It's Today!

For those of you who've forgotten, I'll be in my classroom today, Wednesday, December 30, from 1:00 until 3:00.

See you then.

More Probability

Getting things organized, and some probability stuff.

Today, we spent the first 10 minutes normally spent on a topic of Mr. Max choice, getting our old tests handed back. That's always a highlight of the class.

Then, Mr. Max gave us some things to think about:

1. Accelerated Math is now "out of" approximately 54 objectives. Where are you?

2. Pre-Test for upcoming Thursday test was handed out last week, have you done all questions?

3. NO CALCULATOR for thursday's test.

Here's what you should know how to do for Thursdays test:

Trig 1 (circular functions)

Transformations

Trig 11 (identities)

Logarithms/Exponential Functions

Perms & Coms

Conics

Now, here's what's left to learn:
1) Probability

2) Geometric Sequences

Here's what we learned today: Probablity

We first learned about sample spacing, which is basically creating a picture about the situation your given. Using ordered pairs, or a chart, or a T chart.

Then we learned about the two different event types, Simple and Compound.

A simple event is when the sample space can not get any simpler then what it is.

A compound event is two or more simple events are considered at once.

Then there's the 2 laws of probablity, first being the addition law, "or". Basically if your trying to find the probability of A or B, you will have to add. The formula is a little more complicated, but its on our formula sheets.

The second law is the multiplication law, "and". When your trying to find the probability of A and B, you will have to multiply. This formula is a little harder to understand, with the whole P(BA) part to it.
We basically spent the rest of the class doing examples.

So, do not forget!!!

Thursday = Test. (Forget your calculator at home kids, we don't need 'em.)

January 14th, 12:30 pm = Pilot exam (10 days after we get back from Holidays.. oh boy. Merry Christmas to us!)

Thats about it! Good luck everyone! Study it up!

Sketching Standard Form Hyperbolas (Conics)

Sorry for the lateness of this posting. I had a blonde moment. The video was showing up as text but I guess when I publish it, it shows up as normal. Anyway....



Today we learned how to graph hyperbolas.....YAAAAAAAY!, Lots of fun.......but nowhere as much fun as trig, because it's awesome.


The first slide is just showing you the equation of a hyperbola. A great way to look at it is that either the x coordinate or y coordinate is negative. Also, the coordinate that is positive is the way that the hyperbola opens. So, if you have +x and -y, the hyperbola is going to open left-right because your x coordinate is positive. Furthermore, your 'a' value always goes with your positive value (x or y); However, unlike circles, your 'a' value (transverse axis=2a) is NOT always bigger than your 'b' value (conjugate axis=2b).

Moving on, the next slide asks you to sketch the hyperbola.
The first thing that I do is find the centre of the hyperbola. This is the same as finding the centre for a line. It's just the opposite sign of what's inside the brackets. In this case it is (-2,1).
Next, I find the values for 'a' and 'b'. For this, you just have to square root the denominator to get your values. Now, while keeping in mind which way the hyperbola opens, you can trace your transverse and conjugate axes. I recommend drawing a box for your axes, it really helps. From there you can find your vertecies which are the endpoints for transverse axis. The vertecies are your starting points for your hyperbola.
Now, we have to find the equation of the asymptotes to make our sketch complete and where to draw our hyperbola. This is just finding the equation of a line: y=mx+b, going back to grade 10 if you remembered which I didn't. So, you know your x and y values, they're just the values of the centre (x=-2,y=1). You know the m (slope), it's (by the way that I look at it) the amount that you move from the centre to the corner of the box. In this case, you're moving two up and three over making the slope 2/3 and -2/3. The part that you don't know is b (different from b at the start), but since you know everything else, it is just a matter of putting everything in and solving. After you find your b values, you can draw your asymptotes equations (with dashed lines) and then draw your hyperbola, remembering that your hyperbola follows very close to your asymptotes.

The next slide is just another example.

Last, you are given the graph and it is just a matter of finding the equation.
What I do first is find out whether your x or y is negative by looking at which way the hyperbola opens.
Next, I find the centre of the graph; In this case it is (-4,1),making it(x+4)-(y-1).
Now, you have to find the value of 'a' and 'b'. Without aid of the box or it telling you what your endpoints of the conjugate axes are, you can't find your 'b' value. Since you know the vertecies of the graph, you can determine the value of 'a', but you have to put it in terms of a^2 as opposed to 2a. In this case, you're given a box so you can find the 'b' vale.
Last, you have to find the equation of the asymptotes which is the same as before since you're given the box.

