Pre-calculus mathematics 30S

formulas??

2006-06-19T15:19:00.000-05:00

I was wondering on the formula sheet for the exam is everything on there?? Or is there any specific ones we need to memorize?? Like the ones from the consumer unit and other ones too.
Just wondering. Thanks.

GO FOR GOLD ANSWER KEY!!!!!

2006-06-08T07:59:00.000-05:00

MC:

1) D 2) A 3) D 4) C 5) B 6) B 7) C 8) A 9) A 10) B 11) C 12) C 13) A 14) B 15) B 16) D
17) C 18) C 19) A 20) D

SA:

1) a) 9x^2 - 6x + 3 b) 9x - 4

2) x = -2 (-7 is not a solution when checked)

3) AT = 5 cm; angle TRD = 90 deg.; TD = 10; arc RD = 90 deg.; angle CTS = 70 deg.

4) check your TI-83

5) a) (4, 1) b) x = 4 c) D = (-inf, +inf) R = (-inf, 1] d) x = 3, 5 e) check TI-83

6) f_inv = (x-3) / 2

7) ommitted

8) a) 87 mills b) $6525

9) C = (-4, 5) r = 11

10) a) in class b) 2 c) 2

11) theta = 80.4 deg, 279.6 deg

12) x = (7 +- root(57)) / 4

LA:

1) ... 3x^2 - 33x - 78 = 0
3(x - 13)(x + 2) = 0
x = 13 (-2 is an extraneous root)

2) (tan x - 3)(tan x + 2) = 0

x = 71.6 deg, 116.6 deg, 251.6 deg, 296.6 deg

3) a) x = 1; y = -1; z = -2
(1, -1, -2)

b) (+- 9/4, +- root(175/16))

4) [-1,5] .... number line as well.

5) b = 51.4, 5.4

6) > lengths of sides are equal. (dist. formula)
> opposite sides have equal slopes. (slope formula)
> diagonals bisect each other. (mid point formula)

7) f(1) = 0

therefore..(x-1) is a factor

...after synthetic division....(x-3) & (x-2) are other factors

...roots = 1,2,3, y-int = -6 graph as well

8) Sales: (6000 - 200; Profit: (20 + x)

Max = (6000 - 200x)(20 + x)
...
...
= -200x^2 + 2000x + 120000
* h = 5

P = 20 + 5
= $25.00

9) Assume 2 = 1; given 3 = 1
1 + 2 = angle ABC
3 + 4 = angle ACB
1 + 2 = 3 + 4

therefore: angle ABC = angle ACB

...but given ABC not = angle ACB

therefore 2 does not = 1

10) a) * shift 2 right, up 1 of y = 1/x; VA: x = 2; HA; y = 1; graph using TI-83

b) VA: x = 3, -3 HA: none; y-int: x = -1/9; sketch graph using TI-83

11) r = 7 using (distance from pt to line formula)
C = (1,5)

* therefore equation: (x - 1)^2 + (y - 5)^2 = 49

*

Way to go!!!!

2006-05-05T10:55:00.000-05:00

Way to go Vince!!!!! Fantastic double scribe post. Hopefully everyone can take a look at this and match your efforts in following posts. Keep up the great work!!!!

Today's post

2006-05-04T16:14:00.000-05:00

MAY 4, 2006

COUNTER EXAMPLES

Today's lesson is COUNTER EXAMPLES. It is about proving the CONJECTURE (conjecture: statement made by examples) wrong by providing true statements, examples, diagrams, equations. This lesson is somehow easy because all you do is prove that the conjecture is wrong. However, such conjectures may lead to sudden confusion because sometimes it gives just a small possibility of proving it wrong. You REALLY have to think about it too.

Here are some examples:

1) The sum of 2 prime numbers is always even. (CONJECTURE)

examples that verify this conjecture:

3 + 5 = 8 even

7 + 9 = 16 even

11 + 13 = 24 even

examples that contradict this conjecture:

7 + 2 = 9 odd

11 + 2 = 13 odd

13 + 2 = 15 odd

2) The expression 1 / x will always be less than 1 for any value of x.

examples that verify the conjecture:

x = 2

1 / 2 = 0.5

x = 4

1 / 4 = 0.25

examples that contradict the conjecture:

x = 1

1 / 1 = 1

x = 0.5

1 / 0.5 = 2

That's today's lesson. The homework is Exercise 48.

MAY 2, 2006

VENN DIAGRAMS

LESSON LAST TUESDAY:

EXAMPLES

A upside down U B = m, u

( B \ A ) ' = o, p, s, m, u, r

A \ B = o, p, s

( A U B ) ' = r

Scribe responsibility

2006-05-03T11:41:00.000-05:00

Just a quick note folks. I purposely haven't said anything in class about our scribe responsibilities over the past couple of weeks because there shouldn't be a need to. This is an enriched university entrance level course where standards and expectations are high. Everyone fully realizes that each day that there is a new lesson someone needs to be the scribe for that day. Even if it helps 1 or 2 people that night who might have been confused during the lesson then it's definitely served its purpose. Remember you are all responisble young adults and if the routine of scribing continues like it did for most of the course so far, everyone would only have to be the scribe every 21 classes (almost an actual month). Not a lot to ask for something which might make the world of difference to a few of your classmates, or possibly even people around the world taking the course. I've also noticed that it's the same group of 4 or 5 people who are always the "fill-in" scribes. The honus should be on everyone to put in their small part. Have a good day / night folks!!!

Logic and Proof

2006-05-01T19:33:00.000-05:00

Logic and Proof: Inductive/Deductive Reasoning

If the same result occurs over again, we conclude it will always occur. This is called inductive reasoning.

ex. multiply and odd number by an even number
9 x 8 = 72
11 x 12 = 132
7 x 10 = 10

From this, we can conclude that the products are all even. <-- We made a statement that is called a conjecture. A conjecture is something like an unproved theorem. Basically, we made a conjecture that says that all product of odd/even numbers are even.

If two points are placed on the circumference of a circle and joined by a chord, you'll get two regions.

*note: Having beyond 5 points made it difficult to count the regions, so I only did the first three examples.

*Conjecture: The number of regions follow the pattern 2¹, 2², 2³,... 2ⁿ

The 6-sided polygon proves our conjecture wrong. Using our pattern, we should get 32 regions, but the 6-sided polygon only has 31 regions.

*In deductive reasoning, we make a statement that we are absolutely certain is true based on TRUE STATEMENTS.

ex. Add 2 odd numbers.
1 + 3 = 4
5 + 7 = 12
9 + 11 = 20

Sum is always even?

Odd numbers:
2x + 1, 2y + 1

2x + 1 + 2y + 1
2x + 2y + 2
2(x + y + 1)

* 2 is a factor, therefore sum is always even.

ex

That's it for today's lesson. Homework is the Inductive and Deductive Reasoning worksheet and Exercise 45.

Consumer / Exp. Growth test results

2006-04-28T10:53:00.000-05:00

Hey folks,

The class average for the test was 64%. The marks were very skewed; 8 people received marks from 80% - 100%. 8 people also received marks from 20% - 51%. The other three were somewhere in the middle. Not bad over all, with the most common error being that some people who did not do so well forgot the compound interest formula or just didn't use it when they should have. Have a great weekend and see you on Monday, where we begin our unit on Logic & Proof.

Post?

2006-04-28T08:12:00.000-05:00

I forget why I have to do this again, but it seems we get marks for it. So heres my before test post. I didn't know consumer work would involve so much work and understanding. Probably the unit with the most writing in it in my opinion. But its been straight forward so far hasn't it? Well not for me.. Good luck today on the test. ( I prayed this morning and asked if I could pass of the test haha )

...

2006-04-28T00:01:00.000-05:00

I need help... This thing is driving me nuts... well the fact that I wasn't able to get the pre test does make me go nuts...

Is it Gross Income- Union Dues= Taxable Income??

Well that's all i need to ask since that's the only thing that's messing me up...

LOL...

Have a great day guys

Consumer / Exponential Growth Pre-test answers

2006-04-27T09:15:00.000-05:00

Part I - Short Answer:

1) $1788.46

2) $209.70

3) $7919.36

4) $2576.92

5) $1123.75

6) $423.50

7) a) $473.12 b) $1178.89

8) $11.00 / hr

9) $12916.42

10) $700.00

11) $1300.00

12) $25.90

13) $41.42

14) $1664.00

Part II - Group Problem Solving:

1) $129.57

2) $1722.36

3) a) $2220.37 b) $2228.92 c) $2233.30 d) $2236.25

...as for the estimated budget you're just trying to end up with some money left at the end of the month and making sure everyting is in the right place.

