I will be your host for tonite's Math meal

Check out this picture this is the same one from the white board Mr.Kuros kindly showed us and we instinctively learned like students what the meaning of this mis mash is about!







****too explain in detail.... without too many words ill put up pictures but in groups of 3 for the definitions*****


So check out what a line Segment, a chord and a Diameter are...



How about what the heck is a Secant, Tangent, or ooo a Radius!



Just because I like you Ladies and Gentleman I'll define to you what the heck is are Major and Minor Arcs, what is meant by being bounded by something, what's a Sector and what an INSCRIBED ANGLE is.... although this will be the last circle for tonite

(Aww... too bad actually I'm actually getting tired of making these circles)



Well to end off this night peoples Im going to say there is this assignment similar to that of today's work where it covers all what I summarized for you Ladies and Gents.

Oh yeah Exercise 30 is on this material also.


I hope this meets Textbook/Mr . Kuro's standards just kidding but I do wish that it hasn't been a waste of your time to read this peoples. '

Give yourselves a hand for taking your time to learn.


And to on a final note Gambete Mina (work hard everyone) '''

*bows and waves*

nExT scriber is D'Lan
*clap clap clap and passes baton while running away*

Pre- Test on Analytic Geometry

Hello!! everybody.....

Today in class we did our pre-test but before that, we work on Analytic Geometry #8

here's the question...

8. The total number of known moons around Saturn, Uranus, and Neptune is 43. The total number of moons around Saturn and Neptune is 9 more than the number of moons around Uranus. Saturn has 2 moons more than twice the number of moons around Neptune. Find the number of moons each planet has.


To solve this problem you have to do the following:

• first, you have to find the equation
• Then when you have the equation, set it up so it will be easy for you to solve it.
• Then when solving it, you can use elimination or substitution

here's the equations..

S = # of moons around Saturn
U = # of moons around Uranus
N = # of moons around Neptune

S + U + N = 43
S + N = U + 9
S = 2 + N

(click the images to make it big)

some question that we did on our pre-test..

next scribe is KimC...

Analytic Geometry

Hi there,

Today in class first we did the home work from yesterday and here it is.

Step 1.to substitute -2x+5 in the spot of Y, to get

Step 2. Simplify the equation and add the like terms.

so you will get,

Step 3. Find the roots by using the quadratic formula.

The roots are

x1=5

x2=0.6

Then plug the X roots to find the Y values.
Y1=-2x+5

Y1=-2(5)+5

Y1=-5

Y2=-2(0.6)+5
Y2=-1.2+5

Y2=3.8

After that we did some review questions and here is the first one that we have to find the centre and the radius of a circle.

1) (x+1)^2+(y-4)^2=13

• To find the centre, find the additive inverse of the X and Y. Which is -1,4. There fore the centre is (-1,4).


• To find the radius, square both side and you will find the that answer is the square root of 13.

2) The second Question was a little bit different than the first one because we have to find the distance between a point and equation of line. (3,4) to the line 2x-5y=7


after you plug the numbers into the equation you will find that the distance is

Next Blogger ...Larlyn014

Chae: le six novembre 2008

Haii! (:
I decided to do the blog today since I happened to check on the blog and didn't see an update. I hope Dr.Eviatar doesn't mind if I scribe here with "no permission" :)

Well, today we had to hand in #6 of Exercise 24 from the book. There were some volunteers who put the answer for it on the board. I didn't quite know how to do it, so I didn't volunteer.

Here are the answers for #6 a.)&b.)

a.)
3x + 2y = 4 -->equation [1]
x - y = 3 -->equation [2]

Multiply the equation [2] by 3 like this:
3(x - y - 3) = 0
=> 3x - 3y - 9 = 0

Then you subtract the new equation from the first [1] equation:
3x + 2y - 4 = 0
- 3x - 3y - 9 = 0
5y + 5

Find the value of y:
5y = -5
=> y = -1

Substitute the value of y in the equation [1] to find the value of x:
3x + 2(-1) = 4
3x - 2 = 4
3x = 6
x = 2

Answer is: (2,-1)

b.)
2x + 3y = 48
3x + 2y = 42

Same process, but this time you multiply both equations to get rid of y:
2(2x + 3y =48)
=> 4x + 6y = 96

3(3x + 2y = 42)
=> 9x + 6y = 126

Addition-Subtraction Method:
4x + 6y = 96
- 9x + 6y = 126
-5x = -30

Value of x:
-5x = -30
=> x = 6

Substitute the value of x to get the value of y:
2(6) + 3y = 48
12 + 3y =48
3y =36
y = 12

Answer is: (6,12)

After that, we started a new topic. It was sort of the same as our previous topic (Systems of Linear Equations in Two Variables), but this time it is for THREE variables. I'm still in the process of understanding it, so pardon my explanations. Well, all I have right now are some notes from the board (-_-"). It was so much better when we can just look at the slides again in the blog for information.

But anyway, here are the notes:

Independent - 1 Solution (all 3 equations intersect)
Dependent - Infinite solutions -> concide
-> common line
Inconsistent - No solutions

The graph for the "Ordered Triple" example (3,3,3)

The graph was sort of hard to draw, and to understand. As a matter of fact, I don't really get how you draw the points in it. That's why we're not really asked to draw the graph.

Example of how to solve equations with THREE variables:

x + y - z = 2 --> [Equation 1]
x - 2y + 2 = -1 --> [Equation 2]
3x + y - 2z = 4 --> [Equation 3]

Steps:

1st:
[Equation 1] + [ Equation 2]
=> 2x - y = 1 --> [Equation 4]

2nd:
2 * [Equation 2]
=> 2x - 4y + 2z = -2 --> [Equation 5]

3rd:
Equation 3 + Equation 5
=> 5x - 3y = 2 --> [Equation 6]

4th:
3 * [Equation 4]
=> 6x - 3y = 3 --> [Equation 7]

5th:
[Equation 6] - [Equation 7]
=> x=1

Substitute x=1 into [Equation 4]:
=> 2 - y = 1
=> y = 1

Substitute x=1 and y=1 into [Equation 1]:
=> 1 + 1 - z = 2
=> z = 0

Answer to the equations: (1,1,0)

Well, that's all the notes that was for today. Things for homework are Exercise 25 #1, and the problem she wrote on the board.

In case you weren't in class or didn't get to write it down, here is the problem (Parts of it are already summarized/put into an equation by Dr.Eviatar):

w + l + h = 80cm
Length is 10cm less than twice the sum of width and height and twice the width exceeds the height by 6cm. Find the width, length, and height of the box.

l = 2 (w + h) - 10
2w = h + 6

That's practically what we did for class today. I hope I helped in the new topic in some way. I'm not going to pick who's going to be the next scribe since I sort of just barged in the blog to put what we did today [kind of saying that I scribed with "no permission", haha.] I have to do the homework myself. I'll at least "try" to do it (-_-") I'm still trying to understand the given problem.

Till next time my fellow classmates,
Charizze ^-^