Thursday, September 30, 2010

Section 2.8 Vertical Angles

Today in class we learned about vertical angles. Vertical angles are formed when two lines intersect and are opposite from each other. Two angles are vertical angles i the rays forming the sides of one angle and the rays forming the sides of the other angle are opposite rays, or rays that are collinear and have a common endpoint. You can conclude this in the following proof:(sorry if it is a little hard to read.)







That's what we learned in class today. Sorry it took so long to post this. I had a tennis match until seven and didn't get home until 7:30.

Jacob

P. S. The post time is wrong, I posted this around 8:45.

Wednesday, September 29, 2010

2-6 Multiplication and Division Properties

Today in class we learned about multiplication and division theorems.


Two important theorems are:

1.  if segments (or angles) are congruent, then their like multiples are congruent

2. if segments (or angles) are congruent, then their like divisions are congruent  (shown below)




Next, we proved one of the theorems:
                       ~                                                                                              ~          
Given:     AB   =   CD                                                               Prove:    AM   =  CN
                                   
               M midpt. AB
                                   
               N midpt. CD
This is the flow proof (sorry it's a little blurry):


Using the Multiplication and Division Properties in Proofs:

~ Look for double use of word midpoint or trisect or bisect
~ Multiplication property is used when the segments or angles in conclusion are greater than those given in the info
~ Division property is used when the segments or angles in conclusion are smaller than those given in the info


So thats what we discussed in class today
Hope it helped,
Natalie U.

Monday, September 27, 2010

Congruent segments theorem

The congruent segments theorem states that if the same segment is added to 2 congruent segments, then the resulting segments will also be congruent.
    
           A ___________________________B
                             C                    D

If AC is congruent to DB, then AD must also be congruent to CB for the reason that you are adding the same segment to congruent segments. This is easily represented by the Addition Property of Equality;

if AC= 2, DB= 2, and CD = 5, then that must mean that AD= AC+CD, which substituted in makes AD= 7
You can also see that for the same reason CB = 7, which means that AD and CB are congruent. No matter what the lengths of the lines are, if AC and DB are congruent, when CD is added, they will still be congruent. Also, if anybody reads this, it took way too long because my cat thinks the keyboard is a pillow.

Tuesday's quiz

Hello everybody,

Due to Thursday's modified schedule, I will be available during X-block tomorrow morning.

-Mr. Wilhelm

Sunday, September 26, 2010

Section 2.4


On Friday we learned about the congruence supplementary theorem.
Given:
1 and 2 are supplementary
2 and 3 are supplementary
3 and 4 are supplementary
4 and 1 are supplementary

You can conclude:
1 3
2 4

You can also conclude this with complementary angles.

Saturday, September 25, 2010

Quiz Tuesday!!

We'll be having a QUIZ (through section 2.4) on Tuesday. 
Spread the word.

Have a good weekend.

-Mr. Wilhelm

Thursday, September 23, 2010

Section 2.2

Today in class we learned about;
-Complementary
-Supplementary

Two angles are complementary if and only if their overall sum is equal to 90'.



Two angles are supplementary if and only if the overall sum of the angles is equal to 180'.



You can also show complementary and supplementary angles with proofs;



Thats what we learned in class today and i hope it helps. -Tyler Rogers



if the video doesnt work sorry if it doesnt work click on the link below
http://www.youtube.com/watch?v=EFHQOosKHFA

Wednesday, September 22, 2010

Section 2.1

Hi! Today's section was on perpendicularity.
perpendicular- lines, rays, or segments that intersect to form right angles
intersect- having at least one point in common. (though the intersection may not be labeled as a point)

__       __
AB _I_ CD
This means line AB is perpendicular to line CD. Sorry the perpendicular sign is a bit rough. It should look like this:

*The relationship "is perpendicular to" is symmetrical! (but not reflexive or transitive)*


You can NOT assume perpendicularity from a diagram.


                         As you can see, these lines appear perpendicular. However, because there is no indication that they are, you can not assume anything!
                           This is the exact same picture as before, except now there is a square to indicate a right angle, meaning that these lines definitely are perpendicular!
A couple more things about perpendicular lines:
Perpendicular --> RIGHT ANGLES, not 90 degrees!
2 lines are perpendicular IF AND ONLY IF (<-->) they intersect to form right angles.
If 2 lines intersect and DONT from right angles, they are not perpendicular.

Lastly, we talked about a few important fromulas for the coordinate plane.
First was the slope formula:


This can also be decribed as rise over run or delta y over delta x
Perpendicular lines have oppsite slopes such as y=-2x and y=1/2x

The next formula was the midpoint formula, which is:
or the average of the x coordinates over the average of the y coordinates

The final formula was the distance formula:
Here are some facts about horizontal and vertical lines:
horizontal line:
slope: 0
equation: y=a

vertical line:
slope: undefined
equation: x=b

And thats what we learned! Sorry for the rough pictures, I couldnt figure out the fancy math helper thing...
Bye! Katie <3

Saturday, September 18, 2010

Review 9/17/10

In class on Friday we spent the majority of the time reviewing previous chapters before the test. Here are some of the problems and concepts that we went over.

Probability Concepts:

Combinations: When the order of selections doesn't matter.
You can find the total number of outcomes in a combination using nCr, nCr means the number of n items chosen r at a time. You can find the number of outcomes by setting nCr equal to
                                                                                                             

Example:  5C3 =  gif.latex.gifgif.latex.gif=gif.latex.gif= 10, 10 is the number of possible outcomes for the situation
Permutations: When the order of selections matters

Review Problems:












































































































P(point chosen lands in white)














































































































































































































































































Area(white)= 1
π - 4π




















































































































































































































































































































































































































































































































































































































































































































































































            
                    =12pi
12
π / 36
π = 1/3  (Area of color/Area of white)




































































































































P=1/3
Area of Square= 36
Area of Circle= 9
π







Area of Light Blue= 36-9
π






P(within 5 of C)= 9/12
                           = 3/4

(Sorry that the end of the problem got cut off but point B is at 15)
Thanks for reading my blog post and I'm sorry for any inconveniences in the lineup of problems and equations! And I don't know why the circle problem got all spaced out but I can't fix it sorry.
                           -Corin Murphy