Hoped this helped you out if you were having troubles with this.

Conics- major / minor axis ( from formulas) - Transverse / conjugate axis

For ellipses, you have two axis, a major and a minor . the major axis is always the longer of the two ( see video, starts on ellipses).





For Hyperbolas, you also have two axis, a transverse axis and a conjugate axis. ( see video, after ellipses)





Transverse axis


-connects verticies


-not always longer than conjugate axis


- length is alwas 2a units (a+a)


- there for length of conjugate is always 2b units (b+b)

More Conics

Today in class we talked more about conics. Last class we found out how to change from general form to standard form. This class it was the complete opposite, we learned how to go from standard form back to general form. The reason we convert from general form to standard form is to be able to graph. I didn't quite get the concept at the start of the conics lesson but I did some accelerated math and I am starting to understand it. Also the test has been changed to next Thursday.

More conics

Assignments: exercises 37, 38, 39. This was assigned previously, but after today's lesson, we can do more.

Mr Max expanded our knowledge on conics. The central idea of today's lesson is that we can change general form conics into standard form conics.


Conversion between general form conic and standard conics.

Ax^2 + Bxy + Cx^2 + Dx + Ey + F = 0


Remember how to complete the square?

Take: x^2 + 4x + 4

We can take a square and
(x+2)(x+2)
(x+2)^2

or:
How do we find x? Well, take b/2, and square the result. sqroot(3/2)







If we try to enter (X^2 + y^2 + 6x - 8y) = 11 into graphimatica directly, it fails to map the equation because it cannot parse the data. So, we use this form after we complete the square

(X^2 + y^2 + 6x - 8y) = 11
(x^2+6x+9)(y^2-8y+16) = 11 + 9 + 16
(x+3)^2 + (y - 4)^2 = 36

Using the general form:

(x-h)^2+(y-k)^2= r^2
(h,k) = center; r = radius
we enter (x+3)^2+(y-4)^2= 36
and get:








c)4x^2 - y^2 - 8x - 4y - 16 = 0
4x^2-8x -y-4y = 16 + 4 -4
4(x^2-2x+1) -1(y^2+4y+4) = 16 + 4 - 4
(4(x-1)^2 / 16) - ((y+2)^2 / 16) = (16 /16)
(x-1)^2/(2^2) - (y+2)^2/(4^2) = 1
(x-h)^2 / a^2 - (y+2)^2/b^2 = 1








The final is only ~27 days away. I'm going to start studying today! Yes, that sounds crazy, but the exam is difficult, and this class can be interpreted as being preparation for the exam.

Work Time

Next Test: Thursday, December 10

Although today is Mr. Maks' birthday (Happy Birthday!), he ended up giving us the gift of having the class to do work. Personally, I really enjoyed this to have time to get some work done and give myself some time to process this Conics stuff and allow it to stick.
I'd suggest going through the practice test this weekend at some point so you'll know what to expect and feel more confident when the test comes around. Also, probably check out those Exercises Courtney posted just below mine.
Along with our assignments, I find one of the links we have on our blog really helpful, The Online Version of This Course has a lot of really great examples and practice questions.
Link straight to the Conics section:
http://www.bmlc.ca/Math40S/Pre-Calculus%20Math%2040s%20-%20Conics.html
Everyone just keep on working hard, you're doing great! Have an awesome weekend!

"When we do the best that we can, we never know what miracle is wrought in our life..."
-Hellen Keller

Conic Sections and Other Cool Mathematicalness

That was the title of our class today.
Today was our introduction to conics and we learned there are 4 sections to conics.

1. circle
2. ellipse
3. hyperbola
4. parabola

This is a conic: x^2 +/- y^2
General form of a conic is: Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
*A, B, C, D, E and F are elements of the set of reals. B cannot equal 0
Conics are always squared

A circle happens when the A value and the C value are the same.

(x^2 + y^2) = 1 is a small circle
(x^2 + y^2) = 7 is a bigger circle...etc

Ex. Given the general form of the conic equation,

a) Identify the conic
b) State A, C, D, E, F

given... x^2 + y^2 - 8 = 0
a) So this is a circle because both co-efficients are 1.
b) A = 1, C = 1, D = 0, E = 0, F = -8

given... 2x^2 + 2y^2 + 4x -2y -32 = 0
a) This is a circle, because A and C values are the same
b) A = 2, C = 2, D = 4, E = -2, F = -32

An ellipse happens when A and C values are the same sign, but A cannot equal C. An ellipse is a squashed circle, an ellipse is also a subset of a circle.

