<body><script type="text/javascript"> function setAttributeOnload(object, attribute, val) { if(window.addEventListener) { window.addEventListener('load', function(){ object[attribute] = val; }, false); } else { window.attachEvent('onload', function(){ object[attribute] = val; }); } } </script> <div id="navbar-iframe-container"></div> <script type="text/javascript" src="https://apis.google.com/js/plusone.js"></script> <script type="text/javascript"> gapi.load("gapi.iframes:gapi.iframes.style.bubble", function() { if (gapi.iframes && gapi.iframes.getContext) { gapi.iframes.getContext().openChild({ url: 'https://www.blogger.com/navbar.g?targetBlogID\x3d14084555\x26blogName\x3dPre-Cal+30S\x26publishMode\x3dPUBLISH_MODE_BLOGSPOT\x26navbarType\x3dBLUE\x26layoutType\x3dCLASSIC\x26searchRoot\x3dhttp://pc30s.blogspot.com/search\x26blogLocale\x3den_US\x26v\x3d2\x26homepageUrl\x3dhttp://pc30s.blogspot.com/\x26vt\x3d931551856370134750', where: document.getElementById("navbar-iframe-container"), id: "navbar-iframe" }); } }); </script>

Wednesday, November 30, 2005

Double Acrostic

Well guys I tried and I had to get a little creative to complete the "Double Acrostic". So just bear with me.

Circles have many parts of their anatomy like a chord, or an arC
Having a radius perpindicular to a chord bisects the chord (meaning the line segment on each side of the intersection should matcH.)
One of the theorems that we learned in this Circle Unit was the Parallel Chord Theorem: it states that the arcs between parallel chords form a 1:1 ratiO.
Radii and chords should be measured with a ruler where as angles should be measured with a protractoR.
Diameters are chords that pass through the center and is equal to the value of the radius, doubleD.

The Next Scribe... Hahahaha?

Well Robert told me who the next scribe is lol. Since um... Blogger isn't working for him in some strange coincidence. Don't ask me why... I'm having problems too... lol. Well here's the next scribe... (Robert did make this. He sent me it.)

Yeah it's me. lol. What a strange coincidence... Hahahahahahahahahahahaha

My Scribe

Hello to everyone!  Today’s class started off with two investigations.  Investigation number nine and investigation number ten, which had to do with circle geometry.  

Investigation 9:

A cyclic quadrilateral is a quadrilateral all of whose vertices lie in the circumference of a circle.

First we had to construct a polygon with all four points on the circumference.  After that, for that quadrilateral we measured the sum of the opposite angles and found out that they were equal to 180°.  We then did those same steps two more times to confirm that it wasn’t just for that one type of quadrilateral.  The last step for this investigation was to articulate a general rule.  We found out that “If two angles are opposite each other in a cyclic quadrilateral, then the angles are supplementary.”

Investigation 10:

The diagonals of convex polygons are drawn at a single vertex.  Complete the table of values showing the relation between the number of sides in the polygon and the number of triangles formed.


In conclusion, a convex polygon with n sides can be divided into   n - 2    triangles.


You have learned that the sum of the interior angles of a triangle is 180°.  Complete the following table.


Refer to the preceding table.  Describe in words the relation between the number of sides in a polygon and the sum of its interior angles.

The sum of the interior angles is found by the number of sides (n) subtract two, then
Multiplied by 180°.

Again, referring to the table, write a formula that describes this relation.
The sum of the interior angles of a polygon with n sides = ( n - 2 )( 180° )

After all of that we then went into our dictionaries to take some notes.  Here they are:

Perpendicular Bisector Theorem:

Part I     If a line bisects a chord and passes through the center of a circle then it is      perpendicular to the chord.

Part II     If a line is perpendicular to a chord and passes through the center of a circle then it bisects the chord.

Inscribed-central Angle Theorem:

Part I     If a central angle and an inscribed angle are subtended by the same arc then the central angle is twice the inscribed angle.

Part II     If an inscribed angle and a central angle are subtended by the same arc then the inscribed angle is half the central angle.
A.K.A. “The star trek theorem”

Special Case:
     Inscribed-diameter (Thales) theorem
An inscribed angle subtended by a diameter is always 90°.

