How to Study for Finals

This is going to be somewhat of a general post since the topic is a tad bit wide. Anyways, I think (and hope) that you are all aware of the fact that we have finals this week. After searching a couple of articles about how you can study for your finals and what makes studying easier, I think I have somewhat of a gist of it.


DOs
1. SLEEP
a. You perform better when properly rested

2. Reread your notes, practice on whatever subject you struggle on.

3. Have confidence

4. If you’re studying for a new concept (Chapter 6) then practice on a single section and repeat it until you get all the answers correct.

5. Prepare yourself mentally.

6. Eat healthy
a. Junk food can damage your concentration and memory

7. Choose where you study wisely
a. A quiet, orderly place is most likely the best
b. A peaceful environment will help your concentration




DON'TS

1. Do not just re-read the entire text book.

a. It’s a waste of time
b. You’ll most likely not retain that information
c. Read the chapter summaries
d. Re-read notes
e. Look at charts, pictures, diagrams, and other visual tools.

2. Do not procrastinate.
a. Start studying early and do it often.

3. Do not stress yourself.
a. Take time to rest.
b. You’ll learn more efficiently with breaks in between your studying.


I also have two other things to share~
the first one is a step-by-step how to that shows how it’s possible to pull an all-nighter before a test but still do well.

http://media.tumblr.com/tumblr_liw1rb3Lzj1qbolbn.jpg
~I don't recommend this, I just thought it was interesting.

And the second one is a list of common mistakes that can lower your test scores.

http://homeworktips.about.com/od/schooltests/a/testmistakes.htm



I don’t pick a scribe right? Darn it >//<.
~The dos and the don'ts were in separate charts that were...easier to read, but it all disappeared when I moved it away from the Word document D: ~

6.4 Inequalities for one triangle

THEOREM THAT COVERS ALL THE CONCEPTS IN 6.4:

-If the angles of a triangle are distinct, then you can draw conclusions based on the side lengths.

CONCLUSIONS SHOWN BELOW


(diagram is not to scale)

GIVEN: m^A=40; m^B=75; m^ C=65









CONCLUSIONS:

~m^A < m^C < m^B

~a < c < b ->because of the theorem If one angle of a triangle is larger than a second, then the side opposite the first angle is longer than the side opposite the second angle.


BASICALLY:

~the smallest angle is opposite the shortest side

~the "middle" angle is opposite the "middle" side

~the largest angle is opposite the longest side







REMEMBER!!!!!!!!!!!!!!!!!!!!

In order for a triangle to exist, the sum of any 2 sides MUST always be GREATER than the 3rd side. (also called the Triangle Inequality Theorem)



Example:

If two sides of a "triangle" are 7 cm long and the third is 16 cm long, then the "triangle" is NOT a triangle.

WHY???

-7+7= 14 < 16

-7+16=23 > 7

The sum of the second pair of side lengths IS greater than the third side, BUT the sum of the first pair of side lengths is NOT greater than the third side, therefore, the "triangle" is NOT a triangle.



ANOTHER EXAMPLE:

If two sides of a "triangle" are 8 cm long and the third side is 14 cm long, then the "triangle" IS a triangle.



WHY???

-8+8=16 > 14

-8+14=22 > 8

The sum of the first pair of side lengths IS greater than the third side, and the sum of the second pair of side lenths IS greater than the third side, therefore, the "triangle" IS a triangle.





COROLLARIES:

-The perpendicular segment from a point to a line is the shortest segment from the point to the line.

-The perpendicular segment from a point to a plane is the shortest segment from the point to the plane.



HELPFUL VIDEO:
-http://www.winpossible.com/lessons/Geometry_Triangle_Inequality.html
-http://www.youtube.com/watch?v=e1prq6UFxJE&noredirect=1 (this is a YouTube video so you gotta watch it at home, but its about Indirect proofs and all of 6.4)

Chapter 6.3 Indirect Proofs(You might want to read this before reading David's scribe because he spoils the..... next...... Scribe!!!!!!!!!!!!! So read this first!!!!This is possibly the longest title ever, of all time)

Indirect Proofs!!!!!!!!

A long time ago, (since September, 2011) in this class, we have been using proofs. Now they turn out to be direct proofs. At times, they are difficult to find a direct proof so it's possible to reason indirectly. To use a indirect proof, you assume temporarily that the desired conclusion is not true. Since I don't want to explain how to write a indirect prof, I'll write it in steps so... everyone... is... happy!!!

