Pre-Cal 30S (Fall '06)

The Adventure Continues ...

2007-02-05T14:19:00.000-06:00

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 ...

2007-02-05T14:13:00.000-06:00

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. ;-)

From Natnael .... To Pre-cal30sf06

2007-01-30T20:43:00.000-06:00

Hey guys,

This is my last scribe post on this blog.

I just wanna say a couple of things before i leave from Winnipeg. I really loved this class, and all of you guys. This was such an awesome experience for me, an dam sure for a lot of you. It was such a short time, and this class really kinda got us together on math. I've rarely done math with friends, i usually do it just individually, which i don't think i will continue doing, because of what i saw in the class.

Oh about the postcard... guys ... really THANK YOU V.V.V.V.V.V.MUCH ... and Sam... i will try not to find friends better than you ;[]:...

This class was such an amazing class i think. we all cooperated and worked together till the last minute (be4 the exam) Everyone was blogging, and sharing what he/she knew.. it was awesome.

Well, guys, wish if i had staid with you guys longer.. at least till grad, but u know i have to move.. i guess if God wills we will get together sometime (theres a .009 chance..).

Mr. K. you've been a wonderful teacher of the course, i really liked the way you would put examples on the board.. the pretest... the quizzes.. almost every thing.

Hope you guys enjoy your upcoming academic years. AND EVERYONE IN THAT CLASS SHOULD STICK TO PRE-CAL, and i really do wish everyone good luck!!.. and am sure all of u will.

Well its so sad, but.. this might be my last time talking to you guys...so good bye... in case anyone wanna add me to his/her msn this is ma email nathaniel_td@hotmail.com

Bye

Natnael

The year's come to an end ...

2007-01-29T16:46:00.001-06:00

Autograph that final exam and your life with EXCELLENCE!

2007-01-28T17:53:00.000-06:00

Hi there Pre-Cal 30s!

Thank YOU so very much for those thank you posts!! You beat me to the punch! I wanted to be the first to thank YOU for allowing me to hang around and learn from you--

AND

To wish you the very best!!
Lani

A BIG THANKS TO LANI!

2007-01-28T14:23:00.000-06:00

Dear Lani,
Since our semesters coming to an end, our class and myself personally would like to thank you for your contribution to our blog. You've definately added to our learning experience and your advice will be with us always. To think you took time out of your busy schedule to mentor a small group of Pre-cal students in Winnipeg, Manitoba. Mr. K and our class appreciate your time and sincerity. THANKS LANI!! Good day, and have fun!

THANK YOU Lani!

2007-01-28T11:31:00.000-06:00

Thanks Lani for giving us your support. We gladly appreciate it. You've always been there cheering us on, encouraging us to move forward. I'm sure the class has a lot more to say, but nonetheless, we feel like we owe you a big thank you...

THANK YOU!

~ -Zeph

GO PC30SF06 GO!

Graphing Functions Notes

2007-01-24T22:27:00.000-06:00

As promised, here they are, the full set of notes for graphing polynomial and rational functions. These notes must be copied by hand into your math dictionaries. Do not hand in the printed pages. There is a good reason for this ... if you really want to know ask me in class. ;-)

Learn Hard!

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Scribe Post

2007-01-24T19:50:00.000-06:00

Omg sorry i got mixed up with natnael and melissas post because natnael posted after melissa but yeah i'll do monday, tuesday, and today. Once again sorry for keeping you guys waiting.

Okay so lets start with Monday. On Monday we started off the class by dividing:

Divide. Indicate any remainders as a fraction.

So question 1 had no remainders but number 2 did have a remainder of 2.

The quotient of question 2 can also be written as 71^2/3.

Question 3 had no remainder but the quotient can be simplified to 2x²+5x+3=(x+1)(2x+3).

Question 4 has a remainder like question number 2. it can be written as x³-4x²+2x-1=x²-2x-2-5/x-2

Instead of doing division the long way, we learned how to do synthetic division. This is what synthetic division looks like:

On Tuesday we started off the class by finding intercepts.

Find all intercepts
f(x)=4x³-7x+3
*The degree of the polynomial determines the maximum roots that the graph will have. Something like 4x⁴ could have up to 4 roots.
The first thing you do is find all the possible numerators (you can do this by finding all the factors of the constant in the equation):
±1, ±3
The next thing you do is find all the possible denominators (you can do this by finding all of the factors of the leading coefficient):
±1,±2,±4
From that you can find all the possible roots:
±/1, ±3/1, ±1/2, ±3/2, ±1/4, ±3/4

The easiest way to test if these are roots is by synthetic division:

If there is no remainder, which in this case there isn't, you take the linear factor (x-1) and add the x² and the x where they belong then multiply the 2 to get the equation of the line.
f(x)=(x-1)(4x²+4x-3)
You can factor this further:
f(x)=(x-1)(4x²+4x-3)
f(x)=(x-1)[4x²+6x-2x-3]
f(x)=(x-1)[2x(2x+3)-1(2x+3)]
f(x)=(x-1)(2x+3)(2x-1)
From here you conclude that there are roots at (1,0), (1/2,0), (-3/2,0) and the y-intercept is at 3 because if you substitue 0 in for x in the equation, you are left with 3.

To draw the graph of this function, you must do a sign analysis. You must substitute numbers between the roots into the equation to check if they are negative or positive. The end result for this equation looks like this:

*The only way a root can change its sign is by passing through zero.

The graph of this function would look something like this:

On Wednesday we started by sketching the graph of f(x)=-2x³-7x²2x+3.
These are the steps you take to graph this polynomial:
1.Factor everything
2.Find the y-intercept
3.Find the roots (You may use the roots theorem)
4.Do a sign analysis
5.Sketch the graph

This polynomial is easy to work with in this form.
The y-intercept is the constant which in this case is 3.
Find the roots by finding the possible roots and use synthetic division to find a root:

From this, you can also find points on the graph by using the possible root as the x-coordinate and the remainder as the y-coordinate. Here you have points at (1,-8), (3,-57), and a root at (-1,0)

You can factor the polynomial to find more roots by using the linear equation x+1 and multiplying it by the equation you got by using synthetic division:
f(x)=(x+1)(-2x²-x+3)
f(x)=(x+1)(-2x²-6x+x+3)
f(x)=(x+1)[-2x(x+3)+1(x+3)]
f(x)=(x+1)(x+3)(-2x+1)
From this you get roots at x=-3,-1,1/2 and you can sketch the graph of the polynomial. The graph would look something like this:

Now something like y=1/x would have a vertical asymptote on the y-axis and a horizontal asymptote on the horizontal axis. The graph would look like this:

The next scribe is eeDce

The steps to graphing a function like f(x)=1/x-2 would be:
1.Factor everything
2.Find the y-intercept
3.Find the roots
4.Find the vertical asymptote
5.Find the horizontal asymptote
6.Do a sign analysis
7.Sketch the graph

f(x)=1/x-2 is already factored
The y-intercept would be at y=-1/2
There would be no roots
The vertical asymptote would be at x=2
The horizontal asymptote would be found like this:
lim_x®¥ 1/x-2
lim_x®¥ (1/x)/(x/x-2/x)
lim_x®¥ (1/x)/(1-2/x)
=0/1-0
The horizontal asymptote would be at y=0 on the y-axis. An easier way to find the horizontal asymptote is by taking the leading coefficient of the denominatior and divide it by the leading coefficient of the numberator but only if they have the same degree.
The sign analysis would look like this:

The graph for this would look like this:

THE NEXT SCRIBE IS eeDce

Factor and Remainder Theorem Homework

2007-01-22T15:35:00.000-06:00

Tonight's homework.

Functions

2007-01-21T23:13:00.001-06:00

Function: Functions are set(s) of rules that turn one mathematical object into another.

FUNCTIONS + RELATIONS

-All functions are relations.
-Not all relations are functions.

Four ways of presenting functions: [Using the function f(x) = x + 1]

1) Verbal
...the input changes to an output after adding one unit to it.
2) Graphically: Graph

3) Numerically: Table of Values

4) Symbolically: Equation
f(x) = x + 1

FLAVORS OF FUNCTIONS

1)Many-to-One Function

This is when a function provides one output for two or more inputs.

in the above example; when the input is -2 the output is 3.5, and when the input is 2 the output is 3.5 again.

2) One-to-One Function:

This is a kind function, where there's is only one output for one input.
E.g f(x) = x - 3

FUNCTION COMPOSITION

Function composition dimply means: feeding a function into another.
for e.g.
Note: We can use all the operations on function compositions.

TEST FOR FUNCTIONS:
This is done by the vertical line test, which ensures that there's only one output for each input.
e.g. 1\ Passes 2\ Fails

Baby play + Parent clean up = Functions?

2007-01-21T16:44:00.000-06:00

Hello!

Friday's class Mr.K wanted to tell us a story. He told us a story about his daughter playing in a play room, where she would disorganize everything. Then Mr.k and his wife would put everything back in place afterwards.

At first most of us thought it had nothing to do with functions.

This chart helped us understand more about inverse functions.

Inverse Functions: undoes the input of the function (Exchanging the domain and range)

We then moved on to some examples.

f (x) =2x-3 (inverse will be to divide. Changing the order of operations)
y=2x-3

Inverse functions of y=2x-3.
x=2y-3
x+3=2y
x+3/2=y or 1/2x + 3/2=y

We can really understand it more when we look at the table of values.

Another example:

f-1(x) = 3√5x2-2/6

x = 3√5y2-2/6

x3 = 5y2-2/6

6x3 = 5y2-2

6x3+2 = 5y2

6x3+2 = 5y2

6x3+2/5 = y2

√6x3+2/5 = y

The next scribe for tomorrow is Crysta!

BOB - Logics

2007-01-17T07:17:00.000-06:00

I'd have to say that this unit was quite different from the units that we previously worked on. Some parts were easier for me, so I hope I'll be able to boost up my mark through this. The same goes with everyone else. To be honest, I found it confusing at first, but since it related to real life, I began seeing the concept in a different perspective where I can understand it better. It definitely got my mind going though. So, hopefully this test will be smooth sailing for everyone.

Good luck everyone!

Hey BoB .. you've got anything to say?...no? ...y?

2007-01-17T00:58:00.000-06:00

Hey guys,
i couldn't get to class today. well, i had a doc. appointment, and then felt like i needed rest. so, anyhow...this unit i would say was pretty interesting and sometimes mentally challenging (well, not quite).
Logic + Statistics are my favourite branches of math, cuz they dont just relate to real life. They are real life lol. And in this unit, i would say i tried my best, and it went fairly good, accept that i missed the last day. anyways, I'm hopin that i didnt miss that much, and am looking forward for a good grade on this test.

Hope this test will boost up everyones average!

Nati

BOB TIMEE - LOGIC

2007-01-16T22:53:00.000-06:00

My view on our logic unit was a fair stretch of our mind. I think we may have all found it difficult only because our state of mind and way of thinking differs from dealing with triangles and square roots. Some particular parts of this unit was very plain and simple (i.e: the venn diagrams). Others required more thought and time (i.e: deductive/inductive reasoning). Although everything was possible to solve with either a formula or a broad set of rules to be followed. I hope we all do well on this test. I know its getting harder to stay consitent with our motivation to do well because the course is so close to over. I know for myself, I find myself slacking because I'm just so ready to be done. I know it's not smart on my behalf and I can already see the consequences by suffering of overload before exams. I just need to straighten out and prepare myself with a logical matter! Okay. GOOD LUCK GUYS

I had a fun time BOBing. Hey guys its our last one!

Blogging on blobbing

2007-01-16T22:09:00.000-06:00

I personally think that this logic unit was somekind of a very...very ... long stretch for my mind. I think that i struggle for the most part of this unit. Perhaps , the easiest thing in this unit was the venn diagrams and valid and invalid arguments. I think that I really have to study very hard tonight or else i'm done. Well thats it I hope that I didnt miss too much information from todays class. Study hard and keep your body away from sickness!! Good night

Direct and Indirect reasoning

2007-01-16T21:03:00.000-06:00

Good evening everyonethis is m@rk scribing for yesterday's class. I've been banged up by high fever that is why i was unable to go to class today.

To start yesterday's class we proved using direct reasoning that the product of 2 even numbers is even too.First of all an even number is something that is divisible by 2 and has a factor of 2.

Let 2x=an even number
2a=another even number

2x * 2a =4ax
2x * 2a= 2 (2ax)

∴ Any product by both even numbers is also even as seen from above.

Then, we proved using direct reasoning that the product of 2 odd numbers is odd too.

