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All right. I tried it on 2, I surely didn't expect a precise and exact answer to that but I got something, and if you square this, you'll find the answer kept pretty darn close to 2.
好，我试试求2的平方根,我当然不希望得到一个完全准确的答案了,但是我得到了一个近似值，试试将这个数平方一下,你会发现结果和2相当接近。
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Let me imagine that I've been lucky enough to get you out of the way-- you're fighting him you can't even look at me, but I can do that.
假设我运气够好,把你从这条路上推开了,你在和他搏斗,不可能看得到我,但我可以这样
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So this first little piece of code right here says, ok you give me 2 points, I'll create another 1 of these lists and I'll simply take the x, sorry I shouldn't say x, I'm going to assume it's the x, the x-values are the two points, add them together, just right there, the y-values, add them together and return that list.
好，为了来认识到这一点,让我们来看一个简单的小例子，在你们的课堂手册上，你可以看到我写了一个小程序,它假设我得到了，这些点中的一些，我想对它们做一些操作，例如我想把它们加到一起，那么这里的第一小片，代码的意思是，好给我两个点，我会再创建一个数组。
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The reaction that salespeople would get from women was, you're giving me some probabilities or something, you're asking me to-- it sounds like you're asking me to play in some gambling thing where I win if my husband dies; it doesn't sound right to me.
推销员从女性那里得到的回答是,你跟我讲了一通概率之类的东西,然后劝我...,这听起来似乎是在让我去赌博,如果我丈夫去世了,那我就赢了,这太不对劲了
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That means, when I take two derivatives, I want to get a, then you should know enough calculus to know it has to be something like at^, and half comes from taking two derivatives.
也就是说,当我想求二阶导时,得到了a,你应该有足够的微积分知识,才能知道必须有类似at^的项,而这个1/2则是因为求了两次导数
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So you can actually say, how do I know which methods are associated with the class? For that, we can call dir. And what it does, is it gives me back a listing of all the things, dir all the methods, that are associated with it.
我们开始建立类了，我们得到了这些方法,因此实际上你可以说，我怎么知道，哪些方法是关联与这些类的？,为了解决这个问题，我们可以调用，这个方法的作用。
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So I can concatenate techs plus ivys and assign that result to univs, and then when I print it you'll notice I just get a list of five strings.
和ivys数组用加号串联运算,并把结果赋值给univs数组，接下来我显示下univs的结果,你注意到我得到了一个。
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