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First of all, it's clear from the Pythagoras' theorem that a is the square root of ^2 + ^2.
首先,根据毕达哥拉斯定理,勾股定理在西方被称为"毕达哥拉斯定理"
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and I think she's going to end up with 13 GCSEs, probably all As or A+s.
我想她会考十三门课程，可能都是A或者A+，
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It turns out, and we're going to get the idea of shielding, so it's not going to actually +18 feel that full plus 18, but it'll feel a whole lot more than it will just feel in terms of a hydrogen atom where we only have a nuclear charge of one.
结果是我们会有,屏蔽的想法,所以它不会是完整的，但是它会比原子核电荷量,吸引力要大很多,只有1的氢原子的。
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It's going to be the same temperature V+dV as before but the volume is V plus dV now.
将升温到跟路径1的结果一样,但是现在的体积是。
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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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Because the B+ or A- people, B- you know it, they'll hire B's and B minuses.
原本想要B+或者A-的人，实际能力评分只有B或者。
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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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You have to discount something that's one period in the future by dividing it by 1+r.
你必须把未来一段时间内的资金通过,除以1+r来贴现
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r+ And let's say that sodium has a radius, r plus, r- and chlorine has a radius, r minus, when r is very large in comparison to the radii of the ions, I don't need to draw them this way.
让我们假设钠有半径，是，氯也有半径，是，当r比离子半径大很多的时候，我不需要这样来描述。
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If you are describing a particle with location R, the vector we use typically to locate a particle, R then R is just i times x + j times y, because you all know that's x and that's y.
如果你要描述一个位移为 R 的质点,这个矢量一般用表示质点 R 的位移,R 可表示为 i ? x + j ? y,显然你们都知道这段是 x,那段是 y
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like you see on a map. That's where GMT is, GMT+0 is at that point.
好像你看一幅地图。那里就是格林威治标准时间，GMT+0就是在那一点上。
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Cv+R=Cp Cv is equal, oh Cv plus R is equal to Cp it's a relationship that we had up here that we wanted to prove.
我们就得到了,我们一开始,想要证明的。
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Again we have the charge of the nucleus on plus 2, +2 but let's say this time the electron now is going to be very, very close to the nucleus.
对于我们的氦原子，我们有一次得到了原子核电荷量为,但是我们说这次电子。
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If you had chosen Beta we would have all gotten B+'s but I guess not.
如果大家都选β都能得B+,但不太可行
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The median will fall somewhere in the B+'s.
分数中位数会在B+的附近
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It's times 4, plus PR times 2.
即4+Pr*2
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Then, to find the meaning of b, we take one derivative of this, dx/dt, that's velocity as a function of time, and if you took the derivative of this guy, you will find as at+b. That's the velocity of the object.
接下来,为了弄清b的含义,我们取它的一阶导数,dx/dt,得到速度作为时间的函数,如果你对它求导的话,你会得到at+b,这就是物体的速度
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There are no A+'s; it sounds a little bit like college, but not quite.
没有所谓的A+,这听起来有点像大学,不过不大一样
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I personally don't like swings that much and it's the B-/B+ range, so I'd much rather prefer that to a swing from A to C, and that's my reason.
我不喜欢成绩波动很大的,比如B-/B+这个范围,所以我还是喜欢像A到C这样小点的,这就是我的原因
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So it'll be 1 plus 1/4 times 4, 1/4 times 4 is 1, so 1 plus 1 is 2, so Player II's best response in that case will be 2.
这回就是1+4/4,即1+1=2,所以此时参与人II的最佳对策是2
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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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This point here then is the present value of your income; it's Y today + / .
直线与x轴的交点表示收入的现值,它等于
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That's clear. If I add them, I'll get a vector parallel to i with lengths Ax + Bx.
这很明显 如果把他们加起来,就得到一个平行于 i,模长为 Ax + Bx 的矢量
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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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There's a very natural quantity that you can call A + B.
有一个很自然的量可以记做 A + B
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