*** One more thing folks, exponential growth will also be covered on the test, so be sure to be comfortable with that. Good luck!!!! If you haven't made your post yet, you should do it now if you want your mark on the test. If you have no idea what I'm talking about then scroll down and read my post on the last test results.

...

2006-04-27T02:07:00.000-05:00

I just have a question on the take home unit.

Do you have to hand anything in or show Mr. Malandrakis any of the work that we did??

Well..ya thats all i wanna know, and does this post count for my mark? Haha.

...

2006-04-27T01:56:00.000-05:00

this is nothing just seeing if im doing this right.

Yeah another pointless post

2006-04-26T21:49:00.000-05:00

Consumer Math, I never thought I'd find another type of math that I'd have problems with but apparently I was wrong. It's not that I don't understand it I simply find it frustrating when I get confused half-way through and suddenly lose track of where I was going with a question.......I am doomed when Friday rolls by.

Post haha

2006-04-26T17:59:00.000-05:00

Hey guys! To be honest, I'm having trouble with consumer?! haha. I don't know why. Probably because I don't like reading. It's embarrassing I can't do consumer and I can do pre-cal. HAHA! As Carla said, we did consumer booklet today. I forgot mine so I had to share with John. I was supposed to be the scribe for today but there's nothing really to scribe for. No lesson... no exercise review either since most of us understand Interest. Well, it is pretty easy. Property tax confused me however. Budgeting takes a long time. I'll never want to be a parent! Geez. Okay... I'm done. See you tomorrow!

Post

2006-04-26T16:17:00.000-05:00

This isn't really a scribe post. Lol. That's what Mr. Malandrakis said. I can just post. So here. This won't really make sense.

I've taken Consumer Math before, so i'm hoping that i'll be able to ace the test. But i'll still study.

We did the Consumer Math booklet today.

I think i'm having problem with the conversion. I'll just ask Mr. Malandrakis tomorrow.

So, that's it. I told you this is nothing.

Scribe =\

2006-04-25T19:32:00.000-05:00

Well it seems I'm the scribe =\, I thought I already went this cycle.. oh well.

Today, we did some learning ..

Simple Interest Formula:
i = prt , where..

i is interest
p is principal (starting amount?)
r is rate
t is time

Another formula to remember is..
Amount = principal + interest

Compound Interest formula:
A (amount) = P(1 + r)ⁿ

Revised formula:
A = P(1 + r/s )^ns

Well, sorry that this scribe post is soo short and relatively late. I totally forgot that I was scribe today.. grr John. lol. Well I have no clue who to pick for tomorrow's scribe.. I guess I'll pick..Vince. :\ sorry

Tonight's homework is Exercise.. ermm I'm just guessing but probably Ex. #44. Also, there is a test on friday.. and mm remember to do your consumer math booklet. T.T gtg do mass ELA homework now..

Scribee Postee

2006-04-24T20:38:00.000-05:00

Hey everyone,

I'm going to make this scribe post as short and detailed as possible.

EXPONENTIAL GROWTH

For values to grow exponentially means to increase by a different amount each time.

The formula for an exponential growth: f(x) = ab^x
Restrictions for the formula are: a & b ∈ R, a ≠ 0, b > 1

The simplest exponential function is : f(x) = 2^x

ex. The growth of a $7000 RRSP (RRSP means Registered Retired Savings Plan)

Time (years)	Value ($)
0	7000
1	7630
2	8376
3	9005
4	9881
5	10770

i) estimate the time needed to reach $14000. There are several steps:

Press the STAT button in your graphing calculator.
Plot out the Time variables on the L1 column, and the Value variables on the L2 column.(Note: make sure Plot1 is on. You can enable this by going to the Stat Plot menu on your graphing calculator. [2nd func, stat plot, press enter on plot 1, set it on.])
You will now see dots on your calcuator. We must now find the exponential expression. You can do this by pressing the STAT button your graphing calculator, going the to the CALC menu, pressing 0 or finding the ExpReg on the categories. Press ENTER twice.
To plot it as an equation on the Y= menu, go to Y= menu, press VARS, then press 5 or pressing ENTER on the Statistics selection. Then go to the EQ menu, select option 1 (RegEQ). You should now have the equation on your Y= menu. Now you can graph it and find the solution to the problem.

Doing this, you should have 8 years as an answer.

ii) Find the value of the RRSP after 12 years:
You might want to change your window settings for this one. Set the Xmax value on your graphing calculator to 15. You should get $19630.28 as an answer.

NO HW for today except for that worksheet we finished in class.

Next scribe is Tim-Math-y.

Circle Geometry class average

2006-04-23T18:29:00.000-05:00

Hey folks,

The class average for the test was 70%, on the button. A job well done and this was not exactly a "walk in the park" test if you know what I mean. Starting next unit, being tomorrow, which will be only a few days discussing some consumer math topics along with exponential equations, we are going back to the daily scribe routine. I noticed that we kind of haulted it during the circle geometry unit and that's ok, but this time it will be worth one mark on your next test, being on Friday. This means that EVERYONE must contribute at least one POST before Friday's test in order to receive this one mark. As the final exam is just around the corner, it will become more and more important to have multiple sources of review for different topics that are covered in class. Make sure you share this with anyone else in the class that you think has not read the blog in a few days.

Pre-test corrections

2006-04-20T21:29:00.000-05:00

Hey folks,

Sorry for the inconvinience but there are a few corrections on the pre-test answers. I was in a rush this morning and I quickly posted the answer key that was provided to me from the teacher that gave me the test. Unfortunately there was a few errors. Thanks to Vince and his time and efforts those errors have been corrected. Please take a look at his comment that he left me on the answers for an update. Thanks again to Vince. Good job!!!

Circle Geo. pre-test answers

2006-04-20T10:40:00.000-05:00

MC:

1) d 2) b 3) c 4) b 5) b 6) d 7) a 8) c 9) b 10) b

LA:

1) ABC is an impossible triangle, so no solution.

2) Using the pythagorean thm... OE = 4.9

3) vertex = (2, 4)
eq'n of line in std form: x - y = 0

distance = sqt(2) / 2

4) angle 5 = 45 deg (insc. angle subtented by same arc)

angle 1 = 180 - 45 - 75 = 60 deg

angle 2 = 35 deg (insc. angle subtended by same arc)

angle 4 = 180 - 45 - 60 - 35 = 40 deg (opp. angles of cyclic quad supp.)

angle 3 = 180 - 75 = 105 deg (opp. anles of cyclic quad supp.)

5) angle 1 = 15 deg (cong. triangles - SSS)

angle 2 = 180 - 15 - 90 = 75 deg (tang/chord thm)

angle 3 = 90 - 75 = 15 deg

angle 4 = 180 - 90 - 15 = 75 deg

angle 5 = 90 deg (tang/chord thm)

6) angle 2 = 180 deg (subtended by diam.)

arc AD = 60 deg

therefore...angle 1 = 110 / 2 = 55 deg (central / insc. angle subtended by same arc)

angle 3 = 60 / 2 = 30 deg (insc. angle subtended by same arc)

arc BC = 180 - 60 - 50 = 70 deg

therefore...angle 4 = 120 / 2 = 60 deg (insc. angle subtended by same arc)

angle 5 = 70 deg (cent. angle subtended by same diam.)

7) using pythagorean thm.... OD = 5.29 cm

BONUS:

3

Test #4 Results

2006-04-07T11:52:00.000-05:00

Hey folks,

The tests are marked and the class average is 61%. I realize that some of you might be dissapointed because this means no ice cream in June (for the time being...until the next test). I want you guys to understand a couple of things. First of all this is a very tough, intense, and challenging unit, arguably the toughest of the course. Given that and the fact that we flew through this week covering 5 days of work in only 4 days, a 61% class average is still impressive for the nature of this course. The class average, as a whole for the course so far, is 67%, which is on pace to achieve one of the highest class averages in a pre-cal class (any grade level) in this school in a while. The quality of work I'm seeing from most of you is still extremely impressive and I'm very proud.
Here's another interesting fact, not fiction. The current pass rate for pre-cal students in this school is 63%. This means that in any given pre-cal class in any grade 63% of students, on average, will pass the class. In our class we have 21 students and 17 are currently passing. That's an 81% pass rate, which is unreal. Here's an even more amazing fact. Three of the four students who not passing this course (at the current time) have marks between 46% - 49%, which means we have the potential to have over a 94% pass rate in this class, with a small chance as well of everyone in the class passnig, after the exam of course. Just to let you know, to my knowledge, there has never been a math class (pre-cal, consumer, or applied) in any grade in recent history where everyone passed. This goal is within your reach. We'll talk about this more next week. Enjoy the gorgeous weather this weekend.