How to write Indirect Proofs!!!!!!!!!!!
Steps
1. Assume temporarily that the conlusion is not true!
2.Reason logically until you reach a contradition of a known fact!
3.Point out that the temporary assumption must be false, and that the conclusion must then be true!

Here is a example of using indirect proofs from the book.


Given: n is an interger and n squared (I don't have microsoft word on my computer so i wrote squared) is even.
Proof: n is even
Assume temporarily that n isnot even. Then n is odd, and
n squared = n x n
               = odd x odd x = odd
But this contradits the given information that n squared is even. Therefore the temporarily assumption that n is not false. It follows that n is even.

Now is time for the epic question of the day!!!








Are you ready?





















Did you know that provig the contrapositive of a statement is related to indirect proofs?


Confused? Here's an example from the book!

If you want to prove the statement " If p, then q," you can prove the contrapositive " if not q, then p." Or you could write an indirect proof- assume that q is false and show that this assumption implies that p is false.


Now comes the epic videos of the day!!




http://www.5min.com/Video/How-to-Write-an-Indirect-Proof-516909808


This is the only video that I could find that was not on youtube. Somehow, searching for a video on indirect proofs can lead you to Call of Duty Black Ops?


So that is chapter 6.3!!!!!! Prepare for more epic questions and videos of the day!!! Also, expect a epic picture of the day if there is any picture to show.

Now the next scribe is...
Not Jimmy Bob
















Not John Bob













Not me




But it's



























Whoever David picked!!!!!!!!!!

6.5 Inequalities For Two Triangles

The Next Publisher is Phil J

Theorem

If two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second triangle.

Given: Seg. HA is congruent to Seg. KC; Seg. HB is congruent to Seg. KD; m Angle K is greater than m Angle H.

Prove: Seg. CD is greater than Seg. AB

Proof:

Draw Ray KL so that m Angle LKD = m Angle H. On Ray KL take point M so that KM = HA. Then either M is on Seg. CD or M is not on Seg. CD.

In both cases Triangle MKD = Triangle AHB by Hinge Theorem, and MD = AB.

Case 1: M is on Seg. CD

CD is greater than MD (Seg. Add. Post. and Prop. of Ineq.)

CD is greater than AB (Substitution Prop.)

Converse of Hinge Theorem

If two sides of one triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer than the third side of the second triangle, then the included angle of the first triangle is larger than the included angle of the second.

Same diagram as shown above

Given: Seg. KC is congruent to Seg. HA; Seg. KD is congruent to HB; CD is greater than AB.

Prove: m Angle K is greater than m Angle H.

Proof:

Assume temporarily that m Angle K is not greater than m Angle H.

Then either m Angle K = m Angle H or m Angle K id less than m Angle H.

Case 1: If m Angle K = m Angle H, then Triangle CKD is congruent to Triangle AHB by the Hinge Theorem. (CD = AB)

Case 2: If m Angle K = m Angle H, then CD is less than AB by the Hinge Theorem.

http://vimeo.com/m/17642248

http://www.teachertube.com/viewVideo.php?title=Lesson_23_Topic_3&video_id=39518
http://www.schooltube.com/video/c8315510b6fb421f9b0b/L23T3

6.2 Inverses and Contraposistives

Class Blog 6-2 By Matthew Jones

If Then Statements!!! AGAIN!!!!

The Statement - If P then Q

Converse or P and Q flipped - If Q then P

NEW INFO ALERT

Inverse or the Not statement – If NOT P Then NOT Q.

Contropositive or the Not Converse – If NOT Q Then NOT P

Logical Equivalence
- A statement and it’s Contrapositive are Logically Equivalent
- A statement is NOT Logically Equivalent to it’s Converse and Inverse.
- Similar to an = and the biconditional
- Results of either
Both True
Both False


Examples of everything above :
True conditional : If two lines are not coplanar, then they do not intersect.

Inverse of it : If two lines are coplanar, then they intersect. (false)
Contropositive : If two lines intersect, then they are coplanar. (True)
Venn Diagrams!!!

You use Venn diagrams to represent conditionals. You Use Venn Diagrams/Truth Tables to determine Logical Equivalence.