Let (2x+1) =an odd number
(2a+1) = another odd number

(2x+1) (2a+1) =4ax+2x+2a+1
(2x+1) (2a+1)=2 (2ax+x+a) 1

∴ Any product by both odd numbers is also odd as seen from above because the product does not factor perfectly with 2.

Now, from the previous class we proved that √2 is irrational using indirect reasoning.

First, assume that √2 is rational.

√2=a/b

Note: a/b is in reduced form meaning they have no common factors.

2= a²/b² Multiply both sides by b²
2b² =a²

a² must be an even number and a must be even too so, let a=2c.

2b² = (2c)²
2b²=4c²
b²=2c²

b² and b are both even numbers so, let b=2d. This is a contradiction! Both a and b cannot have any common factors because of our assumption that they are in reduced forms in the beginning.

∴ √2 is irrational

Then, we prooved that √3 is irrational using indirect reasoning.

First, assume that √3 is rational. THis problem is similar to the the previous one.

√3= a/b

Assume that a/b is in reduced form.

3= a²/b²

3b² = a²

On the other hand lets look at square numbers first.
4= 2*2
9=3*3
16=2*2*2*2
25=5*5
36=2*3*2*3
49=7*7
64=2*2*2*2*2*2
81=3*3*3*3
100=5*2*5*2

Can you see the pattern that there is always a pair of prime numbers in square numbers?

So let b²=3c

3(3c)² =a ²

3* 9c² =a²

BUt it cannot happen because a and b will share the same factor which is 3. Therefore contradicting with our first statement that a and b is in reduced form.

∴√3 is irrational

Lastly, we talked about pardoxes. A paradox is a statement that cannot be true but also cannot be false.

Example:
This sentence is false

In this case you cannot decide whether it is true or false.

Thats it for my scribe for tonight. I dont know if im going to be able to go to class tomorrow. It all depends if im feeling better . LIke i told Mr. K the scribe will be Natnael. The homework is the Unusual school play

BOB

2007-01-16T20:25:00.000-06:00

Another unit test coming up. For me, logic was a lot harder than I thought it would be. Even if it's not that easy, I personally enjoy the challenge of finding the solutions. I've had experience with these kinds of problems in junior high, so I have a good starting idea as to how to approach the question. I just hope that we all do really good on this test, and hopefully, I get the mark I've worked to get. 'Night!

BOB

2007-01-16T19:33:00.000-06:00

This logic unit was interesting. It was cool getting back to the Venn-diagrams, bringing back the memories of arithmetic's in elementary. The good old days, everything was so simple. This unit got more difficult when we started the inductive and deductive reasoning. I'm still having trouble with that.

That's all, good luck to everyone tomorrow!

BOB v.6 The LOGICal Bob

2007-01-16T07:04:00.000-06:00

Well I can't say the logic unit was straight forward since we now know about the paradoxes. It was fun though doing those paradox assignments on "A Usual Day at an Unusual School." Venn Diagrams were easy to do and so were the matrix problems. It felt like doing jr.high stuff but more advance. It was awesome learning logic in math class where we discuss things that barely have to do with numbers as compared to the other unit we're doing - consumer math. We discussed confusing things suchas "the inverse is the converse of the contrapositive" and the existence of counterexamples. Learning logic was fun and very interesting and the countdown begins for the logic test to show.

-Zeph

BOBing

2007-01-15T21:49:00.000-06:00

This unit was alright, logic. It was difficult at first because I wasn't sure how to start solving the problems. I like doing problems where you figure out if it's a deductive information or inductive information. I also like doing problems where you verify if it's a valid argument or invalid argument.
What I found most difficult is when I try to figure out if the problem is a paradox or not. Proving that the square root of some numbers are irrational is also something that I find difficult as well. The consumer math stuff was alright. There were times where I didn't understand it, that was when the question asked about stuff I didn't know yet. My parents helped me, because they deal with this stuff all year round. Overall, consumer math was good and logic was quite difficult. Well, I'll see you all in class tomorrow fellow classmates and I hope you all study hard!

-SAMUS :D

BOB for Logics test in 1day and 16 hours

2007-01-15T21:36:00.000-06:00

In logic we had done logical work like indirect and direct reasoning, inductive and deductive reasoning and that is what I am having trouble in. I don't understand the difference between the 2 reasoning in each set. I don't understand the definition on inductive and deductive reasoning.
Something that I am getting the hang of the is the Venn-diagram questions. They seem to be the easiest parts of logic.
In conclusion good luck everybody on the test that is happening in 1 day and 15 hours and 15 minutes from when I post this BOB :-)

Signed,
Benvolio (Ben)

blogging on blogging - logic

2007-01-15T21:13:00.000-06:00

Well, this unit has gone by pretty fast. Of course that's not including the two weeks break we had. So not including that, it hasn't really been that many classes. Logic has been one of the more interesting, fun topics. Venn Diagrams, once you understand the method of how you put them together, they're pretty straight forward. And personally, if i know how to do something, it always makes it fun because i understand it. As for inductive and deductive, that's okay. i had some trouble at first, but nothing i couldn't fix. And what we did in class today was fun as well as our homework. So this unit isn't too bad.

Honestly, i can't remember if Mr. k said to do a blog for this unit or not? A lot of things lately have just been a dream, haha, so I'm not too sure. But better to do this than not to! Well, good luck everyone.

Usual Day At Unusual School

2007-01-15T15:45:00.000-06:00

We talked about proof by contradiction in the last couple of classes. Today we proved that √2 is irrational using proof by contradiction (which is also called, in latin, reductio ad absurdum). You can also review direct proof here (use the pop-up menu to review indirect proof and other topics as well).

Tonight's homework is the play below. Make a list of all the Braves and Brights. We'll discuss it in class tomorrow. ;-)

Page 1 of 4 Page 2 of 4 Page 3 of 4 Page 4 of 4

Challenge

2007-01-15T07:47:00.000-06:00

It's a New Year, with new tests, new units, new learning! And how will you meet those challenges?

2007-01-13T16:05:00.000-06:00

Well like yesterday to start things off we had another quiz. The quiz was not on valid and invalid questions but learning how to draw venn-diagrams based on word problems. I thought the quiz was a little easier than yesterdays. Well after going over the quiz we got off topic and had an interesting discussion about this new device called a smart board. Well the smart board is a school board and a computer in one. It's basically an interactive white board. It's sound really cool and I hope Mr.K gets it in time for us to use it next semester.
Besides chatting about this cool smart board we actually did do some work. We learned about Direct and Indirect prove.

Direct Prove would be Thales' Theorem. We know if an inscribed angle in a circle is subtended by a diameter, then the angle is 90 degrees. Also that 2x+2y=180 degrees or that simplified x+y=90 degrees, as shown below.

Indirect prove is where you begin assuming the opposite of what we want to prove.

So here was the example that we worked on in class
Given triangle ABC with point D on line BC
Prove: If Line AB = AC
and angle BAD doesn't equal angle CAD
then line BD doesn't equal line CD
(so we start we assuming the opposite
line BD = line CD) "a contradiction"

That was the only example that we got to but the next scribe will be able to scribe on the good stuff as Mr. K described the ones we didn't get to yet.
Well the next scribe is m@rk so have fun ...........
Remember our go for golds due Wednesday before exams
and one of our two consumer worksheets on Monday

2007-01-11T20:36:00.000-06:00

This was ok but I had some confusion. I still need a little work on the valid, invalid stuff. I hope we all do well on this test. This unit was a short one but easy. Sorry my BOB is so short and I promise I will put my scribe post up tomorrow morning. I am having problems with my computer.

2007-01-11T19:46:00.001-06:00

Hey guys exams are coming.
Well to start things off we had quiz. The quiz was of course on logic, inductive, deductive, valid and invalid questions. A inductive statement is where you have examples and make a generalization and deductive statement is where you have a generalization and produce an example. The part I thought some of us had problems with was determining whether a statement was valid and invalid. Anyways tomorrow might be the logic test so remember to have your BOB's up just in case. Besides the quiz we had a lot of writing to do in our math dictionaries some new and old things.

MATH DICTIONARY

Some definitions

Set: A collection of objects (could be people, things, numbers etc.)

Subset: A collection of objects that consists of some, all or none of the objects in a given set
Example: Consider set A :(1,2,3)
All of the following are subsets of A : (1), (2), (3), (1,2), (1,3), (2,3), (1,2,3),
Note:is called the "empty set" or "null set" but is still a subset of every set

Universe (U) : all objects are being considered

Notation: Union U A U B means gather all objects in set A with all objects in set B

Notation:Intersection
means gather all the objects that are both in set A and B

- or ' compliment: (reads as " the compliment of A" or "A compliment") means everything outside A

exclusion: means everything that is in set A but not set B

Counterexample: Given any logical argument, theorem or hypothesis, if you can find only one case where it is not true, then the theorem or argument is proven false . This is called a "counterexamlple"

Conditional statements: Any statement of the form" If...then..." is called a "conditional" or "implication"

Hypothesis: The first part of the conditional

Conclusion: The second part of a conditional
Example: If(hypothesis) Then(conclusion)
If a triangle is isosceles Then it's base angles are congruent

Arguments of the form"if..then..." and related statements

A sentence that says only one thing is called a "declarative sentence"

Well that's all for this scribe and I'll scribe on Friday since I couldn't get my post up in time.
Homework is Ex. 48
Flickr assignment due this Friday
Go for Gold due Wednesday before exam
and our 2nd consumer packet

LOGIC

2007-01-10T22:53:00.000-06:00

Salutations fellow classmates,

Today in class, for all you naughty kids that skipped (haha jk), we kind of had a work period. Mr. Kurropatwa had to leave for some meeting, but before that, he had handed out the Go For Gold assignment, which is to be due in about 2 weeks from now.

Rules for Go For Gold:
1) only two options for marks; zero or one hundred.
2) you may talk to other students.
3) you can ask Mr. K ONLY one question.
4) you must complete it and hand it in on time. (it IS a good way to prepare for the exam)

He also handed out the second consumer math booklets. We only got one worksheet to fill in for now and the other worksheets will be handed to us tomorrow. Tomorrow's not a good day to skip then ;)

He also told us that our other project is up and running, the wiki. There will be enough questions for at least everyone to do. Here's the link to it: www.pc20sf06.pbwiki.com
Our photo trigonometry projects are due today and we received our rubric on how we're being marked.

Throughout the class, we all worked on our Go For Gold. Hopefully everyone does well on that, and we all get 1 hundred! Remember to work on the wiki and to go to class tomorrow because we get our worksheets (and who doesn't like to get an early start on work? lol). Yeah, and that's about it. Our next scribe for tomorrow will be

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LAURESSA!!! :D

Peter Donnelly

2007-01-08T16:57:00.001-06:00

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 will learn the math he's talking about here in your grade 12 Pre-=Cal class ... probability.

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

New Year With Continuation of Last Year's Class

2007-01-08T15:56:00.000-06:00

Hi and happy new year everybody and I hope you all had a great holiday. Today we continued on what we left off on the last day of school before the holidays. We worked on some word problems that had to do with Venn- Diagrams. These are the questions:

1. Recently several DMCI students were surveyed about their favourite music groups.
The results were:

22 like Hole
25 like U2
39 like Third Eye Blind (TEB)
9 like U2 & Hole
17 like Hole & TEB
20 like U2 & TEB
6 like all three
4 like non of these three

a) How many students were surveyed?
b) How many students liked U2 only?
c) How many students like exactly 2 groups

2.Every year over 9 million tourists visit Turkey. Last year 9.8 million tourists vistited Turkey.

60% visited only Istanbul
20% visited Izmir
Half of the tourists who visited Izmir also visited Istanbul
2% visited Istanbul, Izmir, and Antalya
The number of tourists who visited only Antalya was half again as much as the number of the tourists who only visited Izmir
Tourists who visited Antalya either saw only Antalya or Antalya, Istanbul, and Izmir
The rest of the tourists visited other regions

How many tourists visited other regions?

Solutions
1.So first we fill in a three circle venn diagram because there are 3 different groups.
I'll explain the diagram, first we work our way from the middle and we know that there are 6 people who like all three groups. Now we pick one of the next 3 overlaps and find those. Lets use U2 & Hole as an example, we know from the clues that 9 people liked Hole & U2 so we subtract 6 from the 9 because the 6 represents 6 people that also like Hole & U2. So we have 3 in that section. Once those sections are complete we work on the outer sections. Like the overlaps we take the total number of people that like that group and subtract the number of people that like that group and other groups also.

a)This question asks us to find the number of students survey, so we just add all the numbers up and we get 50

b)This question asks us to find the number of students who only liked U2 and the answer is 2 according to the graph

c)This question asks us to find the number of students who like only 2 groups and the answer is 28. We get that by adding the numbers that are in the section of 1 overlap which are 3, 11, 14.