Analytic Geometry test outline

2006-04-06T08:09:00.000-05:00

Know how to determine the center and radius of a circle given its equation in expanded form.
Know how to determine if a given point is a solution to a system of equations
Know how to find the solutions, graphically, to a system of linear / non-linear equations using your TI-83.
Know how to solve a 2 x 2 or a 3 x 3 system of equations algebraically given a straight question or a word problem where you might have to set up a system of equations, like your worksheet.
Know how to solve non-linear inequalities: quadratic, rational, absolute value...algebraically and understand that to solve these inequalities means to leave your answer in interval notation unless asked to express as a number line.
Know how to graph a system of linear inequalities using your TI-83.

Yesshterrday's Scribe!

2006-04-05T21:56:00.000-05:00

Hey guys. So I was the scribe yesterday but I just had the chance to do it today because of some internet problem I had on my computer yesterday. But anyway, here's my post..

Systems of Non-Linear Equations

[+] CONIC is a word that describes a graph that has a 2nd-degree equation.
-> These are some examples of Conics:

[+] A QUADRATIC-LINEAR system refers to the equation of one conic & a line.
-> A Quadratic-linear system can have:

-> Consider:

Ex): Solve:

-> A Quadratic-quadratic system can have:

Ex): Solve:

And that's all guys! I hope that my post had refreshed your memory about what we had learn yesterday, and again sorry for the late post.
OH yeah before I forget! Don't forget to do our Analytic Geometry Pre-test oryt guys (:
Sheeeyah!

Fill-in scribe

2006-04-05T21:23:00.000-05:00

Hey folks,

Jamilyn, who was supposed to be the scribe the last couple of nights, has had some computer trouble. I know it's after spring break and all but let's please stick with the "fill-in scribe" routine because as you can see some of these lessons during class are getting quite detailed and some of our classmates might be in need of some extra assistance or a refresher.

Post: Graphing Inequalities

2006-04-05T20:54:00.000-05:00

Equations without an equal sign is an inequality...

An inequality only has a > (greater than), < [less than],≥ [greater than or equal to], and ≤ [less than or equal to] sign...

a graph of an inequality does NOT appear as a line...

if it has the greater than sign.. it means that all the values above the line is your solution...

if it is less than.. all values at the bottom of the line is your solution..

REMINDER!!!: The solutions for these do not include the points of your line... so the line you graph should be a broken line

if it is greater than or equal to, or less than or equal to same rules apply but the line of the graph becomes solid

AND- Intersection

OR- Union Intersection...

it is where the two graphs both have the same points for their solutions Union- all possible solutions for both graphs

HOW TO SOLVE USING YOUR TI-38:

Step 1: Enter the two points...
/Y₁= x
/Y₂= -x

Step 2: Put your cursor in the first character(** the slash thing) and press enter...
/ <-- I mean this one... it should look like a right triangle... one with the right triangle below means less than... one with the right triangle above means greater than... EX: y≤ x AND y≤ -x

Gray part and the Black part is the solutions for y= x and y= -x or the UNION...

The Black part is the INTERSECTION...

For QuAdRaTiC FuNcTiOnS...

EX:

y≤ x^²

I am not quite sure about this since i didn't understand it that well...

So the next scribe is.........

hmmm....

???..

Fevie!!!..Fevie!!!..Fevie!!!..Fevie!!!..Fevie!!!..Fevie!!!..Fevie!!!..

Answer key for analytic geometry pre-test

2006-04-05T11:30:00.000-05:00

Part I - MC:

1) a 2) c 3) d 4) d 5) c 6) b 7) c 8) a

PART II - SA:

1) using distance formula... r = 5.8

2) using substitution, elimination or graphing and finding the pt of intersection on your calculator
sol'n: (-4, -11)

3) 1 solution (as seen by graphing circle and straight line)

4) using distance (from pt to line) formula... d = 5 (root 13) / 13...(sorry about the lack of mathematical symbols, I never had a chance to download the software that produces them.

PART III - LA

1) Graph all lines on paper and find all 4 vertices (2 are y-intercepts, the other two are points
of intersection.
You must then find the equation of each diagonal line (using pt-slope formula from S2)...
The equations of the diagonals are: y = x - 2 and y = -(5/6)x + 4.
The point of intersection of these two diagonals is (3.3, 1.3).

2) You substitute all three points, (10, 1500), (20, 2000) and (30, 1500) into the equation and
you end up with the following 3 x 3 system:

1) 100a + 10b + c = 1500
2) 400a + 20b + c = 2000
3) 900a + 30b + c = 1500

2) - 1): 300a + 10b = 500 4)
2) - 3): -500a - 10b = 500 5)
4) + 5): -200a = 1000

...therefore a = -5

....after more substitution... b = 200, c = 0

...therefore sol'n = (-5, 200, 0)

.... h(t) = -5t^2 +200t

Assignment #2 Results

2006-03-23T21:14:00.000-05:00

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.

Assignment

2006-03-22T21:55:00.000-05:00

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

hello

2006-03-22T19:59:00.000-05:00

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)

2006-03-22T17:23:00.000-05:00

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:

M_MN = M_BC 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

2006-03-22T09:13:00.000-05:00

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!!!

OMG I'm the scribe today

2006-03-21T18:38:00.000-05:00

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

2006-03-20T20:14:00.000-05:00

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

Test results

2006-03-17T15:53:00.000-05:00

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!!

2006-03-17T11:57:00.000-05:00

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.

Solving Absolute value & Rational Equations

2006-03-15T19:53:00.000-05:00

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!

SOLVING RADICAL EQUATIONS

2006-03-14T20:55:00.000-05:00

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.

2006-03-14T18:03:00.000-05:00

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?

Solving Other Non-linear Equations

2006-03-13T20:15:00.000-05:00

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...

2006-03-13T19:47:00.000-05:00

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...

2006-03-13T19:25:00.000-05:00

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

The Discriminant

2006-03-09T18:23:00.000-05:00

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!

stuck..

2006-03-08T22:01:00.000-05:00

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.

...

2006-03-08T21:13:00.000-05:00

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

2006-03-08T18:54:00.000-05:00

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)

2006-03-08T00:49:00.000-05:00

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
2²i²* 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

HeRe Is WhAt I dO...

2006-03-07T22:33:00.000-05:00
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 ]

2006-03-06T23:02:00.000-05:00
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...

Test results

2006-03-03T12:02:00.000-05:00
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

2006-03-03T10:14:00.000-05:00
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.

Test Outline

2006-03-02T18:49:00.000-05:00
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

2006-03-02T18:14:00.000-05:00

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:θ_r = 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 = 5sinθtanθ + 2tanθ

0 = 5sinθtanθ + 2tanθ

*Then, factor out tanθ.*

0 = tanθ (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!

Review day

2006-03-01T20:33:00.000-05:00
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

Fill-In Scribe Post

2006-02-28T21:56:00.000-05:00
Since this is just a fill-in scribe post, I'll just make it as straightforward as possible and since most of us understand how to do this.

MORE ON AMBIGUOUS CASE TRIANGLES

ex:
How many possible triangles?

* 1 triangle
* the angle is opposite the longer side

ex.
sin Y / 5 = sin 42 / 3

sin y = 1.1

* no triangles.
* /sin x/ > 1.

ex.

Find all possible values for side c.

5 / sin 28 = 6 / sin B

5 sin B / 5 = 6 sin 28 / 5

sin B = 6 sin 28 / 5

angle B = 34.3 degrees

To find the related angle:

B_r = 180 - 34.3
B _r = 145.7 degrees

There are two possible answers for angle C:

angle C = 180 - 28 - 34.3
angle C = 117.7 degrees

for the related angle:

angle C_r = 180 - 28 - 145.7
angle C_r = 6.3 degrees

Since there are two possible answers for angle C, we organize our work into a table.

when angle C = 117.7 degrees:
c / sin 117.7 = 5 / sin 28
c sin 28 = 5 sin 117.7
c = 9.4

when angle C = 6.3 degrees:
c / sin 6.3 = 5 / sin 28
c sin 28 = 5 sin 6.3
c = 1.2

last example: Find all possible angles for angle C.

