查询词典 multiply...by...

Multiply on the subway, if be like,drive megalosaurus of iron of one bar iron, in subterranean freely gallop hire, it is quick and convenient really.

乘上地铁，如似驾一条钢铁巨龙，在地下纵横驰聘，实在是快捷方便。

The scientists were Jerry Merryman, James Van Tassel and Jack Kilby. Their first device could add, subtract, multiply and divide. It had twelve bytes of memory -- close to nothing compared to today's powerful calculators.

他们最初的机器只能进行加减乘除的运算，并且只有12比特的记忆容量，这根本无法与当今强大的计算机相比。

The answer is forty. The scientists were Jerry Merryman, James Van Tassel and Jack Kilby. Their first device could add, subtract, multiply and divide. It had twelve bytes of memory -- close to nothing compared to today's powerful calculators. And it weighed more than a kilogram.

答案是40年以前，这三位科学家分别是Merryman，James Van Tassel和Jack Kilby，他们首次发明了可进行加，减，减和除的设备，它拥有十二字节存储器，与今天功能强大的计算器来说跟本就不算什么，并且它的重量有1公斤多。

Experts believe that this should be a kind of mullet, which in the last 80 years the world settled in this region, and open to multiply.

专家们认为，这应该是胭脂鱼的一种，它们于上世界80年代在这一地区安家落户，并且繁衍开来。

For if the objection be raised, that no being can exist separated from itself or show forth local distances between its various selves, the sophism is readily detected; for multilocation does not multiply the individual object, but only its external relation to and presence in space.

如果异议被提出，没有被可以存在脱离本身还是显示了地方之间的距离，其各个守土有责，诡辩，是很容易发现； multilocation并不乘以个别对象，但只是其对外的关系，并在空间。

In order to find the effective length of the multiplier and multiplicand, we gathered detailed information of multiply operations from studying of the SPEC95 benchmark programs in advance.

根据此特性，我们提出了侦测前端零位元检测器和零加法器所构成非同步乘法加速器。

Op1 + op2 The addition operator will add two numbers. op1 - op2 The subtraction operator will subtract two numbers. op1 * op2 The multiplication operator will multiply two numbers. op1 / op2 The division operator will divide two numbers. op1 % op2 The modulus operator will return the remainder of the division of two integer operands. op1 xx op2 The exponentiation operator will raise op1 to the power of op2.++op1 The pre-increpment operator will increase the value of op1 first, then assign it. op1++ The post-increment operator will increase the value of op1 after it is assigned.--op1 The pre-decrement operator will decrease the value of op1 before it is assigned. op1-- The post-decrement operator will decrease the value of op1 after it is assigned.

op1 + op2 对两个数值做加法操作 op1 - op2 对两个数值做减法操作 op1 * op2 对两个数值做乘法操作 op1 / op2 对两个数值做除法操作 op1 % op2 求两个整型数值的余数 op1 xx op2 求幂操作：求 op1 的 op2 次幂++op1 前加操作： op1 的值先增加，然后将值赋给自身 op1++后加操作： op1 的值先赋给自身，再增加值--op1 前减操作： op1 的值先减少，然后将值赋给自身 op1—后减操作： op1 的值先赋给自身，再减少值

Multiply that by 40 and you can see what he lost.

乘上40你就知道他失去的是什么了。

If you multiply this number by this one... No,I give up.

如果你用这个数乘以这个数，不行，我做不下去了。

I asked."Multiply by 2 and see what happens."

我问道。"乘2，然后看看发生了什么。"

London based fashion photographer Matt Irwin might be is the next big thing in fashion photography.

总部位于伦敦的时装摄影师马特欧文可能会是下一件大事的时装摄影。

Based on IMC system,the response would be better if the model matched more satisfactorily.

由于系统基于内模控制设计，故模型匹配度越高，系统响应越好。