BUT HOW!!?! You might ask. Well I’m getting to that. It’s easiest to use

EXAMPLES :
G: all runners are athletic.
P: Leroy is a runner.
C: Leroy is an athlete.

HOW DID I JUST DO THAT!

Well it is quite simple you use the Venn Diagram ( Since Ms.Hunt doesn’t want us using the Truth tables since they will confuse us).

The P is runners and the Q is Athletic. So then what is Leroy? Leroy is a runner as given so he goes in the Runner section or Q. and Q is inside of P. Since in the given.

Then it is quite obvious that Leroy is an Athlete since he is also in the P section.

Or you can look at the given info and see that runners are athletes and Leroy is a runner making him an athlete. TRANSITIVE PROPERTY!!

MORE EXAMPLES : With INVERSES

P: Lucia is NOT an Athlete
C: Lucia is NOT a runner

The conclusion is true. We are still using the same given as the previous Question/ Example.

The reason Lucia is not an athlete is because She is neither in the runner circle nor the Athlete circle so she is a lame-o.

And that’s all folks! Here is one helpful video one funny video and an amazing link to tons of awesome info!

http://www.blinkx.com/watch-video/g1d-converse-inverse-and-contrapositive/z_3OtHw6mD_ALGPXFjtUZQ

http://vimeo.com/15367677

http://accgeo435.blogspot.com/

And the NEXT SCRIBE IS!!!!!!!!!!!













Well uh....









Almost there....








If Not Phil J, then it is not Ms.Hunt

If not Ms.Hunt, then is is not David.

If not David then it is not Francis

If not any other person but one, then it's..............


KATIE HAHAHAHAHAH!!!!

6 - 1: Inequalities

It's a new chapter and we're getting to the inequalities.

First things first, Ms. Hunt wasn't in class today and we had an assignment that I presume needed to be finished at home. We both had to take notes on chapter 6-1 and we had an in-book assignment to do. The notes are self-explanatory, but if you forgot the in-book assignment:

Page 206: #1-13.

Ok, getting back to the chapter . . .


Example 1:



Given: AC > AB; AB > BC
Conclusion: AC ? BC

AC > BC because of the Properties of Inequality.







What are the properties of inequality? Well, they're a group of properties of . . . inequality. You can use these for proofs, and you would just say Prop. of Inequality as the reason.


Properties of Inequality:

If a > b and c >/= d, then a + c > b + d.

If a > b and c > 0, then ac > bc and (a/c) > (b/c).

If a > b and c < 0, then ac < bc and (a/c) < (b/c).

If a > b and b > c, then a > c.

If a = b + c and c > 0, then a > b.


Example 2:


Given: AC > BC; CE > CD
Prove: AE > BD







Statements: Reasons:
1. AC > BC; CE > CD 1. Given
2. AC + CE > BC + CD 2. Prop. of Inequality
3. AC + CE = AE; BC + CD = BD 3. Segment Addition Postulate
4. AE > BD 4. Substitution Prop.


Example 3:


Given: Angle 1 is an exterior angle of Triangle DEF.
Prove: m Angle 1 > m Angle D; m Angle 1 > m Angle E.







Statements: Reasons:
1. . . . 1. Given
2. m Angle 1 = m Angle D + m Angle E. 2. The measure if an ext. angle of a triangle = the sum of the measures of the two remote int. angles.
3. m Angle 1 > m Angle D; 3. Prop. of Inequality
m Angle 1 > m Angle E


This specific example proves the following theorem:


Theorem 6-1: Exterior Angle Inequality Theorem

The measure of an exterior angle of a triangle is greater than the measure of either remote interior angle.

OKAY, so I couldn't find a video about Inequalities in Geometry because it's too broad of a term and all that showed up were different types of inequalities - not even from geometry. So I specified my search to "Properties of Inequalities Geometry" and that came up with nothing except for a 30-minute video of this one guy teaching his class about inequalities, but half of them weren't related to this lesson. Then I just tried to find ANYTHING having to do with this specific lesson and nothing came up. So, what I decided to do, was to give a link to main points of EACH section in Chapter 6. It's basically a link to notes for the whole chapter. I figured this would be helpful to understand mainly what the chapter is about.