19 / sin 19 = 23 / sin B
19 sin B = 23 sin 16
sin B = 23 sin 16 / 19
angle B = 19.5 degrees

Solving for the related angle:

angle B_r = 180 - 19.5
angle B_r = 160.5 degrees

Solving for angle C:

when angle B = 19.5 degrees:
angle C = 180 - 16 - 19.5
angle C = 144.5 degrees

when angle B = 160.5 degrees
angle C = 180 - 160.5 - 16
angle C = 3.5 degrees

And that's the end of today's lesson. I hope that helps.

Er.. do I have to pick a scribe for tomorrow? If I have to.. well, I pick:

rus-L. You can do it!

Scribe Responsibilities

2006-02-28T21:34:00.000-05:00
Hey guys,

Remember the "scribe responsibilities" that we talked about before. It's been two days now with no-one filling in as the scribe and noone taking that responsibility.

ASSIGNMENT #3

2006-02-26T22:12:00.000-05:00

Here's #3. Is this what you're talking about? So, this is how I drew my diagram, hoping it's correct.
You find the sides 'c' and 'a' by using sine law.
You have to add all of the sides together to get the total distance that the person traveled.
I hope this helps. If it's wrong, I'm sorry. HAHA!

The homework

2006-02-26T19:22:00.000-05:00
Can anyone help me on # 3?? Do I use Tan or Sine to figure out the side?

oh my goodness ....

2006-02-26T16:08:00.000-05:00
HELP HELP HELP HELP!!

I am having the hardest time with the assignment that is due tomorrow. I have literally been staring at it, and I have no idea what to do, or where to start. Can someone help me? If not, I know I'm going to fail......yikes..

I'm super frustrated....

First Weekend Scribe Post

2006-02-24T22:54:00.000-05:00
My first time doing the scribe post and thankfully I got the weekend job. It's not going to be a very long one since we did have the quiz in the middle of class.

After the usual mental math we started to correct the homework sheets that we had for homework the previous night, we also got some hints for the hand-in assignment that's due on Monday at 9:00:

(My computer's been acting nutty with any kind of pictures lately so you'll just have to bare with my descriptions for now, sorry guys.)

#5) For the diagram you drew one large square and a smaller one in the center of the first. Inside the lengths are of course 30 and 40 meaning the area is 1200. on the outside the width and length is representing by X with the area being 984. The equation to use is y=(40+2X)(30+2X).

#6) Replace TAN and SIN with X and Y:

Y=2XY^2-2XY-12X
Y=2X(Y^2-Y-6)
0=SX(Y-3)(Y+2)

Since the question is worth 5 marks that meant that there are 5 answers, so all you have to do now is solve the equation.

And that's all I can remember from today, oh yeah as for the quiz the only thing I can say is that it's not hard but then again it's not easy, can't give you much more then that since I don't remember much of it, it's how my mind works once a test or semester is over my mind gets wiped blank except for the things that were really pounded into my head, it only starts to come back when I'm back in class.

The next scribe will be, Jonathon!

LOST: episode 2

2006-02-24T12:01:00.000-05:00
I am SCREWED! I'm okay with the lessons I missed on Tuesday since they were fairly easy; also with wednesday's topic - about related angle -- sin, cos, tan and stuff, then the quadrants. Those are some easy stuff. But wut are these 'bquadrantrent' and so forth?! I'm totally lost! And I'm missing that quiz! I am screwed! Then that assignment is pretty hard. I don't get the last two questions. I amy be slow at understanding Dino's post but it's hard to understand especially if I missed the class. It would be easy for those who actually witnessed how things happen. Someone help me! UH! I hate getting sick!

Question... again....

2006-02-24T00:07:00.000-05:00
Mr. M... is there any way to solve for sine, cosine or tangent manually?...

Also, i wanted to ask if:

tan=cos/sin or tan= sin/cos

Which one is right??

I think this actually has something to do wwith the assignment... since if tan= cos/sin... i could cancel the sin and simplify the problem...

this post was made late coz i went to work...

Solving Trig. Equations

2006-02-23T17:54:00.000-05:00
To Solve Trig. Equations

There are two simple rules:
Find values which satisfy the equation.
The domain will always be from [0°,360°], unless stated.
Here are some examples to help you:
ex. sinø = 0.78615 Solve!
To solve this question, you first take your (TI 83) calculator and press the (2nd) button, then the (sin-1) button, and type in the number 0.78615. then the calculator will give you a number of 51.83° (round of to two decimal places). Then you need to find out which quadrants have a sin value that is positive (Wednesday's class), which is bquadrantrent 1 quadrantrent 2. So you already hquadrantrent 1 done, it is 51.83° , so to figure out the degree valuequadrantdrent 2 you take 180° ( 180° is the maximum degree value you can have in Quad. 2) and subtract 51.83° from it . The value for Quad. 2 is 1that's°, thats it your done.
sinø = 0.78615
ø= 51.83°
ør = 180 - 51.83 = 128.17°
ex. cosø = .43214 ( for (cos) use the same method as above but press the (cos-1) button instead of the (sin-1) button)
ø = 64.40°
ø= 360 - 64.40 = 295.6° ( you have to subtract it from 360° , because cosine is positive in quad.1 and quad. 4).
Now try this one on your own. (answer at bottom of page)
2 sinø-1=0 ( hint move the 1 to the other side)
For homework do exercise #9 and ask Mr. Malandrakis for a sheet for review, there is also a quiz tommorow on quad. Equations and trig. equations. So study and Good luck.
Answer to question uptop is ø = 30° , 150°
Tommorow's scribe is Jason.
By: Berardino Petrelli (Dino)

Angles in standard position, reference(related) angles, and the like.

2006-02-22T19:26:00.000-05:00
Helloo fellow scribes ...

I am your scribe for today, and hopefully this will start the scribe program up again, and no one will get confused and mxed up again.
Today in class we learned about Angles in the standard position. We started off with this recall about quadrants: the top right section of a coordinate plain is the first quadrant. Moving counter clockwise from the first quadrant makes the top left section of the coordinate plain the second quadrant, and so on.

When the postive x axis is the first side of an angel, that is called the inital side or arm. When a ray is drawn through the origin between 0 and degrees, this ray is called the terminal side or arm. The angle that the initial and terminal sides make is called an angle in standard position.

When an angle is in standard postion:

1) it's initial side is the x axis, and the terminal side is formed by a ray extending out from the origin.
2)the angle is measured in counter clockwise direction

Consider:

Remember : r= the square root of x^2 +y^2
if (x,y) is any point on the terminal side of an angle in standard position:
The SIN of the angle is y/r
The COS of the angle is x/r
The Tan of the angle is y/x

This diagram lays out just what you have to do to find the angle you are looking for.
Remember : unknown angles are usually expressed with the greek letter theta.
Some angles have special relationships with each other. This relationship could be referred to as a Reference/related angle.
A related/reference angle is the acute angle formed between the terminal side of the angle and the nearest x axis.

We were also given a summary of reference angles, and basic facts about SINE and COSINE graphs. This I do not need to type here because I am sure you all have it. Also, there is a project due on monday, about quadratic functions, trigononmetry, and a little bit of mental math. You can work with anyone, just make sure you hand it in on monday!
Homework for tonight is Exercise #8....
If anyone has any questions, or comments, feel free to contact me. I may have made a mistake on something, so feedback is always appreciated. I am kind of in a hurry, but I hope this post does justice to all the ones that have come before it!
And the scribe for tomorrow is....Dino!
Good luck man, I know you will do well! Keep up the good work guys!
Chow.

sine and cosine graphs...

2006-02-21T22:54:00.000-05:00

This is something I got from my trigonometry class last year in the Philippines...

I hope this can help most of you guys in figuring out how the graphs work...

sin graphs always start at zero (only if there is no vertical shift)
+ cos graphs start at the MAX point
-cos graphs start at the MIN point

These graphs have zero vertical shift and zero horizontal shift... so it's just the basic form of the graphs...

Also...

sin -360= 0
sin -270= 1
sin-180= 0
sin -90= -1
sin 0= 0
sin 90= 1
sin 180= 0
sin 270= -1
sin 360= 0

cos -360= 1
cos -270= 0
cos -180 = -1
cos -90= 0
cos 0= 1
cos 90= 0
cos 180= -1
cos 270= 0
cos 360= 1
I know this isn't that much of a help since these are obvious data in periodic graphs...

y= asinx + k y= acosx + k

k= vertical shift
a= amptitude... (did i spell it right?) or also known as the stretch factor of a periodic graph

Well... that's how i understood the stuff...

