Sample letter for use in validating your mortgage (but with modification can be used for any debt) debt as described by your lender or your mortgage servicing company. As always; I recomm…Full description
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Tutorial on programming the Raspberry micro computer. Includes unpacking, description of onboard components, connecting peripherals, and writing code. This is a great introductory document for some...
Book of proof
The undersigned officer of the Internal Revenue Service, a duly authorized agent of the United States in this behalf, being duly sworn, deposes and says that:Full description
About Pi
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Descripción: Proyecto de Ingenieria
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This is a reference of some of the most common proof marks that are found on firearms. These allow a person to trace its history, what its proofed for and what repairs were done and by who.
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The Mathematics of the Vesica Vesica Piscis. Piscis . For every angle we get two triangles as seen here here..
As you can see, these triangles have a very interesting relationship, for every angle the radius (r) h ! " h#.
The symmetry of these triangles is shown $elow.
For every angle the radius (r) h ! " h#.
The %osmic &dentity. This mathematical identity wor's for A num$ers (*).
The %osmic Triangle. +e can lin' the cosmic identity with pythagoras theorem to get the following cosmic triangle. u$stituting * -/ and !* /- as shown $elow. $elow.
0ecause of the nature of the cosmic identity, identity, the triangles a$ove will wor' with any value * and /.
1sing the cosmic triangle a$ove in a vesica piscis, where $oth circles have radius (r) -/ " /-.
As seen $elow the hypotenuse of the $lue triangle 2/
The Angle of Perfect ymmetry. ymmetry. At 567 this special angle, the chord (c) r " h !.
The Pi Triangle. As we can see a$ove, it seems the following two triangles must $e correct.
2/ (-/ " /-) " (-/ 3 /-) /# (-/ " /-) 3 (-/ 3 /-) &f we use pythagoras theorem and the $lue pi triangle to calculate pi (/), we get the following value.
& have $een told that this is not a mathematical identity $ecause it does not wor' for >ero (?), lets e*plore. Assuming that * is always @ ! and !* is always !, and also that the reciprocal of >ero (?) is infinity (B).
As you can see, these varia$les always cancel each other out. u$stituting !* ? (>ero) and * B (infinity) .
CDvery e*plicit duality is an implicit i mplicit unity.E unity.E Alan +atts +atts