Ex. Given the general form of the conic equation,
a) Identify the conic.
b) State A, C, D, E, F

given...(1) x^2 + 49y^2 - 49 = 0
a) Ellipse because both A and C are positive, but the values are different.
b) A = 1, C = 49, D = 0, E = 0, F = -49

given... 4x^2 + 9y^2 - 3x + 2y +0
a) Ellipse because both A and C are positive, but the values are different.
b) A = 4, C = 9, D = -3, E = 2, F = 0

A hyperbola happens if A and C have opposite signs.
*If the x is positive, than the hyperbola will open horizontally, but if the x is negative the hyperbola will open vertically.

Ex. Given the general form of the conic equation,
a) Identify the conic
b) State A, C, D, E, F

given... 9x^2 - 4y^2 - 36 = 0
a) Hyperbola because the A value is positive, and the C value is negative.
b) A = 9, C = -4, D = 0, E = 0, F = -36

given... -3x^2 + 3y^2 +2x - 12y + 2 = 0
a) Hyperbola because A is negative and C is positive.
b) A = -3, C = 3, D = 2, E = -12, F = 2

If one of A or C is the value of 0, than you have a parabola.

Ex. Given the general form of the conic equation,
a) Identify the conic
b) State A, C, D, E, F

given... y^2 - 4x = 0
a) Parabola because A value is 0
b) A = 0, C = 1, D = -4, E = 0, F = 0

given... 3x^2 - 2x + 5y - 3 = 0
a) Parabola because A ihas a value bigger than 0, and C's value is 0.
b) A = 3, C = 0, D = -2, E = 5, F = -3

The exercises we were assigned are:
Exercise 36 #1-7
Exercise 37 #1-11
Exercise 38 #1-9

Dont panic because there is no way you wil be able to do all these questions with just the information we learned on conics yesterday. You need more information.

Communicating using technology, Japanese Vending Machines....and some more Binomial Theorem

We started the class with a discussion about how beneficial (or not) a ban on cell phones would be in schools, and how some people have lost the connection with people around them, by only talking to there "friends" on their phone or computer. Just something to think about...Back to math class we did mental math today. We did more examples of using the Binomial Theorem and had a matherific time! Remember that smoking and meatbelts are stupid.

Assignment is Exercise #35 Questions 3,4,5,6,7

Binomial Theorem

Binomial theorem is defined as (a+b)^N where N equals a natural number.

See these terms??? (look up), well, look at the cool patterns within them (look down)

Here is an example showing how to use the binomial theorem:

Here is an example showing how to find a certain term within an expansion:

Here is another example showing how to put the binomial theorem to use:

This nifty picture down here is Pascal's Triangle. What you'll notice is that if you add any two numbers beside eachother, their sum is the number found directly below it. Example: 6+15=21

Homework:

Exercise 32, Questions 1-10

Exercise 33, Questions 1-10

Exercise 34, Questions 1-10

Below is a calendar of our next few weeks of school before the end of the semester.

-November 29th to December 1st we will be learning BINOMIAL THEOREM
-December 2nd to December 8th we will be learning CONICS
- December 8th and 9th we will be given time to review
-December 10th will most likely be our test
-December 11th to December 17th we will learning PROBABILITY
-December 18th we are off for Christmas Holidays!

-January 4th to January 11th we will be learning SEQUENCES

-January 12th to the 19th we will be doing review, but remember PILOT EXAM IS JANUARY 14th, and our FINAL EXAM IS JANUARY 20th

KRISTIN HANSFORDS TEST QUESTION

Question: Simplify: (log210)(log48-log3)

Solution: (log10/log2)(log16)
=(1/log2)(log2^4)
=4log2/log2
=4

Test Question

1st question (2 marks)

Convert expression into a single expression logarithm:
½ln(x) - 2 ln(x-1) - 1/3 ln (x²+1)
=ln √(x) - ln(x-1)² - ln ³√(x²+1)
= ln (√x)/ [ (x-1)² * ³√(x²+1) ] <--- Log Law #2.
surprisingly, thats all you have to do for that question (simplify)

2nd question (1 mark)

Express log 6 1/216= -3 in exponential form
6^-3= 1/216

3rd question (1 mark)