2.This quest is a doozy because we have to go back and forth with the clues.

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If the pictures are going too fast, feel free to pause or replay.

The first picture shows inserting the info from the first clue and from the clue it tells us that 60% of the total tourists visited only Istanbul. We can put it on the venn diagram because the clue states that 60% visited only Istanbul.
The second picture shows the third clue and with the second clue in mind we can find the Istanbul- Izmir overlap because the clue states that half of the number of tourists that visited Izmir also visited Istanbul. So half of 20% is 10% and that leaves us with 10% for Izmir.
The third picture shows the addition of the middle section with the forth clue.
The forth picture shows the sixth clue which states that the people who visited Antalya only visited Antalya or saw all three places and not 2 places. So we know that the Izmir-Antalya and Istanbul-Antalya sections have 0%'s. And now we can find how many people visited Izmir and it is 8% because before we had 10% and due to the forth cluwe are left with 8% and that 8% go into the Izmir section.
The fifth picture shows the fifth clue and it states that the people that visited Antalya is equal to 1.5x the number of people who visited Izmir and 8 x 1.5= 12. So Antalya is 12%.
The final picture shows the addition of the last clue which states that the remainder of the tourists visited other regions.

To answer the question we need to know the number of people who visited Turkey and the percentage of the tourists who visited the other regions. To find the number of people who didnt visit Istanbul, Izmir, and Antalya we take the total number of tourists multipied by the "other" percentage in decimal form.

After the explanation of those two questions we began with a new logic subject.
Mr. K told us to take out a new piece of loose leaf and draw the biggest circle you can draw on that paper and make a dot on the circumference. Then he asked us how many different flat areas there are in the circle and there are 1. He told us to make a second dot and connect the dots and asked us how many areas are there and there were 2. Then a 3rd dot and then connect that dot with every other dot and then do the same as before and there were 4 areas. Now look at the graph.
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So lets think of an equation to helps us find the number of areas if the number of points is 457.
As a class we discovered that the equation 2^(n-1) where n is the number points will work. So we tried the equation with 6 as the number of points and using the equation we predicted the number of areas will be 32. Now continue with the diagram with 6 dots and count the number of areas. Oh no there is only 31. That is situation called a counterexamples and that is when there is a theorem made and there is one example that doesn't fit the theorem. That just makes the theorem a theorem that is not true. Tomorrow will be the continuation about counterexamples.

Okay that was our first class back from the holidays and it was kind of tiring to go back to school after a 2 week break. Homework (homework*sigh*) is exercise #48 and it is about counterexamples.

The next scribe will be SAMUS!!!
Good Night :-)

VEry ...very late post

2007-01-07T22:22:00.000-06:00

Ok, everyone I hoped you had some very good days of rest. Now, let's start with the math. I know that this should have been posted a long time ago but my computer sort of meltdown.First of all I need to thank my cousins for fixing my computer and making me recovery disks.

Last two weeks we reviewed the questions from our circle unit. The first question was:

Given:
line AP and line AQ are tangents
Arcs BS and RE are congruent
Angles 2 and 5 are congruent

Prove that lines BC and ED are congruent

Statement and Reason
Lines AB and AE are congruent Tangent Theorem
Angles 3 and 4 are congruent Inscribed Angle Theorem
Triangles ABC and AED are congruent Angle-angle-side
LInes BC and ED are congruent From above

The second question was:

Isosceles triangle ABC is inscribed in a circle with a diameter 12 cm. Side AB is a diameter. Determine the length of BC.

Let X=the length of lines AC and BC.
Then use the phythagorean theorem
x^2+x^2=12^2 2x^2=144
x^2=72
x=√72
x=6√2

The third question was:

Quadrilateral PQRS is inscribed in a circle. Side PQ is parallel to side SR. The diagonals intersect at T. Prove that triangles TSR and TPQ are isosceles.

Statement and Reason
Arcs PS and QR are congruent Parallel Chords Theorem
Lines PT and QT are congruent Congruent chords theorem
Lines ST and RT are congruent Congruent chords theorem
TRiangles TSR and TPQ are isosceles From above

The fourth one was:

The circle with centre O has 4 points of tangencyA,B,C and D.Angle AEB =30°, angle ADC=90, and Angle DCB=85.FInd the measure of angle CDG.Statement And Reason
Angle BAD=95 Cyclic Quadrilateral Theorem
LInes EA and EB are congruent Tangent theorem
Triangle AEB is isosceles From above
Angles EAB and EBA=75 BAse Angles
LInes HD and HA are congruent Tangent theorem
Angle HAD =10 Supplementary angles
Angle HDA=10 base angles
Angle CDG=80 Supplementary angle

Then after we reviewed the test we practiced on more stuff on Logic

The first question was: A survey of 80 students showed that 45 like rock music, 25 liked country and 10 liked both. How many liked neither?

The second question was:

140 homes are surveyed 90 have a stereo, 70 have a DVD player and 40 have both. HOw many have neither?

The last question was:
150 people were surveyed on their reading preferences
90- mystery
70- history novels
45-sci-fi
30-mystery and history
25- mystery and sci-fi
22-history and scifi
10 -reads all 3

How many read
1. At least one type of book
2. Exactly one type
3 Exactly two types
4. Do not read any of the 3 types

That's it for this edition of my scribe. Again, i'm sorry for the images because my paint and RFXsoftware is not working properly eversince last year. I must admit that i need a new computer.
REMINDERS
Flicker Assignment (trigonometry)
DElicious (Put at least one link for every unit we finished)

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.)

Scribe news

2006-12-22T18:22:00.000-06:00

Hi it's me benofschool and m@rk will be finishing up his scribe post over the holidays and he has told me that I will be the next scribe. So therefore I will be the first scribe in 2007 for our pc30sf06 blog. Until 2007 Merry Christmas and happy holidaze!!

Don't Let This be You ...

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

Have an excellent holiday ... but try to do some math too ... you don't want this to be you do you?

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.

First things first? One day left to go--

2006-12-21T11:20:00.000-06:00

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

BOB v.5 BcircleB Circle Geometry

2006-12-20T12:49:00.000-06:00

When I first thought about Circle Geometry, I first didn't know what the unit will be about, but as we continued on with the unit, I found out it was about the angles and lines which make up theorems. Doing the investigations were easy and straightforward because all we had to do was follow the instructions...but in the end, we had to memorize them all! Yeah, so circle geometry was really straightforward. The only difficult thing about this unit is where you have to prove something which can take a LOT of time to do and that's the only part I'm worried about. My goal is to get at least a 70% or above on the test (depending on how many "proof" questions there are) which is coming up in about...less than two hours!! XD So I'll cya all in 6th period!

" L O G i C " - cont'd

2006-12-19T17:35:00.000-06:00

To start off today's class, we were given a few reminders:

FLICKR.com RUBRIC: Reminder to students that this is for your benefit. Putting in your voice nto this document will affect how your flickr assignments are graded. This document should be about how you want your assignments to be marked, and how the assignments should be done. Get crackin'!

NEXT FLICKR ASSIGNMENT: The next flickr assignment will be to take a picture of TRIGONOMETRY. Be unique and creative and find a photo no one else will have. This won't be due until the first tuesday after winter break. We'll be given a great amont of time for this assignment, but don't procrastinate!

Today, we added in some more notes into our math dictionary:

Here are a few examples of some arguments:

Example 1 and 2 both are true and valid statements. The reason being the premises of each example makes sense with what the conclusion says. The conclusion makes sense of what has been stated and flows naturally. This is called a sound argument.

SOUND ARGUMENT: an argument that is both valid and true.

AN EXAMPLE OF A FALSE ARGUMENT:

"All men are mortal.
Mr. K is a man
... Mr. K wears glasses."

That was a false statement because, even if the three statements alone are true, together as an argument, it's not. The conclusion doesn't flow naturally with the premises, and it doesn't make sense.

**************************************

We also discussed different TYPES OF REASONING that we go over in this unit of Logic:

Induction: When we observe several particular examples that identify a pattern and conjecture that it must always be that case.

EXAMPLES:
QUESTION: Will the sun rise tomorrow?
ANSWER: It has every day before, so it will rise again tomorrow.
*That's the answer because through all the years we've been living, a reoccurring pattern of the sunrise has always taken place.

NOTE: To view more examples of inductive reasoning, look over all the investigations we worked on in our circle geometry unit!

Deduction: When we argue from basic, unarguable truths, to a valid conclusion.

An example of deductive reasoning is the process of proving THALES' THEOREM.

***************************************
In logic, we also look at relationships between different sets, and compare them in Venn Diagrams.
In the following images, we see how sets are made, and how we can compare two different sets using a venn diagram.

***************************************

Weeell ..

TOMORROW'S THE DAY! CIRCLE GEOMETRY UNIT TEST

I HOPE YOU GUYS ENJOYED MY SCRIBE..

blogger was being a pain in the behind and erased my

first draft and I had to start all over again!

BUT, I DID IT AGAIN. YET ANOTHER SCRIBE POST BY ME!

TOMORROW'S SCRIBE WILL BE ...

M@RK.

(just cuz he told me to pick him!)

NiGHT!!

Circle Geometry Review

2006-12-19T16:01:00.000-06:00

We're gearing up for our test tomorrow. Here's a bunch of online quizzes you can use to help get ready.

Have you added any really good links to our del.icio.us box lately? ;-)

You can find circles in the most unusual places ...

Proof of Congruence (5 questions - refresh the page for more quizzes)

More Congruence (5 questions - refresh the page for more quizzes)

Circles (5 questions - refresh the page for more quizzes)

Arcs &Angles (5 questions - refresh the page for more quizzes)

Arcs & Chords (5 questions - refresh the page for more quizzes)

Inscribed Angles (5 questions - refresh the page for more quizzes)

Tangents (5 questions - refresh the page for more quizzes)

Polygons (5 questions - refresh the page for more quizzes)

Whew! More quizzes than you can shake a stick at! ;-)

New Unit: Logic

2006-12-18T20:47:00.000-06:00

So today we started logic which is a system of reasoning or the way you present an argument.

Mr.K asked us if the sun is going to rise tomorrow and if i'm not mistaken almost everyone thinks that the sun is going to rise tomorrow based on the fact that the sun has been setting and rising every single day of our lives so far. That brings us to inductive reasoning and deductive reasoning

Inductive reasoning is when you come up with a hypothesis based on examples or observations to make a general rule more specific.

Deductive reasoning is when you use fundamental trues to create a rule. An example of deductive reasoning is Thales' theorem.

Something like this would be inductive:

Draw the next two.
With the next two it would look like this:

Something like this would be deductive:

Draw the next figures.
There really is no right answer for this question, you can look for a pattern but you must explain 3 things: the square, the curve, and the dot.

Here is a true or false questions:
1. The square of a number is larger than the number.
- this is false because the square of 1 is not larger than 1 (1² = 1)

A sentence like "This sentence is false." is what we call a paradox. It is true and false because it refers to two things at the same time and it is a self-reference.

valid means it flows natuarally from the premisis.
invalid means it does not flow naturally from the premisis.

All men are mortal
Mr.K is a man
Mr.K is mortal
^^ This is what we call a sound argument because it is both true and valid

A few examples of unsound arguments are:
All men are squibbles
Squibbles are pink
All men are pink
^^ This is unsound because it is valid but it is false.

All men are mortal
Mr.K is a man
Mr.K wears glasses
^^ This is unsound because it is all true but it is also invalid

All men are green
Mr.K is a man
Mr.K is green
^^ Something like this is also unsound because it is valid but most of it is false

SO THAT IS WHAT WE DID TODAY. HOMEWORK IS EXERCISE 45
&&
CONSUMER MATH - WAGES ASSIGNMENT **due on friday

NEXT SCRIBE IS LAURESSA

BOB; circles

2006-12-17T23:14:00.000-06:00

Circle Geometry was by far the hardest unit yet. It was alright at first because we went through it as a class, step by step. Although, I didn't always understand the investigation because of my poor measuring skills. But, when all ten investigations were done I finally understood it all. I find it more easier for me to understand things when I'm told of what to look for and practicing it, rather than finding it out myself. Most of the investigations, I didn't know what I found because my measuring skills aren't so good.
I'm happy that I reviewed before the pre-tests and quizzes because that's where it came in place, piece by piece I began to understand what was going on and why different angles are the same or two times larger. This happened on Thursday I believe. It was like someone had replaced a light bulb to the lighthouse in my head. However, I wasn't prepared enough. So, my options are study harder or study harder (haha).
The up side is, I understand the theorems and I look for different triangles and angles. The down-side is, sometimes I doubt myself and have second thoughts about some things, which makes me to change my answers (which were right in the beginning). But, most of the things, I understand. Well, I hope that everyone does well and studies hard. See you all tomorrow fellow class mates!