Inscribed Angle Theorem:

If two (or more) inscribed angles are subtended by the same arc then they are congruent.

Parallel Chord Theorem:

The arcs between parallel chords are congruent.

Congruent Chords Theorem:

Congruent chords subtend congruent arcs.

The next scribe is: (

Tuesday, November 29, 2005

My Acrostic

Hey, I feel sort of stupid for picking the shortest word so I had tried to make a double acrostic...well if you can find a word that ends in h and one that ends in o kootoe's to you :D.

Chords are made up of a straight line that joins two points on the circumference of a circle and creates an arC.
Having a radi bisect a chord creates a perpendicular and a right angle for those who remember their matH.
One other fact is that two congruent chords are equidistant from the centre of a circle, this will also work with more then twO.
Right now we have three chord theorems to study: Perpendicular-Biscetor Theorem, Parallel Chords Theorem, and Congruent Chords Theorem which are all under the chord theorem scection to make it easier to remembeR.
Drawing the centre of the circle can be done by using two chords that are not parallel, a ruler, and a pencil that is sharpeneD.

My Double Acrostic

Circumference is a part of each and every arc.

Inscribed angles that are subtended by the same arcs have equal angles. Like angles BAI and BCI.


Right angles are made when inscribed angles are subtended by a diameter.

Circles are comprised of a 360 degree arc.

Lens flare is an example of circles that gradually go from large to small.

Every tangent line touches a circle once.

If I you already used the one I used for "I" then I'll change it, but I can't remember what you used this morning Mr. Kuropatwa.

bee-logging pee-rompt

Hi everyone I thought I'd get started early so no one has thing as me. Anyways, here is my acrostic:

If it's blurry just click the picture. ;)

My Angle on Circles

Hello everybody! Well today in class we started off with a quick presentation by Mr. Kuropatwa on how to get a del.icio.us account. If you need any more info on that look at Mr. Kuropatwa's previous post on it. We then moved on to our first project of the course. There are options for what you can do but the general project is when you answer a question that he handed out then you solve it, communicate and present your results the best you can. He said that with this project it is easy to get an 88%. This is because all you need to do is do the math correctly then present it and communicate in a little above average manner. But if you want the 90's or the fabled hundred then you got to work. Some people have made a webpage, dioramas, videos and things like that. But if you do make one of those it doesn't garauntee anything. If you make a video and it's not decent you can still fail. If you don't want to do the question you can also talk to Mr. Kuropatwa about making a instructional video or something along those lines (your deadline will also be extended).

After he explained the project we went on to our Circle Theorums. We did two today. The Tangent Radii Theorum and The Tangent Chord Theorem.

The Tangent Radii Theorem

"If two tangents are drawn from a common point, exterior to a circle, then they are equidistant."
"If a radius intersects a tangent line, then where they meet it is a right angle."

This is a diagram of that:
All of the angles where the tangent lines and the radii meet are 90 degree angles.

The Tangent Chord Theorem:
"If an angle is formed between a tangent line and a chord then, then on the opposite side of a chord if an inscribed angle is made then those angles are congruent."



Angle RAB = 148.3 degrees
Angle SAB = 21.7 degrees
Angle ACB = 148.3 degrees
Angle ADB = 21.7 degrees
Angle MNY = 34.7 degrees
Angle XNM = 145.3 degrees
Angle NLM = 145.3 degrees
Angle NWM = 34.7 degrees

That concludes my summary on the class. The next scribe is:

It's Robert!!!!

Monday, November 28, 2005

Circle Acrostics

Kudos to Mrs. Armstrong for turning me on to the idea of acrostics in math.

Blogging Prompt
Your task is to create an acrostic "poem" that demonstrates an understanding of circle geometry related to any one of these concepts:

CIRCLE
INSCRIBE
ANGLE
CHORD
TANGENT
SUBTEND
CONGRUENT

As an extra challenge (worth an additional bonus mark) try to make a Double Acrostic, that is, each line should begin and end with a letter of the word you are working with.

