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Luc Armant proposed the same idea before I did: leaving only the centreboard in the water augredelair.fr (36MB in French with drawings) and he's a member of the Syroco team. Best wishes to the project!

Reporting suggests it was just another publicity stunt to help launch his own television network farther to the right than Fox News. He never expected to win or even go ice farther than the primary. We all apparently underestimated the stupidity and gullibility of the American voters.

Where is the well regulated militia when you need it?

Here is an image "from a "pre-created list of values" from y=x^2... Do the numbers "going up" or going "down" "travel" towards infinity towards that 0 in the center??

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Weak decays are. e.g beta decay in nuclei. Not alpha, or gamma emission.

Dear Joigus, I have been thinking and working at this topic and I have done some progresses! I have extended your intuition to systems with more equations. So if we have a system of 3 unknows and 1 parameter of which we want to find the higher value that makes the system solvable, we can use the parameter as an unknows and add a 4th equation. To write the 4th equation we can consider the gradient of the other 3 equations. We can do \( \vec{v_{1}} = \boldsymbol{\nabla}f_{1}\wedge\boldsymbol{\nabla}f_{2} \), so \( \vec{v_{1}} \) is the vector tangent to the curve created by the intersection of \(f_{1}\) with \(f_{2}\). In the same way we can do \( \vec{v_{2}} = \boldsymbol{\nabla}f_{2}\wedge\boldsymbol{\nabla}f_{3} \) to get the vector tangent to the curve created by the intersection between \(f_{2}\) and \(f_{3}\). Now to have just 1 solution of the system, we need that those 2 curves are tangent between them, therefore we can impose \( \vec{v_{1}} \wedge \vec{v_{2}} = 0 \) as 4th equation of the system. I have tried with many systems and it works!!! Let me know if what I have done until here is clear... because now I would like to do the same with more equations in the system, but it looks quite difficult! Do you have any idea how to continue?

My opinion of your guess is very low. I think you see a pattern you want to see. Humans are dependent on pattern recognition to the point where we often force its use. Prophecy and astrology are examples of this. Vague descriptions are easily assigned, and if they're even close we can convince ourselves of a match.

In other words, Post are "Integrated." Kind of which i was trying to figure out when i was asking if 4 is a coefficient or a natural number😉 What your saying makes perfect sense now... I agree, especially when speaking on ""incompleted theories"" that have not been previously published and or on its final edit process by the distribution publishers... This science forum should make a "special note" on that as there are many people whom are prematurely excited to share their ideas with the world without their copyrighted information being "actually published" released and authenticated in news papers or other publicity measures. This really is a serious issue..

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Yes joigus, I think we start to be on the right way. So, as you already understood, my system becomes: \[ \begin{cases} −x^2 + 10x − 1/2y^2 + 6y − K = 0 \\ −2x^2 + 20x + 3/2y^2 − 18y = 0 \\ (−2x+10)−(−4x+20)=0 \end{cases} \] And like you said, from the 3rd equation I get \(x=5\). Substituting this in the 2nd equation, we get \(y=4.367\) or \(y=7.633\). Finally substituting x and y in the 1st equation, we get \(k=41.667=\frac{125}{3} \) . So for this example imposing the 3rd equation as the difference of the derivates respect x, works. You can try to do the same with the second example, but in this case you have to derivate respect y to obtain the 3rd equation. Solving the system, you will get: \[x=1.6626 \\ y=6.0 \\ k=45.0\] that is exactly again the maximum value of k that we were looking for. The reason is that the 3rd equation adds the condition of tangency between the 2 intersection of the surfaces represented by the first 2 equations with the plane xy. So , because they are tangent, the solution of the system that is the intersections of the previous curves, can only be 1. I show you both cases in the following picture: So, they case on the left that has 2 solutions is for \(k<125/3\) and the case on the right with just 1 solution (tangency between the 2 curves is \(k=125/3\). If we have \(k>125/3\), the 2 curves doesn't intersect and there isn't any solution.