-SAMUS

B.O.B.

2006-12-17T19:23:00.000-06:00

Once again, a unit finished.
This semester is almost over too. (tear*)
The circle geometry unit was a lot of fun. I really liked this unit, mostly because it involves looking at a real image and trying to figure out measures. I like how everything all connects, when you find one part you can also find another. The trouble I have is remembering the theorems. I'll just have to look over my math dictionary more. That's all for now, good luck!!

BOB

2006-12-17T14:55:00.000-06:00

I actually found this unit to be the easiest out of all the other units we've worked on. I have the whole concept down on what Circle geometry is all about. Remembering the names of the theorems is probably the only thing I have to get used to. Other than that, I think I'll finally be able to get this test down and get a well rounded mark from it. I hope it's the same with everyone else. Study hard, and Good luck!

Circle Geometry: Theorem Summary

2006-12-16T16:11:00.000-06:00

Okay, so I was responsible in making the first post dedicated to Circle Geometry and now I'm responsible in making the last post dedicated to Circle Geometry. Just like the first post, the following information is the "Dictionary Information."

The usual human brain can only remember at max 7 items, according to Mr. K. In remembering a phone number, we chunk the 7 digits into 2 groups so its easier to remember. Just like remembering a phone number, we chunk all the theorems into 4 categories.

next scribe is crysta! go crysta! i'm sure you'll find some new innovative way for learning just like what u did with that bubbleshare thingy! =)

BOB

2006-12-15T00:35:00.000-06:00

Well, I think that this unit is pretty much straight forward for me. I think that I can do better in this test if we only practice more or at least spend one more class using the theorems.The name of the theorems doesn't really stick on my mind so I devise a naming method myself. Hopefully we can all have an excellent mark in this test.
******BREAKING NEWS******
THe test is postponed FOR MONDay.
*********************************

BOB - Anatomy of Circles

2006-12-14T22:48:00.000-06:00

Well well well. Another test, another BOB. I find myself saying that in every BOB entry I write. Anatomy of Circles... well it hasn't been my favourite unit but more of a reason to look foward to our test tomorrow (so it'll be the end of it). I don't know why but I guess this whole theory of circles is something I find hard to understand and process in my brain by myself. We'd be in class and Mr. K goes over it step by step and it seems to be okay. I understand the rules and different theoroms when I'm in class. However when I get home and is presented a question- I sometimes dont even know where to start. I guess it is the fact there is a broad number of ways to solve any given question. Well, hopefully we all pull through the test tomorrow. As for myself I don't expect myself to do incredibly well :(, since it is one of my weaker units.

WELL GOOD LUCK AND GOOD NIGHT :)

11 days until christmas!

BOB - circle geometry

2006-12-14T22:37:00.000-06:00

Yet another unit test. I was supposed to have two tests tomorrow. One for pre cal, and one for chemistry. Luckily! my chemistry has been postponed to monday. haha. Well, this unit was really confusing for me. I found it difficult at first to prove everything. And i still haven't memorized all those theorems. :\ Well, lots of cramming tonight! Anyway, i'd like to become like, pro at this. so i'm going to keep practicing to keep myself familiar with it. Besides that everything is fine. I've spent my whole evening besides my brothers christmas concert on math, so it's time for my break. Happy holidays guys! And again, good luck tomorrow on the test. =)

Circle Geometry - Pretest

2006-12-14T22:35:00.001-06:00

Today we did our pre test in class for Circle Geometry. More or less i think the whole class found that the long answer question and the last multiple choice question were the hardest, so i'll explain those. =)

LONG ANSWER
Determine the measure of ےECB, ےBDC, ےBAD, and ےDBE, where E is the centre of the circle.

So first of all we always start with a table that has STATEMENT on one side and REASON on the other.

It's obvious at this point that BE and CE are radii, therefor triangle BCE is an isosceles triangle. This also means that ےEBC and ے BCE are the same.

NOTE: CLICK ON BUBBLESHARE BOX TO SEE FULL EFFECTS. =)

We now know the exact amount of ےBCD and according to cyclic quadrilateral theorem ے BCD's and ےBAD's sum should equal 180°. So in order to find ےBAD, we subtract 30° and 40° from 180° and come up with 110°. Therefore ےBAD is 110°.

Another thing we looked at were the two triangles that are subtended by the same arc; triangle BCE and triangle BCD. Since all of us know of the central inscribed angle theorem ( AKA the star trek theorem ) we can easily point out that ےBDC is half the amount of ےBEC. As a result ےBDC = 100°/2 which equals 50°.

We only have one angle left! Angle DBE! So if we look closely at triangle BCD you'll realize that we only have one angle left to find, ےDBE. We also know that every the angles of a triangle sum up to be 180° and by subtracting every angle that we have we'll be able to come up with the value of ےDBE! :D

Which in this case in 20°! =)

Once you've found your information and proofs, your table (including "Statement" and "reason") should look something like this:

MULTIPLE CHOICE
In the diagram below, R, S, T, U are points on the circle with RS = RU = UT and SU = ST. Find the measure of ےUST.
a) 31°
b) 35°
c) 36°
d) 40°
e) none of these

The first thing you should always do is mark down what is given to you. Have a wishful thinking colour, a colour used for given facts and a colour used for things you have proven! ( it comes in handy :D )
I'll use purple as the facts that are given, blue for what we're looking for and orange for the facts i prove. =)

There's no particular order to do this in, but what i looked at first was the isosceles triangles. We can easily point out that ےRSU and ےRUS are the same and ےSUT = ےSTU.

Now if we look at RS, RU and UT they are all the same length. Which means that the angles subtended by this chord length will all be the same. SOOO.. ےRST = ےRUT = ےUST.

In order to find the exact degree that angle is we make three equations. (this is like going back to substitution in linear equations). Here are the three equations (let ےSRT = w):
1. w + z = 180
2. 3y + z = 180 --> z = 180 - 3y
3. w + 2y = 180 --> w = 180 - 2y

If we substitude 2 and 3 into 1, we can find out what Y is, which is ےUST.
180 - 2y + 180 - 3y = 180
180 + 180 - 180 = 2y + 3y
180 = 5y
36 = y

And there you have it!
c) 36° is the correct answer. =)

That's all for tonight! Sorry for the late entry, i had to go and watch my brother's christmas concert! haha. =) Well, no homework tonight, just study! Test tomorrow. Good luck guys!

HAPPY HOLIDAYS! =]

And the next scribe will bee..... -Zeph! Enjoy. =)

Blogging on blogging

2006-12-14T19:47:00.002-06:00

Well the circle geometry was pretty fun till we had to start proving what was true, and that I think was the only thing in this unit that is still a little fuzzy. I tried my hardest on this unit but I really need to find a trick to memorize that darn theories. This unit I think was definitely one of the better ones, but good luck to all on the test.

B0B Natnael

2006-12-14T18:48:00.000-06:00

Well, this unit as Cherieeee said was quite easy, and the thing that made it easy, i think, is that its pretty straight forward. I mean, right now we now whats on the test (at least the layout).

I think everyone will do very good on this test...and some excellent(its not u Lauressa..jk)

goodluck on the test everyone! ;D

BEEE - OHHH - BEEE (.. spells BOB)

2006-12-14T18:36:00.000-06:00

Ahhhhhh! Another unit test!
Oh well, nothing to worry about. I honestly think that this unit, for me, was easier to understand than the ones we've had in the past. There's just something about circles, triangles and geometry that make them easy to learn. There were times when I hated this unit because of all the PROVING we needed to do. But I guess, the more proving we did in class, the easier it seemed, and my hatred lessened. Today we had the pretest in class, and I don't think I did my best. Although, I did put in my first effort, and got about 70% of the questions right! "Sweeeeeet!" Anyway, good luck to everyone on the test tomorrow!

-- Cherrrieeeee

Test Prep Exercises - Questions 1, 2 and 8

2006-12-13T20:15:00.000-06:00

hey guys. Sandy and I had an agreement to switch days, if she chose me tomorrow.
So yes, its me EEDCE.

(1) In the diagram below, AB and CB are tangents to a circle with radius 10. If angle AB find x, the shortest distance from B to the circle.

cos50°=10/h
h = 10/cos50
h = 15.557units

x = h - 10
x = 15.557 - 10
x = 5.557 units

(2) Given: Circle with centre C
- Angle 1 = Angle 2
Prove AB=CD

statement proof
Angle 1 = Angle B given
BC ~= CD ~= AC radii
Triangle ABC is isoscles from above
Triangle ACD is isoscles from above

Therefore, AB cannot be congruent to CD.

8. Complete the proof:
Diagram clarification: E, D, C are collinear
Given: EA is tangent to the circle at A
AB EC

Statement Proof
AD // BC given
EA is tangent at A given

Angle 8 = Angle 7 Tangent-Chord Theorem
Angle 7 = Angle 5 Alternate Angles
Angle 6 = Angle 8 - Angle 9 Tangent-Chord Theorem
Angle 1 = Angle 2 Triangle ABC ~ Triangle AEC

Okay. thats it folks.
homework is to finish the rest of this pre-test if not yet done so.

SANDY will take it away tomorrow.

Circle geometry

2006-12-13T00:02:00.001-06:00

Salutations fellow math students,

Today, for all you naughty kids that skipped (haha jk), we worked on five different circle problems that Mr. K had put up for the class to do. They were:

1) A chord that is 10cm long is 12 cm from the centre of a circle. Find the radius.

2) Line AB and line CD are two parallel chords in a circle. Line AB is 10cm and Line CD is 14 cm. Find the radius if the chords are 3cm apart.

3. a) O is the centre of each circle. Find the measures of the indicated angles. Justify your answer.

3. b)

4) Given:
-Line AC is tangent at B
- Angle 1 is congruent to angle 2

Prove: Line BD // Line FC

So yeah that's what happened in class. Homework was posted up on the blog by our very own math teacher, Mr. K. And I choose........................................................................................SANDY cheeks to be our scribe for tomorrow's class.

BOB for the Circle Geometry Test

2006-12-12T21:29:00.000-06:00

Hi everybody and in a few days it will be our circle geometry test and I'm almost ready. All I need to do is to study and get down all the stuff I miss about 2-3 weeks ago. I need to practice the recent work which is to prove somthing using the given info in a question which is not that bad but since it's new it doesnt hurt to do stuff. I need to get my theorems straight and finish the new review. That's all and good luck on the pre-test and test. =)

Circle Geometry Test Practice

2006-12-12T15:32:00.000-06:00

As promised, here are 8 problems to help you get ready for the test.

Remember, as I said in class, if you start working on one of these questions and get the horrible feeling in the pit of your stomach ... STOP. Go on to the next problem and try it out. If that happens with all 8 problems take a break and come back to them later ... lather. Rinse. Repeat. ;-)

We'll go over all the questions in class tomorrow.

Math Dictionary, Circle Geometry

2006-12-11T18:46:00.000-06:00

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I will add the last bits of info. left in the dictionary later...for now, the next scribe will be Samus.

Investigation 10. Angle sum of Polygons

2006-12-11T16:33:00.000-06:00

Did you know?

2006-12-10T09:51:00.001-06:00

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?

Circle Geometry - Investigation 8

2006-12-06T19:09:00.000-06:00

Good evening everyone. Today Mr. K talked to us about our flickr photo assignment at the beginning of class. The due date is now on Monday. Just a little reminder for you all.

We did investigation 8 today. Here were the instructions.

1.

(a) Construct a chord AB in the circle.
(b) Construct a tangent, RS, to the Circle at A. The tangent should not be perpendicular to AB.
(c) Measure the angle of RAB.

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You can plot down the points anywhere you'd like. Just be sure that your chord and the tangent line aren't perpendicular. I measured angle RAB and ended up with 35° with mine.