Hope this can actually help...

Sorry if the post was made this late coz i was finishing some other stuff...

What happened..

2006-02-21T22:16:00.000-05:00
I'm getting really anxious. Where's the scribe post? I have no idea what's going on. I and Vince are worried. If this goes on...

Mr. M, I'm going to miss tomorrow's class too because I have to host games for the junior high kids.

Scribe responsibilities

2006-02-21T21:23:00.000-05:00
This is a quick comment based on what I've seen the last couple of days. Yesterday Jojo, who was the assigned scribe, didn't get a chance to make a post because his internet wasn't working. That's no problem but he assigned Andrew to be today's scribe and it appears that he perhaps forgot to make a post and Candace made an open post as you can see wondering what hapenned in class today.

Keep in mind folks, the main reason this blog exists is FOR REASONS LIKE THIS, when people miss class or are not understanding what went on in class. I have a challenge for, from what I can tell so far to be a very mature, intelligent, and motivated class (that's you guys!). If one of you dosen't see a fresh post by let's say around 8:30 or so at the latest, please take the mature and responsible initiative, for your fellow class members, and voluntarily be the "fill-in" scribe for the night. If it even makes a difference to one or two people, like Candace tonight, then it's all worth while.

Remember you're only "required" to post on class lessons once every 25 classes or so, that's once every month and a half or so. So please adhere to this very miniscule responsibility as some of your classmates truly depend on the scribe to do their job as it just might make the difference for them.

today??

2006-02-21T21:03:00.000-05:00
Hello everyone...
ok, well I was just wondering what happened in class today?? Also who is the scribe for today??

??

2006-02-21T20:45:00.000-05:00
Are TI-89 calculators the same as the calculators we use?? or even somewhat similar...

Sunday Funday

2006-02-20T01:49:00.000-05:00
Well, well, well, it's the end of another weekend, and it's the start of a new week. Why don't we start a new week with games, hey? (I got the idea from Mr.K ",)

This is a game from last semester's Pre-Cal 20S blog. This is a hard one, I tell you. It took me weeks to figure it out. First person to get out of the room gets a free pack of gum from me. =)

Scribe # 4: Weekend Scribe

2006-02-18T22:08:00.000-05:00

SORRY FOR THE ERRORS. BLOGGER.COM HATES ME! HAHAHA..

Scribe #4 : Weekend Scribe
.
.
.
Hey guys! So.. a test on Monday, right? I'm actually kind of panicking right now. I might not do so well on the test. Well, before I state what we did on Friday, let's review some important things that we have learned in the past several days. These terms will be essential for Monday's test.
1) Quadratic Function - any function or equation with degree 2
ex.
f(x) = x2 + 2
2) Parabola - the graph of a quadratic function
ex.

3) General Form - equation of a quadratic function written in: f(x) = ax² + bx + c
ex.
y = 4x² + 5x + 15

4) Standard Form - equation of a quadratic function written in: g(x) = a(x - h)² + k
g(x) = 4(x - 3)² + 7

5) Zero Product Property - one of the many methods of determining the roots of a parabola
ex.
y = x² - 3x - 4
0 = x² - 3x - 4 [ make y = 0 ]
0 = ( x + 1 ) ( x - 4 ) [ factor ]

[ solve for x ] 0 = x + 1 x - 4 = 0
-1 = x x = 4

[ state the roots of the equation ] roots = -1, 4

6) Completing the Square - a method used to transform an equation of a parabola from general form to standard form
- shown by Alge-tiles

ex.
y = x² + 4x - 5
y + 5 = x² + 4x [ move 'c' over to the other side (do not forget to change its sign) ]
y + 5 + 4 = x² + 4x + 4 [ add 'c' to both sides by finding the half of 'b' and the square of whatever the value might be (ex. half of 4 is 2... 2² is 4... therefore, 'c' is 4) ]
y + 9 = ( x + 2 )² [ factor ]
y = ( x + 2 )² + 9 [ move 9 over to the right side, this will be your standard form of the equation: y = x² + 4x - 5 ]

7) Characteristics of a Quadratic Function:

a. VERTEX - a point (x, y) directly through the center of the parabola
b. AXIS OF SYMMETRY (a.o.s. - Mr. Malandrakis' version) - equation of a line passing through the vertex
c. OPENING - if a > 0, parabola opens up (has a MIN)
- if a < color="#3333ff">MAX)
e. ROOTS - where the parabola crosses OR touches the x - axis [ x-intercept, zero ]
f. DOMAIN and RANGE - limits of x (domain) and y (range) axis
+ ways to express DOMAIN and RANGE:
Set Notation
Number Line
Interval Notation
g. WIDTH - how wide the parabola is
- depends on the coefficients of 'x'
- if coefficient > 0, parabola is THIN
- if 0 > coefficient <>
8) Vertical Shift - identified by the letter 'c' of the general form
- if 'c' > 0, parabola shifts up from the vertex
- if 'c' <>
9) Horizontal Shift - graph is shifted h (absolute value of 'h') units horizontally

10) 'a' controls the width and the direction of the
11) 'h' controls the horizontal shift
12) 'k' controls the vertical shift
NOW... this is what we did on Friday (that took a while)..
A pre-test for Monday about Quadratic Functions1) An archer shoots an arrow into the air that its height at any time( t ) is given by the function h(t) = -16t2 + kt + 3. If the maximum height of the arrow occurs at time t = 4, what is the value of k?
SOLUTION
h = -k / 2a
4 = -k / 2 9 16 0
4 (32) = -k / ~~-32~~ (~~-32~~)
-128 = -k
-128 / -1 = -k / -1
128 = k

2) [pretty straight forward]

3) What is the y-coordinate of the vertex for the function f(x) = x² - 12x + 7?

SOLUTION: use complete the square or the 'h' formula...
f (x) = f(x) = x² - 12x + 7
y - 7 = x² - 12x
y - 7 + 36 = x² - 12x +36
y +29 = (x - 6)²
y = (x - 6)²- 29

ANSWER: -29

4) Find the exact values of the root of the quadratic equation: x² + 8x + 2 = 0

SOLUTION: complete the square
x² + 8x + 2
x² + 8x = -2
x² + 8x + 16 = -2 +16
x² + 8x + 16 = 14
root of ( x + 4 )² = root of 14
x + 4 = + or - root of 14
x = -4 + or - root of 14

AND THE REST ARE JUST ON YOUR SHEET. I WAS GOING TO WRITE MORE BUT I THINK IT'S POINTLESS BECAUSE EVERYONE HAS THE ANSWERS AND SOLUTION. YES, I DO GET LAZY (sometimes). Well, good luck to you guys! And good luck to me too! HAHA! I'm done for now... I hope this post helped some of you. I'm so tired. It took me 2 and half hours to do this thingy. Perhaps, it's also because my computer is slow.

The NEXT SCRIBE is JOJO! Sorry, man... Just do it...

REYES OUT!
PS.. DONT FORGET OUR HOMEWORK! > exercise 7.. and the sheet Mr. Malandrakis gave us..

HEADS UP!

2006-02-17T20:57:00.000-05:00
Hey guys,

I just found some errors on the answer key on the Quadratic Functions review Mr. Malandrakis gave us. The answer for 1a is the answer for 1b, and the answer for 1b is the answer for 1a. I hope this reduces the confusion.

Have a jolly weekend!

I'm the scribe for Friday.

2006-02-16T22:16:00.000-05:00
So guys, I'm the scribe for Friday. But is it okay to do it sometime on Saturday since it's the weekend? It's because I signed up for leadership tomorrow and I won't be home until 9:30. So I'm wondering if it's okay to do it the next day. I don't consider Leadership more important than Pre-cal. But, I didn't know I'm going to be the scribe for tomorrow. So yeah.. can I post it on Saturday?

Scribe # 3 "Applications of Quad. Functions"

2006-02-16T20:37:00.000-05:00
Okay lets get down to what we learned today in class. Most of what we did today in class was based on what we did this whole week with completing the square and changing expanded form into standard form using the skill of completing the square. All we did today was applied it to real life situations.
We were given a sheet and Mr. M walked us through it.

Before I get to the questions I'd like to point out some of the instructions that were given to us to help us understand. Such as read the question carefully. We were also instructed to look for any words which could help make the question clearer such as MAXIMUM or MINIMUM. And drawing a diagram can also prove useful too.

