Reflection...

EyLoW (:

For this unit, I can say that it was quite easy. But I still have troubles on proving by contradiction (the indirect and direct reasoning thingy). Overall, I think that I am capable of answering the questions on the test. Yet, I am still quite nervous because I don't guarantee a perfect mark on it since I found some part of the unit a little difficult. I hope that everyone would do well on the test tomorrow! ^-^

-CharChar

It's 2009! :D Scribe post for January 05, 2009

Happy New Year

&

Welcome Back To School!

Hii Everyone! ^-^

First of all, I hope that you all had a wonderful winter break :] That 2-week hibernation and fun was great, but now it's time to get up and have ourselves back on track. As what Mr.K said, "We only have 3 weeks, almost exact, before the exam. After Logic, we have one more unit to do." (Well, it was something like that :P) To warm up our brains, we had a Pre-test on Logic today. Our real test will be on Wednesday (January 07, 2008), since Mr.K said that we'd have one more day to review Logic a little more.

Here are the questions & answers/corrections if you've missed the Pre-test (They can also be seen in the slides):

(1) In a class of 50 students, 18 take music, 26 take art, and 2 take both art and music. How many students are not enrolled in either music or art? (a) 6 ; (b) 8 ; (c) 16 ; (d) 24

to explain it in the form of a Venn Diagram...

(2) What is true about the statement "If two angles are right angles, the angles have equal measure" and its converse "If two angles have equal measure then the two angles are right angles"?
(a) The statement is true but its converse is false.
(b) The statement is false but its converse is true.
(c) Both the statement and its converse are false.
(d) Both the statement and its converse are true.

- Converse is false because two angles that have equal measure doesn't necessarily have to be right angles. For example, an isoceles triangle has two equal angles but they are not right angles.

(3) The inverse of the converse of a conditional statement is the contrapositive.

(4) Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8}, B = {1,3,6,9}, and C = {1,3,4,6,8}. List the members in the set: ( A ∩ C )'

to explain in the form of a Venn Diagram...
(click to enlarge)

Therefore, the members in the set ( A ∩ C )' are 1, 2, 3, 5, 7, 9.

(5) Prove that the difference of square of two odd numbers is always divisible by 4.

=
=
=
=
This has a factor of 4 and is divisible by 4.

I think I explained them as well as I could. Tomorrow, we will have time to review Logic with the class once more. Homework for today was to try to prove if is rational or not. Look at the slide that was on December 19 for the explanation on because it is similar. Also, we have to finish other unanswered exercises that we were assigned including Exercise 50.

That is all for today! ^-^

The next scribe will be Randall :]

- CharChar

Reflection

Hiya (:

I find the Circle Geometry unit easier now than the first time that I saw it. The drawing part was fun, but understanding the theorems was a quite hard. I had difficulties on how to state and reason, and up until now, I still use the "paragraph/sentence" thingy. But I know that I will get used to the more formal one. The theories, as I said, were hard to know which one to describe with. Thanks to the summaries of the theories, I already know which theory to use. The extra days that we took to review helped me understand more. The pre-test today made me confident that I was ready for the actual test. I'm just a little nervous about the test tomorrow, though... But I hope that we all do well on the test tomorrow. Good luck, my fellow classmates! ^-^

--CharChar :D

Welcome back, Mr.K! :)

Allo~! ^-^

Here are the topics that I don't quite understand:

• Circles and Radius in Analytic Geometry
• Ambiguous Triangles
• Systems of Linear Equations in Three Variables
• Absolute Value Inequality

They're pretty much the same as the others -_-"

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 ^-^

Char's posting for September 10, 2008 ^-^

Hello there! (:

For today's class, we continued on with Graphing Quadratic Equations. We were to graph the equations: y=x² and y=x²+1.

Here are my Table of Values & Graph:

y=x²

y=x²+1

Since sketching a graph and making the Table of Values seemed to have taken a lot of time, Mrs. Ingram taught us how to use our graphic calculator.

How to use the graph in a calculator:

Click to enlarge*



Notes:

• Quadratic Equation in the form of y=ax^2+k

If k is positive the graph will move upwards k units

If k is negative it will move down k units

We spent almost half of the class learning how to use the graphing calculator. It made graphing a lot easier and more accurate than a sketched one. But remember that you should know how to sketch a graph on a paper too.

For the rest of the class, we just worked on our homework.

I'm sorry for posting late. Pardon my explanations if they ever confuse you.
Don't forget that the homework for today is EXERCISE 2 on PAGE 3; QUESTIONS 1-8.

The next random scribe is ...

k a r L e e n .