We now introduce the %c{N+} substitution, which evaluates to part N, N+1, N+2, ...
我们现在介绍%c{N+}替换,它将被计算为N, N+1, N+2…
We first introduce methods of finding the sum of the first n terms for an arithmetic sequence of higher order.
我们首先介绍了高阶等差数列前n项和的求解方法。
So if we work out the chemical potential, it's just one over N times A.
如果我们计算出化学势，它就是N分之一乘以A
So it looks a lot less messy if we just draw our Lewis structure like this for h c n, where we have h bonded to c triple bonded to n, and then a lone pair on the nitrogen there.
这看起来整洁了不少,如果我们把氰化氢的路易斯结构画成这样的话，这样我们就有氢与碳之间的单键和碳与氮之间三键，然后还有一对孤对电子在氮这里。
N: No, we can just keep it intimate10 and do it when all of the family is here.
诺拉：不用，我们可以让它保持私人一点，在我们所有家人在场的时候进行。
So when we count, generally, again, we start from zero, we go to N minus 1.
当我们计数时，一般的，再次强调，我们从0开始，到N-1结束。
So, if we took the case of nitrogen, if we add an electron to nitrogen and go to n minus, we find that the change in energy is 7 kilojoules per mole.
如果我们以氮为例，如果我们给氮增加个电子令它变成-1价的氮，我们会发现能量的变化是，7，千焦每摩尔。
So we know that we're in the n equals 5 state, so we can find what the binding energy is here.
我们知道,我们在n等于5的态，我们可以找到结合能是多少。
With the N-Tier improvements for Entity Framework, we want to address some of the same problem space as DataSet, but we want to avoid the primary issues with it.
随着对EntityFramework进行N层改进，我们想解决一些相同的问题空间，例如数据集，但要避开它一些主要问题。
So let's start with a case when we have some number of the equations say n equations and n unknown.
所以让我们从一个例子入手，当我们有一些方程式，假定n个方程，n个未知数。
PROFESSOR: Good. So, it's going to be in 3, because that's the shortest energy difference we can have, 3 those 2 are inversely related, so it must be n equals 3.
教授：好，是3，因为它的能量差最小，红色的是我们能看到的，所以一定是n等于。
And unlike n, l can start all the way down at 0, and it increases by integer value, so we go 1, 2, 3, and all the way up.
不像n，l可以从0开始取，然后每次增加一个整数，所以我们可以去1，2，3，一直下去。
It's going to weigh too much, we don't need to look at it. But it'll still be order 2 to the n.
我们没必要去看它，但是它还是要遵循2到n的顺序。
x So the mole fraction at any time is n minus x divided by the total number of moles, x which we just calculated as n plus x.
任何时候的摩尔分数是n减,除以总摩尔数，就是我们计算的n加。
We receive shipments o n Tuesday and Saturday, which means that we have different items in the store at least twice a week.
我们通常在周二和周六安排发货，这就意味着一个店面的在一个星期内至少更新两次。
Are they going to take us seriously as an enemy if they think we eat Cap 'n Crunch for breakfast?
要是他们认为我们早餐吃的是‘克朗奇船长米粉’，他们还会继续把我们真正当成敌人来看待吗？
N what can we see through financial ratios?
n透过财务比率我们还能看到什么？
So, let's try another example here, and let's try a case now where instead of dealing with a neutral molecule we have an ion, so we have c n minus.
那么，让我们来试一下另外一个例子，这次不是一个中性原子,而是一个离子，氰离子。
We know what you need to do is take all the molecular partition functions, the transitional ones, and to the N factor. The number of particles.
只要把所有分子平动配分函数相乘,对平动部分,幂次是N，粒子总数。
我们也是那个时候认识的。
We said that output redirection using n> usually overwrites existing files.
使用n>的输出重定向通常覆盖现有的文件。
Our oldest son, Justin, was enjoying the cookies n cream ice cream we got the kids as a treat while we were at the water park.
我大儿子，Justin在享受美味的奶油曲奇雪糕，这是我们在水上乐园给孩子们的奖赏。
We want to write a function that generates accumulators-- a function that takes a number n, and returns a function that takes another number i and returns n incremented by i.
我们需要写一个函数，它能够生成累加器，即这个函数接受一个参数n，然后返回另一个函数，后者接受参数i，然后返回n增加（increment）了i后的值。
Divided by the number of moles initially, which is n. That's what we want.
除以开始时的摩尔数，也就是n也就是我们想要的。
So it's a more specific version of the equation where we have the n final equal to 2.
当我们令n等于2时,使这个这个方程变成更具体的版本。
We might then more simply create n-ary subtrees of "as many plains as possible."
然后，我们就可以更简单地创建“plain尽可能多”的n - ary子树了。
The 'p' command does what it always does, explicitly telling sed to print out the line, since we are in '-n' quiet mode.
因为是处于' -n '安静方式，所以'p '命令还是完成其惯有任务，即明确告诉sed打印该行。
Once we multiply it by a very large number n, it becomes very much greater than kT.
一旦我用一个很大的n与它相乘，它就会变得比kT大得多。
Once we multiply it by a very large number n, it becomes very much greater than kT.
一旦我用一个很大的n与它相乘，它就会变得比kT大得多。
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