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Examples
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Instead of x= + or -b, times the square root of 4ac, all over 2a, it is now simply =+squareroot.
Tassel Hassles Jen 2009
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Assuming that the parity operator is hermitian, meaning it is observable, means that its eigenvalues can only be real, meaning there are only two possibilities: 1 or -1. (squareroot of 1 is either 1 or -1). which leads to the conclusion that in this simplified case of two particles in one dimension there are two types of wavefunctions, one we will call bosons and the other fermions.
Thanksgiving Sean 2008
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Problem 149650 solved by Alan3354 webmaster@algebra. com: This problem was solved by Alan3354: Solve x = Arc tan (- squareroot 3)
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Problem 149650 solved by Alan3354 webmaster@algebra. com: This problem was solved by Alan3354: Solve x = Arc tan (- squareroot 3)
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(x) squared/16 - (y) squared/25 = 1. i have the focus = squareroot of 41 and asymptotes = +or -
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(x) squared/16 - (y) squared/25 = 1. i have the focus = squareroot of 41 and asymptotes = +or -
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(half of the length of the base) = squareroot of (4 - height squared), so that length of base = 2 x squareroot of (4 - [height squared]). en Español
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t = 2pi [squareroot (l/g)] he exhibits a similar discernment.
The Approach to Philosophy Ralph Barton Perry 1916
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