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Plus it makes it, as I talked about last time, much easier to think about distributing the vaccine around the world because shots require skilled medical personnel, whereas, an oral vaccine could be self-administered.
而且,这会使得,就像我上次谈到的,这会使得疫苗更容易在世界范围内分发,因为注射需要专业医疗人员,但是,口服疫苗自己就可以服用
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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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And using the text plus these archive notes,I think you will be able to piece together what you need.
使用这个课本，再加上存档笔记，我想,你们已经能组合出你们需要的了。
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dU=dq+dw Well from the first law, du is equal to dq plus dw, and I wrote down everything I knew at the beginning here.
第一定律“,前面我们已经,看到了。
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J++ Now here's the semicolon, J less than N, where N is this, J plus, plus; so what am I doing?
现在这里是一个分号，J小于N，N是这个，那么我在做什么？
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When I started this company back almost nine, ten years ago, I started to recruit talents, recruit engineers, and I wrote in the job description that requires five-plus years of related experience.
大约是9到10年前，公司成立初期，我们开始招聘人才，招聘工程师，职位介绍由我亲自执笔,要求具有五年以上的相关工作经验。
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What I want is survival plus something else.
我希望拥有的是带有其他条件的存活。
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The claim is, I can write any vector you give me as a real scale i plus a real scale j.
也就是说,我可以把任何你给我的矢量,写成 i 和 j 的实系数线性组合的形式
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, okay. So 1 plus 1/4 of 0 is 1, so if Player II chooses 0, player I's best response is to choose 1.
是1,因为1+0/4=1,参与人II选0时I的最佳对策是1
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Three plus five. I can take something to a power, double star, just take three to the fifth power.
+5，我也可以,求一个数的次方，例如求3的5次方。
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The power of linearity is F=k1+k2 if I come across f of x, y, z equals k1 plus k2, if it is a linear equation, I don't have to go and solve it all over again.
线性的威力是,一个方程，如果它是个线性方程，那么我就不用再去解他了。
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So what I'm saying is that du through path 1 plus du going through path 3.
于是1,因为能量。
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In which I said, you know, I didn't like the fact that things like plus are overloaded, because you can use plus to add strings, you can use plus to add numbers, you can use plus to add floats.
那个玩笑是这样的，你们懂得，我并不喜欢,加法被重载这样的事情，但是你可以用加法，来对字符串进行合并，你可以用加法来加数字。
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I want to then do, I need to find the square root b squared plus h squared, right?
我接下来要求b的平方和h的平方,的和的平方根对不对？
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But here's a new and improved, I think, version of increment; ++ returns nothing, takes nothing, but it does perform plus plus, but I did something stupid.
但这里是一个新的，改进了的increment版本;,没返回值，没输入值，但它执行的是，但是我做了件愚蠢的事情。
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Being and void, being and void, so now I've got my plus two little projectile coming in, and plus two zooms right through.
有和无，有和无，现在带两个正电荷的发射体进来了，两个正电荷正好穿过。
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It takes the number whose Fibonacci memo I want plus a memo.
它将我想要的斐波那契数,的值加上了参数。
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That means starting at the first one, I'm going to do something to it. And what am I'm going to do? I'm going to take that character, convert it back into an integer, and add it into some digits. And I've done a little short hand here, which is I should have said some digits is equal to some digits plus this.
这意味着从第一个字符开始，我要对他们进行一些操作,我要去做什么呢？我要取得这个字符，然后把它转换为整数,然后加到某些数上面去,我在这里用了一些缩写,我本来应该写一个数字等于这个数字。
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Therefore, the vector A that you gave me, I have managed to write as i times Ax plus j times Ay.
这样,你给我的矢量 A,我已把它写成 i ? Ax + j ? Ay的形式
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Just swapping them, right? I temporarily hold on to what's in the i'th element so I can move the i plus first one in, and then replace that with the i'th element.
交换他们，对么？,临时的保存下第i个元素,然后把第i+1个元素移进来，把i+1的位置替换为第i个元素。
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Initially, its location as a function of time is equal to i times x plus j times y.
初始时刻,它的位移作为时间的函数等于,i ? x + j ? y
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It says, in either case in general, t of b-- and this is where I'm going to abuse notation a little bit but I can basically bound it by t, 12 steps plus t of b over 2.
我可以用一个,比12+t的数代表,这里有点不准确的地方,具体的步数依赖于奇数偶数，但是你们可以看到在两个case中。
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I'm going to just put that in, and it's the cosine of the number times i plus sine of the number times j times R.
我要在式子加入这个量,这个式子就等于这个值的余弦乘以 i,加上这个值的正弦乘以 j 再乘以常数 R
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You can now collect the whole expression as something something i plus something something j.
你们现在可以推导出完整的表达式,用 i 和 j 相关的式子来表示
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There I have it Cp is equal to Cv plus R, right?
所以Cp=Cv+R，对吗？
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So similarly, I would find that ?2 2 equals 1 plus B S1 and this is the best response of Player II, as it depends on Player I's choice of effort S1.
同理可得?2等于1+B*S1,?2是参与人II的最佳对策,因为它与参与人I的策略S1有关
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And the way I remember that is I put a little cross here, and then I make that the plus end.
我还记得要,在箭头尾部打个叉，然后让加法完结了。
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So, I've got this tiny volume with, in the case of gold 79 plus of charge, and I've got some electrons out here somewhere, and the vast majority of the atom is nothing.
我认为这个小体积里面，比如金的79个正电荷，电子在外面的某些地方，原子里面大部分是空心的。
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So, I've got plus twos, zooming in at high energy against a wall of positive charge.
两个高能的正电荷,在正电荷壁垒时会遇到阻力。
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So I take 2 plus 3 b steps to go through this loop OK. So if b is 300, it takes 902 steps.
好，如果b是300的话就是902步,如果b是3000那就有9002步。
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