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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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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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But it really wasn't until, I would say, the arrival of Java that object-oriented C++ programming caught the popular attention.
但是知道Java的诞生，才吸引了大众的关注,而后Java和。
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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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s1+s2 Instead I can write with impunity the solution will be s1 plus s2.
而我会写下解就是,毫无风险。
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+ Because I can just use plus, plus.
因为我可以直接用。
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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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One thing I can guarantee from past experience in this class, is that the median grade will be a B+.
不过从教学经验看,中间成绩应该是B+
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+3b I've said the product was is 2 plus 3 b.
我们已经说了答案是。
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If you want to call something as A + B, it means that I did A and then I did B.
如果你想某个量称为 A + B,它表示我先沿着 A 走,然后是 B
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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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Normally, if I tell you 3x + 4y = 6, and 9x + 6y = 14, you know how to solve it.
一般来说,如果我给你两个方程 3x + 4y = 6,和 9x + 6y = 14,你们肯定会解
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And then on the right hand side I++, we saw that in pseudocode with our socks example last week that just says increment I.
然后是最右边的i++，我们上周在socks例子的伪代码中,曾经见到过，不是么？只是代表i加1而已。
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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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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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But this is the same vector we are calling i times Ax plus j times Ay.
这和矢量 i ? Ax + j ? Ay 是一样的
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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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If I told you 3x + 2y = 9 and 4x + 6y = 6, you certainly know how to solve for x and y, right?
如果我给出 3x + 2y = 9,和 4x + 6y = 6 这两个方程,你们肯定知道怎么解出 x 和 y,对吧
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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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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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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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See, I've defined f of x to be a function x=x+1 that takes a value of x in, changes x to x+1, x and then just returns the value.
我定义了f是一个函数,输入x，让，然后输出。
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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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It's looking at the i'th and the i plus first element and it's saying, gee, if the i'th element is bigger than the i'th plus first element, what's the next set of three things doing?
他在观察第i个元素,和第i+1个元素,如果第i个元素大于第i+1个元素，接下来的要做哪三件事情？
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The point is the arrow A, somebody has chosen to write in terms of i prime and j prime as Ax prime and Ay prime.
解决问题的关键在于矢量 A,可以用这样的形式来描述,i' ? Ax' + j' ? Ay'
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For the x part, I'm going to write x = x0 + v0 t.
对于 x 分量,有 x = x0 + v0 ? t.
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That would come by saying, "I'm taking i times Ax + j times Ay + i times Bx + j times By, " and I'm trying to add all these guys.
那就这样做,"i ? Ax + j ? Ay + i ? Bx + j ? By",然后把它们都加起来
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I ask when y=0, then I say 0=15+10t-5t^.
我问何时y=0,然后得到0=15+10t-5t^
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if I say a particle's location is i times t^2 plus j times 9t^3, for every value of time, you can put the numbers in and you can find the velocity by just taking derivatives.
一个质点的位置,i ? t^2 + j ? 9t^3,在每一个时间点,你可以把数值代入,并通过求导得到速度
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