Evaluate ln ( 1/ 4√e)
*multiple choice here guys..
a) ¼
b) -¼
c) -4
d) 4

Thursday Test Question

This is my test question:
Solve for x: ln6+ln(x−2)=−2
(Express your answer correct to 3 decimal places.)
This question is worth 3 marks as well
Here is the method I used to get the answer:

ln[6(x-2)]=-2
ln(6x-12)=-2 (1 mark here)
(e^-2)=6x-12
6x=(e^-2)+12
x=(e^-2)+12/6 (1 mark here)
x=2.022 556
x=2.023 (1 mark here)

(that's where the marking guide said the marks should be awarded)

I would suggest to everyone that you go over all the test questions mentioned on here. This way you know what you will have to answer and how to do it. Good Luck to everyone!

My Text Question

http://www.edu.gov.mb.ca/k12/assess/archives/pc_mg_june_06.pdf

page 58.

(3 marks)

Transformations Questions

Questions: #13 - 2 marks (1 per letter)
#14 - 3 marks (1 per letter)
http://www.bmlc.ca/Math40S/Pre-Calculus%20Math%2040s%20Standards%20Test%20-%20Transformations%20QUESTIONS.pdf
Answers Link:
http://www.bmlc.ca/Math40S/Pre-Calculus%20Math%2040s%20Standards%20Test%20-%20Transformations%20ANSWERS.pdf

Test Questions

Here are my questions for our test tomorrow:


1) Point P is a point on the terminal arm of an angle Θ in standard position. If cscΘ<0 and sin theta=-15/17, deterine the value of the other five trigometric functions.

2)P(Θ) is a point on the unit circle. If secΘ<0 and tan(theta)<0, and the coordinates for P are (x, 2root6/7)

a. Find the value of x
b.State the coordinates of P(Θ+pi)

Solutions:

1)step 1: Determine quadrat-(quadrat III)
step2: Find sinΘ=1-(-8/17)squared=-15/17
Step3: Determine other 4 trig. values
sec=-17/8
csc=-17/15
tan=15/8
cot=8/15

2) a.
step1: determine quadrat-(quadrat 2)
step 2:1-(2√6/7)²=-5/7
b.step1: determine quadrat-(quadrat 4)
step 2: determine integer values for x and y: (5/7, -2root6)

My question.

These questions are pulled from the Mickelson book.The solutions can be found on pages 198 and 199.

Page: 179

Number:16)

Letter: c)

Evaluate(worth one mark):

10! 4! / 8! 6!

= (1*2*3*4*5*6*7*8*9*10)(1*2*3*4) /

(1*2*3*4*5*6*7*8) (1*2*3*4*5*6)

= 9 * 10 /

5 * 6

90 / 30

= 3

Second question (page 180, number 23) (worth two marks):

How many different groups of four letters can be made from the letters A, B, C, D, E, and F if the letters can only be used once?

P(6, 4) = 6! / (6 - 4)! = 6! / 2! =

1 * 2 * 3 * 4 * 5 * 6 /

1 * 2

= 3 * 4* 5 * 6

= 360

We can create 360 groups of four letters from A, B, C, D, E, and F if the letters can only be used once.

TEST QUESTION!

HERE WE GO GUYS...QUESTION ON THE FUNDAMENTAL COUNTING PRINCIPLE (FCP)

This is out of the Mickelson (hmm i always thought it was nickleson) book. pg 176.

There are 4 roads between cities A and B, and 3 roads between cities B and C.
a) How many ways can a person travel from A to C through city B??
(4)(3) = 12
b) How many ways can a person make a round trip from A to C and back to A, through city B?
(4)(3)(3)(4) = 144
c) How many ways can a person make a round trip from A to C and back to A through city B, without using any road twice?
(4)(3)(2)(3) = 72

Alright, hope that helped and you all do great

Brett's Test Question

Question:

1. a) How many 6 letter arrangements can be made using the letters M A T T E R ? Give your answer as a whole number. (1 mark)

b) How many 6 letter arrangements end with T T? Give your answer as a whole number. (1 mark)

c) How many 6 letter arrangements can be made if the letters M T T R must be together? Give your answer as a whole number. (2 marks)

Solutions:

1. a) 6!/2!= 360


b) _ _ _ _ T T
4!= 24


c) (3!) (4!/2!) = 72

**3 groups, 4 consonants, 2 t's


GOOD LUCK THURSDAY!