q1 So let me just rewrite that as, I just want to divide by q1 everywhere.
分子分母同时除以,现在我写的这些。
Therefore, when you divide by the mass to get the acceleration, the response of different bodies is inverse to the mass.
因此,当你拿力除以质量得到加速度时,不同物体的加速度与质量成反比
By tradition, they divide by 360, not 365, and so you understand that dealers of Treasury bills tend to quote "discounts."
习惯上他们除以360,而不是365,国库券的经销商,习惯于谈论"贴现"
He saw that if you take the atomic mass of chlorine, 2 add it to the atomic mass of iodine, divide by two, you get something that is really close to the atomic mass of bromine.
他看到,如果你将氯的原子质量,加上碘的原子质量,除以,得到的数将与,溴的原子质量很接近。
I, somehow if, I want to walk through some sequence of data structures, gathering up or doing the same thing, adding ages in until I get a total age and then divide by the number faculty.
我想,以某种方法,来遍历,一些数据结构,把它们相同的属性加到一起,就是一直的把年龄加到一起一直到,得到了一个年龄总数,然后除以员工的数目。
So if you divide by $10 million, that's a lot of money for each employee.
如果将1000万平分,每个人都能拿到一大笔钱。
- And then I divide by an int -- that's okay because so long as you have a float involved at some point early on, you're okay.
然后我除以一个整型数-,那是可以的因为现在为止,我们在之前已经,引入了一个浮点数,你是对的。
We know we need to divide by 266, 266 but what we need you to help us with is to figure out this top number here and see how many particles are going to backscatter. So if the TAs can come up and quickly hand out 1 particle to everyone.
知道背散射的概率就可以了,我们知道要除以,但还需要你们来搞清楚,分子上的数是多少,有多少个发生背散射的粒子,助教们请过来,把这些球分给同学们。
I have 3 q1* is equal to a - c over b; and finally divide by 3 q1* is equal to a - c over 3b.
q1*= /b,最后两边同除以3得,q1*=/3b
Take the difference of the two velocities and divide by the difference of the two times, and you've got the acceleration.
求出速度的差值,除以时间的差值,这样就能得到加速度
Then you say, well why didn't they divide by 365, because I know there are 365 days in a year?
也许你们会问,为什么他们不除以365,因为一年有365天
They had to divide by hand and so they didn't like the number 365, so they thought, let's just round it to 360.
他们必须手工计算,所以不太喜欢365,所以他们就近似成360
So, what's the conservation of mass if we divide by the atomic masses here?
那么质量守恒是什么呢?,如果我们在这儿按原子质量划分?
So, when you divide by the mass to get the acceleration, you get the same answer.
因此,当等号两边除以质量来求加速度时,结果是相同的
You can take one full circle divide by the time and you will get this.
你可以先转一整圈,再除以时间就得到这个
In the log case, it's divide by an amount.
而在对数算法中。
If you have a bond with an interest rate of 4.375% -that's not an easy one to divide by two but you would get half of that every six months until maturity.
如果你持有一种利率为4.375%的债券,这个数除以2有点难算,而在债券到期前,每半年你能得到4.375%一半的券息
And we want to divide all of that by our wavelength, and to keep our units the same we'll do meters.
我们想用波长除以所有这些,而且保持单位统一我们采用米。
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。
Then to get the monthly payment, you take the square bracketed thing and you divide by the mortgage balance and that's the monthly payment; that's how it's calculated.
如果要算每月偿还额,你用贷款余额,除以这个方括号的值,这样就得出每月偿还额,这就是每月偿还额的计算方法
You cannot say to me, "Take a force, due to a spring, and see what force it applies and divide by the acceleration and get the mass, " because we haven't defined force, either.
你不能说,"给弹簧施加一个力,看看它能提供多大的力,然后除以加速度,得到质量",因为我们还没给力下一个定义
He could divide all of that by the absolute value of the charge of the positive particle, all over the mass of the positive particle.
他可以把这整个,除以正电粒子的,电荷绝对值除以,正电粒子的质量。
The result is that when you divide an int by an int, the answer no matter what is going to be an int.
当你把一个整型数除以一个整型数时,无论如何答案将会是一个整型数。
Instead divide an int by a floating point value.
我们用一个整型数除以浮点数。
Because the square root of 10,000 is 100, whatever the standard deviation of the portfolio is, you would divide it by 100 and it would become really small.
因为一万的平方根是一百,无论这个投资组合的标准差是多大,当除以100后就都变得很小很小了。
I need to divide it by three to get the average.
我应该先除3得到平均数
What are you going to divide by to get the acceleration?
你会拿它除以多少来得到加速度呢
first So what do I do? I find the midpoint by taking last minus first, divide by 2, and add to start.
要怎么做?我把last减去,除以二,加到起点上去。
Specify the quantum number n and divide by Z.
只需说明n的具体值,并用Z去除就行了。
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