Question #1

What is the maximum rectangular area that can be enclosed by 120 m of fencing, if one of the sides of the rectangle is an existing wall?
*NOTE: the word "maximum" was used

Work and Explanation

First we must come up with a base equation.
since we know the perimeter is 120 and that the sum of the fenced sides must equal 120 we come up with equation: 2w+l= 120

Now we make a statement for l.
we do this by moving the 2w over.
l= -2w+120

Now to find w we let y = area.
y= (-2w + 120) (w)

Convert into Standard form.
To do this we will use the skill of completing the squares to convert this into standard form.
y=-2(w^2-6Ow)
-1800=-2(w^2-6Ow + 9oo)
=-2(w-30)^2 + 1800
v=(30+1800)

Understanding the vertex.
The "k" value in the vertex is the area of the rectangle.
Therefore, max area = 18OO m squared

The homework for tonight is Exercise 6. The scribe for tommorow is vincent.

The Scribe List

2006-02-15T22:20:00.000-05:00
Now Mr. Malandrakis has implemented the "scribe" posts, we should have a Scribe List. I'm making this, and I will update it everyday so that we will not lose track of who will be the scribe. If you see someone's name crossed off on this list you CANNOT pick them as scribe again, unless we start a new cycle. We start a new cycle when everyone has been picked to be a scribe.

You can access this by the Links section, and you may want to take a look at the scribe list before you pick a scribe.

Cycle 2

rus-L
~~John D. - #12~~
Jojo
KAT L.
Jamilyn G.
Jason
Kristin_R
~~tim-Math-y~~
A.Z
StevenF
sacramento
Carla
miles_23
Jason
candace
Dino
vincent
Rick
albert

Scribe Post #2

2006-02-15T16:56:00.000-05:00
Well hello everyone.. I'm Tim-Math-y your scribe for today. I'm going to try and do this quick so everyone can read it and so that it won't take up too much of my time.

Today, we started off class with Mental Math which included 11 questions. Next we did a 10-15 minute review of last night's homework: Ex. 3 (questions 5a, 5b, 7, 8, 4c, and 11).

Now onto today's lesson!

TRANSFORMING GENERAL TO STANDARD FORM

*This proccess is referred to: completing the square.

When thinking about this, we think back upon "Alge-tiles".

---Sigh.. unfortunately, it says error when I try to add images.. so please bear with me on this Scribe post holding no visual pieces... (-.- )---

y = ax² + bx + c ---> General / Expanded Form

>>>Consider:
- a square with length of (x+1) and a width of (x+1)
- the area would be expressed as (x+1)(x+1) --> x^2 + 2x + 1

>>>Consider:
- x^2 + 2x + _?_
- to complete the square, the unknown is equal to 1.

>>>Consider:
- x^2 + 4x + _?_
- to complete the square, the unknown is equal to 4.

Do you see how to find the unknown value of C by looking at what's givin?

*In a quadratic func in general form:
C = (b/2)^2 ... is used to complete the square

>>>a perfect square trinomial is where a trinomial has 2 equal factors.

Example: x^2 + 6x + 9 = (x + 3)^2
Example: x^2 + 8x + 16 = (x + 4)^2

>>>convert from general form ----> standard form by completing the square.

Example:
y = x^2 + 4x - 5
Step (1.) y + 5 = x^2 + 4x
Step (2.) y + 5 + 4 = x^2 + 4x + 4
Step (3.) y + 9 = (x + 2)^2
Step (4.) y = (x + 2)^2 -9
V = (-2, -9)

Example:
y = 3x^2 - 12x + 16
y - 16 = 3x^2 - 12x
y - 4 = 3(x^2 - 4x)
y - 4 = 3(x^2 - 4x + 4)
y = 3(x - 2)^2 + 4
V = (2, 4)

Now that those two are done, Mr. Malandrakis decides to blow our minds with the so called "friendly" fractions..

Example:
y = -2x^2 - 5x + 3
y - 3 = -2x^2 - 5x
y - 3 = -2(x^2 + 5/2 x)
y - 3 - 50/16 = -2(x^2 + 5/2 x + 25/16)
y - 98/16 = -2(x + 5/4 )^2
y = -2(x + 5/4)^2 + 49/8
V = (-5/4, 49/8)

__________ / / __________

*In any quadratic function in standard form:
h = -b/2a
K = 4ac - b^2/ 4a

Example:
y = x^2 + 4x - 5

h = -4/2
h = -2

K = (-2)2 + 4(-2) - 5
K = 4 + 8 - 5
K = -9

V = (-2, -9)

And that is all folks! Wow.. I can smell dinner from downstairs.. I'm sure HUNGARY :). But before I am done, have to mention some things ^^.

Mr. Malandrakis: "It is very important to take some time to read what's on the blog."
Tomorrow's Scribe: (sorry no drum roll :P) It is.. MR. "StevenF" because he said it's fine by him ^.-
Homework!: (which is essential for this course) EXERCISE #5

And for sure now, I think that is all.. see you guy tomorrow! I'm going to eat dinner :P

Valentine's Scribe

2006-02-14T18:39:00.000-05:00

Hello!

It's true that we learn new stuff everyday. Today in Pre-Cal, we learned about determining characteristics of a quadratic function algebraically, or in other words, without you using a graphing calculator.
________________________//_____________________________

DETERMINING CHARACTERISTICS OF A QUADRATIC FUNCTION ALGEBRAICALLY

First of all, we now know what is the standard form of a quadratic function. Now on this lesson, we used the expanded form mostly.

The expanded form of a quadratic function is:

y = ax² + bx + c

1) a represents the width and direction of the opening of the parabola.
2) b creates an oblique translation (a translation of a line that is neither parallel nor perpendicular)
3) c is basically the y-intercept of the parabola.

________//________

Recall: Find the roots and the vertex of f(x) = 6x² + x -12 using zero product property.

Factoring the expression:
f(x) = 6x² + x -12
f(x) = (2x + 3)(3x - 4)

Ah.. this involves fractions. So this is how we set it up to find the roots:
2x + 3 = 0 | 3x - 4 = 0
(note: ALWAYS organize your solutions.)

Solving for the roots:
2x + 3 = 0 | 3x - 4 = 0
2x = -3 | 3x = 4
x = -3/2 | x = 4/3

So the roots of the equation f(x) = 6x² + x -12 are:
roots: x = -3/2, 4/3.

* the x coordinate of the vertex a.k.a horizontal shift is found by finding the midpoint of the roots:

h = (x1 + x2) / 2

To find the horizontal shift, we just need to substitute x values in the equation with the roots.

h = (-3/2 + 4/3) / 2

Ah... dissimilar fractions. Now we need to find the LCD of the equation so we can solve them. LCD = 6.

h = (-9/6 + 8/6)/ 2
h = (-1/6) / 2
h = -1/12

Now we have found the x coordinate a.k.a horizontal shift of the vertex. We just need to find the y coordinate a.k.a vertical shift so that we can find the vertex. To find the horizontal shift, we substitute "h" back into f(x).
[*Note: This involves ugly fractions for this particular solution, so please just bear with me.]

And oh, don't make it hard for yourself, just use the factored form of the expanded form of the parabola since it's easier. This is one handy tip coming from me. ;)

f(-1/12) = [2(-1/12) + 3][3(-1/12) + 4]
f(-1/12) = [-1/6 + 3][-1/4 + 4]
f(-1/12) = [-1/6 + 18/6][-1/4 + 16/4]
f(-1/12) = [17/6][17/4]
f(-1/12) = -289/29

Now we have the y coordinate of the vertex a.k.a horizontal shift. Therefore, we can state that:
vertex: (-1/12, -289/29)

ex. Find the roots, vertex, axis of symmetry of f(x) = x² + 4x -5

Solving for the roots:
x² + 4x -5
0 = (x + 5)(x - 1)

x + 5 = 0 | x - 1 = 0
x = -5 | x = 1

Therefore, we can say that the roots: -5, 1.

Finding the x coordinate of the vertex a.k.a horizontal shift:

h = (x1 + x2) / 2
h = (-5 + 1) / 2
h = (-4) / 2
h = -2

Now that we know the x coordinate of the vertex a.k.a horizontal shift, we can also state the axis of symmetry which is:
x = -2

Finding the y coordinate of the vertex a.k.a vertical shift:

f(-2) = (-2 + 5)(-2 -1)
f(-2) = (3)(-3)
f(-2) = -9

And there you have it! We can now state the coordinates of the vertex:
v:(-2, -9)

We also came up with the rules of finding the roots, axis of symmetry, and vertex of a quadratic function in expanded form.

