calculate when the population would reach 1 million.

We'll be using the same formula, but change it up a little to relate with the question.

A=Ao(model)^t

1,000,000=828,500(1.022)^t

1.2070=(1.022)^t (divide A by Ao to isloate the model and t, then go in terms of ln)

ln1.2070=(t)ln1.022

t=ln(1.2070)/ln(1.022)

t=8.6455

It will take 8.6455 years for Calgary to have a population of 1 million.

5) On the Richter scale, the magnatude, R, of an earthquake's intensity, I, is defined as follows:

R= log (I/Io)

When Io is an arbitrary minimum intensity used for comparison. The Mexico City earthquake of 1978 had an intensity of 10^7.85 times Io. What was it's magnitude on the Richter scale?

R=log(I/Io)

=>log10^7.85(Io)/(Io) (cancel out the Io's)

=>log10^7.85 (because 'log' is naturally known as log(base10), therefore log(base10) of 10 cancels out and the exponent is brought down therefore...)

=>7.85 on the richter scale.

The following refers to the online assignment sheet Mr.K posted up.

2)The population of a colony of Pre-Cal 40S math students grew from 3(10^5) to 4(10^5) during the period from noon to 2pm. (These guys really love math!) At what time will the population be 6(10^5)?

Po=5(10^5)

P=4(10^5)

t=2

When P=6(10^5), t=?

4(10^5)= 3(10^5)(model)^2 the (10^5)'s cancel out, and the 3 is taken out from the right side while the left side is multiplied by (1/3)

(4/3)^(1/2)=((model)^2)^(1/2) (to get rid of the "square" we give both sides a multiple of (1/2))

ln(4/3)=2ln(model)

(1/2)ln(4/3)=ln(model)

e^k= model (let k = that beast we call (1/2)ln(4/3))
6(10^5) = 3(10^5)e^(kt)
2=e^(kt)

ln2=kt
ln2/k=t
t=4.8188
The time would be 4 o'clock and 0.8188 hours (Im kind of clueless right now how to convert that to minutes >_> oh shame to me.)

Here, Mr.K just threw some exponential problems and showed how in different situations exponents might pop up in nasty places can be solved.

3^(2x-1)+1=2
3^(2x-1)=1 (bring the 1 over)

3^(2x-1)=3^0 (that is whats awesome about the number 1, it can be any base you want to the power of zero! That's great! Its like a candy store, with all these candies and the best of all, they're all candy...? No...? Never mind.)
2x-1=0 (both having the same base the exponents have a little party of their own in the form of an equation!)
2x=1
x=(1/2)

5./([4^3]-2)/[8^(x+1)]) = 16
([(2^2)(3x-2)]/(2^3)(x+1))^(1/5) = 2^4
((2^(6x-4))/2^(3x+3)^(1/5) = 2^4
(2x^(3x-7))^(1/5) = 2^4
2^(3x-7) = 2^20
3x-7 = 20
3x = 27
x = 9

Well, that's about it for the questions. Then Mr.K had popped up the questions: "Who here owns a suit?" And well, a few raised their hands. Then he said: "Well, what if I had 2 pants and 3 shirts. How many combinations of outfits can I get from that?

Introducing: Combinations and Permutations. A.K.A Counting.

Now back to the million dollar question: How many combinations could there be? Well given this picture we break up the combinations into visual "paths."

You see we give each 'P' their own 'trees' because, well... you can't wear two pants at once! And then what branches out are the three shirts.

Simply, u can see hear, counting up all the little roots on the right hand side there are 6 possible combinations.

Well, great. 6 combinations. Now what if we wanted to....Ok, we'll add two ties to the bunch. How many combinations could there be? Well we'll jsut have to draw out the trees again, this time. Including the ties.

It gets big easily. So, now that we've added the ties, u can see there are 12 combinations possible.

Easy huh? Well after that, Mr.K decided to grab 4 students out of their chairs and stand infront of the class. He then explained how many combinations would there be for 4 students to sit in 4 chairs? Given that once one student had sat down, there would only be three combinations left. And so forth. Then he explained how to calculate this easily he wrote simply this on the board: 4!

No, not 4! as in I REALLY ment it. But 4! as in 4*3*2*1.

So what if it was 5!, then it would be 5*4*3*2*1, remember that it stops at 1, because if it reaches 0 it all becomes 0, and if its a negitive number, well it wouldn't make sense.

Variably it would be: n!=n(n-1)(n-2)(n-3)....3(2)1

Well if you've reached this far, you'll see a light at the end of the tunnel. This concludes the behemoth of a scribe post! Because Mr.K had been called down... to take a picture o_o;; Well, that and my fingers are getting really sore from typing all this. Ooh ooh, speaking about pictures being taken. After all that reading, enjoy this short, even thought it's about mother's day I still think it makes sense for this umm, current rambling. Enjoy, ScubaSteve out~!