Примеры использования Sqrt на Английском языке и их переводы на Русский язык
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Sqrt square root.
The latter shows I{\displaystyle{\sqrt{I}}} is itself an ideal.
One exact solver runs in time n O( k){\displaystyle n^{O{\sqrt{k.
It is known that 2{\displaystyle{\sqrt{2}}} is irrational see proof.
This is the mathematical coincidence π≈ 4/ φ{\displaystyle\pi\approx 4/{\sqrt{\varphi.
Sqrt(double val) The square root of val number where val must be more or equal to zero.
To calculate the square root is used function Sqrt() from class Math.
The shift by 2 n{\displaystyle{\sqrt{2n}}} is used to keep the distributions centered at 0.
The inclination of a side face for such an ideal pyramid is\( arctg\ sqrt{ 2}=\) 54 44.
The factor of 1/ 8 π{\displaystyle 1/{\sqrt{8\pi}}} simplifies a number of equations in general relativity.
The eccentricity of a rectangular hyperbola is 2{\displaystyle{\sqrt{2.
Then K a b c d{\displaystyle K={\sqrt{abcd}}} since opposite angles are supplementary angles.
For brevity, it is also useful to introduce the variable p 0 2 m|E|{\displaystyle p_{0}={\sqrt{2m|E|.
In the alternate tileset, the long edges have 1+ 2{\displaystyle 1+{\sqrt{2}}} times longer sides than the short edges.
The best known algorithm approximates it within a factor of O(|U|){\displaystyle O{\sqrt{|U|.
However,⌊ n⌋{\displaystyle\lfloor{\sqrt{n}}\rfloor} is not necessarily a fixed point of the above iterative formula.
Every locally k-chromatic graph has chromatic number O( k n){\displaystyle O({\sqrt{kn}})} Wigderson 1983.
Improving the bound O( x){\displaystyle O({\sqrt{x}})} in this formula is known as Dirichlet's divisor problem.
The first group of vertices belongs to the one-dimensional domain corresponding to the SQRT operation.
An ex-tangential quadrilateral ABCD with sides a, b,c, d has area K a b c d sin B+ D 2.{\displaystyle\displaystyle K={\ sqrt{ abcd}}\ sin{\frac{B+D}{2}}.} Note that this is the same formula as the one for the area of a tangential quadrilateral and it is also derived from Bretschneider's formula in the same way.
It suffices to check that each prime gap starting at p is smaller than 2 p{\displaystyle 2{\sqrt{p.
Solving for x yields x R 2+ r 2- r 4 R 2+ r 2.{\displaystylex={\sqrt^{ 2}+ r^{ 2}- r{\ sqrt{ {4R^{2}+r^{2 Fuss's theorem, which is the analog of Euler's theorem for triangles for bicentric quadrilaterals, says that if a quadrilateral is bicentric, then its two associated circles are related according to the above equations.
This contradicts our initial supposition, so we are forced to conclude that 2{\displaystyle{\sqrt{2}}} is an irrational number.
This tiling can be seen as the union of two perpendicular hexagonal tilings,flattened by a ratio of 3{\displaystyle{\sqrt{3.
The optimization problem permits approximation and is approximable within a O(| V|/ log| V|){\displaystyle O\ left(|V|/{\ sqrt{\log|V|}}\right)} approximation ratio.
Wagner VI is equivalent to the Kavrayskiy VII horizontally elongated by a factor of 2{\displaystyle 2}⁄ 3{\displaystyle{\sqrt{3.
The carry-select adder is simple but rather fast,having a gate level depth of O( n){\displaystyle O{\sqrt{n.
In the uniform case, the optimal delay occurs for a block size of⌊ n⌋{\displaystyle\lfloor{\sqrt{n}}\rfloor.
With high probability, this process produces a graph with independence number O( n log n){\displaystyle O{\sqrt{n\log n.
Even stronger, for any fixed H, H-minor-free graphs have treewidth O( n){\displaystyle\scriptstyle O{\sqrt{n.