1) To find roots:
- factor y = ax² + bx + c using zero product property

2) To find axis of symmetry, find the midpoint of the two roots. MAKE SURE to always write it like this:
x = h

3) To find vertex, substitute h for f(x) to get the k value. MAKE SURE to enclose the points of the vertex in round brackets as it is a point in the line.
vertex:(h, k)
_______________________________________________________

And there you have it! That's today's lesson. Sorry it took a long time to post. Homework for today is Exercise 4, and tomorrow's scribe is..... (drumroll please)...... tim-Math-y. Haha.

Have a great evening people! Get out there and commit heinous acts of kindness.. Cheers!

the quizz! :(

2006-02-12T22:22:00.000-05:00
well...so the quiz..yeah i didn't do so well on that..hahh. i knew how to like do all the other stuff. that was easy, but the last page screwed me over. i like didn't know how to do the last three or four questions, which is yenno...bad.hahah...well i hope when we go over it i'll be able to understand...this is my first real blog, if this is a blog. i don't know what i'm doing or if it is right, so can someone tell me how i'm supposed to used this....hahh i'm a BIG noob. hah yupp..well see you guys at school tomorrow! bye:D
-romina

Quiz correction

2006-02-11T01:58:00.000-05:00
Hey there folks,

A minor correction from Friday's quiz, which we'll go over in class on Tuesday (Monday is the assembly) anyways but I'll give you a heads up...it was the last question on the quiz where you were asked to write the equation of the graph of the parabola. You were asked to leave your answer in expanded fom, and the question was supposed to ask you to leave it in expanded form. If you left it in standard form you received full marks. Just a little "FYI" for everyone who checks out the blog on the weekend. Also another "FYI" the marks (overall) on the quiz were great. A 72% class average, which is rare and outstanding. Great job overall and keep up the fabulous posting.

Also...in case you guys haven't realized, if you look in the links section on the side, you'll find some great sites that really explain quadratic functions in full colour and worked out examples. You should check them out if you're still having trouble.

Blog, Quiz

2006-02-10T22:42:00.000-05:00

BLOG

* I think this blog is cooler than i thought. I just regret that I didn't view it the day before the quiz (yesterday). Man, seriously. How can you guys spend so much time on this?! I used to update my AA page like every time. But I got lazy. I know this is not that much different form those pages but I think I lost my touch on web designs. You guys did a good job on your posts.

* I have a question. Do we have to write like this in here every time? proper spelling, capitalization, and punctuations? coz i wanna write like this... its so much easier..

* Anyways, I decided I'm going to go here every night to see the updates that might help me dig deeped into the topic.

QUIZ

* That quiz SHOOK me! I thought it was pretty easy even though I kind of screwed up on the last page. I didn't understand the EXPANDED FORM. Is it this one? :

y = ax² + bx + c

* I didn't know what to do then. I just left my answer in standard form:

y = a(x - h)² + k

* I think I will get that wrong.

* Also, about Zero Product Property... Are you supposed to enclose the points in BRACKETS? I didn't do that actually. I thought it would be okay just to leave it the way you factored it. AHHHH! I hope I did well on that quiz.

* I think QUADRATIC FUNCTIONS is easy overall. I just have to study more. Pay attention to little details since they might play an important role on the answer/equation.

Look I'm Back...

2006-02-10T22:39:00.000-05:00
Hi fellow bloggers. If you don’t know my name… My name is Richard. If you don’t know why I’m here… Then… I don’t either. But anyways. The main reason I’m like on your blog is that I’m here to help you guys. Yes as impossible as it may seem. I’m here on the blog. So live with it. I actually passed this course last semester if that’s so hard to believe. I know it is for me. But in other cases I’m just around to help the fellow bloggers blog and to give at least a tiny bit of incite into every thing you do on the blog. Yes I’ll be commenting allot and allot and that means allot. I’ll also be posting up allot of blogging tips and crazy help from myself the remnants of the first semester bloggers. Who knows… A study group may get back together again. Hahaha. Well anyways. I just wanted to say hi and bye. I’ll be back…

Oh yeah. Here’s a tip for Mr. Malandrakis… Make a chat box. Because it’s actually a good study tool and it creates a classroom outside a classroom. ;) And... If you want to read the course... Then.. Read PC30S first Sem. It's all there. There's some intresting posts there too. You know... Pop goes the weasel... Hahahahaha

Richy Out.

Some Blogger Help

2006-02-10T21:07:00.000-05:00
Hello there!

Now, making a post with a lot of detail takes time, like the one I just did. Thankfully, I got some help from the post that I found in the last semester's Pre-Cal 30S class of Mr. K. It was created by Graeme and Richard and it sure was a lot of help. This post contains a lot of stuff you can do in Blogger e.g. creating tables, making a marquee, stuff like that. I hope this helps.

Have a good weekend!

Edit: Mr. Malandrakis, when can we have a chatbox? It'll be really helpful and marvelous to have one. It'll encourage student-to-teacher and student-to-student interaction a lot. And a del.icio.us tag can help us too. In case you don't know, when we are bored, we go to Google to find some sites that may be able to help us out in the course. Cheers!

Remember: You can find everything in the Internet. If you can't find it, you are not looking hard enough. ;)

And I found another post that is really important. NO, NO, NO, NO, NO! We don't ever do this. Never. And I'm sure that some of us from Mr. K's class can remember this. =D happy times

Quiz

2006-02-10T17:25:00.000-05:00
The quiz was actually longer than i expected. I'm used to doing quizzes with only 5 questions that's usually out of 14 marks. But i guess i did pretty good. I just had trouble with question no. 3. But i got it after i handed in my paper. I was surprised that i answered all those questions. I guess i understand it now.

my first post

2006-02-10T14:54:00.000-05:00
This is my first post, just seeing if it works. Great job guys!

Scribe Post?!

2006-02-09T21:02:00.001-05:00
Hi everyone!

I just decided to make a scribe post about what we learned today and I have nothing to do. Still concerns quadratic functions, it's turning out to be alright.

Anyway, we started the class with Mental Math. It wasn't that bad.. Then we moved on to homework. I asked how to solve number 11 on Exercise 2. Everyone didn't get that question, so we went into small groups trying to figure out how to solve it. The question is:

Jim and Kim each have money to buy ice-cream cones. Unfortunately, Jim is 24 cents short of the price of a cone and Kim is 2 cents short. They decide to pool their money and buy a single cone, but they discover they still don't have enough money. What is the cost of an ice cream.
We set it up like this:

let x = cost of ice cream
(x-2) = Jim's money
(x-24) = Kim's money

Then, we combine similar terms.
(x-2)+(x-24) = 2x-26

This is not a linear equality, this is a linear inequality as it is said in the problem that when they pool their money, they are still short. So we set it up like this:

2x-26 < (6)

We solve the equation like a linear equation.
Putting the xs on one side and the constants on the other side:

2x-x < (6)

Combining like terms:
x < (6)

Now, if Kim has (x-2) cents (24 cents) and it's not enough, then one ice cream cone must be 25 cents. Let's see if that's true:

If x=25;
(25-2) = 23 cents (Kim's money)
(25-24) = 1 cent (Jim's money)
Total: 24 cents

Therefore we can say that one ice cream cone must be 25 cents. This may become useful in future encounters, so take note.
_____________________________________________________________

Today's lesson: More on Quadratic Functions

The standard form of a quadratic function is:
y = a(x-h)² + K

1) The vertex of this equation is: (h,k)
2) Axis of symmetry: x = h
3) If a > 0, width is thin.
If a < style="color: rgb(204, 51, 204);">a controls width and opening.
4) "h" controls horizontal shift.
"k" controls vertical shift.
ex (Sorry for the poor graph, I'm a noob at MS Paint)

Here, we plot a parabola with the coordinates (3,0), (1,4), (0,3) and (-1, 0). We need to
state all eight characteristics of this parabola and create an equation(y = a(x-h)² + K ).
_______________________________________________
Recall: The eight characteristics of a parabola are:

1) Vertex
2) Axis of symmetry
3) Domain
4) Range
5) Roots
6) Opening
7) Max/Min
8) Width
_________________________________________________________

Now we state the characteristics of the parabola above.
1) Vertex: (1, 4)
2) Axis of symmetry: x = 1
3) Domain: (-∞, ∞)
4) Range: (-∞, 4)
5) Roots: (3, 0), (-1, 0)
6) Opening: down
7) Max/Min: max
8) Width: none, since it's the base width (note: we can either write a dash (-) or none if it's the base width. If a = 1, it's the base width.

