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m Inelegant but effective brackets group the expressions compared, with lozenge denoting modal qualification (possibility) of nonequivalence of the expressions. Alternatively, exclamation symbolizing negation and square denoting modal qualification (necessity) of the expressions' equivalence.
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<math>[div(\vec{F})|_{\mathbf{x_0}=(x,y,z)}=div(\vec{F}|_{\mathbf{x_0}=(x,y,z)})=\lim_{V \to 0} \frac{1}{V}</math>{{oiint}}<math>\ _{S(D)} \vec{F}\cdot\mathrm{d}\vec{S}]
<math>[div(\vec{F})|_{\mathbf{x_0}=(x,y,z)}=div(\vec{F}|_{\mathbf{x_0}=(x,y,z)})=\lim_{V \to 0} \frac{1}{V}</math>{{oiint}}<math>\ _{S(D)} \vec{F}\cdot\mathrm{d}\vec{S}]
!\square=[\nabla \cdot \vec{F} = \frac{\partial\vec{F}(x,y,z)}{\partial x} + \frac{\partial\vec{F}(x,y,z)}{\partial y} + \frac{\partial\vec{F}(x,y,z)}{\partial z}]</math>
!\square=[\nabla \cdot \vec{F} = \frac{\partial\vec{F}(x,y,z)}{\partial x} + \frac{\partial\vec{F}(x,y,z)}{\partial y} + \frac{\partial\vec{F}(x,y,z)}{\partial z}]</math>

<math>
pV=\frac{1}{2}m|\vec{v_{total}}|^2=\frac{3}{2}Nk_BT
</math>

Revision as of 22:17, 2 July 2023

\oiint

\oiint