Tuesday, April 21, 2009

Aspartame Withdrawal Appetite Loss

Problem Solved

SQUARES AND STICKS


------------ --------------------------------------------------
---------------------- Look at the sequence of figures below.


1) Draw the two figures that follow in sequence.

2) Calculate the number of sticks needed to draw the figure N ° 10

3) Noting some regularity, calculates necessary to draw stick figure ocuupa place 100 of that sequence.

4) If we call n the figure number, write a formula to calculate the sticks in n function.

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

POSSIBLE SOLUTIONS

told The manner in which sticks found two different solutions but turned out to be two different ways to write the same formula. See ...

In the figure below, drag the slider and see how it follows the sequence.


















Sorry, GeoGebra Applet can not start. Make sure you have Java 1.4.2 (or later) installed and activate JavaScript in your browser ( Click here to install Java now )




This fellow was as follows:

Four sticks to the first set and then added three sticks for each of the following figures.

Move the slider to confirm. Have been painted a different color sticks that are adding figure to figure.

See if there is any relationship between the amount added 3 and the figure number in each case. To see more clearly, it should complete a table like the following. Remember to leave raised the operations.


The number of times that sum 3 is one less than the figure number!

That is:

sticks Qty = 4 + 3. (Figure number minus 1) If you call

n the figure number and p the number of sticks, then, to 4 must be added ( n - 1) times 3.
The formula here is:


ANOTHER WAY OF TELLING STICKS
That led to another formula. See ...



















Sorry, GeoGebra Applet can not start. Make sure you have Java 1.4.2 (or later) installed and activate JavaScript in your browser ( Click here to install Java now )



This companion aims to have the sticks like this:

Three sticks for each a stick figure at the end to close.

With what the formula here is:

Monday, April 20, 2009

Best Slideshow Software For A Mac



ACTA No. 47 - Resolution No. 2 of CODICEN

I recommend reading Chapter VI, which describes the Advisory Council Teaching ( CAP). Remember that in a few days should elect their representative!
The first five leaves are the circular signed by the Code and then starts the Statute. I leave you a summary of the chapters for easy reading.

CHAPTER 1 - General Principles - Articles from 1 to 9

CHAPTER 2 - From the substantive scope - Article 10

CHAPTER 3 - From the charts and study centers meetings - Articles from 11 to 13

CHAPTER 4 - Responsibilities of the student - Articles from 14 to 18

CHAPTER 5 - Among the responsibilities of officials, including pertains to the implementation of this Charter - Article 19

CHAPTER 6 - The Educational Advisory Board - Articles from 20 to 22

CHAPTER 7 - Disciplinary Proceedings - Articles from 23 to 30

CHAPTER 8 - Of the corrective measures - Articles from 31 to 36

CHAPTER 9 - Transitional provisions - Articles 37 and 38. Student Status
at Scribd Publish or explore others:
This videito

Wednesday, April 15, 2009

Wrestling Bedroom Ideas

PYTHAGORAS STUDENT STATUS AND DONALD DUCK

Walt Disney tells the story of Pythagoras and the Pythagoreans. Many
contributions made by the Pythagorean sect to mathematics and even though you do not believe it, the music too. If you watch this video would be good to discuss what the teacher of music, sure you will love.


Video Playlist
I made this video playlist at myflashfetish.com




PYTHAGORAS THEOREM

The square of the hypotenuse is equal to the sum of the squares of the legs.


visual demonstration
Move the slider shown in the figure and is worth every step.


















Sorry, GeoGebra Applet can not start. Make sure you have Java 1.4.2 (or later) installed and activate JavaScript in your browser ( Click here to install Java now )


Monday, April 13, 2009

Difference Between Magnums Regular

Discovering regularities. Powers

DISTRIBUTED PROBLEM No. 2
I leave, down three problems solved in class this week.
also left them in the photocopier. To print directly from the blog to click the button that says "MORE" and finding the option to print.


generalizacion1


entry in second year Mathematics: Solved the problem is solved No. 1.

Wednesday, April 8, 2009

Marriage Poems In Tamil

The launch of Planck just around the corner

No final decision yet for the launch day of the Planck satellite, which will be released along with the telescope Herschel on board the Ariane. The forecast was a launch in mid-April, in particular the 16th, but no date was postponed
final. The reason seems to be the complexity of Herschel, according news page of the ESA :
The Herschel telescope mirror, the largest ever to Be Launched in space, is a novel and advanced concept using silicon carbide petals brazed 12 together Into a single piece, it is one of the major Technological highlights of the mission. The Complexity of the structure and Its uniqueness Means Must Be taken great care to Ensure That stress exerted on it DURING launch Are well understood.
Over the next Few Days, a panel of independent experts led by the ESA Inspector General and Arianespace will carry out a final cross-check of the documentation to Demonstrate That the required safety margins for the telescope Are met.

The new launch date Will Be Shortly defined.

We listen. Late Information: Herschel and Planck launch

campaign.

Tuesday, April 7, 2009

The Molecular Structure Of Clf2



Afshar's experiment is controversial because the author says he refuted the principle of complementarity. This states that two descriptions, the wave and corpuscular, are needed to understand the quantum world, but are complementary: when it's one the other is not valid, and vice versa. This conclusion can be questioned as discussed below.
To understand the experience part of the experiment usual double-slit.



