During Monday's class, we pretty much steered our own course for preparing for the exam by either doing accelerated math or practice exams, so not much went on. We did, however, fill out a self-assessment for the class blog. I wanted it printed on green paper and Megan wanted it printed on pink, but we ended up with purple in the end.
I also found that the Math40S website is a very good resource for studying for the exam, so those of you in class today should check it out. It has a list of questions and answers for each subject covered in the course, great if you are having trouble with a certain part of the course. This beats having to go through old exams to find certain questions, and the questions on the Math40S site are questions off old exams anyway.
As far as accelerated math, I think we have until the 29Th to finish so that leaves you with 10 days.
I would like to close by saying that it has been a fun class and good luck on the exam tomorrow morning!
Tuesday, January 19, 2010
Friday, January 15, 2010
Pilot Exam (Not flying) review and discuss
after yesterdays pilot exam we all felt it was a little hard, but after todays class i think we all felt a little better after Mr. Max explained some of the harder questions.
after the discussion we had the rest of the class to work and learn.
Above is some pics to brighten your day
Wednesday, January 13, 2010
Pilot tomorrow!
Today we were able to once again review, work on accel, and go over some old exams. I'm really glad the course has been wrapped up for awhile which has given us some time to just review everything. I personally haven't reviewed as much as I probably should have... as much as I wish I have more, it will give me a good idea as to what needs to be looked at again and what I'm able to do.
I'd suggest to just do some light review and try your hardest to not get super stressed; eat some cucumbers, they're supposed to help relax you. And like Mr.Maks said, have a good supper and a good sleep tonight! Just be sure to be in the room tomorrow at 12:30 at the latest.
"I've got a theory that if you give 100 percent all of the time, somehow things will work out in the end."
-Larry Bird
Good luck everyone, you'll do great!
And as for that long haired 4 year-old boy, I personally think someone being suspended for their hair length is ridiculous. It actually really irritates me that people think they have the right to suspend someone for their chosen hair style.
Link to an article:
http://thechronicleherald.ca/World/1162049.html
I'd suggest to just do some light review and try your hardest to not get super stressed; eat some cucumbers, they're supposed to help relax you. And like Mr.Maks said, have a good supper and a good sleep tonight! Just be sure to be in the room tomorrow at 12:30 at the latest.
"I've got a theory that if you give 100 percent all of the time, somehow things will work out in the end."
-Larry Bird
Good luck everyone, you'll do great!
And as for that long haired 4 year-old boy, I personally think someone being suspended for their hair length is ridiculous. It actually really irritates me that people think they have the right to suspend someone for their chosen hair style.
Link to an article:
http://thechronicleherald.ca/World/1162049.html
Tuesday, January 12, 2010
Monday, Monday, Monday
This is a blog for monday, as you probably gathered from the title "Monday, Monday, Monday". We had a work period, where we got a chance to ask Mr. Max questions about accelerated math, and the practice exams.
*Reminder that the pilot exam is on Thursday this week and we need to be here @ 12:30 pm.
Mr.Max also made an announcement that we can decide if we want the pilot exam to be marked as a test or not. But it's up to us individually. Thank you for that:)
And thats about all, good luck!!
*Reminder that the pilot exam is on Thursday this week and we need to be here @ 12:30 pm.
Mr.Max also made an announcement that we can decide if we want the pilot exam to be marked as a test or not. But it's up to us individually. Thank you for that:)
And thats about all, good luck!!
Thursday, January 7, 2010
Partial Sum Examples
At the beginning of class we were shown a blog post that was about a teacher's view of smartboards and sympodiums and how he believed they were a waste of money. Here is a link to that blogpost if you'd like to check it out: http://teacherleaders.typepad.com/the_tempered_radical/2010/01/wasting-money-on-whiteboard.html
*Our assignment for today is to do questions 1-9 on Exercise 47.
Today we did 3 more example dealing with partial sums. Here's the slides that have the examples:
*Our assignment for today is to do questions 1-9 on Exercise 47.
**Reminder: Exercises 48, 49 and 50 are the last 3 exercises left. They are all review so you should challenge yourself by doing all 3 completely without using the answer key/solutions. This is really good practise for the upcoming pilot exam next Thursday!!
Tuesday, January 5, 2010
Work Class!
Today, we handed in our calendars for studying, in hopes of creating and sticking to your study goals.
Also, we did our 8th, and final mental math. The marks have been added to Edline.
Then some students had the brilliant idea to catch up on accelerated math for the rest of the class. Mr. Max agreed and let us. YAYYY!
Monday, January 4, 2010
Wednesday, December 30, 2009
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.
See you then.
Wednesday, December 16, 2009
Tuesday, December 15, 2009
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.
Now, here's what's left to learn:
1) Probability
Here's what we learned today: Probablity
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.
Thursday = Test. (Forget your calculator at home kids, we don't need 'em.)
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.
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!
Thursday, December 10, 2009
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.
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.
Tuesday, December 8, 2009
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)
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)
Monday, December 7, 2009
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.
Friday, December 4, 2009
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.
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
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
Wednesday, December 2, 2009
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.
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.
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