This is a blog for my second semester pre-calculus 30s students. Have a blast!

Thursday, March 23, 2006

Assignment #2 Results

Hey folks,

I hope your evening is going ok!! If not I have some good news for you. The results are in for assignment #2. The class average is 81.4%. What else is new though? You guys just continue to raise the bar whenever you get the opportunity. Five people received 100%, while another five people received marks in the 90's. SIMPLY OUTSTANDING!!! Keep up the amazing efforts guys. Have a great evening.

Wednesday, March 22, 2006

Assignment

Just wanted to check and see if my answer was correct. For number 4 for does k=10??

hello

Hey guys, i was jsut wondering if someone would be able to give me a hand with this hand-in assignment...i can't seem to do the questions :S...im not sure what to use from my "tool box"!!...

SCRIBE (O.o)

Hey guys! I'm Tim-Math-y your scribe today..

Today, we started off class with our usual MENTAL MATH. Moving that aside, we reviewed last night's homework which I believe was Exercise #22.

Now today's lesson was Proving Assertions Using Coordinate Geometry which was on a handout.

With this lesson, "our goal is to USE and MANIPULATE (control) the formulas any way we can in order to prove certain assertions and/or theorems in coordinate geometry (about shapes on a graph)".

With the first example, Given A=(-1,3), B=(0,5), and C=(-2,6)

a.) Verify that ABC is a right triangle
We start by drawing a diagram which should be something like..








Now with that, we find the lengths of the triangle so that we can prove that this triangle is a right triangle with the Pythagorean Theorem.

Using the distance formulas d=root(x1-x2)2+(y1-y2)2, we find that distance AB=root(5), distance BC=root(5) and distance AC=root(10). With these values, we plus them into the pythagorean theorem which is a2 + b2 = c2.

root(5)2 + root(5)2 = root(10)2
5 + 5 = 10
10 = 10

This means that the triangle is indeed a Right Triangle.

b.) Is ABC isosceles? justify your assertion
This should be a simple answer like this:

Yes, because AB=BC. :)

c.) If M is the midpoint of AB and B is the midpoint of AC, prove that MN is parallel to BC.
Start by drawing a simple diagram.








With this, we first must find the coordinates of N and M by finding the midpoints of AC and AB. We do this by utilizing the formula:
mpt=(x1+x2/2 , y1+y2/2)

We find that the coordinates are N(-3/2 , 9/2) and M(-1/2 , 4).. yess eeew to fractions.

With that, we now prove that the two lines are parallel by finding both of their slopes. We do this by using the formula:
M(slope)=(y1-y)/(x1-x)

To cut this short, the two slopes are equal values which were -1/2. With that we conclude that they are parallel with a statement:

MMN = MBC therefore, MN is parallel to BC

Now comes the hard questions which I must apologize because my brother must use the computer for his own homework purposes thus booting me off :( .

Homework for today is Ex.#23 i think.
There is also a Quiz this friday.

TOMORROW's SCRIBE IS .... hmm.. I'm looking on the scribe list and I know mr. Jojo was scribe yesterday so... the list must be old but I pick the next one in line who is..... JAMILYN G.!

Have a great.. rest of day. lol

Quiz #4: Analytic Geometry outline

Hey folks,

For your quiz this Friday be sure to be comfortable with the following:

  • Finding the center and radius of a circle, given its equation in expanded form.
  • Finding the shortest distance from a point / line to a line.
  • Solving a 2 x 2 system of linear equations algebraically, either by substitution or elimination (this is Thursday's lesson).

Good luck!!!

Tuesday, March 21, 2006

OMG I'm the scribe today

Wooow who picked me. Everyone knows I'm no good at scribe posts. Oh well.

We started a new lesson today about the distance between points and lines. It's easy to find distances vertically and horizontally. I'd make a diagram but I'm too poor to afford a good photoshop application :) Anyways now another way is to find the perpendicular distance which will always be smaller than the shortest distance of what you're given. The formula is..




The X and Y in this formula represents a point on a line.