Remember, this is a bit of a race. Your answers have to be posted to the blog in the comments to this post. If someone has already used a word or phrase in their acrostic you cannot use the same word or phrase. i.e. It gets harder to do the longer you wait. ;-)

Here is an example of an acrostic that Mrs. Armstrong wrote:

Always in 2 dimensions
Region between the boundaries
Entire surface is calculated
Answer is in units2

Be creative and have fun with this!!

Scribe of Thales

Hey guys, I guess it's my turn to put another chapter into our "textbook" so I decided to begin with Thales of Miletus.Now, by looking at this picture, you can probably tell he was, as Mr. K. would say, "A very miserable human being". Apparently, he tricked his best friend by saying his son was dead just to show him why he wouldn't have a family. As well, he made a farmer put sponges on a donkey's back to make it carry more weight. Actually they are all very smart ways of looking at things and proves an intellegent point, but he has cruel methods. But the 'Father of Mathematics' was a genius and he found out about the first investigation we corrected in class today, The Inscribed Angle/Diameter Theorm. This states that,"If an inscribed angle is subtended by a diameter, then it is a right angle."
How did Thales come up with this? What was his proof? He must be a genius!
Well, here's the proof.

So as you can see it is quite simple, the angle subtented by the diameter creates two isoceles triangles when the radius is connected to the inscribed angle. This is because the radius of a circle is always the same so the three congruent lines are all radiuses or radii? Anyways, then the two base angles are congruent and since the three angles of a triangle add up to 180 degrees, then a plus b equals 90 degrees.

The second investigation we corrected (#4) was to do with Inscribed and Central angles. It is the Inscribed angle/Central angle theorm and it states that,"If an inscribed angle and a central angle are subtended by the same arc, then the inscribed angle is half of the central angle."

The angle AOC is done to show what it means by "subtended by the same arc."
We haven't learned
the proof yet and Mr. K. has put it up as a challenge to prove this or any of the other investigations so I won't try to tell you.

Also, Mr. K. should be putting up th acrostic list as an alternative for pre-test posting and also I hope everyone's ready to see their test mark because he said he'd have them done by tomorrow.

One last thing before I go eat, I choose


As the next scribe. See Aichelle, I can do it too.

Bye!

Sunday, November 27, 2005

Sunday Connected-Slide Funday!



ConSlide puzzles are a new type of sliding block puzzles invented by M. Oskar van Deventer... The pieces of a ConSlide puzzle move like regular sliding block puzzles, but some of the pieces have sections connected by bars of various heights. This means pieces can pass over and under one another as long as the bars and posts don't run into one another. The goal of each of these puzzles is to move the red block to the upper left corner. Play it here.

(Once again, thanks to Think Again!)

Saturday, November 26, 2005

The Editor's Initiative

Instead of posting a pre-test reflective comment on your progress in this class you may undertake The Editors' Initiative. Here's how it works:

  Step 1: Scan through the previously posted Scribe Posts on the blog. Find one that has one or more errors.

  Step 2: Discuss the error(s) and what you think the correction(s) should be with me. If I agree with your editorial proposal go to Step 3.

  Step 3: Discuss the editorial change with the author of the post. The author will chose to proceed in one of the following two ways.



3a3b
The Editor is briefly allowed administrative privileges on the blog. They will edit the post to make any necessary corrections. They then sign the post at the bottom:
Edited by: [name] on [date]
The author will edit the post in consultation with the editor who will vet the author's changes until they are correct. The author then signs the post at the bottom:
Consulted editor [editor's name] on [date]

Students may chose to make more than one edit. Each additional edit will earn them a bonus mark on the next test. Your mark on the previous test determines the maximum number of edits/bonus marks available to you.

Mark on Last Test / Max Edits Allowed
> 90 / 1

80-89 / 2 (1 bonus mark)

70-79 / 3 (2 bonus marks)

60-69 / 4 (3 bonus marks)

50-59 / 5 (4 bonus marks)

40-49 / 6 (5 bonus marks)

30-39 / 7 (6 bonus marks)

20-29 / 8 (7 bonus marks)

10-19 / 9 (8 bonus marks)

0-9 / 10 (9 bonus marks)

You may also assume the role of Content Consultant to earn marks as outlined above. Here's how it works:

  Step 1: Scan through the previously posted Scribe Posts on the blog. Find one that doesn't provide enough detail or leaves out too much information. Decide what additional content should be added.

