英文字典中文字典


英文字典中文字典51ZiDian.com



中文字典辞典   英文字典 a   b   c   d   e   f   g   h   i   j   k   l   m   n   o   p   q   r   s   t   u   v   w   x   y   z       







请输入英文单字,中文词皆可:

enveloped    音标拼音: [ɛnv'ɛləpt]
pp. 包围

包围

enclosed \enclosed\ adj.
surrounded or closed in, usually on all sides. Opposite of
{unenclosed}. [Narrower terms: {basined}; {capsulate,
capsulated}; {closed, closed in(predicate)}; {coarctate};
{confined, fenced in, penned}; {embedded, fixed}; {embedded,
surrounded}; {encircled}; {enveloped}; {fogbound};
{self-enclosed}; {surrounded, encircled}]
[WordNet 1.5]


enveloped \enveloped\ adj.
enclosed or surrounded completely; as, the fog-enveloped
city.
[WordNet 1.5]


Envelop \En*vel"op\, v. t. [imp. & p. p. {Enveloped}; p. pr. &
vb. n. {Enveloping}.] [OE. envolupen, envolipen, OF.
envoluper, envoleper, F. envelopper; pref. en- (L. in)
voluper, voleper. See {Develop}.]
To put a covering about; to wrap up or in; to inclose within
a case, wrapper, integument or the like; to surround
entirely; as, to envelop goods or a letter; the fog envelops
a ship.
[1913 Webster]

Nocturnal shades this world envelop. --J. Philips.
Envelope


请选择你想看的字典辞典:
单词字典翻译
enveloped查看 enveloped 在百度字典中的解释百度英翻中〔查看〕
enveloped查看 enveloped 在Google字典中的解释Google英翻中〔查看〕
enveloped查看 enveloped 在Yahoo字典中的解释Yahoo英翻中〔查看〕





安装中文字典英文字典查询工具!


中文字典英文字典工具:
选择颜色:
输入中英文单字

































































英文字典中文字典相关资料:


  • Reflexive Generalized Inverse - Mathematics Stack Exchange
    Definition: G is a generalized inverse of A if and only if AGA=A G is said to be reflexive if and only if GAG=G I was trying to solve the problem: If A is a matrix and G be it's generalized inverse then G is reflexive if and only if rank (A)=rank (G)
  • Let $G$ be a group, $a \\in G$. Prove that for all $g \\in G$, $|a . . .
    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it This scores points for you and for the person who answered your question You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?
  • Conjugacy Classes of the Quaternion Group $Q$
    Disclaimer: This is not exactly an explanation, but a relevant attempt at understanding conjugates and conjugate classes
  • Functional equation $x^2 (f (x)+f (y))= (x+y)f (yf (x))$ [closed]
    I think there's an interesting math question from a new user here, but unfortunately it was received much worse than it deserved @Jack costaux You can edit the question to improve it In particular: (1) Your original question asked how to find a solution of this equation If it stays same, I suggest to remove the solution-verification tag and the part "if i made any mistake clarify", so that
  • Let $a \\in G$. Show that for any $g \\in G$, $gC(a)g^{-1} = C(gag^{-1})$.
    Try checking if the element $ghg^ {-1}$ you thought of is in $C (gag^ {-1})$ and then vice versa
  • Prove that $o (a)=o (gag^ {-1})$ - Mathematics Stack Exchange
    Your proof of the second part works perfectly, moreover, you can simply omit the reasoning $ (gag^ {-1})^2=\cdots=e$ since this is exactly what you've done in part 1
  • Subgroup that contains all Sylow $p$-subgroups
    You could "copy" the proof of Frattini's Argument: let $\;g\in G\;$ and let $\;P\;$ be some Sylow subgroup of $\;G\;$ , so $\;P\le H\;$ But also $\;P^g:=g^ {-1}Pg\;$ is a Sylow subgroup and thus also $\;P^g\le H $ Note that all the Sylow subgroups of $\;G\;$ are also Sylow subgps of $\;H\;$ , so by Sylow's Theorems $$\exists\;h\in H\;\;s t \;\;P^g=P^h\implies P^ {gh^ {-1}}=P\implies gh
  • Centralizer : center subgroup ~ normalizer : normal subgroup . . .
    The normalizer of 𝐴 in 𝐺 is the largest subgroup of 𝐺 that contains 𝐴 and in which 𝐴 is normal -> in which 𝐴 is normal to the normalizer correct?
  • Difference between a group normalizer and centralizer
    Let H is a Subgroup of G Now if H is not normal if any element $ {g \in G}$ doesn't commute with H Now we want to find if not all $ {g \in G}$, then which are the elements of G that commute with every element of H? they are normalizer of H i e , the elements of G that vote 'yes' for H when asked to commute Hence, $ {N_G (H)=\ {g \in G: gH=Hg }\}$ | Now Centralizer of an element $ {a \in G
  • Proving that $gHg^ {-1}$ is a subgroup of $G$
    $1) $$ (gag^ {-1})^ {-1}=g^ {-1^ {-1}}a^ {-1}g^ {-1}=ga^ {-1}g^ {-1}$ $2)$ $ ga (g^ {-1}g)bg^ {-1}=g (ab)g^ {-1}$ I'm stuck at this point, Is it correct so far? is





中文字典-英文字典  2005-2009