[ Example ]

Given the equation 2x-2y+6=0 and a point at (4,0) find the shortest distance from the point to the equation.

After substituting you're formula will look like this..



After doing the math you will be left with..



You could simply root 8 but it's all the same. Since you cannot have a root in the denominator you multiply top and bottom by root 8 or 2 root 2 whichever you prefer and you'll be left with 14 root 8 over 8 which is you're answer.

I hope this helps you guys so don't worry if this is not as good as everyone elses :)
I don't know who hasn't been scribe yet so I choose you, Timmy!

- OMGitsCLouD

Monday, March 20, 2006

HI:)every body I’m the scribe for today
Today we start a new unit (the forth unit), and it’s called:
ANALYTIC GEOMETRY:

--> consider a fixed point. A circle is:

A circle is the set of all points (x, y) that are equidistant from a fixed point called the centre of the circle. The distance, r, between the centre of the circle and any point (x, y) on the circle is the radius.

--> What if the center of the circle is (h,K)
(x-h)^2+(y-k)^2=r^2 this is the Std. form.

-->expanded form:
(x-h)^2+(y-k)^2=r^2 ... where d,e,f E real #.

Ex.:

For the circle:

x^2+y^2-8x+12y-35=0 {Rewrite with like variables together and constant on the right.}

i) Find center:

1. (x^2-8x)+(y^2+12y)=35 {Group the like variables and begin to complete the square of each group.}

2. (x^2-8x+16)+(y^2+12y+36)=35+16+36 {Take half the middle term and square it to complete the square. Add to both sides.}

(x-4)^2+(y+6)^2=87

center= (4,-6)

ii) r=squ. root of 87


Ex.: Wirte the eqn. of a circle with center (-1,-2) and passing through (3,4)
first we have to find the r using the distance formula:
then place it back at the eqn.



Ex: Find the eqn of a circle with diameter AB, with A(1,2) & (-3,-6)

first we have to find the mid. point , in order to find the r, because half of diameter is =to r. then use mid. point. with one of the point from que. and solve.

Ex.

find the eqn. of the circle with center (0,4) and circumference of 8pi.

area of cirucm=2pi(r)

8pi=2pi(r)

8pi/2pi=2pi(r)/2pi

4=r

(x-0)^2+(y-4)^2=4^2

there for x^2+(y-4)^2=16

Note:

the home work for tomorrow is ex#21:)

and the scribe for tomorrow is jojo

Friday, March 17, 2006

Test results

Hey folks,

Guess what? Yes, that's right the tests are marked already. The class average was 74%.
Great job as that is still a staggering class average for any pre-cal test. The highest mark was 99% as well. Have a great weekend!

Long weekend post!!

I hope the tests went well on Thursday, while I was away sick. I haven't really looked at them yet, but by Sunday afternoon at the latest I will try to post the class average of the tests on the blog again.
Another thing I have to mention to you guys again is that over the first month and a half of the course so far I feel that we have gone fairly quickly, to some respect. On the other hand though I think that most of you guys have been right there with me the whole time. Let's face it, this is a very tough, challenging and time demanding course and so far I must say that I'm very impressed with what I've seen so far. I'm referring to your work ethic, the types of questions (and lots of!!!) that you guys are asking every day from the previous day's homework, and just the general interest and hard work that I'm seeing on a daily basis. You guys might not realize this but over 90% of students in highschool don't ask any questions even when they have them. Asking questions makes the WORLD of difference in your learning. You guys probably ask the best, and the most of, questions than any other class I've had.
Overall, very impressive work so far, and I'm proud of everyone for your efforts up to this point and I'm sure this trend will continue. Have a great weekend and see you on Monday!!

* One other thing folks - in the "Links" list there's a new link called "The equation of a circle". On Monday we begin our fourth unit of the year, "Analytic Geometry," where we begin with coming up with the equation of a circle on a coordinate plane. So if you have some time before Monday, take 10 or 15 minutes and check out the site. There are some great interactive tutorials there too where you can...well you'll see if you do. Trust me Monday's lesson will seem A LOT easier if you check this out first. If not, then check it out on Monday or Tuesday after the lesson and it can help you then too.