(d) Construct a point, C, on the circle, on the opposite side of chord AB from angle RAB.
Some of us where have a few problems plotting down C. We did plot C opposite of the chord AB, but had it plotted down inside the circle. The question asks to plot it on the circle, meaning on the circumference. What we did was plot it down in the circle which was different from what was being asked. So, just look out for those small things and read the question carefully.

(e) Construct and measure inscribed angle ABC. What do you notice?
From looking at the slide below, you will notice that angle ACB is the same as RAB.

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(f) Measure angle SAB.

As you can see, angle SAB measures 145°.

(g) Construct a point, D, on the circle, on the opposite side of the chord AB from angle SAB.
(h) Construct and measure inscribed angle ADB. What do you notice?

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Angle SAB and ADB are the same.

Here is sort of a final drawing on all the angles that were measured.

THIS HERE IS CALLED THE TANGENT-CHORD THEOREM.

(i) Can you articulate a general rule for what you have found? Your sentence should begin:
"If an angle is formed between a tangent line and a chord, then...."

The rule here is simply that "If an angle is formed between a tangent line and a chord, then the inscribed angle subtended by the opposite side of the chord is congruent."

RULE
IF AN ANGLE IS FORMED BETWEEN A TANGENT LINE AND A CHORD, THEN THE INSCRIBED ANGLE IS SUBTENDED BY THE OPPOSITE SIDE OF THE CHORD IS CONGRUENT.

And yes, I did mean to repeat that, and repeat it in a way that will be easily read. It's so that you can remember the rule better. Through repetition. Hopefully it works.

The HOMEWORK for tonight is to Re-do the questions we had trouble with in EX. 33. And also EX. 34.

REMINDER: There will be a quiz maybe tomorrow, or the next day after.

I chose the next scribe to be Natnael.

Good Night~

2006-12-04T20:07:00.000-06:00

Today's class we started off by handing in our projects. I hope everyone gets over 80% just like Mr. K said we should.

After that he showed us some examples of the 40s student's trigonometry pictures, so have a better clue of what to do with our quadratic function pictures. Make sure it has a lot of details a quadratic function should have.

*Reminder it should be uploaded on flickr by Thursday.

We went over some question in our exercise #32 which were # 10, 12, 13

#10 was a question that didn't really fit in with any unit we did so far but it was a fun question to solve.

#12 this question we had to find the ticket price that will yield maximum profit.
This is a good review for our Analytic Geometry unit.

revenue = (# ticket) (cost/ticket)
let x = 1 discount of $3.00

R (x) = (1000 + 100x) (60 - 3x)

roots @ x= -10 x= 20

We ended up with the price $45 per ticket.

#13

A) y= x and y = -1/2x+2

B) x=-1/2x+2
2x=x+4
3x=4
x=4/3

(4/3, 4/3)

C) 8/3

After going over some homework we started on our Investigation #7

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"If two tangents are drawn from a common point, exterior to a circle, then...the tangent lines are the same measure"

"If a radius intersects a tangent line, then...the angle is 90 degrees"

After all of that the next step is to go back and construct a new point S exterior to the circle and repeat steps b, c and d for point S.

Today's homework is exercise #33
Make sure to attempt every question. The ones you are really stuck on should be looked at in class.

And most of you may notice that the scribe list is now on our fourth cycle!! that's a lot of scribing!. I just wanted to say I'm really proud of every one's work. Look how far we came with our scribing, we're using a lot more tools to make our post explained to the best of our abilities.

Okay enough of that...the next scribe for Wednesday (Tuesday afternoon NO SCHOOL!) is... MERIAN!

GOOD NIGHT!

Circle Geometry!

2006-12-01T16:17:00.000-06:00

To start off today's class we all congratulated Crysta on her Fantastic! scribe post. Once again, great job Crysta! Mr. K was also trying to explain to us the meaning of how much one million (1000000) really is. If we broke it down to how long it would take to reach one million by receiving one penny each second. When we finished calculated it, it took 11.5 hours. Then we decided to calculate how long it would take to reach a billion (1000000000). We all found out that it would take 32 years to reach that. We also realized that we haven't even lived for a billion seconds yet AND one billion penny's will fit into 5 school buses. Quite interesting, don't you think? If you're wondering how Mr. K knew all this, he went to this website

http://www.kokogiak.com/megapenny/

Anyway, once we got back onto to track we discussed questions that we had trouble with from last nights homework and we did Investigation 5.

(a) Choose 6 different points on the circumference of the circle. Do not space them out evenly. Label these points A, B, C, D, E and F.
So this part is pretty straight forward. We start with a blank circle and add 6 points anywhere on the circumference.

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(b) Construct angles ACB, ADB and AEB.
For this part of the investigation, we're pretty much making chords, but these are to make angles.

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(c) Measure the angles formed at C, D and E.
So here we take the angles and measure them. It's best to write them down that way we can keep track of them. instead of just adding it to the picture.

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When I completed this step in class, i got this:
ACB = 13°
ADB = 13°
AEB = 13°
Notice a pattern?

(d) Construct angles EAF, EBF and ECF.
Here, we will repeat what we did for (b).

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(e) Measure the angles formed at A, B and C.
Here we'll do the same as (c).
EAF = 60°
EBF =60°
ECF =60°

(f) What do you notive about all these angles? Can you articulate a general rule for what you have found? Your sentence should begin:
"If two (or more) inscribed angles are subtended by the same arc, then..."
Have you seen the pattern to these? Well, If you have you've noticed that all the angles come out to equal the same. To complete the sentence can be said in different ways.
i) ... they have the same measure.
ii) ... are congruent
iii) or in your own words!

So there you have it! investigation 5! I don't know if it's just me but these things are super fun to do. haha. (:

Anyway. Homework for the weekend is:
1. Attempt to complete Investigation 6. (but if you get frustrated with it, then leave it be)
2. Exercise 31.
3. Project (due monday)
4. Picture of a parabola put onto Flickr. (due thursday)
REMEMBER TO DO YOUR HOMEWORK, GUYS. =)

As for the next scribe.. i think** the only person left is MELISSA! .. haha. if i'm wrong, correct me, otherwise, Have a great weekend!

2006-11-30T20:49:00.000-06:00

Okay so i didn't get to do my post when i was scribe last time cause my net wasn't being very nice. Today in class we did investigations 3 and 4.

Investigation 3:
1.(a) Use what you know about chords to find the centre of the circle. Erase all construction lines once you're done.

This is how you do question 1.(a) (If you do have any questions, this question is explained in the previous scribe):

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(b)Construct diameter AB.

Heres what you do:

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(c)Choose 4 different points on the circumference of the circle. Do not space them out evenly. Label these points C, D, E and F.

This questions is self-explanatory:

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(d)Construct chords AC and BC.
(e)Measure the angle formed at C

Draw lines from A to C and from B to C and measure the angle C:

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(f) Repeat steps (c) and (d) for the points labeled D, E and F.

It's pretty straight forward:

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(g) What do you notice about all these angles? Can you articulate a general rule for what you have found? Your sentence should begin:
"If an inscribed angle is subtended by a diameter then...."
If an inscribed angle is subtended by a diameter then.... the inscribed angle is always 90°.

So that was Investigation 3. We started Investigation 4 but did not get a chance to go over it in class. THE SCRIBE FOR THE WEEKEND IS .. SANDY !

Dictionary

2006-11-29T20:28:00.000-06:00

For those of us who missed out on our math dictionary I'll fill you in.
Well it's actually just the end of what we couldn't finish yesterday.

Being able to find the root sum and root product of f(x)=2x^2-6x+7

A=2 Root sum= -b/a Root product= c/a
B=-6
C=7 -(-6)/2 = 3 7/2

Find a quadratic function whose sum is -3 and whose product is 5/2

Root sum= -b/a Root product= c/a
-3/1 = -b/a 5/2= c/a

-6/2 = -b/a both must have a common denominator 5/2= c/a

a=2 f(x)=2x^2+6x+5
b=6
c=5

Circle Geometry

2006-11-29T18:28:00.000-06:00

Well where should I start. In class we were given our first assignment in class, it wasn't hard but it was relatively new to some of us. Investigation 1 and 2 were our given assignments and here are completed ones in case anyone didn't quite get it .

INVESTIGATION 1
A- Construct 4 different chords (a straight line joining two points on the circumference of a circle), AB ,CD, EF, and GH

But none of them should be diameters (longest possible chord in a circle that passes through the center) nor should they be parallel. The 4 different chords are in black

B- O is the center of the circle. Construct lines from the center that are perpendicular to each of the chords. AB, CD, EF and GH. Label the new points where the new lines meets W, X, Y and Z , The perpendicular lines will be in red . The perpendicular lines are also known as bisectors, because they cut the chord into two.

C- Measure the lengths of the AW and BW. Do so with the rest so every chord has two measurements on either side of the perpendicular line

D- Now after your picture is drawn you should see a relationship between a chord and their perpendicular line.

A LITTLE HELP: "A line perpendicular to a chord, that passes through the center of a circle, the chord."

Here is an example of what your drawing should somewhat look like

INVESTIGATION 2
A- Like the A in Investigation 1 construct 4 different chords AB, CD, EF and GH

Remember none of them should be parallel to each other nor be diameters

B- Construct perpendicular bisectors for each chord. Each bisector should be a chord.

KEY: the way to find the bisector is to measure the chord, and find the midpoint. At that midpoint line up a triangle from your geometry set to make a 90 degree angle and draw the perpendicular.

C- What do all the bisectors have in common? Well once your done they should all intersect at the same point.
D- What can you say about any perpendicular? You could say that the perpendicular bisector of any chord will always dissect the center of the circle.

This diagram should give you an idea of how it should look

Example: chord AB. Length of chord divided by two = point where bisector should cross

To complement our learning we also learned some definitions to go along with our new unit
Bisect- to cut into two exact equal halves
Subtend(subtended)- to hold arms of an angle open by an arc

Well that's it thats all folks
Oh yea before I forget

Share the kindness and leave some kind comments on our friends blog in Scotland
If you forgot the address its exc-el.org.uk/blogs/s3srcibeposts, I checked it out and its pretty neat!

Ex 30 for homework.
NEXT will be ......................Crysta!

Working Harder or More Effectively?

2006-11-29T17:40:00.000-06:00

Greetings Pre-Cal 30S! From your BOBs, it appears you are really working hard and feeling that to be more successful, you need to "practice more", "study more", and "need more practice". My sense is you may be practicing and studying now; I wonder if you think you could be more effective in your efforts? I know from lots of experience (60 years worth) that habits can ensure my reaching my goals or not. And often, I've studied more, and practice, but not really effectively. Given that, I think I've found 7 habits, that if I've worked at them, have helped me be more effective! Have you seen these 3? Do you think these habits can be of value to you too? Or are these a habit for you already?

--------------------------------------------

I think this short autobiography is a good introduction to Habit 1.

There's A Hole In My Sidewalk: AUTOBIOGRAPHY IN FIVE SHORT CHAPTERS

by Portia Nelson
Chapter 1
I walk down the street.
There is a deep hole in the sidewalk.
I fall in.
I am lost.... I am hopeless.
It isn't my fault.
It takes forever to find a way out.
Chapter 2
I walk down the same street.
There is a deep hole in the sidewalk.
I pretend I don't see it.
I fall in again.
I can't believe I am in the same place.
But, it isn't my fault.
It still takes a long time to get out.
Chapter 3
I walk down the same street.
There is a deep hole in the sidewalk.
I see it is there.
I still fall in.... It's a habit.
My eyes are open.
I know where I am.
It is my fault.
I get out immediately.
Chapter 4
I walk down the same street.
There is a deep hole in the sidewalk.
I walk around it.
Chapter 5
I walk down another street.

Habit 1: Be Proactive®
Take responsibility for your life.

--------------------------------------------------------------

Does this cartoon say it all for Habit 2?

Habit 2: Begin with the end in mind.®
Define your mission and goals in life..

-----------------------------------------

Finally, Habit 3: Rocks, Pebbles, Sand, Water—And Calculus is which?

A time management specialist was asked to give a presentation on her specialty. She decided to do a demonstration. First she asked her assistants to bring a big bucket and put it on the table in front of the audience. Then she asked for large, grapefruit-sized rocks and filled the bucket with them.
"Is the bucket full," she asked?
"Yes!" said the crowd, but she asked for more to put in anyway. This time her assistants brought in pebbles. She poured the pebbles in the bucket and it held a surprising number in the space between the big rocks.
"Now is the bucket full?" she asked.
"Yes!" "No!" "Yes!" "No!" said various persons in the crowd. Some people were uncertain; some were getting suspicious. The time management specialist asked for more. This time the assistants brought her sand. She poured sand in the bucket and it filled the spaces between the pebbles.
"Now is the bucket full?" she asked.
"No!" they answered. By now, everyone was suspicious. So she asked for water and poured in quite a lot. Now no one could think of anything else that could fit in that bucket.
"What does this process demonstrate?" asked the time management specialist.
One member of the audience spoke up: "No matter how busy you are, you can always fit in one more thing."
"I can see how you might think that was my point, but it is not," said the specialist. "I was trying to show you that if you don't put the big rocks in first, you'll never get them in at all!"