In this experiment we know that if both slits are open it shows the wave behavior in the form of interference on the photographic plate to the right. If however we are able to determine the beginning and somehow the way that a particle took (slot S1 or S2), then the board does not appear to be interference and shows us corpuscular behavior.

Now imagine that after the slots first put a grid. This grid is such that openings correspond to the interference peaks. This means that for the case of the two open slots in the grid does not attenuate the signal no end to the photographic plate as it is letting through the correct photon peaks. Returning to the experiment

double ranuda usual, above the grille. Consider now that we slots behind a lens, which acts as follows. When only slot 1 is open, the lens sends the photons that pass through it to the detector 1. When only the slot 2 is open, the lens sends the photons that pass through it to the detector 2.

Now imagine that both things together, the grid behind the lens after slots and the slots.


This
Afshar's experiment. The result is that there is no signal attenuation in the photographic plate. That is, there is interference between waves passing through each of the slots.

Asfhar concluded that these two things happen at once: there is interference due to the lack of attenuation and there is a determination of the path of each photon due to the use of the lens. This is the principle of complementarity, as in a single experiment, the photons have wave and particle characteristics shown.

However, this conclusion need not be correct. What seems quite beyond doubt is that there is interference. But that does not seem clear is that the lens actually is determining the way in the case of two open slots. What is proven is that the lens sends the photons that pass through the slot 1 to detector 1 for slot 2 being closed, and vice versa. Conclude from this that there is a determination of the road for two open slots for an extrapolation is probably unacceptable.

To illustrate consider an electron in a superposition of the projection of its spin on the z axis ( is in the state mentioned say anything about the position during the trip. We can not ask which slit the electron passed since the system is not in an eigenstate of the basis on which we project.

The difference in this experiment is that the final projection of the particle state is not about all possible vertical positions on a photographic plate, but as Ruth Kastner (in
Why the Afshar Experiment Does Not Refute Complementarity
) calls "slit-basis"




In the double-slit experiment, the system prepared in a particular state can be projected on all possible positions on the vertical axis using a measurement of position with a photographic plate. In this case, using lenses, the system can be designed only on L 'and U'. However, these two basic states do not correlate one to one with the states L and U.
More information:
http://en.wikipedia.org/wiki/Afshar_experiment



Monday, April 6, 2009

Which Currency Is The Best To Invest In?

Afshar's experiment conditions for the paradox of Olbers


Olbers's paradox tells us that in a static and infinite universe the night sky should be completely brilliant free regions devoid of light or dark. Out the terms of the paradox precisely help us understand better.
Specifically, the mathematical formulation of the paradox is to calculate the flow of light we receive to be located in the origin of a spherical coordinate system. To begin with we establish two hypotheses. The first hypothesis is that light sources are distributed homogeneously in space. The second is that space is static. first consider the flow of light that reaches us emitted from a source at a distance r. In a static space this flow is proportional to the inverse square of distance, f ~ 1 / r ². Then consider the amount of light sources in a spherical corona. This number n increases with the radius squared, n ~ r ². Therefore, the sum of the flow of light from all light sources located in a spherical shell at any radius is F ~ f n. This value is a constant as r dependence cancels. How many layers summary, we have found an infinite stream combining four hypotheses:

1. The light sources are distributed homogeneously in space
2. The space is static
3. The universe is infinite in extent
4. The light sources are timeless We are going

indicating these assumptions made at a time.

First, the homogeneity of the light sources. If the source distribution is not homogeneous paradox need not be. Benoit Mandelbrot, the creator of the fractal, proposed a cosmology that solved the paradox of Olbers precisely that. Specifically, in a universe in which the light source distribution of fractal dimension is less than two, the paradox does not occur even if you have infinite sources in an infinite universe whose existence is eternal.

The reason is that it no longer n ~ r ² and there will be no cancellation f ~ 1 / r ². What is true in general that for a fractal dimension D, n ~ r ^ {D-1}, making the paradox is resolved if D

second hypothesis, the static space. If space is not static and is not satisfied that f ~ 1 / r ², but in general for an expanding universe flow is more diluted. In some models this resolves the paradox. This was the solution of the stationary model of cosmology of Hoyle and Burbridge was based on a space-time de-Sitter.

third hypothesis, the infinite expanse of space. If space is not infinite and has an edge (or say that the source distribution Light has an edge), then the sum over spherical crowns ends at a certain radius. Thus the sum of the flow is finite.

But beware, if space is finite but unbounded (and the rest of the following assumptions remain valid, namely the old infinite light sources), then the light emitted in more remote past makes one or several times around the universe to reach us. Since the age of sources is infinite the number of crowns to consider again is infinite and the paradox is not solved. Fourth

hypothesis, age infinite light sources. This is obvious: If the light sources do not exist from the infinite past, but from a time T, then because the speed of light c is finite, the number of crowns to consider is finite, in particular to a radius R = c T. < 2. La justificación de un modelo cosmológico así viene del hecho que a cierta escala las fuentes de luz parecen agruparse en distribuciones fractales en el universo - aunque según me consta a mí los estudios no son concluyentes y no son extrapolables a muy grandes escalas.

is wonderful to see how this simple, everyday paradox confronts us with an unprecedented cosmological mystery and becomes a cornerstone of cosmology, as it shows the depth of the possible solutions to the paradox. Any model of the universe must necessarily deal with it.