Wednesday, March 15, 2006

Solving Absolute value & Rational Equations

So this is my first scribe duty and what your going to see won't nearly be as creative as the other posts...but i'll try my best. In today's class we learned how to solve Absolute Value & Radical equations and if your a bit lost, or didnt go to the class...hopefully this will help.

Absolute Value Eq'n

* The Absolute Value of a number can be thought of as the DISTANCE FROM ZERO on a number line.
* Any positive or negative number can have a POSITIVE DISTANCE from ZERO.

--> The absolute value of any number is:

a = a, if a<0
a = -a, if a<0

--> 5 = 5
--> -3 = 3

EX- Solve x² - 3x = -4x + 6

x² -3x = -4x + 6 x²-3x = 4x - 6
x²+x-6 = 0 x² -7x +6 = 0
(x + 3)(x - 2) = 0 (x - 6) (x - 1)

x = -3, 2 x = 6, 1

Check: (-3) : 18 = 18 (correct)
(2) : -2 = -2 (In-correct, can't be negative)
(6) : 18 = -18 ( " " " " " )
(1) : -2 = 2 (correct)

______________________________________________________________________

RATIONAL EQ'N

* A Rational Eq'n is where the variable (x) is part of the denominator of a fraction

There are 2 rules :

1) Factor
2) Find LCM (lowest common multiple) and Multiply each term by it.

Ex - Solve: 3x/x-2 = 2 + 6/x-2 LCM: x-2

3x = 2(x-2) +6
3x = 2x-4+6
x = 2 <--- Restriction "can't = 2"

And i'll give you one more example because i have to study guys ! (much needed studying time !!!!!!)

Ex. Solve. 3x/x+1 = 12/x²-1 + 2 LCM: (x-1)(x+1)
3x/x+1 = 12/(x-1)(x+1) + 2
3x(x-1) = 12 + 2(x²-1)
3x² -3x = 12 +2x² - 2
x²-3x-10 = 0
(x-5)(x+2)

Therefore x=5, -2

So that pretty much was the lesson of today...sorry if it didn't help, i tried my best to make it look good

The next scribe will be...STEVEN!

















Tuesday, March 14, 2006

SOLVING RADICAL EQUATIONS


MARCH 14, 2006
SOLVING RADICAL EQUATIONS

- a radical equation is where the variable (x) is under a root or part of a radical expression.

Here are some examples: [ I wish I could have made the images bigger so you guys can easily see. ]

++ sorry for the delay.. computer froze many times.. haha. also, i want everything to be clear. that s why i cant just say "square root of" or "to the power of". anyway. homework is Exercise 18.

++ for all those Coin Hunt people, for clue set number 3, first question, mr. k said he's gonna reclarify that question.

++ KENNY is the next scribe!

Help with Exercise question.

I need some help with question 2 ALL of the questions for number 2, specifically the methods to finding the answer or something along those lines, help....please?

Monday, March 13, 2006

Solving Other Non-linear Equations

let's just have some review!
let's solve for: x2-2x-24 by completing the square, factoring, and using the quadratic formula. Take note...we must get the same answers for all of this.

Completing the square:
x2 -2x-24=0

x2-2x=24
x2 +1=24+1
square root of (x-1)(x-1)=square root of 25
x-1=±square root of 25
x-1=5
x=6,-4

Factoring:
x2-2x-24= 0
(x-6) (x+4)
x= 6,-4

Quadratic formula:


Solve for: y= x4 -2x2 -15

Let p= x2

y= p2 -2p-15

p2-2p-15=0

p2 -2p=15

(p-1)2 =15 +1

square root of (p-1)2 =square root of 16

p-1=± 4

p= -3, 5

substitute p back to x2

x2= square root of -3

x= ± i square root of 3

x= ± square root of 5

Solve for: y= x8-1

Factor x8-1

(x4-1) (x4+1)=0

Factor (x4-1)