Habit 3: Put First Things First®
Prioritize and do the most important things first.

What do you think?
Best,
Lani

And by the way, these come from 7 Habits of Highly Effective Teens by Sean Covey.

Zeph's Notebook/Dictionary: Algebra

2006-11-29T03:49:00.000-06:00

just to make things a bit organized, this scribe post is divided into two post. one section will be the circle geometry notes we took and the other post will be the algebra notes. this is so that this scribe post doesnt end up having the labels, circle geometry and algebra, at the same time.

Note: there is still more algebra notes to come from the next scribe!

REMINDERS:

~ FLICKR/NUMBER ASSIGNMENT - GET MORE PICTURES!! TOTAL OF 5 ALTOGETHER
~ "LOST IN MATHLAND" ASSIGNMENT DUE THE DAY AFTER TOMORROW!!
~ NO HOMEWORK
~ LAURESSA'S SCRIBE
~ GEOMETRY SET

Zeph's Notebook/Dictionary: Circle Geometry

2006-11-29T03:31:00.000-06:00

just to make things a bit organized, this scribe post is divided into two post. one section will be the circle geometry notes we took and the other post will be the algebra notes. this is so that this scribe post doesnt end up having the labels, circle geometry and algebra, at the same time.

CIRCLE GEOMETRY

Note that the small "o" stands for "degrees" not to be confused with exponent zero...

REMINDERS:

~ FLICKR/NUMBER ASSIGNMENT - GET MORE PICTURES!! TOTAL OF 5 ALTOGETHER
~ "LOST IN MATHLAND" ASSIGNMENT DUE THE DAY AFTER TOMORROW!!
~ NO HOMEWORK
~ LAURESSA'S SCRIBE
~ GEOMETRY SET

Setting A Circular Mood ...

2006-11-28T16:40:00.000-06:00

We began our study of circle geometry today. I thought this might be a good way to "set the mood" for this next unit of study... ;-)

And another test finished!

2006-11-27T18:30:00.000-06:00

And yet another test accomplished!
Crysta set me to scribe on Thursday/Friday, but there was just a pre-test.

HOWEVER,
Today was the day we had our Analytic Geometry unit test. How do you think you guys did? I'm positive to say I did my best on the whole test, except my mind fell apart on the last two questions :

Anyway,
There isn't much to write for today's class, but I sure hope everyone did well on it, despite the difficulties of the questions.

A few reminders:
1. FLICKR ASSIGNMENT -
In class we discussed the assignments, and the pictures involved with it. According to Mr. K, Flickr will not 'activate' (for lack of a better word) our accounts unless we have a minimum of five (5) pictures. So remember to take four more pictures (not of anyone's faces, thanks), of random things, and upload it into your flickr account.

2. MATH PROJECT #1
Don't procrastinate! Get started on your math project now! Reminder: Due Friday, December 1^st, 2006. (Oh wow, DECEMBER ALREADY?!?!?!) You have a choice of working by yourself, or with a partner. No outside help, but from books, the internet, or Mr. K! AND also remember that the minimum and expected mark we should all get is an 88%. Let's make them good!

I hope this is a good scribe for today's class, and helps people in reminders. I choose the next scribe to be ....

" -Zeph "

Late BOB

2006-11-26T21:38:00.000-06:00

Hey guys,
well, this unit, i guess, have been tough on you guys, and I really hope u guys be fine on Monday. But for me, this unit, have been really easy, and I think I'm doing fine in this unit. One of the reasons, I think, is that at first I tried to get the concept...and i was so interested in the 'dimension' stuff. So, I tried to get the main idea, and once i got it...i dunno, but it wasn't that hard.

good night

BOB - Analytic Geometry

2006-11-26T20:28:00.000-06:00

I found this unit to be quite fun with some challenging points to it too. Like people have mentioned already on their BOB's, I tend to make the smallest errors which really screws things up. I hope I'll be aware of the mistakes I make on the test so that I could correct them. I think I'll need to do more reviewing as well so that I can really get the concept down.

BOB - Analytic Geometry

2006-11-26T19:48:00.000-06:00

Well, for myself I found this unit of "Analytic Geometry" much like a bad shopping trip. There you are, searching and searching for the perfect outfit, but no success. Although it only takes awhile to enter that one store and see that perfect top you've been looking for hanging on display. Finding that perfect top would be when you finally find the equation(s) thats being asked to solve. Understanding what the questions being asked for is the hard part, but once you get that resolved and figuire out the equation(s).. its a breeze from there, much how its a breeze to just pick up the top and walk to the cashier. Although I realized you have to be very aware while solving. Be aware of little errors you can make which will screw the whole process up. Much like shopping, be aware of the price and be sure you can actually afford the item.
- Other then that. GOOD LUCK AND GOOD NIGHT.

don't mind my metaphoric BOB entry, we just finished poetry in english =) Haha.

BOB!

2006-11-25T15:21:00.000-06:00

Another unit done. Analytic Geometry was a hard, yet fun unit. The distance between a circle and a line, I'm pretty sure I have that down. I'm still having trouble with the linear equations. The word problems are difficult also, but I always seem to mess up on the smallest things. Well this weekend I'll be studying for this test. I need to practice more.

'so good luck on the test! :)

BOB v.4 An Analytical Analysis

2006-11-25T12:00:00.000-06:00

well that was weird... I had to re-register back to blogger.com but anyways i got my account back and i'm still registered at this blog. so thats good news for me!

i wont be coming to school on monday cuz i found out that i have a viral infection. all the clinics were closed and my appointment was scheduled til the second week of december. i'd probably be dead by then. haha. anyways, all the clinics that my parents suggested on me going to were all booked and on friday, my parents sent me to the hospital. there were so many people at the hospital. you have to die to get your way in...literally. what i meant is that there were so many patients waiting in line that one of them already died. the paramedics were there and the mother was yelling "I WANT MY SON BACK!" and was freaking out. Well i was very lucky because we found a clinic near the hospital that wasnt all booked and the doctor said that i have a viral infection. the medicine that he prescribed to me had strange side-effects such as nausea, headaches, bad dreams, and excessive sleep & sleepiness. b4 u no it, i'll be back at school and i'll be alrite.

in analytic geometry, word problems are still an obstacle thats just in the way, but once i find the equation everything seems to be straightforward. solving the system of equations are easy for me but one mistake anywhere in the long process can equal death. graphing the equations is the topic that i needed to focus more on and other than that, analytic geometry seemed ok.

~ zeph

BOB

2006-11-25T08:46:00.000-06:00

Okay, so this is my BOB for the unit analytic geometry. I think out of all units this is the one that i found the hardest. I missed one day of class and i was so confused for the following classes. I think I've finally gotten the hang of it and realize what is asked in the question. I'll just have to study some more over this weekend to be completely ready for the test on Monday. Other than that, i think everything is okay.

Hope you guys are having a great long weekend! =)

BOB for the Analytic Geometry Test

2006-11-22T17:01:00.000-06:00

Hi everyone and this is my BOB. I get some of the stuff on the unit of analytic geometry but I need a bit more practice especially on the inequality questions. I seem to screw up a lot so I'm guessing I need a bit more work. I'll just study to get the other stuff into my head. About the multiplying with the fingers blog was cool. It works and it's kind of fun doing it also.

It's All In Your Hands

2006-11-22T15:58:00.000-06:00

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

Don't forget about your flickr assignment.

Cheers!
Mr. K.

blogging on blobbing

2006-11-22T11:26:00.001-06:00

I think that this unit about analytic geometry is difficult for me. This unit started off really easy with all the equation of the circle and distance between a point and a line. Then it got a little bit confusing with the systems of linear equation with three unknowns and word problems. I think that I'm average at everything but struggles at doing systems of linear equation with three unknowns and word problems. I wish that we can do a little bit more of word problems. I hope that everyone do well for the test on Monday. That's it for my BOB.

B-O-B SPELLS BOB!

2006-11-22T06:56:00.000-06:00

So another unit done, and another test to come. (Ooh, that rhymed!).
In this unit, I was struggling quite a bit in the beginning. Finding distance between two points is simple, but sometimes when it comes to distance from a line to a point, depending on the question, I have a little difficulty. However, I have been working on review questions from past exercises and I think I finally got the hang of it! Well we won't know for sure until the test! Aw, and I miss the pretest today, because of jazzfest : (. But I hope everyone does well on this pretest! See ya guys on Thursday!
♥ ♥ Cherrieeeee

B-B-B-O-B

2006-11-21T23:59:00.000-06:00

The unit went by pretty quick. I understand the point to line stuff. I had little difficulty, the only thing that messes me up sometimes would be the number line because I forget what I'm looking for and if its true or false. That is all i have to say about this unit. I hope everyone does well!

-SAMUS

2006-11-21T16:40:00.000-06:00

Hey everyone, I'll be scribing for both today and yesterday's class.

Yesterday we had a what seemed like a shorter class, because Mr. K had to leave early.

In the beginning of class Mr. K talked about a new project we should start on. Which involves our favorite number, and why it's so special.

* Dont forget to sign up to flickr.com so we can post up our pictures to the tag PC30SF06

We had a review sheet that day on "Quadratic & Rational Inequalities"

Some of us had a hard time solving the questions, so Sam found our good friend Mr.Clark. He came and helped us with question 1 (h).

2x² -7x +3 >= 0
(2x-1)(x-3) >= 0
e.p. (3,1/2)

Solution
(-oo, 1/2] U [3, oo)