TURN OFF YOUR SOUND BEFORE GOING TO THIS LINK, THERE ARE ADDS IN THE SIDEBAR AND THEY'RE LOUD.

http://www.scribd.com/doc/522464/Geometry-Notes-Chapter-Six-Inequalities-in-Geometry

I REALLY hope this link is aloud past the filter.



So, it's that time of the day . . . who will be the next scribe.....?












Yeah, Matthew, you should have known it was going to be you from the beginning. HAVE FUN!

Kites

PROPERTIES OF KITES:
-Diagonals are perpendicular
-When shown as in above diagram:
--The two upper sides are congruent
--The two lower sides are congruent
--The left and right angles are congruent
--When we add the diagonals:
---The horizontal diagonal is bisected by the other diagonal
---The vertical diagonal bisects the two angles it goes through
---Two Isosceles triangles are created
-These properties apply to ALL kites, but sometimes rotated.

Kites are so weird, Khan Academy didn't have a video for them... That was annoying.
So then, I looked at vimeo, because apparently that's the thing that you use at school to get videos because they blocked youtube. I got nothing. Well, I got this video... http://vimeo.com/22253704 but it's not very educational. It's funny, but not educational, unless you're like, six. We're not six. Plus, it's like twenty five minutes... not fun.

Now, apparently I'm not going to post a video, because kites are so bizarre that no one who makes videos cares about them. I have no other ideas, so I'm just gonna write a poem, because things that rhyme tend to be helpful. They just do. Please don't mock my excessively fail poetry.

with perpendicular diagonals
kites can look like pterodactyls. (sort of.)
one diagonal bisects the other
and they form isosceles triangles like brothers.
one diagonal bisects angles
and bisects the diagonal that is the base of the two triangles.
two pairs of sides are congruent
but not opposite sides, 'cause that'd ruin it.
there's a pair of angles that are congruent
they are opposites, but it didn't ruin it.
this poem is really bad, but I don't care
'cause it rhymes, and it sticks in your head, SO THERE.

NEXT SCRIBE hopefully can do better than me... (shouldn't be too hard)
DMITRI! Good luck with that.

Trapezoids!

Today in class, we learned about trapezoids, and how they associate with chapter 5.

All trapezoids have one pair of parallel sides, and they are called the bases. The other two sides are called the legs. The bases are often labeled b1 and b2.

------------------------------------------------------------------------------------------

Isosceles Trapezoid:

They carry many of the characteristics that a regular trapezoid does, but in this kind;

• the legs are congruent
• the base angles are congruent
• the diagonals are congruent
• the angles created by a median (below) are congruent
  

------------------------------------------------------------------------------------------

Median:

• Is a segment that extends from one leg of a trapezoid, to the other.
• A median to a trapezoid divides each leg into congruent segments.
• The formula for finding the median of a trapezoid is [(b1+b2)]÷2
• The median is parallel to b1 and b2.
• If you have an isosceles triangle with a median,than all four segments created are congruent.

And now for an informational video!

http://www.youtube.com/watch?v=d8DAYzbmuBE

And the next student scribe is...................











Hollee! Good luck!

-Love Emma, 11/28/11 (my Birthday!!)

Rectangles, Squares, and Rhombuses

Rectangles, Squares, and Rhombuses (Special Parallelograms)

Prop. of Parallelograms:

• both pairs of opposite sides are parallel
• both pairs of opposite sides are congruent
• the diagonals bisect each other
• both pairs of opposite angles are congruent

Rectangles:

• 4 right angles
• diagonals are congruent
  
• prop. of a parallelogram

Rhombuses:

• all 4 sides are congruent
• the diagonals are perpendicular to each other
• the diagonals bisect opposite angles
• all properties of a parallelogram
  

Squares:

• all properties of a rhombus
• all properties of a rectangle
• all properties of a parallelogram

Always, Sometimes, Or Never

1. If an angle of a parallelogram is a right angle then is it a rectangle?
2. Is a square a rectangle?
3. Is a rectangle a square?
4. If a diagonal bisects 2 angles then is it a square?
5. If the diagonals of a parallelogram are perpendicular to each other then is it a rectangle?

1.always 2.always 3.sometimes 4.sometimes 5.sometimes

HERE ARE SOME HELPFUL VIDEOS

http://vimeo.com/15571270 <- the teacher in that one is a little wacky, but it helps

http://www.5min.com/Video/How-to-Use-the-Properties-of-Special-Parallelograms-516909821

AND THE NEXT SCRIBE ISSSSSSSSS..................

