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• More than 15 million Americans are out of work, and 7.2 million jobs have been lost since the recession began in December,2007.

VOA: standard.2009.10.03

• So, once we figure out what our probability of backscattering is, we'll just raise that to the 1/2, and we'll multiply that by 12 . 20 centimeters.

所以一旦我们,知道了散射概率，我们开根号。

麻省理工公开课 - 化学原理课程节选

• The way that we can figure this out is using something called bond order, and bond order is equal to 1/2 times the number of bonding electrons, minus the number of anti-bonding electrons.

我们可以用叫做,键序的概念来弄明白它，键序等于1/2乘以成键电子,数目减去反键电子数目。

麻省理工公开课 - 化学原理课程节选

• So I want to figure out the best response of Firm 1 as a function of the price chosen by Firm 2.

我想找出公司1的最佳对策,并用公司2价格的函数表示

耶鲁公开课 - 博弈论课程节选

• If you look at solve 2, it's going to solve2 run through the same kind of loop, printing out all of the answers.

我会让你们看一看,如果你在看，那将运行相同的循环。

麻省理工公开课 - 计算机科学及编程导论课程节选

• Sometimes there are completely new emphases and the story can come out to be a very, very different story. 1 and 2 Chronicles are a retelling and a reworking of much of the material from Genesis through 2 Kings, and it cleans up a lot of the embarrassing moments.

复述的故事会有全新的侧重点，可能和原来的故事之间,存在很大的差别，《历代志》上下就是,对从《创世记》到两部《列王纪》中大量材料的复述和,再加工，它清除了很多令人窘迫的故事。

耶鲁公开课 - 旧约导论课程节选

• In case 2, we're taking the 3 p out of the neutral atom, whereas in case 3, we're taking it out of the ion.

在第二种情况中，我们要从中性原子中拿走，3，p，电子，而在第三种情况中，我们要从这个离子中拿走它。

麻省理工公开课 - 化学原理课程节选

• So, in particular, let's start working out what share of the votes you'd get if you chose position 1 or position 2, against different positions the other guy can choose.

先来看看如果你选择立场1和2,在你对手选择不同立场时,你分别能获得多少选票

耶鲁公开课 - 博弈论课程节选

• So this is going to be the choice of Player 1, and this is going to be the choice of Player 2, and what I want to do is I want to figure out what this looks like.

这个表示参与人1的策略,这个表示参与人2的策略,接下来我想要知道,这个函数究竟是什么样子的

耶鲁公开课 - 博弈论课程节选

• Now, with this in hand, what I want to do is, and in Firm 2 I have a similar expression, what I want to do is, I want to figure out the Nash Equilibrium of this game.

知道它了以后,因为公司2的表达式也是相似的,接下来我们就可以尝试找出纳什均衡了

耶鲁公开课 - 博弈论课程节选

• So let's look at another example, Li2 let's take lithium 2 and see what we can figure out here. In lithium 2, we have two atoms of lithium, each have three electrons in them.

让我们看另外一个例子，我们看一看,看我们能得到什么，在Li2中，我们有两个Li原子，每个带有3个电子。

麻省理工公开课 - 化学原理课程节选

• If you see a problem that asks you, for example, the third ionization energy versus taking a second electron out of the 2 s in a photoelectron spectroscopy experiment, those are two very different things.

如果你遇到一个题目问你的是，比如说，是第三电离能,还是在光电子能谱实验中从，2，s，轨道中,拿走第二个电子，这可是两个完全不同的问题。

麻省理工公开课 - 化学原理课程节选

• The highest occupied orbital is now the 2 s orbital, 1 s 2 2 s 1 so we're going to end up with boron 2 plus 1 s 2, 2 s 1, plus the electron coming out of there.

现在最高的被占据轨道是，2，s，轨道，因此结果应该是正二价的硼，再加上一个出射的电子。

麻省理工公开课 - 化学原理课程节选

• And what I want to point out that we just figured out for molecular orbital theory, is that o 2 is a biradical, because remember, the definition of a radical is when we have an unpaired electron.

我要指出的是,我们刚利用分子轨道理论，指导了O2是二价自由基，因为记住，自由基的定义是,有个未配对的电子。

麻省理工公开课 - 化学原理课程节选

• In both cases we're taking an electron out of the 2 s orbital.

在这两种情况下，我们都拿走了一个，2s，电子。

麻省理工公开课 - 化学原理课程节选

• So, for example, 1s1 again we see hydrogen is 1 s 1, helium we say is 1 s 2, or 1 s squared, 1s2 so instead of writing out the 1 s 1 and the 1 s 2, we just combine it 1s22s1 as 1 s squared, lithium is 1 s 2, 2 s 1.

简化符号所以举例来说，我们又看到氢是，氦是1s2或者1s的平方，所以不是将它写为1s1和1,而是我们将它合并为1s的平方,锂是。

麻省理工公开课 - 化学原理课程节选

• But there's something you'll note here also when I point out the case of the 2 s versus the 2 p, which is what I mentioned that I would be saying again and again, which is when we look at the hydrogen atom, the energy of all of the n equals 2 orbitals are exactly the same.

但是这里有一些事情你们也会注意到,当我指出2s和2p的情况，我之前提过,我会一次又一次的说，我们在观察氢原子的时候，2层轨道的所有n的能量,是完全相同的。

麻省理工公开课 - 化学原理课程节选

• So what we're going to do is we're going to figure out Player 1's best response quantity to each possible choice of Player 2, and then we're going to flip it around and figure out Player 2's best response quantity to each possible choice of Player 1, and then we're going to see where those coincide, where they cross.

下面我们就需要表示出,参与人1对于2不同产量下的最佳产量,然后反过来写出,在参与人1的不同产量下,参与人2的最佳产量,然后再来看看这两者在哪里相交

耶鲁公开课 - 博弈论课程节选

• So what I need to do then is I need to figure out what is Player 1's best response for each possible choice q2 of Player 2.

所以我们要先找出,在参与人2的每个可选策略q2下,参与人1的最佳产量

耶鲁公开课 - 博弈论课程节选

• And if we talk about electrons or photoelectrons, again we can describe it in terms of energy, we can talk about velocity, and from there, of course, you can figure out the energy from 1/2 m v squared, and actually we can also describe the electron in terms of wavelength.

如果我们谈论电子或者光电子，我们又可以从能量的角度来描述它，我们可讨论速率，从那里，当然，你可以计算1/2，m，v2得出能量值，而且事实上我们也可以,从波长的角度来描述电子。

麻省理工公开课 - 化学原理课程节选

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