  Step 2: Discuss the new content you think should be added with me. If I agree with your editorial proposal go to Step 3.

  Step 3: Discuss the editorial change with the author of the post. Together, you will chose to proceed in one of the following two ways.


3a3b
The Content Consultant will add a new post to the blog inserted at the appropriate time and date. They then sign the post at the bottom: Additional Content by: [name] on [date]The author will edit the post to include the additional content provided by the consultant. Additional content will appear under a heading "Additional Content". The author then signs the post at the bottom: Additional Content Provided by [consultant's name] on [date]

Students may chose to make several additional content contributions for bonus marks according to the table above.

After discussing the issue with all my classes I have decided that you cannot edit your own scribe posts. When it's your turn to be scribe try to write a post that is so excellent no will be able to edit it. ;-)

Friday, November 25, 2005

Circles R Us

Today we started class by Mr. K. putting up questions on the board. This was the first question:


To find line AO we used the Pythagorean Theorem (a² + b² = c²). If you plug in the numbers you get:
5² + 3² = c²
25 + 9 = c²
34 = c²
√34 = c

You know that line AB is 10 because line AC and line CB are congruent; and you know they are congruent because of the perpendicular bisector theorem.

This was the second question:


We know that line OE is a radius and we know that the radius is 23. So we subtract line CE from OE and get 10 (23 – 13 = 10). We know that line OC is perpendicular to line AB because of the perpendicular bisector theorem. In order to get line AB we must use the Pythagorean Theorem to get line CB. Once you know line CB you also know line AC because they are congruent. If you do everything properly you find that line CB is √429 or 20.71 if you round it off. Then you know that line AB is 2√429 or 41.42.

This was the third question:


To find line OC we used the Pythagorean Theorem and we got √203. Since line OC is perpendicular to line AB we know that line AC is congruent to line CB because of the perpendicular bisector theorem. The answer is 22 (11 + 11 = 22).

Mr. K. showed us a different way of solving these questions by using a chart with two columns, one that says statement and the other that says reason. This is how it would look like for the first question:


This is how it would look like for the second question:

This is how it would look like for the third question:


The fourth question was:
A chord that is 10 cm long is 12 cm away from the centre of a circle. Find the length of the radius.To solve this question we drew a picture and this is how our picture looked like:


We got that from the information given to us and by using the statement and reason chart.


The last question was:
The diameter of a circle is 20 cm. A chord is 8cm from the centre. How long is the chord?

For this question we also drew a picture, we went fairly quickly with this question so here is the picture:


If you are still confused, maybe this can help you. We drew what was already given to us which were a chord that was 8cm from the centre and a 20cm diameter. Then we drew a radius connecting to point C, which was 10cm. Then we used the Pythagorean Theorem to find line BC and got 6. Once you know that it’s 6, you also know that line AB is , then you add 6 and 6 together and get 12.

After all of that Mr. K. gave us two investigations to do and showed us how to construct perpendicular bisectors. The two investigations are for homework if you didn’t finish them in class and Exercise 31 is also for homework.

*Reminders*
*You only have to use the chart when you are asked to justify your reasoning.
**Be careful about putting exact answers and/or approximations.
***You can use your statement as a reason after you have proved it.
**Use different colours with your circles.
*When drawing a radius imagine an infinite number of radii coming out from the centre and see which one connects with another point.


Lastly, the next scribe is:

Mr. Dixon's on eBay - How Trusted is He?

Mr. Dixon is in my classroom right now and he has a question he needs your help with.

His trust rating on eBay is 99.2%. Out of 122 transactions he's had, 121 people said he treated them well and he can be trusted.

Mr. Dixon wants to know when will his "trust rating" reach 99.3% and he needs your help.

(1) Can you write an equation he can use to solve this problem?
(2) Can you solve the equation and tell Mr. Dixon how many more favourable exchanges he needs to have to raise his trust rating to 99.3%.