EMMA! (your welcome)

AND THATS IT FOR TONIGHT THANKS FOR READING SPECIAL PARALLELOGRAMS!!!!!

by, Jessica Young

5.2 Proving Quadrilaterals are Parallelograms

Today Ms.Hunt showed us 4 theorems that we can use to prove that quadrilaterals are parallelograms.

Review on Parallelograms: Opposite angles are congruent, opposite sides are congruent, diagonals bisect each other, and both opposite sides are parallel.

Theorem 5.4- If both pairs of opposite sides in a quadrilateral are congruent, then the quadrilateral is a parallelogram. Here is a proof problem that will prove this.


Given: Segment Ts is congruent to segment QR; segment TQ is congruent to segment SR.



Prove: Quadrilateral QRST is a parallelogram.

Statement Reason

1. 1.Given

2. Segment SQ is congruent to segment SQ 2. Reflexive Prop.

3. Triangle QST is congruent to Triangle SQR 3.SSS postulate

4. Angle 1 is congruent to angle 2; Angle 3 is congruent to angle 4 4.CPCTC

5.Angle 1, angle 2, angle 3, and angle 4 are alternate 5. Def. of alt. int. angles

interior angles

6. Segment TS is parallel to QR; Segment TQ is parallel to segment SR

6. If 2 parallel lines are cut by a transversal and alt. int. angles are congruent, the the lines are parallel.

7. QRST is a parallelogram 7. Def. of parallelogram

Here is another theorem

Theorem 5.5- If one pair of opposite sides of a quadrilateral are both congruent and parallel, then the quadrilateral is a parallelogram.

Given: Segment TS is congruent to segment QR; Segment TS is parallel to segment QR.


Prove: Quadrilateral QRST is a parallelogram.

This one was for homework, so I can't write the answer down.

Theorem 5.6- If both pairs of opposite angles are congruent, then the quadrilateral is a parallelogram.





Given: Angle A is congruent to Angle C: Angle B is congruent to Angle D.


Prove: Quadrilateral ABCD is a parallelogram.


This one was on the homework too, so I'll leave you guys to solve that.


Theorem 5.7- If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.


Given: Segment AM is congruent to segment MC; Segment DM is congruent to segment MB.
Prove: ABCD is a parallelogram.

Another one on the homework! Good luck with this one!

So let's review...
The five ways to prove that a quadrilateral is a parallelogram is to show that...

• Both pairs of opposite sides are parallel.


• Both pairs of opposite sides are congruent.


• One pair of opposite sides are congruent and parallel.


• Both pairs of opposite angles are congruent.


• Diagonals bisect each other.

Our homework for today is Page 174 numbers 1-22, 23, and 24, with 25 being optional for a challenge.

Here is a link for more detailed information, as well as extra practice!

http://www.onlinemathlearning.com/parallelogram.html

http://www.sonoma.edu/users/w/wilsonst/courses/math_150/theorems/Parallelograms/default.html

The next scribe is....

Not Sean, Dmitri, J.D., Zahra, Phil K., Phil J., David, Matt, Katie, Krista, Kerryann, Ms. Hunt, Francis, Emma, myself, Hollee, or Olivia.

It's....

Ashley! Good luck on your next scribing!

Have a nice day!

Review for Chapter 4

Tonight's review was pg. 160 #1-20
Tomorrow we will be reviewing more

Ways to Prove Triangles Congruent

• SSS
• SAS
• ASA
• AAS
• HL (Only use with right triangles)

How to prove parts of two triangles congruent

1. Prove the two triangles are congruent by using one of the theorems/ postulates that proves them congruent
2. Then state that the two parts are congruent by using CPCTC ( Corresponding parts of congruent triangles are congruent)
   

Theorems/ Postulates

• Theorems/ Postulates used to prove triangles congruent (SSS, HL, ASA, etc...)
• Isosceles Triangle Theorem: If two sides of a triangle are congruent, then the angles opposite those sides are congruent.
• Converse of Isosceles Triangle Theorem: If two angles of a triangle are congruent, then the sides opposite those angles are congruent.
• If a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment.
• Converse: If a point is equidistant from the endpoints of a segment, then the point lies on the perpendicular bisector of the segment.
• If a point lies on the bisector of an angle, then the point is equidistant from the sides of the angle.
• Converse: If a point is equidistant from the sides of an angle, then the point lies on the bisector of an angle.

