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∂(z3) ∂z = 103z2 = 1 3z2 Hence for every (x,y,z), divF≥ 1 > 0 Example 49 Find the values of n for which ∇2(rn) = 0 Solution We have r = p x2 y2 z2 and so from Example 45, ∂(rn) ∂x = nxrn−2 Therefore, ∂2(rn) ∂x2 = nrn−2 nx(n−2)xrn−4 = nrn−4 r2 (n−2)x2, and because of the symmetry in r with respect to x. 1/27/17 · n ɒɂ C N X f ڋ֎~©2chnet ꒷ ̕ȂɊ̐S ́g  J R h ߂ g p i Ȃ̂ 1day ̎g 񂵂Ȃ̂ 킩 Ȃ ĈӖ Ȃ ȂƎv B. @ w T y X A o X E ւ̖ x ̋ O ̑ q b g ɂ Đ ܂ꂽ ̂悤 ȃ o ł̃p t H } X ɂc u c B b h c F b y ̃T E h ő q b g w T y X A o X E ւ̖ x ȍ~ ̋Ȃ𒆐S ɉ t Ă ܂ A ȊO ɁH t 肵 Ă ܂ B ͂ b N X ̗ǂ ˔ ~ W V ̃ C u p t H } X ́A h ȃX e W O ʼn t U Ń Y ɂȂ Ă ܂ ł A ̃ C u ł́A t p t H } X ō ̃ x ł B X e B @ C ̃t @ G C h A @ f @ O ̃t @ ɂƂ ẮA e X ̃\ y ߂ Ƃ p ӂ Ă ܂ A w T y X A o X E ւ̖ x ȍ~ ̃z C g X l C N ̑ \ Ȃ͂قƂ lj t Ă ̂ŁA f B b h J @ f.

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P{Z≤ z} = y≤zx f XY(x,y)dxdy= ∞ x=0 zx y=0 e−xe−ydydx F Z(z)= ∞ 0 e−x(1−e−zx)dx=1− 1 1z,z≥ 0 f Z(z)= 1 (z1)2,z≥ 0 F Z(z)=f Z(z)=0forz. H Z C N z H S ݓX 戵 V f ځ@19 N3 x ԗD ǃV b v ܁@ H ̂ ȉ i E h t @ X i z I ׂ 4 _ Z b g/ L P X/ z/ p X P X/ R C P X/ h Z / ƋL O/ Əj /. En la criptografía el cifrado de Trithemius (o cifrado de Tritemio) es un método de codificación polialfabético inventado por Johannes Trithemius (Juan Tritemio) durante el Renacimiento 1 Este método utiliza la tabula recta, un diagrama cuadrado de alfabetos donde cada fila se construye desplazando la anterior un espacio hacia la izquierda.

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