All of my classes are working on this. The first student in each class who solves Mr. Dixon's problem correctly gets a chocolate bar. ;-)

Thursday, November 24, 2005

Greatful

Hey Mr. K.

Thank you very much for your advice. I downloaded Firefox and now I release that yes, it is the best browser. It's helped me a lot, I just put up the diagram in my post.

Oh yeah I meant to ask for our tests because some of the people from class are asking ME if YOU will mark the tests by the end of the week so this is for them and me.

Wednesday, November 23, 2005

Scribing.

WELL I'm very sorry about the late post. I didn't time earlier.

First in class we went over the Circle Geometry - Investigation #1. And we learned that the relationship between a chord and the perpendicular is that the perpendicular to the chord that passes through the center, bisects the chord.
As you can see here:



BUT..
Then we realized that when we drew the perpendicular on some of the chords it DID NOT cut the chord in half. We figured out that the circle wasn't a circle because the radius wasn't the same throughout the entire circle! But that was just a mistake the computer did so no biggy.

Then we did Circle Geometry - Assignment #2.
AND AGAIN, the circle wasn't a circle (good job figuring that out kasia!) because the lengths of the diameter weren't the same throughout the circle. So we crumpled up the paper and threw in the recycling box. Mr. K ran out and got us fresh sheets with REAL circles.
This assignment was also similar as Assignment #1 but we learned that all bisectors become the diameter of a circle and we know this by measuring. And we can say that the perpendicular bisector of a chord passes through the center of a circle.
Like this:


Then Mr. K talked to us about 4 ways of getting bonus marks.
If you are getting greater than 90's on your last tests you are only allowed to get 1 bonus mark. Then if you are getting 80-89's you can get 2 bonus marks.
70-79 / 3 bonus marks
60-69 / 4 bonus marks
50-59 / 5 bonus marks
40-49 / 6 bonus marks
30-39 / 7 bonus marks
20-29 / 8 bonus marks
10-19 / 9 bonus marks
0-9 / 10 bonus marks

Here are the ways:
1) making an acrossticks
2) Editing someone else's post
- But you have to go to Mr. K first and tell him which post and what you would like to do with it.
3) Editing your own post
- And again you have to go to Mr. K and tell him what you want to edit on your post.
4) Add most content to old posts
- You must go to Mr. K again to tell him what content you believe you missed out on that you want to add.


And that is all for today folks.

Tomorrows scribe is AICHELLE =) .

Tuesday, November 22, 2005

Proof, but from a different prospective.


Basically, use the formula about opposite angles and the method of solving systems (you should all be familiar with it):

d=a+b          |  L1
c+d=180*      |  L2
____________________

c+(a+b)=180 |  L3: sub L1 into L2
.: a+b+c=180



* angles c and d are supplementary

simple right?

untitled

haha .. well I'm not gonna try to "prove" the thing about the triangle. But I wanted to say that RICHARD i like yours haha . It's so simple and logical =)

Proof that Proof Exists...?

I can't prove what's allready been proven yet
there is no proof that we have proven anything.
So I'll prove that I proven everyone wrong with
this really simple thing. HAHAHAHA











Look's strange but it's so true. Make a triangle
and cut off the corners. Then put them together.
You make a line. And a straight line is 180º. Now
that's proof that YOU can do at home. Hahahaha.

Oh yeah. It was taken from this site.

Proof By Disproving all other Angles

Proof that the total of the degrees cannot exceed 180 or go below 180


Here is a right triangle with the two adjacent sides both being three units long

(A and C). In this triangle C + B + A = 180 degrees. Now if the triangle had 185 degrees…

As you can see the line off the 50 degree angle at c does not meet up with the endpoint of line C (And if you extended line C then the angle at the top would no longer be 45 degrees but 40 degrees. So it would still be a 180 degree figure). This would be true for any angle above 180 degrees. Then when it is below 180 degrees line B cuts through line C instead of meeting it at the endpoint. The angle where that meets it ends up being b – c = a (Assuming that you had b as the largest angle). Then if you input any of those into the formula a + b + c = 180 it works. (b – c) + b + c = 180 then you input the largest angle for b any angle for c that doesn't bring the total to 180 into that formula and you will get 180 unless your angle b is 180 or greater or sixty or less (sixty or less and b is not the greatest angle). Then you get a closed figure with three angles, a triangle. If you put anything greater than 180 for the total you get something like 180 = 181. You put something less than 180 than you get something like 180 = 179 which is completely wrong. So by proving that the angles in a triangle cannot be anything greater than 180 degrees or less than 180 degrees and that 180 degrees works, we can state that all triangles have 180 degrees.

