<body><script type="text/javascript"> function setAttributeOnload(object, attribute, val) { if(window.addEventListener) { window.addEventListener('load', function(){ object[attribute] = val; }, false); } else { window.attachEvent('onload', function(){ object[attribute] = val; }); } } </script> <div id="navbar-iframe-container"></div> <script type="text/javascript" src="https://apis.google.com/js/plusone.js"></script> <script type="text/javascript"> gapi.load("gapi.iframes:gapi.iframes.style.bubble", function() { if (gapi.iframes && gapi.iframes.getContext) { gapi.iframes.getContext().openChild({ url: 'https://www.blogger.com/navbar.g?targetBlogID\x3d14084555\x26blogName\x3dPre-Cal+30S\x26publishMode\x3dPUBLISH_MODE_BLOGSPOT\x26navbarType\x3dBLUE\x26layoutType\x3dCLASSIC\x26searchRoot\x3dhttp://pc30s.blogspot.com/search\x26blogLocale\x3den_US\x26v\x3d2\x26homepageUrl\x3dhttp://pc30s.blogspot.com/\x26vt\x3d931551856370134750', where: document.getElementById("navbar-iframe-container"), id: "navbar-iframe" }); } }); </script>

Monday, October 24, 2005

ESS-SEE-ARE-EYE-BEE-EE [S-C-R-I-B-E]

I’m the scribe for today and I don’t know if I can explain everything well but I’ll do my best. Today we started off the class by Mr. K. telling us what we’ve learned this unit and what we learned was:

Quadratic Formula
Discriminant
Imaginary Numbers
Sum and Product of Quadratic Roots
Solving Equations-Radical / Absolute Value / Rational

He then put two questions of the board.
√-12 and (i-3) (i+2)
He gave us some time to solve them and once you solve them you get:

√-12
= √12(-1)
=√12 √-1
=2i√3 [We wrote it like that instead of 2√3 i so you can tell which numbers are under the square root.]

(i - 3) (i + 2)
= i² - i – 6

= -1 – i – 6
= - i – 7

Then Mr. K. asked a few people their favourite numbers from 1-10 and he came up with a sum and product of quadratic roots. From the sum and product we had to find an equation. We started by finding a common denominator which is 18. After that we wrote down what a, b and c were, then we plugged in the numbers for the equation.

sum=2/9
prod=3/6

-b/a=2/9
-b/a=4/18

c/a=1/2
c/a=9/18

a = 18 b = -4 c = 9

y = 18x² - 4x + 9

Mr. K. put up another question where the roots were x = 2 + √3, x = 2 - √3 and we had to find the equation. So to find the equation we found the root sum and the root product. From that we found what a, b and c were. Then we had our equation.

Root Sum= 2 + √3 + 2 - √3
-b/a = 4

Root Product= (2 + √3)( 2 - √3)
c/a=4-3
c/a=1


a = 1 b = -4 c = 1

y = x² - 4x + 1

Then Mr. K. asked if you can find a different equation and Graeme said you could put it in its factored form which is: y = (x - (2 + √3))( x – (2 - √3)). Then we solved it to see if it was the same. Then Graeme said you could multiply everything by 2. Mr. K. said that it wasn’t unique because you could have the same roots and have a different function with a new constant.

y = (x - (2 + √3))( x - (2 - √3))
y = x² -(2 - √3)x - (2 + √3)x + (2 + √3)(2√3)
y = x² - 2x +√3x - 2x - √3x + 4 – 3
y = x² - 4x + 1

f (x) = 2x² - 8x + 2


The last question from the pretest was the next question we went over and this is what we got in class:


I’m not quite sure what the answer is but to get the answer you are supposed to use the quadratic formula.Mr. K. put three more questions on the board and the first one was:

The profit, P, for publishing a book is given by the equation
P(x) = -5x² +400x - 3000
where x is the selling price per book.
a) Is it possible to set a selling price that will earn a total profit of $6000?
b) What range of selling prices allows the publisher to make a profit on this book?

The answer to a) is:
6000 = -5x² +400x - 3000
0 = -5x² +400x - 9000
0 = x² - 80x +1800 [We divided everything by 5.]
After that you use the quadratic formula. You end up getting a negative number as the discriminant so there is no sol’n. The answer is, no, you cannot get a profit of $6000.

For b) you find the roots of the original equation to find the range of the selling price. You end up getting [$8.38, $11.62] as the range.

2) Jason ran 4 kph faster than Gary walked. Jason ran 15 km in the time it took Gary to walk 9 km. What were their speeds?

When Mr. K. was doing the question he said he made a mistake somewhere so the answer was quite unclear.

3) The square root of three less than a number is 12. What is the number?
12 = √ x – 3

144 = x – 3
x=147


That was pretty much everything we covered in class. Mr. K. also said not to worry about the distance equations because there won't be a lot of those on the test. Remember we will have a substitute and the test is on Wednesday, so review your notes from your dictionary and go over the questions you had difficulties with. Tomrrow we will start on circles I think it was. Just tell me if I made mistake;).


***The scribe for tomorrow is Rosel S.

2 Comments:

At 10/24/2005 6:11 PM, Anonymous WAF3R said...

Great Scribe very colorful and very accurate to the way the class went. Great job Aichelle!

 
At 10/24/2005 10:51 PM, Blogger Richy said...

I kinda just relived an hour of my life. lol

 

Post a Comment

Links to this post:

Create a Link

<< Home