(x2-1) (x2+1) (x4+1)=0

(x-1)(x+1)(x2+1)(x4+1)

x-1=0 x+1=0 x2+1=0

x=1 x=-1 x2=-1 x4+1=0

x=± square root of -1 x4=-1

x=± i let p=x2

p2=-1

p=± i

substitute p back to x2

square root of x2=square root of ± i

x=±square root of i

therefore: x= ±1, ± i, ± square root of i

Solve for: y=x3-8

0=x3-8

cube root of x3=cube root of 8

x=2

by: fevie javier


Fill-in Scribe please...

Hey folks,

Albert was the scribe on Thursday's class and he never picked a scribe for Friday's class and now he wasn't in class today. All in all this means that there is no "official" scribe for today's class. As you can see, one of our classmates needs help because he wasn't in class today. Can somebody please continue the great reputation of "fill-in" scribe and make a short summary of today's class? That would be greatly appreciated, not only by myself, but more by Kenny I'm sure. If the "fill-in" scribe could please assign tomorrow's scribe that would be greatly appreciated as well. If you would like to know who hasn't scribed yet the updated list is in the "Links" section on the blog.
Just so you guys know we also have a class member who will be sick for the entire week, she e-mailed me this morning. So she will be depending on the blog to be updated as to what's going on in class. Thank you very much.

Im lost...

So today i didn't go to class because i was late so i didnt get any notes, so if someone could fill me in quickly i'd appreciate it...thanks

Thursday, March 09, 2006

The Discriminant

b^2-4ac in the quadratic equation:

-b+(b^2-4ac)
- _______
2a

(pretend these parenthesis are the square root)
is what you called "the discriminant". The discriminant helps you to find how many real solutions a quadratic equation has without solving the equation or graphing.


ex: x^2 - x - 12 = 0
substitute the value of discriminant from the equation like this:

b^2-4ac
(-1)^2 - 4(1)(-12)= 49

which means it has 2 real no. solutions because the answer is greater than 1.

another example: x^2 - x + 2 = 0

let's substitute it again in the discriminant like this:

b^2-4ac
(-1)^2 - 4(1)(2)= -7

which means it has 0 real no. solution because tha answer is a negative no. or less than 1.

I'll give you one more example and i'll explain things to you more so you can understand(I hope I could do that).

ex: x^2 - 6x + 9 = 0

use the discriminate formula which is:
b^2-4ac
(-6)^2 - 4(1)(9)= 0
which means it has only 1 real no. because the anwer is 0.

Explanation:
The ff. should be keep in mind in order to understand how many roots and what kind of roots(irrational or rational or imaginary) does a quadratic function has:

1.)If the answer to the discriminant formula that we just did is > 0(greater than zero)and it's a non-perfect square then the equation has 2 real irrational roots.

2.)If the answer to the discriminant formula is also > 0(greater than zero) and it's a perfect square then the equation has 2 real rational roots.

3.)If the answer to the discriminant formula is = 0(equals to zero) then the equation has 1 real no. root.

4.)If the answer to the discriminant formula is <0(less than zero) then the equation has 0 real number roots but the roots are called the "imaginary numbers or roots".

That's all...whew!I already made my first post being a scribe tonight.I promise Mr. M to make one this night. Don't forget to study guys we do have a quiz tomorrow morning so good luck!my next scribe is..I'll tell you tom. ok?see yah!

Wednesday, March 08, 2006

stuck..

I am stuck on this worksheet. I thought it would be easier than this. When I looked at #s 3, 4, 6 & 7.. I was like "uhhh.. Am I still on Earth?" HAHA.. whatever. I'll just wait for corrections tomorrow.

...

A correction has been made for our math work sheet this morning....

number 4 is supposed to be x = 4 and not x = 2.93...

I just want help some people add some effects to their blog posts so i'm gonna give my favorite html tag which can be useful to some who would like to make their blogs "alive" and "moving"...