I think that this is somewhat correct. I hope it proves it.





Circle scribe

Hey again

This is post is for today's class
today we wrote in our math journals most of the class and then we did a circle geomerty that we have to finish tonight for homework. This post was edited Nov 23 at 5:45

Circle Geomerty

fundamentals

1) The sum of all angles in a triangle is 1800

2)the angle exterior to a triangle is equal to the sum of
the opposite interior angle

3)Congruent: Means the same shape and size

4)Transversal: A line intersects 2 parallel lines

5)Opposite angles: two angles that are opposite
each at the intersection of to two lines are called opposite angles.
opposite angles are congruent

6)Alternate angles: two angles on alternate sides of a transversal are
called alternate angles. They are congruent

7) Corresponding angles: two angles on corresponding sides of a
transversal are called corresponding angles. They are always congruent.

8) Congruent Triangles: any two triangles can be proven congruent using
one of these four properties .

The Anatomy of a Circle

O - is the centre of the circle


is a chord- A straight line that joins two points on the circumference



is a diameter - chord passing through the center of a circle.



is a sector - a lime segment that contains a chord






is a tangent- a line that intersect, (touches) the circle at one point




radius- a straight line that joins a point on the circumference to the centre



is a minor arc- part of the circumference


is a is a major arc some but bigger



segment AB - part of the area of a circle bonded by a core and a minor arc

Sector FOC - part of the area of a circle bonded by the centre angle and an arc


centre angle: angle whose vertex lies on a circumference of a circle.

Inscibed angle: A angle whose vertex lies on the circumference of a circle

"Arc": the measure of an arc is equal to the measure of the centre angle that subtends that arc

This is all the notes I got I think I Missed a couple on the board sorry. If anyone has the rest please send me a comment and tell me so I can edit the post thanks. Thanks Robert for letting me borrow your notes

Tomorrows scribe will be ....... larisa

does this count as proof?

I'm pretty sure mine is different from Jennie's, so here it is:

is this proof?

http://www.apronus.com/geometry/triangle.htm

i found this site, who knows this might be useful to all of us.

Sunday, November 20, 2005

Blogging Before The Test

Hello everyone! This unit has been the hardest unit for me because it has so much to it. By studying a lot, I feel good about the test on Monday and hope everyone does well. Just a little reminder for the test:
independent - intersects at one point
dependent - intersects at all points
inconsistent - does not intersect/parallel
Good luck to everyone once again.

test-age

Wow, this Analytic Geometry stuff has a lot to it. There's a lot to take in and a lot of practical work that goes with it too. Basically it's like if you don't do your homework in this topic you're grabbing that one way ticket to not understanding what is happening. Well at least that's what it seems like to me, but i'm pretty confident that some people might agree with me. Then again not doing your homework is alreayd a one way ticket to not understanding pre cal at all. Anyway, I think analytic geometry was alright, not finding too many things difficult except for the problem solving stuff, but other than that I think this topic was a good intro to how hard precal acually is. So yeah before I end off my blog, i'd just want to say good luck to everyone tomorrow on the test and don't burn yourselves on doing stupid mistakes that could cost you a pass, ie. labels on the graph. Consider yourselves warned =D

blogging before the test

This unit was okay for me. I think the systems of equations is pretty easy. I also think that solving inequalities are okay. The thing that I have trouble with is using the information when they give you a word problem to solve. Well I'll keep this short and simple and just do my best tomorrow. Good luck on the test everyone.

bloggin' before the big test...

this is my blog before the test... i need that one mark hah =)... well im pretty comfortable with solving equations.. whats really bothering me is the one with the equation and the greater or less sign then you need to find a coordinate, that really confuses me. plus the asymptote, its a funny word but really hard for me to comprehend. well this unit is pretty hard and we all have got to work even harder. well good luck everybody! especially me haha =)

