It's that the pounds of gold plus the pounds 8 of silver plus the pounds of raisins is no greater than eight.
就是金的磅数加上银的磅数,加上葡萄干的磅数不大于,所以我已经找到了。
And so when we say something is exponential, we're talking about in terms of the number of bits required to represent it.
所以当我们说某些东西,是指数增长的我们指的,就是代表它的比特数。
So by parallel we mean - they're either both spin up remember that's our spin quantum number, that fourth quantum number.
所以我们意味着,它们都是自旋向上,记住我们的自旋量子数,是第四个量子数。
So the temperature is intensive, and you can make intensive properties out of the extensive properties by dividing by the numberof moles in the system.
所以温度是强度量,你可以通过,除以系统中物质的摩尔数,来从广延量中导出强度量。
You bought the house for $300,000; you sell it for $270,000 that's exactly what you owe, so you have nothing left.
你以30万的价格买入房子,然后以27万的价格卖出,27万就是你的负债数,所以你什么都没剩下
A lot more of you chose numbers between 20 and 30, so we're really getting into the meat of the distribution.
有很多人选了20至30之间的数,所以我们已经涉及到分布的问题了
4 So even if the correct mathematical answer is 1.4 or whatever, when you divide an int by an int, you only have room in that variable, in the response for an actual integer.
所以即使那个正确的答案是4,或别的数值,当你用一个整型数除以一个整型数,在那个变量的返回值里,只有,存储一个整型数的空间。
So carbon 12. We know that it has the proton number, by definition, is 6. And the neutron number, 6 from 12 is 6. So it has 6 protons and 6 neutrons.
所以碳12,我们知道它有质子数,根据定义,那就是6,而电子数,12减6等于6,所以它有6个中子。
So, Elvis Presley has a Bacon number of two.
所以猫王的Bacon数为二。
So we can have, if we have the final quantum number m equal plus 1 or minus 1, we're dealing with a p x or a p y orbital.
所以如果我们有,磁量子数m等于正负1,我们讨论的就是px或者py轨道。
Here just two, so we changed the number of moles of gas by three. All right, how much did it matter, right?
所以我们将气体的摩尔数,改变了3摩尔,好,它会起多大作用?
The reason there are three quantum numbers is we're describing an orbital in three dimensions, so it makes sense that we would need to describe in terms of three different quantum numbers.
我们需要,3个量子数的原因,是因为我们描述的是一个,三维的轨道,所以我们需要,3个不同的量子数,来描述它。
So we get at the number of neutrons indirectly because we know the proton number here.
所以,我们可以间接得到中子数,因为我们知道质子数了。
So we know that all the numbers between 45 and 30, these strategies were not dominated.
所以可知选择30至45之间的数,这样的策略在原博弈中并不是劣势的
We know no one's going to choose 68 and above, so we can just forget them.
我们知道没人会选择68及以上的数,所以可以忽略它们
He stayed up all night and counted every leaf on the tree and it came very close to what Rituparna said, so he--the next morning-- believed Rituparna.
他彻夜未眠,数了树上的每一片叶子,发现结果和睿都巴若那所言相差无几,所以他在...第二天早晨...,相信了睿都巴若那
So we know no one's choosing any strategies above 45.
所以我们知道没人会选择大于45的数
So this declares an integer, a variable of type int called I, 0 and initializes it to zero.
所以这里声明了一个整型数,一个叫做I的整型变量,把它初始化为。
Yeah. Suppose I choose it right down there I guess 0. Well, the tangent there will not even have an x intercept. So I'm really going to be dead in the water.
好,如果我选猜想数为0,好吧,这个点的切线甚至和x轴都没交点,所以在这儿这一原则,真的不起作用了。
So what data type can we use to actually get more bits of precision than an int?
所以我们能用什么数据类型来表示,比int类型更多位数的数呢?
So, we need to actually add on this fourth quantum number, 1/2 and it's either going to be plus 1/2 or negative 1/2.
所以我们需要加上这第四个量子数,它等于1/2或者负的。
So now we know the Avogadro number and are able to count the quantities accordingly.
所以现在,我们知道阿伏伽德罗数,并且能够以此计算出数值。
So they're unlikely to choose numbers at random.
所以他们并不大可能随机选数
So Rituparna tries to prove to him his abilities and he says, see that tree there, I can estimate how many leaves there are on that tree by counting leaves on one branch.
所以睿都巴若那就试图证明自己的能力,他说,看那边的树,我只需数一根枝杈上的叶子,就能估算出树上叶子的总数
So, you can understand they really felt quite confident at this time that we could explain everything that was going on and in fact, a really telling quote from the time was said by a professor at the University of Chicago, and what he said is, "Our future discoveries must be looked for in the sixth decimal place."
所以你可以理解,为什么当时的人那么,自信的说我们可以解释,这世界上发生的一起事情,事实上,当时有个著名的名言,是芝加哥大学的一个教授说的,他说“我们今后的科学发现,就在于弄清楚小数点第六位后的数,“
to the n, every value in the 1 bit vector we looked at last time is either 0 or 1. So it's a binary n number of n bits, 2 to the n.
从2到n,我们上次看到的,位向量的每个值不是0就是,所以它是n,比特的二进制数,从2到。
So your role in here in this demo is to get an int from Jordan.
所以你在这个演示里面的角色是,从Jordan那里获取一个int数。
And when we make these comparisons, one thing I want to point out is that we need to keep the constant principle quantum number constant, so we're talking about a certain state, so we could talk about the n equals 2 state, or the n equals 3 state.
当我们做这些比较时,我想指出的一件事是,我们需要保持常量原则,保持量子数是常数,所以我们在讨论一个确定的态时,我们可以谈论n等于2的态,或者n等于3的态。
So isn't f already a float?
所以f不是已经是个浮点数了吗?
So, if we have a high intensity, we're talking about having more photons per second, and it's important to know also what that does not mean.
所以,如果我们有一个高强度,我们就是在讨论每秒钟,有更多的光子数,同样理解它不代表什么,也是很重要的。
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