Difference Between Median and Altitude
Median: A segment from the vertex to the midpoint of the opposite side.
Altitude: Perpendicular segment from the vertex to the opposite side.

Reminder
Name triangles so congruent points correspond

Helpful Videos
http://www.khanacademy.org/video/congruent-triangles-and-sss?playlist=Geometry
http://www.khanacademy.org/video/finding-congruent-triangles?playlist=Geometry
http://www.khanacademy.org/video/review-of-triangle-properties?playlist=Geometry

For Chapter Summary pg. 159-160
Test is Wednesday 11/9/11


The next scribe is......... Tim

4.7 Info about segments

4.7 Info about segments

Median: A segment that extends from a vertex to the opposite side

- Segment intersects the side at the midpoint

- 3 per triangle

Altitude: A segment that extends from a vertex to the opposite side

- Creates rt. angles

- 3 per triangle

- Located IN, ON THE SIDE of or OUTSIDE of a triangle.

- If a point lies an a perpendicular bisector of a segment, the point is equidistant to the endpoints of the segment

- Converse is also true

- If an angle bisector contains a point, then the point is equidistant from the sides of the angle (the new segments must be perpendicular to the rays of the angle)

- Converse is also true

VIDEO: http://vimeo.com/16443191

Using More Than One Congruent Triangle In A Proof

Today, we learned about using more than one pair of congruent triangles in a proof. It’s really simple and just added steps to normal proofs. You can use SSS, SAS, ASA, AAS, HL and CPCTC to help through the proof.

Some tips for proofs:

· Remember key steps

· ALWAYS write in 2 column form unless told otherwise

· Make sure your reasons are correct

· Stay neat with proof

· Draw diagrams

· The given is ALWAYS first, the book is wrong, and the point to prove is always lastToday, we learned about using more than one pair of congruent triangles in a proof. It’s really simple and just added steps to normal proofs. You can use SSS, SAS, ASA, AAS, HL and CPCTC to help through the proof.

Some tips for proofs:

· Remember key steps

· ALWAYS write in 2 column form unless told otherwise

· Make sure your reasons are correct

· Stay neat with proof

· Draw diagrams

· The given is ALWAYS first, the book is wrong, and the point to prove is always last

this helps also- http://www.vhstigers.org/ourpages/auto/2006/6/20/1150817470106/Geometry%204-6%20Using%20More%20than%20One%20Pair%20of%20Congruent%20Triangles.pdf

4.4 Isosceles triangle theorems

Today we learned a theorem and its converse for isosceles triangles. We also learned three corollaries and a theorem about perpendicular bisectors.

Theorem: If 2 sides of a triangle are congruent, then the angles opposite those sides are also congruent.
Converse: If 2 angles of a triangle are congruent, then the sides opposite those angles are also congruent.

Corollary 1: An equilateral triangle is also equiangular.
Corollary 2: An equilateral triangle has 60 degree angles.
Corollary 3: The bisector of the vertex angle of an isosceles triangle is perpendicular to the base at its midpoint.

This picture has the vertex angle (C), the base (C), the base angles (B and A), and the congruent legs (B and A). Both sets of A and B are congruent, and both Cs can be bisected with a perpendicular bisector.

Isosceles triangle perpendicular bisector: a line that bisects the vertex angle and the base, and is perpendicular to the base.
Perpendicular bisector theorem: If an isosceles triangle has a perpendicular bisector, then it is perpendicular at the midpoint of the base.

Here is a 5 minute video on the isosceles triangle theorems:


http://www.5min.com/Video/How-to-Use-Properties-of-an-Isosceles-Triangle-516909805


And last but not least, the next scribe is...


NOT Matt, Kerry, Katie, Krista, Francis, Ashley, JD, Sean, David, Tim, Emma, Hollee, Zahra, Phill J, or Dmitri. So that leaves...


ALIVIA! MWA-HAHAHAHAHAHAHAHAHAHA! Happy Halloween! Or not! MWA-HA-HA! I'm not feeling too evil, or I would have picked Phill J. That would be EVIL!


PS If the video has an error (which it probably will), then I sent Ms Hunt an email with the video as well. Bye! -Phil K