Since now we have enough information, we can create the equation.
(y = a(x-h)² + K )

Substituting the vertex coordinates:
y = a(x-1)² + 4

To complete the equation, we need to find a. Any point from the parabola won't matter. I'm picking (0, 3) for this one.

Substituting x and y coordinates:
3 = a(0-1)² + 4

Solving:
3 = a(1)² + 4
3 = a + 4
-4+3 = a
a = -1

Final equation:
y = -(x-1)² + 4

ex If (1, 3/2) lies on this parabola y = (x-2)² + k, find the minimum value.
Substituting the x and y coordinates:
3/2 = (1-2)² + k

Solving the equation (k):
3/2 = (-1)² + k
3/2 = 1 + k
-2/2 + 3/2 = k
1/2 = k
Now we have the minimum value, we can complete the equation.
y = (x-2)² + 1/2

Phew. Now, on to the last example.
Find the root(s) of y = x² - x - 2.

What we would do normally when we see x²-x-2 is factor them out. So we do that to find the roots.
0 = (x - 2) (x + 1)
x = 2, -1

* Note: This is called zero product property.
ex. Use zero product property to find the roots of: y = x² + 5x + 6
0 = (x + 2) (x + 3)
x = -3, -2
_________________________________________________________
End of lesson. Phew, that took a long time. And oh, we have a quiz tomorrow, and homework is exercise 3.

Great Posts!!!!!

2006-02-08T22:46:00.000-05:00
Katrina, that was a fabulous post. I'm not sure if you wrote this blog tonight or back in semester one but it was great. The diagrams were out of this world, excellent job on that. Carla, I hope that helped a bit with your trouble with understanding interval notation.

Just remember that interval notation is just a simpler way of expressing domain and range. You use round brackets when that value is NOT included and square brackets when it is included.

If it still dosen't make sense by tomorrow, remember to bring it up after mental math! Great job again!!!

BLOGGING ^_^

2006-02-08T17:52:00.000-05:00
I'm not sure how this blogger is going to work...
If we`re going to have a "scribe" everyday like what we used to do in Mr. K's class or what.
All I know is that blog thingies like this is, indeed, VERY helpful.
Last Semester, our blog was of BIG help to everyone. I'm sure this one will be too.

Anyways... For those who are having a hard time figuring out how to work this blog thing... You guys can go here:
BLOGGERS'R'US.

This is a post made last semester by one of my classmates in Mr. Kuropatwa`s class.

What do I think so far?
Well, it's almost the same thing like last semester with Mr. K. I guess I just understand it a little better now.

What Mr. Malandrakis said about "not doing homework or putting enough effort" is so true. If you don't do homework/put enough effort you'll have a hard time passing this course. SOOO... ALWAYS DO YOUR HOMEWORK :)

Since some people still get a little confused about the parabolas, I hope this little summary will help you guys understand it a lil` better ;)

QUADRATIC FUNCTION

It is a function degree 2 a.k.a. parabola

DOMAIN: all possible inputs for a given function or relation
RANGE: all possible outputs for a given function or relation

Properties of a Parabola

Vertex: the maximum or minimum value for a parabola. Its coordinates are always written (h,k). The max or min value is always y=k

Roots aka zeroes/x-intercepts: the points where a parabola crosses the x-axis. a parabola can have 0 , 1 or 2 roots.

Axis of Symmetry: The line of reflection through with one half of the parabola is symmetrical to the other half. Its equation is always x=h

How to write an Interval Notation

1.) Write the smallest number in the required interval.
2.) Write a comma.
3.) Write the largest number in the required interval
4.) Use square brackets around the end points that are included in the interval.
5.) Use round brackets around the end points that are not included in the interval.

!! Important !!

i) Infinity always has ROUND brackets.
ii)Interval Notation can only be used with the reals.

ON A NUMBER LINE we use an open circle to the points that ARE NOT INCLUDED and a shaded circle to the points that ARE INCLUDED.

okay... well... this is it for today... I still have to do my homework.

I hope this helped you guys!

HAPPY BLOGGING!

Pre Calculus 30S

2006-02-08T17:22:00.000-05:00
I'm used to blogging so i know what to do. But using it in a clss, no. I guess i need to get used to it. I'm still confused about the Parabolas. Because we didn't cover the Interval Notation. I don't know which to put, either the round or straight brackets. Actually, i didn't understand it at the first time. But when i got home, i really went over it one more time. I guess i kind of understand it now.

I do like math, but Algebra part only. When i listen to our lesson, there are parts that just go through my ears and not understand it. I don't know why. I'm just really kind of confused. My friend said that that's only for first time, that as more time and lessons go by, i'll understand it. I hope so. Because i think i'm going to fail this course if i don't understand it soon.

Thoughts

2006-02-07T21:36:00.000-05:00
So far I really don't have any problems with quadratic functions,it isn't exactly hard but I still end up getting confused a few times with things such as interval notation which was thinking about when I should use a round bracket or a square bracket. Since classes have only just started again I don't really have any further thoughts about this class so there's really nothing else I have to say.

Quadratic Functions, Parabolas, the whole ball of wax.

2006-02-07T21:01:00.000-05:00
Pre Cal 30s is quite interesting so far, I must say..

Luckily, being in Mr.K's class has helped prepared me for parabolas. Mr.K tends to "dip in" to future pre cal experiences, which is really nice, because I am not freaked out when I get to new information.

At first, I really was a little confused by parabolas. I knew what they were, and what they looked like. The thing that is messing me up a little bit right now is using the graphing calculator. I just bought it a few days ago, and I am still getting used to using it. I know I will get used to it in a while, and I will get better and better as I practice, it'll just take me some time.

Some characteristics about parabolas and quadratic functions? Well, we know so far aboput roots, zeros, width, sketch, axis of symmetry, the up or down shift of the parabola on the y-axis, and so on and so forth.
I honestly didn't think that there would be more characteristics of parabolas/quadratic functions.... I really can't even think of a possible characteristic...we've covered so much already.

I am pretty pumped about pre calc, but I am already stressing about the 3 hour exam. Eek, working for three hours straight, and in an EXAM? Eep. Maybe it won't be so bad, who knows.
I really like math for some reason, even though sometimes, I am not so good at it. I am eager to learn this semester, and I can't help but anticipate what is to come.

I'm also really glad we have a blog. I know that some people in the class aren't so excited, but I think they may change their minds when they start to use it more and more. It is a really neat tool in the tool box to use in studying, and learning. Believe me- it really helps!

Well, I guess I should cut this off now. I have talked enough.
But just wait- this isn't the last time you have heard from me!
Keep on learning guys!
Lets make this semester a fun one!
In the words of Mr.K - Cheers!

Quadratic Functions I

2006-02-06T20:34:00.000-05:00
So what do we think about quadratic functions so far? Are parabolas easy to work with? Easy to understand? Make a post with your thoughts on what you've seen so far along with perhaps you might think you'll see in the upcoming classes on parabolas, what more characteristics you might be discovering...

Oh and remember you need to make your own post with your thoughts, not just a comment. You can also practice changing font colours, sizes, and so forth...

First day, new faces, new places...

2006-02-03T18:17:00.000-05:00
So here's the first non-rehtorical post (that means you will have to respond):

Now that you know how the course will run (weekly tests/quizzes) and the rules and regulations, what do you think so far?

ETIQUETTE

2006-02-03T17:59:00.000-05:00

I realize that some of you are avid bloggers already but here are my basic rules for etiquette on the blog:

You are to treat this blog as if it were our classroom. You treat and address everyone with utmost respect.

There is to be abolutely no use of abusive, vulgar, or profane language of any kind for any reason what so ever. There will be severe consequences for anyone who abuses this rule or takes it for granted.

If you are posting in reference to someone else's post then you are commenting in a completely mature and proffessional manor.

Most importantly, this blog's main purpose is as a learning tool for all of you to work together to tackle some of the struggles you might encounter along the way. You know the quote "2 heads are better than 1," well imagine 20 or 30 heads!

2005-12-16T20:30:00.000-05:00

Just remember, when things get really stressful, take some advice from someone who knows what they're talking about.

2005-12-16T19:47:00.000-05:00
Hey there pre-cal 30S students!!!! Guess what? You guys get to officially be the "guinea pigs" of my first blogging experience. For those who don't know, a "blog" is a combination of the words "web log" and it's the biggest technological advance to hit the world since the internet. If you're not already, by a few weeks into this course you will become active bloggers as well. Stay tuned for more coming soon!!