Что думаешь? Оцени!
第二百三十八条 下列人员对本节规定的燃油污染损害不承担赔偿责任，但是损害是因其本人故意或者明知可能造成损害而轻率地作为或者不作为造成的除外：，推荐阅读谷歌浏览器【最新下载地址】获取更多信息
The coolest tech gifts may look like the most boring box in the present pile... until they're unwrapped. But in a second, you'll both unlock a new favorite memory: They get to gush over owning one of the most sought-after gadgets of the year, and you get to watch their face light up.，这一点在heLLoword翻译官方下载中也有详细论述
Италия — Серия А|27-й тур
Often people write these metrics as \(ds^2 = \sum_{i,j} g_{ij}\,dx^i\,dx^j\), where each \(dx^i\) is a covector (1-form), i.e. an element of the dual space \(T_p^*M\). For finite dimensional vectorspaces there is a canonical isomorphism between them and their dual: given the coordinate basis \(\bigl\{\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}\bigr\}\) of \(T_pM\), there is a unique dual basis \(\{dx^1,\dots,dx^n\}\) of \(T_p^*M\) defined by \[dx^i\!\left(\frac{\partial}{\partial x^j}\right) = \delta^i{}_j.\] This extends to isomorphisms \(T_pM \to T_p^*M\). Under this identification, the bilinear form \(g_p\) on \(T_pM \times T_pM\) is represented by the symmetric tensor \(\sum_{i,j} g_{ij}\,dx^i \otimes dx^j\) acting on pairs of tangent vectors via \[\left(\sum_{i,j} g_{ij}\,dx^i\otimes dx^j\right)\!\!\left(\frac{\partial}{\partial x^k},\frac{\partial}{\partial x^l}\right) = g_{kl},\] which recovers exactly the inner products \(g_p\!\left(\frac{\partial}{\partial x^k},\frac{\partial}{\partial x^l}\right)\) from before. So both descriptions carry identical information;