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So if I take away the quotation marks, all I have is the word with no indication that it's a concept.
所以如果我拿掉引号，剩下的就只是这个词而不指示一个概念了。
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It's an historical question, as I say, and I want to devote the rest of the lecture to it.
这是一个历史问题，正如我刚刚说的，我会用剩下的时间讲讲这个问题。
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In Alice in Wonderland, the Cheshire Cat disappears and all we have left, the last thing that disappears, is the smile.
在《爱丽丝漫游仙境》里面,笑脸猫消失不见了而剩下的,最后一个东西,就是微笑
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Just the first one or more generally the address of the first one and then, man, I'll just figure it out from there where the rest of the letters are because by definition of a string, they're back to back to back.
只需要第一个字节的地址,然后，我可以指出之后的,剩下的字母,因为通过一个字符串的定义，它们是紧邻的。
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Moving a stack of size 3, again, if I want to go over there, I need to make sure I can put the spare one over here before I move the bottom one, I can't cover up any of the smaller ones with the larger one, but I can get it there.
移动上面的三个圆盘的话，我在移动最底下的圆盘之前，同样要确保剩下的,是在另外一个柱子上，因为我不能把大的圆盘,放在小的上面。
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So, what we have left in our equation is only one part that n we haven't explained yet, and that is that n value.
现在方程里只剩下一个常数,我们还没有解释，这就是。
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Lithium has three electrons. If it loses two one remains. What is this formula telling us?
锂原子有三个电子，如果它失去两个,剩下一个，那么这个式子告诉了我们什么？
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That leads to our next topic that I want to pursue for the remainder of the class, the reform of poetry and the arts.
这导入我们下一个主题,而我想要利用今天剩下的时间，寻找诗学与艺术的改革。
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It'll probably be the last example I have time for and so I'll go through the rest of them quickly in the next lecture.
因为时间关系,这可能是最后一个例子,下节课我会快速讲完剩下的内容
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I've set these two things equal to each other: I've got one equation and one unknown.
我让两个方程相等,只剩下了一个等式和一个未知数
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If you do that, you will find there is one unknown, which is v0, and we can solve for v0.
这样就只剩下一个未知量了,也就是 v0,我们就能解出 v0
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Before we get to suicide, you might say, the question that's going to entertain us for the remaining few weeks is this.
在我讲到自杀之前，你可能会说，剩下几个星期里让我们有兴趣的,问题是这样一个问题。
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Yup, so one total node, 2 minus 1 is 1, and that means since l is equal to 1, we have one angular nodes, and that leaves us with how many radial nodes?
一个节点，２减去１等于１,因为ｌ等于１,我们有一个角向节点，那剩下径向节点有多少个呢？
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And that then let's me get to, basically, base this code. Having done that, height I simply call base with get float, I call height with get float, and do the rest of the work.
现在让我们看看这段代码，其实已经写好了，我简单的调用下，用来得到一个浮点数，再调用下，来得到另外一个浮点数，然后做剩下的工作就可以了。
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Well, I might move my hands here 'cause this list only has one element left.
我会将手移到这边，因为这个列表中,只剩下一个元素了。
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And where we had left off was we were going to start one example of thinking about now where we have a heteronuclear diatomic molecules, so two different atoms in terms of forming the molecule.
我们还剩下一个,异核双原子分子的例子没讲,这里组成分子的原子，是不同的。
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So that leaves each carbon with only one hybrid orbital left.
这样每个碳原子只剩下一个杂化轨道。
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And I'm going to get minus 2, which came from here: minus 2 and that is in fact negative, which is what I wanted to know.
最后只剩下-2了,2是一个负数,这正是我想要的
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I want to play one more game today in the remaining minutes.
在剩下这几分钟我们再玩一个游戏吧
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I think it depends on whether the list is odd or even in length. Actually, that's probably not true. With one, it'll probably always get it down there, but if I've made it just equal to two I might have lost.
是奇数还是偶数，事实上,这是不正确的，如果最后剩下一个,那可能得到了结果,如果剩下两个，可能错了,所以，首先我们要格外。
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It took me seven comparisons, because I can take advantage of the fact I know I only ever have to look at the first element of each sub-list. Those are the only things I need to compare, and when I run out of one list, I just add the rest of the list in.
进行了7次对比，因为我可以利用我知道的优势：,每次只需要比较每个子列表的第一个元素,那才是我需要进行对比的内容,当一个列表的元素处理完了,只需要将另一个列表剩下的元素直接添加进去。
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So this is a little bit trickier to look at and see what it means, but essentially we have two hybrid orbitals, which are shown in blue here, and then we have one p orbital that's left alone that's going up and down on the page.
看这个图肯能会觉得比较诡异，但本质上，骂我们有两个杂化轨道，这里用蓝色表示，还有剩下一个p轨道,在图中上下方向上。
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And if we hybridize these orbitals in carbon, what we end up with is having two hybrid orbitals, and then we're going to be left with two of our p orbitals that are each going to have an electron associated in them.
如果我们杂化碳原子里这些轨道，我们能得到两个杂化轨道，另外剩下两个p轨道,每个里面有一个电子。
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Take these two equations, put an equal sign between them, replace this PR throughout with X, I'm going to have one equation and one unknown and that even the math phobics in the audience did in high school.
在这两个方程中间加个等号,将这里的Pr换成X,就会得到一个等式和一个未知数,剩下的问题,大家高中的时候就该回做了吧
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And if you just look at the numbers, you can see how it cuts in from one side and then the other side as it keeps narrowing that range until it gets down to the place where there are at most two things left, and then it just has to check those two to say whether it's there or not.
你能看到他是如何不断的，从一个大的范围被拦腰劈开,知道最终只剩下两个数字,然后就只需要,比一比就知道结果了,将它同线性查找比较下。
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