It's not like all the orthopedic injuries I've had where you go get an MRI or an X-ray and you get it fixed.

VOA: standard.2010.02.26

And then from that start point, get the x-value. Same thing, from that instance, get the endpoint, from that end point get the x-value, square.

取得结束点，然后取得结束点的y坐标，然后求差开平方，然后同样求的y坐标。

麻省理工公开课 - 计算机科学及编程导论课程节选

Locally x got bound to 3, 3 I added 1 to it, whoop-dee-doo, 4 I get back a 4. But what's the value of x?

这是我想要的对吧？,在局部程序中x被赋值为，我给它加了1，我得到了？

麻省理工公开课 - 计算机科学及编程导论课程节选

There is no way that you could irradiate a crystal of nickel with a single beam of x-rays and get that circular ring pattern if the electron beam were behaving as a particle beam only.

没有什么其他的途径,可以让镍晶体发射,单束X光来,继而得到环状图案,如果电子束,是以粒子束形式出现的。

麻省理工公开课 - 固态化学导论课程节选

That's a choice, and that choice turns out to be very interesting and really important, because if you connect these two points together, you get a straight line that has to intercept the x-axis at some point.

在这一选择下,我们会发现一件非常有趣，而且极其重要的事,当你把这两点用直线连接起来，你会发现这条直线,将与x轴在某点相交。

麻省理工公开课 - 热力学与动力学课程节选

We know it's going around in a circle because if I find the length of this vector, which is the x-square part, plus the y-square part, I just get r square at all times, because sine square plus cosine square is one.

我们之所以知道它做圆周运动,是因为我求出了这个矢量的模长,也就是 x 的平方加上 y 的平方,我就得到了它在任意时刻的模长平方,因为正弦平方加余弦平方始终等于1

耶鲁公开课 - 基础物理课程节选

And keep going, until the square of one of those integers is greater than or equal to - sorry, just greater than x. OK, why am I doing that? When I get greater than x, I've gone past the place where I want to be.

如果还是比x小的话，跳到3，这么继续下去,直到一个整数的平方大于或者等于,对不起，是大于x，好,为什么我要这么做呢？,让我得到的整数的平方和。

麻省理工公开课 - 计算机科学及编程导论课程节选

If de Broglie is correct, let's irradiate the nickel crystal with a beam of electrons where the wavelength of the beam of electrons is identical to the wavelength of the x-rays that were used and see what we get.

如果德·布罗意是正确的，那么镍晶体放出,电子光束,电子束的波长,和原来讲过的X光的波长是相等的,再看看得到了什么。

麻省理工公开课 - 固态化学导论课程节选