**replace the square brackets with <>

[marquee]text[/marquee] ---> makes the text move side wards...

this tag can be used for both text and pictures

The Nature of Roots

THE NATURE OF ROOTS



Before I start talking about the nature of roots...Let's have a short review of what roots are and how can we get them.

Roots: are where the quadratic "crosses/touches" the x-axis. It's an x-intercept or zeroes.

We can get the roots by:
1.) factoring
2.) quadratic formula

3.) or completing the square.

Review link(video)

Okay, for those who didn't know that... Now you do ;)

and now to the...


The Sum and Product of Roots.


-When a (leading coefficient) is 1, the sum of the roots is the negative or opposite of b (coefficient of x).
-The product of the roots is c (constant term).

EXAMPLE:


b (coefficient of x) is (-6), it is the negative/opposite of the sum of the roots.
c (constant term) is (8), it is the product of the sum of the roots.

THE SUM OF THE ROOTS IS : -b/a

b (coefficient of x)
a (leading coefficient)


THE PRODUCT OF ROOTS IS: c/a

c (constant term)
a (leading coefficient)

EXAMPLE:


okay... this is it!

no exercise homework... just the worksheet handout on nature of roots =)

bye for now!

ANNNND the next scribe is....oliver.

More on Quadratic Functions.. (fill in post)



Well there is no post again...

Though this is too late... It might help...

IMAGINARY NUMBERS are set of negative numbers inside a root...
The imaginary unit
, "i" , is defined as:



i = square root of -1; i2 = -1
therefore for any positive real number "a", square root of -a= i*square root of "a"
Examples:
a. square root of -4 = i* square root of 4
square root of -4 = 2i
b. square root of -25= i * square root of 25
square root of -25= 5i

c. ± square root of -8= ± 2i * square root of 2

** from this part onward i will be using /x /to represent square root.. the number between the two slashes is inside a root sign

CHECK:
(2i/2)2= -8
22 i2 * 2= -8
8i2= -8
8(-1)= -8
-8 = -8












FORMULA FOR FINDING THE VALUE OF X... SHORTCUT FOR COMPLETING THE SQUARE




-b ± /b-4ac/
2a



EXAMPLE:

y=x2 + 4

a= 1, b= 0, c =4

= [0 ± /o -(4)(4)(1)/]÷2(1)

=(/-16/)÷ 2

=4i ÷2

=2i

Example:

y= 2x4 - 5x2 + 2

let p be equal to x2

y=2p2-5p+2

y=(2p-1)(p-2)

y = 2p-1

p= ± /½/

p= ±/2/÷2

y = p-2

p= ± /2/

p= ± /2/

x= ± square root of 2 divided by 2, ± square root of 2

well... i'll end this here since I am too sleepy already...

Hope this helps... :P...

MILES.. MAYORDOMO..




I guess i pick the next scribe so i choose..... KAT

Tuesday, March 07, 2006

HeRe Is WhAt I dO...

Well.. what I do for square roots, I just write the whole word like this:

5 + root of 2

Then, for degrees, you can use either:

1) 900 [ put "sup" (short for superscript) inside these brackets < >; place the desired text in between two sets of these.. the first one should be <> then the last one is < /SUP > ] but when you write this, make sure each letter follows each other..
2) 90 degrees [ just write the whole word down ]

Monday, March 06, 2006

Albert cannot do the post today... He said so since he gave the wrong e-add to Mr. M. again...

Mr. M. he said it was supposed to be
jay_dintch15@yahoo.ca...



Friday, March 03, 2006

Test results

Hey guys,

The tests are marked and I must say I'm in shock. I don't think I've ever had a class score an 80% average, as a whole class, on any test in any subject in any grade let alone PC30E. That's right, that's not a typo, the class average was a whopping 80%. I didn't round up or fudge any marks either, the exact class average was 79.87%, which is obviously 80%. Based on some of the reactions I got before you guys left on Friday, I was under the impression that the opposite would occur. Now I'm doubting myself; Am I making these tests too easy? :) . Have a great weekend and you'll get them back after Monday's lesson.