He also did another example

(x+2) (x-1) / x+3 <= 0
e.p. (-2,1,-3) Nothing else really happened in that class so we continued to work on our review sheet.

~~~~~~~

Today's class we worked on the question Mr K gave us.
Which were: Slove
1. a) x²+3x >= 18
b) 2x+5/x+1 <= x+1/x-1
c) |3-4x| > 9
d) |5x-8| <= 2

2. There are $950 in $5, $10, and $20 bills in the cash regester. There are a total of 71 bills. The number of $20 bills is 7 less than the total number of $5 and $10 bills. How many bills of each denomination are in the cash regester.

We had time to work on them and corrected it with the class.

a) x²+3x >= 18
x=3x-18 >= 0
(x-3)(x+6) >= 0

Solution
(-oo,-6] U [3,oo)

b) 2x+5/x+1 <= x+1/x-1
x can not = -1 , 1
we multiply each side by (x+1) (x-1) then we'll get.
2x² + 3x-5 <= x²+2x +1
x² + x -6 <= 0 (x+3) (x+2) <= 0

Solution
[-3, -1) U (-1, 1) U (1, 2]

c) |3-4x| > 9

d) |5x-8| <= 2

2) # 5 bills = F
# 10 bills = N
# 20 bills = W

equations
F+ N+ W = 71
5F + 10N + 20W = 925
-F - N + W = -7

There are 21 $5 bills, 18 $10 bills and 32 $20 bills.

Don't forget our pre test is tomorrow and our test for this unit will be on Monday.

Tomorrows scribe isss........ Crysta!! have fun!!

** Don't forget to do the review down there\/

Analytic Geometry Review and Reminders ...

2006-11-21T15:34:00.000-06:00

Your first project was handed out in class today. It is due next Friday, December 1. You can use any presentation format you like. Mr. Harbeck and I have started collecting a list of free online tools you can use any time you have a project to do. You can find it here.

Don't forget your first flickr assignments are due on Friday. (Take a picture of an "interesting" (to you) number using your cell phone or digital camera. Upload it to flickr. Include a description of why you chose that number and tag it pc30sf06.)

Below is the review sheet we talked about in class today ... try to finish it for class tomorrow. The test is next Monday.

Solving Inequalitie Equtions & A Splash of Review

2006-11-17T21:40:00.000-06:00

Well well well. Another friday scibe post!
Okay. So lets get this scribe going.

Minutes after the 5th period bell, we all walk into our Pre-Cal class, as we do everyday and by our dismay we spot Mr. K holding a small stalk of papers. Since we're all "smarter then the average bear", we immediately knew a lovely quiz was coming in our direction. So after a brief discussion of wither or not we should include quizes for marks, we all got to work. The quiz was nothing new; a nicely summed up quiz about everything we learned so far in 'Analytic Geomtry'.

After self marking our peer's quizes, Mr. K placed the following questions on the board.

S O L V E:

Question 1:
x²+x<6

Okay, so we haven't seen much of these before and you may be a little puzzled.
BUT- think back to the Mathmatician who set the house on fire
"Turn the problem into something we already know how to solve"

First and foremost- Ignore the "<" for now, think of it as an "="
x²+x=6

And if we move the 6 to the opposite side..
x²+x-6=0

Why what do we have here? A parabola! Not too terrifying is it?

Okay so lets place the "<" back.
x²+x-6<0

Before actually solving this equation, it would help to understand what infact the question is asking. Well what we're trying to do is-
"Find the values of X where this parabola will have negative values"

So first we must factor the equation we came up with
x²+x-6<0
(x+3)(x-2)<0

Once we get the two roots, you must draw the line graph (it will be included for marks for the final answer). Finally pick any point in each of the 3 regions and sub it into the equation.

Answer: (-3, 2)
Remember! we use round brackets "( )" because we know the -3 and 2 are not included because the original equation did not include "≤" (less then AND equals to)

Question 2:
2x-3=5

Remember! The values in the absolute signs "x" can be both negative and positive, leaving us with two cases.

Case #1
2x-3=-5
2x=-2
x=-1 check: it works!

Case #2
2x-3=5
2x=8
x=4 check: it works!

Okay that was simple enough, BUT- what if we had..
2x-2>5

Yikes! Do not fear, remember- ignore the "<" and treat it as an "=" for now.

So we'd continue to solve just like before-
Case #1
2x-3>5
2x>8
x>4

Case #2
2x-3>-5
2x>-2
x>-1

Answer: (-∞,1)U(4,∞)
"U" - Union, that group together with this group

To be finished soon, very soon.
Sorry folks!

Homework assignment: All the questions you didn't get done on the past exercises (up to Exercise 28).
Mondays scribe will be Mel!isa

.
.
Enjoy your weekend!

Solving a System of Linear Inequalities

2006-11-16T20:18:00.000-06:00

Sorry for the late post. Well, here's my scribe.

Solve the Following Linear Equation Graphically

Solve the Following Linear Inequalities
Hint: Answer is always graphically

Answer is the region highlighted ({), since its where the two regions meet.

Notice: that the lines (y < -x +3 , and y > 2x - 3) are dotted. This is because in linear inequalities a point(s) can only be in 3 areas. These are: BELOW the line, ABOVE the line, OR ON the line. In this case "y < -x +3" is below, and "y > 2x -3" is above.

Notice: that the point where the two lines meet is an open circle. This is because the two regions can go infinitely close to that point but they will never intersect at that point.

WHAT IF Y IS ALSO EQUAL TO THE LINE?

This Concept can be applied to any other shape (Circles, parabola, radical, etc.). Lets see Parabolas.

in Parabolas the region is either INSIDE, OUTSIDE or ON the parabola.
Example:
y <>

ONE MORE THING TO REMEMBER

When solving any inequality equation, if you divide/multiply by a negative, don't forget to change the direction of the inequality sign (<,>).

Example. When Solving 2x - y < 3
-y < -2x + 3 Divide both sides by a NEGATIVE.
y > 2x - 3

using the point (2,2)
Using the correct equation.
y > 2x -3
2 > 4 - 3 TRUE

BUT IF WE USE THE WRONG EQUATION.
y < 2x -3
2 < 4 - 3 FALSE Since 2 is greater, not less than 1

FINALLY...THE NEXT SCRIBE WILL BE me...jk....Its eedce. Have fun!

Applications with Systems of Linear Equations

2006-11-15T16:19:00.000-06:00

Hi it's me ben again. Today in class Mr.K created a new rule to get people to participate in class and that's to pair up and let one person speak for the other person. Okay back to the math. Today we started with some Systems Of Linear Equations With Three Variable questions for practice.

a+2b+c=0
3a-b-2c=11
2a+b-c=3

3a+3b=3
5a+3b=11
2a=8
a=4

3(4)+3b=3
12+3b=3
3b=-9
b=-3

4+2(-3)+c=0
4-6+c=0
-2+c=0
c=2

EXPLANATION:
L₁
L₂
L₃
L₄=L₁+L₃
L₅=2L₁+L₂
L₆=L₅-L₄
L₇ Solve L₆
L₈ Solve L₄ w/ a=4
L₉ Solve L₈
L₁₀ Solve L₁ w/ a=4, b=-3
L₁₁ Solve L₁₀

Lines intersect @(4, -3, 2)

After a little work on that question we had to apply our Linear Equation skills we just learned and use them in word problems.

1.When 3 numbers are added in pairs the sums of the pairs are 28, 39, 45. What are the three number?
2Mark sells wildlife photos at the county fair, his revenue, R dollars is determined by the price, p dollars, that he charges for each picture as shown in the equation:
R(p)=ap²+bp+c
At a price of $40, his revenue is $2900
At a price of $50, his revenue is $3000
At a price of $70, his revenue is only $2600
Find the values of a,b, and c

Okay for #1 we need to find out the equation. So it states that the numbers are add in pairs so we need the possible combinations for the 3 equations and they are shown below then now solve.
a+b=28
a+c=39
b+c=45

-b+c=17
2c=62
c=31

b+31=45
b=14

a+31=39
a=8

EXPLANATION:
L₁
L₂
L₃
L₄=L₂-L₁
L₅+L₄
L₆ Solve L₅
L₇ Solve L₃ w/ c=31
L₈ Solve L₂ w/ c+31

Numbers are 8, 14, 31

For #2 we need to get another equation for each price and revenue. Notice that the price is p so just sub the price of each set of pictures and same with the revenues and you should get this:
a(40)²+40b+c=2900
a(50)²+50b+c=3000
a(70)²+70b+c=2600

Now just simplify and Solve
1600a+40b+c=2900
2500a+50b+c=3000
4900a+70b+c=2600

2400a+20b=-400
900a+10b=100
600a=-600
a=-1

900(-1)+10b=100
-900+10b=100
10b=1000
b=100

-1600+4000+c=2900
4000+c=4500
c=500

EXPLANATION:
L₁
L₂
L₃
L₄=L₃-₂
L₅=L₂-L₁
L₆=L₄-2L₅
L₇ Solve L₆
L₈ Solve L₅ w/ a=-1
L₉ Solve L₁ w/ a=-1, b=100

Therefore a=-1, b=100, and c=500

That's what we did today on Applications with Systems of Linear Equations and homework for tonight is EXERCISE #27 *But don't do #'s 1-5*
The next scribe will be M@RK and good night.

Systems of Linear Equations in Three Variables

2006-11-14T21:17:00.000-06:00

Salutations,
Today in class, (for all you naughty kids that skipped, haha JK) we worked on equations with three variables. We did work on equations with two variables too. Well, here are the questions that we worked on:

3x+5y= 17
4x-y= -8
23x= -23
x= -1

4(-1)-y= -8
(-4)-y= -8
y= 4

EXPLAINATION:
L1
L2
L3= L1+5L2
L4 solve L3

L5 solve L2 w/ x= -1
L6 solve L6
Therefore lines intersect at (-1, 4)

4x+3y= 1
3x+2y= 2

-x= -4
x=4

3(4)+2y= 2
12+ 2y= 2
2y= -10
y= -5

EXPLAINATION:
L1
L2
L3= 2L1-3L2
L4 Solve L3
L5 solve L2 w/ x=4
L6 solve L5
Therefore, lines intersect at (4, -5)

-x+y+2z= -3
2x-y+ z= -7
x+2y-z= 7

3x+y= 0
-5x+3y=11
14x= -11
x= -11/14

3(-11/14)+y= 0
y= 33/14

2(-11/14)-(33/14)+z= 98/14
-22/14-33/14+z= 98/14
z= 153/14

EXPLAINATION:
L1
L2
L3= L1+5L2
L4 solve L3
L5 solve L2 w/ x= -11/14
L6 solve L5
L7 found x
L8 solve L4 for y w/ x= -11/14
L9 solve L8
L10 solve L2 for z with x=-11/14 and y= 33/14
L11 solve L10
L12 found z
Therefore lines interesect at (-11/14, 33/14, 153/14)

2x+y-3z= 9
x+y-2z= 7
zx-y+z= -7

3x-z= 0
4x-2z= 2
2x= -2
x= -1

3(-1)- z= 0
-3= z

x(-1)+y-2(3)= 7
-1+y+6= 7
y= -2

EXPLAINATION:
L1
L2
L3
L4= L2+L3
L5= L1+L3
L6= 2L4-L5
L7 solve L6
L8 solve L4 / x= -1
L9 solve L2 w/ x= -1 and z= -3
Therefore, the solution is (-1, 2, -3)

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

(1, 2, 3)
Are these solutions to that equation?

To solve this equation, all you have to do is sub in the numbers 1 for x, 2 for y, and 3 for z! Then, ta-dah there you have it.

3(1)-2(2)+4(3)= 11
2(1)+3(2)-3= 5
1+4(2)-3(3)= 3

As you can see, both L1 and L2 work out. However, the third line does not. Zero does not equal three, well last time I checked.

And, Ta-dah. That's what happened in math class today. Our homework for tonight is the numbers that were omitted from yesterday and number 26. Once again, our homework for tonight is numbers 1 and 2 from 25 and all of 26! Before I leave, I hold in my hand, the name of who will be the next scribe! The next scribe is.................................BENOFSCHOOL! Well, see you tomorrow everyone!

-SAMUS

Systems of Linear Equations in Two Variables

2006-11-13T23:14:00.000-06:00

Ok, today in class we did something like the Friday's class. We practiced doing the different techniques on how to solve systems of linear equations in two variables.

We were given three systems to solve:

x=y+6
3x-5y=7

3x+2y=16
y+7x=19

2x+3y=4
5x+2y=3

Ok, lets do the first one. This first question is best suited to be answered using the substitution technique, because as you can see there is a equivalent for the x value.

x=y+6
3x-5y=7
3(y+6)-5y=7
3y+18y-5y=7
-2y=-11
y=11/2
x=11/2+6
x=11/2+12/2
x=23/2

Therefore, these lines intersect at (23/2,11/2).

Now, lets do the second one.This certain question is best suited to be answered using the elimination technique , because there is no equivalent on any of the variables unlike the first one.

3x+2y=16
y+7x=19

3x+2y=16
7x+y=19
-11x=-22
x=2
7(2)+y=19
y=5

Therefore,these lines intersect at (2,5).

Lastly, before we do some word questions. This question is suited to be answered using the elimination technique just like the second one.

2x+3y=4
5x+2y=3
-11x=-1
x=1/11
3y=4-2x
3y=44/11-2/11
3y=42/11
y=14/11

Here we go again with this word questions....

125 tickets were sold at a basketball game. A total of $337.50 was collected. Students paid $2.50/ticket and other people paid $3.50. How many students attended the game?

Let s=# of student's tickets
p=#of other people's tickets

2.50s+3.50p=337.50
s + p =125
-s =-100
s=100

Therefore , there are 100 student tickets and 25 other people's tickets.

After doing all the warm up things Mr. K started talking about consistent and inconsistent lines.
A consistent line means that there is an answer.It can be break up into a dependant and an independant line. Independant means that there is only one solution. Dependant means that there is an infinite amount of solutions. An inconsistent line means that there is no solution.

I'll be editing this tomorrow because the html codes for line 1 and line 2 doesn't seem to work.

