查询词典 debug statement

I don't agree with your statement, but I give deadly advocation for your rights of speaking.

我不同意你说的话，但是誓死捍卫你说话的权利。

According to a statement, the new role is in addition to his current position as CEO Aegis Media Pacific.

据一份声明表示，这个职责是基于他目前在安吉斯媒体太平洋首席执行官的职位增设的。

Here is the affidavit of support from them and a bank statement of their financial resources.

这是他们的担保证明和银行出具的关于他们的财力证明。

There was an affidavit of support submitted and photocopy of a bank statement.

那是一份表示支持的宣誓书和一张充分的银行证书复印件。

Evidence of my financial resources in the form of a bank statement accompanies this Affidavit of Support.

现随信寄上银行存款支付报告书一份，做为我财源的证明。

First, we introduce and discuss the various methods of multivariate polynomial interpolation in the literature. Based on this study, we state multivariate Lagrange interpolation over again from algebraic geometry viewpoint:Given different interpolation nodes A1,A2 .....,An in the affine n-dimensional space Kn, and accordingly function values fi(i = 1,..., m), the question is how to find a polynomial p K[x1, x2,...,xn] satisfying the interpolation conditions:where X=(x1,X2,....,xn). Similarly with univariate problem, we have provedTheorem If the monomial ordering is given, a minimal ordering polynomial satisfying conditions (1) is uniquely exsisted.Such a polynomial can be computed by the Lagrange-Hermite interpolation algorithm introduced in chapter 2. Another statement for Lagrange interpolation problem is:Given monomials 1 ,2 ,.....,m from low degree to high one with respect to the ordering, some arbitrary values fi(i= 1,..., m), find a polynomial p, such thatIf there uniquely exists such an interpolation polynomial p{X, the interpolation problem is called properly posed.

文中首先对现有的多元多项式插值方法作了一个介绍和评述，在此基础上我们从代数几何观点重新讨论了多元Lagrange插值问题：给定n维仿射空间K~n中两两互异的点A_1，A_2，…，A_m，在结点A_i处给定函数值f_i(i=1，…，m)，构造多项式p∈K[X_1，X_2，…，X_n]，满足Lagrange插值条件：p=f_i，i=1，…，m (1)其中X=(X_1，X_2，…，X_n)，与一元情形相似地，本文证明了定理满足插值条件(1)的多项式存在，并且按"序"最低的多项式是唯一的，上述多项式可利用第二章介绍的Lagrange-Hermite插值算法求出，Lagrange插值另一种描述是：按序从低到高给定单项式ω_1，ω_2，…，ω_m，对任意给定的f_1，f_2，…，f_m，构造多项式p，满足插值条件：p=sum from i=1 to m=Ai=f_i，i=1，…，m (2)如果插值多项式p存在且唯一，则称插值问题适定。

At the basic of the review of traditional intangible assets in the side of affirmance、dimension、amortization and publication, this article analyses the problems of the Traditional intangible assets accounting, which is in the age of the knowledge-based economy, and introduces some concrete steps of enlarging its vange, improving the method of dimension and amortization, then beefing up the content of a financial statement.

本文在对传统无形资产会计核算的确认、计量、摊销、披露四个方面进行评述的基础上，分析了进入知识经济时代传统无形资产核算所面临的问题，并提出了拓展无形资产的范围，改进无形资产的计量、摊销的方法及充实财务报表的内容等具体措施。

Should any part of the Waiver for any reason be judged legally invalid or ineffective under applicable law, then the Waiver shall be preserved to the maximum extent permitted taking into account Affirmer's express Statement of Purpose.

若因为任何原因，「抛弃」的任一部分依据准据法，在法律上会被法院认定为无效，则该「抛弃」在宣告者明示的「目的之声明」中，应在被允许的最大范围内被考量。

Should any part of the License for any reason be judged legally invalid or ineffective under applicable law, such partial invalidity or ineffectiveness shall not invalidate the remainder of the License, and in such case Affirmer hereby affirms that he or she will not exercise any of his or her remaining Copyright and Related Rights in the Work or assert any associated claims and causes of action with respect to the Work, in either case contrary to Affirmer's express Statement of Purpose.

若因为任何原因该「授权」的任一部分在相关法律下被法院裁定为无效，该无效部份不因而使「授权」其余部分亦无效，且在这个情况下，宣告者於此声明他/她将不会行使任何他/她对本著作剩余的著作权及其相关权利或提起任何与本著作有关的相关联诉讼主张，在前述任一情形，本处之声明并不会与宣告者明示的「目的之声明」相违反。

If any person B, in any imaginable situation at any time t', were warranted in believing that at an earlier time t some specified person A sincerely made the statement "My present afterimage is red," and also that A did not immediately revoke or otherwise discredit it, then B would not be warranted at t' in believing that at t A had an afterimage that was not red.

这里排除A所可能发生的"使用和表达错误"，A确实有红色的余象（ afterimage ：视觉刺激结束后持续的图像）；而且这里不仅要表明B没有证据表明A的陈述是错的，而且之所以如此是因为A的陈述意味着被描述为红的是余象而非其他，不是一个反思，不是一个玻璃杯上的红点，不是墙上的红斑。B不能矫正的不是A的陈述，而是A的余象。

Many will continue to choose to live in duality and in conflict.

许多人将继续选择活在二元对立性和冲突中。

I find that students of the University of Physical Education all wear sportswear at first sight.

我发现：体育大学的学生乍一看，都是穿运动衣，大家都一样