Applications

Just a quick reminder folks that those green applications for next year are due on
Wednesday, Mar.8 next week. Please don't forget as you heard Mr.Lukie say that they are very important in insuring you having a good chance of getting into the courses that you want.

Thursday, March 02, 2006

Test Outline

Hey folks, here is a brief and repeated outline for tomorrow's test:

  • Know how the sine and cosine graphs look like and all their key points.
  • Know how to find roots / zeros of sine and cosine graphs on your calculator.
  • Know your four quadrants and all the corresponding trig. ratios for each quadrant.
  • Know how to solve standard trig equations along with related angles.
  • Know how to solve quadratic trig. equations along withn related angles.
  • Know when the ambiguous case occurs in triangles and how to find 2 possible values for a side or an angle for a triangle.
  • Know how to find the vertex and roots of a quadratic function algebraically.
  • Know how to solve max / min quadratic function problems.
  • Know how to manage your time :).

GOOD LUCK!!!!!!!!

Quad./Trig. Equation

This is my first Scribe Post, so just bear with me and I apologize if you do not understand what I’m saying. I’m just not good at explaining things.

Today, the lesson was about Quad/Trig Equations. This lesson involves a lot of factoring. We did a recall about what a quadratic equation is.

  • A Quadratic Equation is an equation with the highest power, which is 2.
    • Example:
      • Y= X2
    • This could be a quadratic equation:
      • f(x) = sin2θ


Example:
Solve: y=3x2 + 2x - 1

  • First, try to see if it can be factored. In this case, it can be.

y= (3x-1) (x+1)

  • Then, change Y into a zero.

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

  • Then, let each factor equal to zero.

3x-1=0

-----------

x+1=0


  • Solve for x.

3x= -1

x=-1/3


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

x= -1



In this next example, we'll do the exact same thing. Just imagine that the sinθs are the x variables.


  • Solve: 2sinθ - sinθ - 1.

y= (2sinθ + 1) (sinθ - 1)

0 = (2sinθ + 1) (sinθ - 1)


2sinθ = -1

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

sinθ = 1

*But this time, we solve for angles, not just numbers.*

sinθ = -1/2


θ = -30
°

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

sinθ = 1


θ = 90
°

*Since the angle is negative,
we have to find the related
angle in the 1st quadrant.*


θ
r = 30°

Q3:
θr = 180° + 30°

= 210
°

Q4: θ
r = 360° - 30°

= 330
°

Therefore, the angles are 90
° , 210° , 330° .

Here's another example:
  • Solve: 5sinθtanθ = -2tanθ

*First, we need to transpose the -2tanθ to the left.*


y = 5sin
θtanθ + 2tanθ

0 =
5sinθtanθ + 2tanθ

*Then, factor out tanθ.*


0 = tan
θ (5sinθ + 2)

tanθ = 0
°

θ = 0
°

*This is a special situation.

We need to find where tanθ
is 0 other than Q1. It is in
the 180 Quadrant and
360 Quadrant.*

θr = 180
°, 360°

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

5sinθ = -2


θ = -2/5


θ = -23.6
°

θr = 23.6
°

Q3: θr = 180
° + 23.6°

= 203.6
°

Q4: θr = 360
° - 23.6°

= 336.4
°


Therefore, the angles are 0
°, 180°, 203.6°, 336.4°, 360°.





If i'm right, we have a test tomorrow. So, good luck to everybody! Before i forget, the next scribe is,
Albert!

Wednesday, March 01, 2006

Review day

Hey guys! I'm scribe for today. Well today we basically did reviews. No new stuff guys so for those who were away today in class, don't worry you didn't missed anything. We did Exercise 12 an exercise for review.

For those people whoe were away and needed help on Exercise 9 here is the solution for #9:


















So for exercise 12 there might be one question that might be still a little tough for some people so here is the solution #2 and #3:



Hope this all help. The next scribe will be........ CARLA