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Thursday, October 06, 2005

Postage from a Scribe

Hello, everyone! I’m not very good at explaining things but I’ll try to explain things and show you what we learned in class. Today we started algebra, exciting isn’t it? Mr. Kuropatwa explained to the class that it’s an in depth analysis of solving equations and it’s about finding roots. We started class with questions from the board, where we had to find the roots, by factoring the equations:

x² + 7x + 10 = 0

2x² - 5x + 3 = 0
2x² + 3x = 2
3x² - 2x = 1

We then factored the equations and these were our answers:

1) x² + 7x + 10 = 0

(x+5) (x+2) = 0
x = -5 x = -2

2) 2x² -5x + 3 = 0

(2x-3) (x-1) = 0
x = 3/2 x = 1


3) 2x² + 3x = 2
2x² + 3x - 2 = 0 For number three and four we balanced the
(2x-1) (x+2) = 0 equation for the second step.
x = 1/2

x = -2

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

x = 1

After that we had to solve these equations:

tan² + 7tanx + 10 = 0

2sin² - 5sinx + 3 = 0
2cos²x + 3cosx = 2
3sin²x - 2sinx = 1

Mr. Kuropatwa said that these equations with tan, sin, and cos (trig) are like the equations above. He said that it was in a pattern of a quadratic but isn’t a quadratic. Then he told us math is all about the science of patterns and how we have to remember patterns in order to solve problems. He also said if it fits the quadratic pattern, then you can solve complicated equations. We solved the trig equations like this:



tan² + 7tanx + 10 = 0
Let tanx = a
a² + 7a + 10 = 0
(a+5) (a+2) = 0a=-5 a=-2
tanx = -5 tan x = -2
X=281.3° x=296.6 °
X=101.3° x=116.6°


You are probably wondering how we got that…well I’ll explain it to you, well I’ll try. So first we let tanx=a and we rewrote the equation as a² + 7a + 10 = 0 and then we factored it and got (a+5)(a+2)=0. The roots we found were a = -5, a = -2 but since we let tanx=a we had to change it back to tanx. The roots were, tanx = -5, tanx = -2. Then we had to find the angles by putting arc tan into our calculators. Since it’s negative, we know that the angles are in quadrants IV and II. You then find out the angles are:


X=281.3° x=296.6 °
X=101.3° x=116.6°


After we solved that question as a class, he told us to do the second question by ourselves. After he gave us some time he went over the question and got:

2sin² - 5sinx + 3 = 0
Let sinx = a(2a-3) (a-1) = 0
a = 3/2 a = z
sinx = 3/2 sinx = 1
D.N.E. x=90°


Since sinx = 3/2 and you know it’s impossible you write D.N.E. (does not exist), Reject, undefined, or no sol’n (no solution).

***Make sure to write, D.N.E. (does not exist), Reject, undefined, or no sol’n (no solution) SO YOU DON’T LOSE A MARK otherwise you will lose a mark if you don’t write it. ***



That may be blurry but it says Do not let TANX=X! Make sure you switch the variable when you are looking for the angle. The bottom says Let TANX equal any other variable other than X.

After that Mr. Kuropatwa went over the other two questions real quickly.

2cos²x + 3cosx = 2
cosx = 1/2 cosx = -2

x= 60°, 300° D.N.E.

3sin²x - 2sinx = 1
sinx = -1/3 sinx = 1
x=340.5° x=90°
x=199.5°


Once we were done all of that, Jacky reminded Mr. Kuropatwa that we had not finished taking down notes in our math dictionaries. So Mr. Kuropatwa finished the notes off by starting it off with the role of parameter B.

Here you go, knock yourself out:


The Role of Parameter B
In this course B=1, always. For more about B take Pre-Cal 40s.

The Role of Parameter C
C is called the Phase Shift or the Horizontal Shift.
C>0 The graph shifts right C units.
C<0 The graph shifts left C units.

The Role of Parameter D
D determines the Sinusoidal axis or Vertical Shift.
D>0 The graph shifts up D units.
D<0 The graph shifts down D units.


The Relationship Between
y=sinx and y=cosx

sinx = cos (x - 90) cosx = sin (x + 90)

After writing our notes Mr. Kuropatwa told us our homework was Exercises twelve and thirteen. Thanks for taking the time to read this and if I made a mistake just tell me=).

So tomorrow's scribe is Aldridge.

5 Comments:

At 10/06/2005 6:50 PM, Blogger Richy said...

Nice Nice Very Nice. Hahaha. Excellent Job. Hahaha. I like how you made the diagrams your own. Great Job

 
At 10/06/2005 9:38 PM, Blogger Craig K said...

Ahhhh! Complete Ownage! Amazing scribe post. You've "Raised the Bar" haha. I wonder how high that bar is now? I think this class has raised it through the roof and knocked Mr. K's socks off many times. Hahaha

 
At 10/06/2005 10:32 PM, Blogger Mr. Kuropatwa said...

Outstanding! I've run out of superlatives. I no longer own any socks. ;-)

 
At 10/08/2005 1:09 PM, Blogger Ionut Alex. Chitu said...

To solve equations graphically and numerically you should try DeadLine. The freeware finds the real roots of an equation, evaluates functions and the first two derivatives extremely fast and accurately, finds extrema of the function.

 
At 10/12/2005 5:41 PM, Blogger Rannell d. said...

Wow Aichelle, tons of colors. Cooool ;)

 

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