Blogging Before Test

This unit was a lot to take in, and I hope that this will be an okay test. Most of the time when Mr. K makes tests, he asks us about it, and then most people say it was hard. But yeah, just read through your notes guys, do your exercises, and everything, and look over the review. And I'm pretty sure it'll be okay! Hope everyone does good. =) You can't really study for this, you either know it or you don't. SO GUYS... know your stuff! =)

blog post before our test

so tomorrow we're having our test for the unit about linear inequalities ang solving equations with 2 and three variables. from all the unit we tackle, this seems to be the easiest for me (but not that much). we already tackle this topic from my former school. the good thing of studying this topics again is that i learned far more than what i already know about it. with this, the transition from highschool math to university math would be smooth. my classmate friends from the philippines, who loves the subject math when we are still in highschool (yes, im supposed to be a 2nd year university student in the philippines but because it is just equivalent here to grade 12, i need to be back in highschool), were having a hard time in university specifically engineering courses. i'm also stuggling in math when i came here because the lessons here are explained more deeply than just knowing how to solve a specific equation and that's it, we can get our grade if we pass our tests. i hope learning more topics about math can give me back my interest in math.

so goodluck to everyone for tomorrows test!

Post before the big test.

This unit is really jampacked with so much information. somethings did get me confused especially with the fractional inequality equations. There were things that I thought were really easy that we learned like solving systems of equations. But I think I will do okay cause everything seemed to sink it well. This is supposed to be one of the hardest tests we will receive and I think it will be because the pre-test was so difficult. Well I should be getting back to studying. Good luck everyone. Study hard!

Fourth Powers


- I hear you have extraordinary powers!

- Not true! What I said to the press was that I have fourth powers at home.

- What do you mean?

- 1, 16, 81, .... That kind of stuff.

- But these are square numbers!?

- Yes, but they are more than that. They are square numbers squared.

- I see. And you have these at home?

- Some of them. Not all of them.

- And what do you do with them?

- Play with them, of course, what else?

- What kind of games?

- This morning I added them.

- That must have been great fun!

- That's what I said to the press, but they didn't believe me.

- How did the game go?

- Not very well. I tried to pick less than nineteen of them, one several times if
neccessary, to get any number in the universe.

- How did you get 20?

- 20 was easy. 1 + 1 + 1 + 16 = 20. I only had to use five of them.

- Which number gave you problems?

- I found a number less than 500 that required nineteen fourth powers.

- So you lost the game?

- Yes, and I can't even remember which number it was.

Thursday, November 17, 2005

Quiz scribe

Hey everyone

So today in class we did a quiz and the corrected it, and I'm not really sure what I should scribe about today. So I just wanted to make a reminder to everyone that tomorrows a pre-test and Monday will be the test. I don't really think there's not going to be much to scribe about until next Tuesday and I'll be scribe on Tuesday and I'll pick the next scribe then.

Wednesday, November 16, 2005

Scribe's Post

It's my turn as scribe again and we did a lot of things today. When we came in class, Mr. K put up five questions on the board. The first one we did was:









Don't forget to make a clear cirlcle if the equation does not have an "and equal to" under the <> sign, because the clear circles mean that it doesn't include the roots


The solution to the equation is : (-∞, ⅓) U (1, ∞)


The second question was pretty simple. All we had to find were the restrictions.













For this problem, we first put both expressions on one side and make a common denominator.










To find out of the equation has any roots we use the discriminate.

Because the discriminate is a negative, it has no real roots and has no solution. This saves us time and work than it would if we were using the quadratic formula.



Then Mr. K took time to explain to the class about asymptotes, while we used out graphing calculators to graph one of the equations. Asymptotes are when the graph gets infinetly close but never touches it. For example, when we graphed the second question we did, we checked for the y-value of x = -3. The y-value always got closer and closer to the graph the more the numbers got lower.














This graph will have asymptotes and will never touch the graph.











Here are the last two questions:






























Okay so that's the end of my scribe post and the next scribe is going to be Jonathon.