The scribe for tomorrow is ............... "SAMUS" .

Sorry SAMUS ,but you're the one that pop into my hEAD first. I was gonna pick Lauressa but she said her computer wasn't working...

Homework for tonight is Exercise 25 excluding questions number 1 and 2.

Solving a System of Linear Equations

2006-11-12T14:39:00.000-06:00

HEY! I'll just cut through the intro and get this scribe started.

On Friday's class we sort of started something new, but we still used some previous knowledge in order to work with the following questions.

We were given two equations.

x + y = 6
x - y = 2

y = -x + 6
x - 2 = y

Question: Do the lines intersect?
_________ Where?

It's better to know more than one way to solve a problem. You have to take on a certain equation in as many different angles as you can.

Mr. K taught us different ways to solve the equation.

There were 4 different techniques that he showed the class.

Graph : to actually draw out the equation on a graph
Comparison
Substitution
Elimination

*Important note: The Lattice Point is a point in the cartesian plane where the points are integers. (where the lines cross)

GRAPH

We've graphed the lines and can easily see where the lines cross (4,2). Graphing it will not always solve the problem because the point where the lines intersect won't always be a Lattice point.

COMPARING

At the Intersection, Y1 & Y2 are the same.

eg.)

y = -x + 6
x - 2 = y

-x + 6 = x - 2
____8 = 2x
____4 = x

This technique is called comparison. It's when 2 things equal to the same thing.

SUBSTITUTION

To figure out the rest, substitute 4 for x.

x + 4 = 6
4 + y = 6
y = 2

Lines intersect at (4,2)

ELIMINATION

OK...hopefully people will be able to understand this..

The Homework is Exercise 24

Next Scribe is: M@rk

2006-11-10T14:32:00.000-06:00

2006-11-09T06:27:00.000-06:00

didn't u read my scribe post yesterday at 11:18pm?

"the next scribe after this lesson was sandy. oh and dont pick me for scribe the next day cuz i was already picked for scribe. its just that my name hasnt yet been crossed off cuz my scribe wasnt yet posted but now it is. so after today's scribe, i think cheriee's next cuz her name hasnt yet been crossed off so she's the next and last scribe. good luck and you'll do a great job like always!!"

Analytic Geometry

2006-11-08T17:33:00.000-06:00

Hello everyone!.
Today in class Mr. K said we were going to learn something new and other stuff we already knew about.

We worked on two questions:

1) If the distance from the line 2x-3y+8=0 to the line 2x-3y+K=0 is 24/√13 units find the value(s) of K.

2) Given:A (-1,3)
B (0,5)
C (-2,6)

a) Is ABC a right triangle?
b) Is ABC isosceles?
c) If M is the midpoint of AB and N is the midpoint of AC prove that MN // BC
d) Find a point D so that ABCD is a parallelogram
e) is ABCD a rectangle?

We were given some time to work on it. Most of the class had trouble with the first question like where to start.

* For the first question we talked about ways to find out if the lines were parallel ( // ).We knew that if the slopes were the same they are //.

some examples shown.

a) 2x-3y+8=0
4x-6y-k=0
* multiplied by 2

b) 2x-3y+8=0
-8x-12y+k=0
* multiplied by -4

-back to solving question #1

1) We changed the first line to be written in Standard Form

2x-3y+8=0 | 3y=2x+8
y=2/3x+8/3

From changing it to standard form we found the y intercept and the x intercept.

y intercept (0, 8/3)
x intercept (-4,0)

We looked at the question again and wrote down all the information we knew.

P (-4,0)
A= 2
B=-3
C=K
D=24/√13

Now that we know all of this information we just replaced them in the formula to find a distance from a point to a line.
D = | Ax + By + C | / √A2 + B2

24/√13 = | 29(-4) -3 (0) +K | / √4+9

24/√13 = | -8+k | / √13

24 = | -8+K |

24= -8+K -24 = -8+K

32= K -16 = K

The answer to this question has 2 Values, which are 32 and -16.

The second question wasn't all answered, but this is what i got.

2)

D = √(x2 -x1)2 + (y2-y1)2

AB=√1+4
=√5

BC =√4+1
=√5

CA =√9+1
=√10

(√5)2 + (√5)2 =√10

a) ABC is a right triangle
b) ABC is not a isosceles

* If anyone has more answers to this question feel free to comment and I'll update it. =)

Today we also wrote in our dictionaries, most of the information we already know.

My Analytic Geometry Tool Kit

The Pythagorean theorem | a2 + b2=c2

Slope of a line (a.k.a. rate of change) | (tri)y/(tri)x = y2 -y1/x1- x2

Midpoint of a line segment | midpoint = (x1-x2/2 , y1+y2/2)

Distance between two lines | D = √(x2 -x1)2 + (y2-y1)2

Distance between a point and a line | D = | Ax + By + C | / √A2 + B2

Equation of a lines

General Form | Ax+By+C=0

Standard Form | y=mx+b

Point-slope form | y2-y1=m(x2-x1)

Parallel lines (//) have equivalent slopes | m1-m2

Perpendicular lines have slopes which product is (-1) | m1-m2=-1

**Homework for today is exercise #23, and last the scribe for tomorrow is......-Zeph!!! Have fun!

Algebra Review

2006-11-07T23:18:00.000-06:00

ALGEBRA REVIEW
LAST SCRIBE POST REVIEW BEFORE THE PRE-TEST!!!

in class, mr.k gave us equations to solve and two word problems.

SOLVING FOR X IN A GIVEN EQUATION

solve for x:

x + 3 = 4 - x

[(2x + 2)/(x^2 + 2x - 15)] + [2/(x - 3)] = 1/(x + 5)

√(w + 5) = √(3 - w) + 1

SOLVING FOR X IN AN ABSOLUTE EQUATION

x + 3 = 4 - x

since we are solving for x in an absolute value equation, x can only equal a positive number because the absolute value represents the distance and distance is always postive. because an absolute value of one side equals the absolute value on the other side, then there is more than one answer meaning that there are 4 equations.

we write down all the possible scenarios and since the absolute value operation means that whatever the number is inside the absolute value brackets, we're only look for its positive. for example, the absolute value of -3 is 3; OR -3 = 3.

mathematics is the science of patterns...

mr. k showed us that theres a pattern when solving for x in an absolute value equation. these are the possible solutions:

first, we remove one expression's absolute value brackets and there we are left with either a positive or a negative of that expression.

here are the possibilitites:

x + 3 = 4 - x;

x + 3 = -(4 - x);

-(x + 3) = 4 - x;

-(x + 3) = -(4 - x);

since we're looking for the positive value, all of these expressions have to be greater than 0.

wow...thats a lot of work just to solve for x but mr. k has lots of neat tricks up his sleeves so he showed us a shortcut.

"x + 3 = -(4 - x)" is the same things as "-(x + 3) = 4 - x" because all whats different is that a -1 was multipled on one side.

"x + 3 = 4 - x" is the same thing as "-(x + 3) = -(4 - x)" because all whats different is that a -1 was multipled on one side.

so really, we only need to solve for x for 2 equations which are:

x + 3 = 4 - x AND x + 3 = -(4 - x)

for x + 3 = 4 - x...

x + 3 = 4 - x
2x + 3 = 4
2x = 1
x = 0.5

for x + 3 = -(4 - x)...

x + 3 = -4 + x
x + 3 = x - 4

notice anything abnormal?

if we subtract both sides by x, then we get 3 = -4 which doesnt make sense.
if we add 4 to both sides we get 7 = 0 or if we subtrace 3 on both sides we get -7 = 0.

THIS IS AN EXTRANEOUS SOLUTION!

so for that equation, x = { }, N/A (not applicable), or "No Solution"

so x can only equal 1/2. BUT! we have to check if 1/2 fits perfectly into the given equation and it makes sense. so we sub in 1/2 for x.

x + 3 = 4 - x
(0.5) + 3 = 4 - (0.5)
3.5 = 3.5

both sides of the equation are equal so x = 3.5

in solving for x in an absolute equation, if the equation was like: x + 3 = 4 - x + 1, we have 4 different equations cuz a -1 isnt being multiplied on both sides. solve it and you'll see what i mean.

its like that because we are not looking for the absolute value of 4 - x + 1. we are solving for x when an absolute value equals another absolute value plus a number.

the next scribe after this lesson was sandy. oh and dont pick me for scribe the next day cuz i was already picked for scribe. its just that my name hasnt yet been crossed off cuz my scribe wasnt yet posted but now it is. so after today's scribe, i think cheriee's next cuz her name hasnt yet been crossed off so she's the next and last scribe. good luck and you'll do a great job like always!!

2006-11-06T20:43:00.000-06:00

oh yea its Lauressa I forgot Mel!ssa is our next scribe

2006-11-06T18:09:00.000-06:00

Hey guys ,

Well today in class we recieved our grades :( Some of us were suprised and for others well there's always room for improvement. Well when Mr.K handed out our tests from AGES AGO! we went over the questions we had problems with.

The units our tests were from were recently algebra, quadratic functions and trig.

There were just some of the questions that we went through together in class.

Quadratic Functions

Long answer :

A farmer with 750 feet of fencing wants to enclose a rectangular area and then divide it into 4 pens w/ 3 fences parallel to one side of the rectangle. What is the largest possible total area of the 4 pens?Step 1: Write an equation to match the question asked

2L + 5w = 750

Step 2: Get one of the variables by itself
2L = 750 -5w

Step 3: Solve for the variable
L =750/2 - 5w/2 ; L = 375 -5/ 2w

Aw = (375 - 5/2w)w Where one of the roots are at 0 (w=0)

1) 375 - 5w/2 =0

2)375 = 5w/2 Get w by itself, so in order to do that you multiply 2 by both sides, so it cancels out

3)750 = 5w > 750/5 = w w=150

750= 2L + 5w > 375 = 5w > w = 75

Im really sorry for the short blog today because we did so much in class but I couldn't get my computer working. Mr. K I need a one on one session on blogging, the programs on my computer weren't working some would't open. SOrry again to the class for that. NEXT time I promise it will be better. Sorry :(

REVIEW / THE ROAD TO A "NEW" DISTANCE FORMULA

2006-11-04T19:54:00.001-06:00

Hey Guys!

Well on Friday's class we had Mr. Clark once again to fill in for Mr. K, and no time was wasted. To kick off the class, Exercise 20: Review 2 was assigned and is expected to be done.

As soon as that was established, our class as a whole had asked Mr. Clark to look over a few questions on Exericse 21: Circles on a Coordinate Plane (which was previously assigned) that we were having problems with.

4. Find the centre and radius, and sketch the graph for the following circles.
(c) x²+y2+6y-12=0
HINT: To solve, complete the square

and put into standard form of the equation of a circle
x²+y²+6y-12=0
x²-10x+25+y²-4y+4=25+4
(x²-5)2+(y-2)²=29
centre: (5,2) radius: √29

9. Sketch the graph of y=x²-2x+5. State the domain and the range.
HINT: The degree is 2, we have a parabola!
y=x²-2x+5
y=(x²-2x+1)+5-1
y=(x-1)²+4

HINT: Vertex is (1,4) and Y-Intercept is (0, 5)
Domain: (-, ∞) Range: [4, ∞)

10. Write a quadractic equation with roots 2±√3
HINT: Find Sum of Roots + Product of Roots
s.o.r. = 2+√3 + 2-√3
=4
p.o.r.= (2+√3)(2-√3)
=4-3
HINT: (√x)(√x)=x
=1
eq'n = x²-4x+1

11. Two sidewalks meet at right angles. At noon, Person A is 12 km north of the intersection, walking south at 2 km/h. Person B is 18 km east of the intersection, walking east at 4 km/h. At what time is the area of AOB a maximum?
HINT: Maximum.. think parabola.. think sign of a quadratic equation.. think vertex
Original Area = Base(Height)/2
= 18 km(12 km)/2
= 108 km²

After X hours A will be 12 km-2x
After X hours B will be 18+4x

New Area = (18+4x)(12-2x)/2
= (216-36x+48x-8x²)/2
= (-8x²+12x+216)/2
= -4x²+6x+108
HINT: lets complete the square to find the vertex
= -4x²+6x+108
= -4[x²-(3/2)x+(9/16)]+108+(9/4)
= -4[x-(3/4)]² + 441/4
Vertex (3/4, 441/4)

So that was all with Exercise 21: Circles on a Coordinate Plane.

Students then asked questions about our next assignment
Exercise 22:Distances between Points and Lines

We were all familar with the line distance formula
distance =√(x₂, x₁)²+(y₂-y₁)²

Mr. Clark then presented to us the "new" distance formula
Below is the points and lines distance formula
distance=ax₁+by₁+c/√a²+b²

i.e) Find the distance from Point (2, -5) and the line 3x+4y+9=0
distance=ax₁+by₁+c/√a²+b²
=3(-2)+4(-5)+9/√3²+4²
=35/5
=7

HINT: Keep the Pythagurian Triples in mind when solving these equations
a²+b²=c²
*3, 4, 5
*6, 8, 10
*5, 12, 13
*12, 16, 20
*8, 15, 17
7, 24, 25

So I guess homework for the weekend is all uncompleted exercises up until..
Exercise 22: Distances between Points and Lines

Have a good weekend and enjoy!

Scribe for Monday will be Lauressa
since you expressed your ready-ness to blog!

P.S. Images will be placed tomorrow if I find the energy, sorry.

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.