addend造句
例句与造句
- Thus, the augend and addend must each be small enough to fit within one memory location .
由此可见，加数和被加数都应足够小，以便能够适合地装入一个存储单元。 - Thus, the augend and the addend must each be small enough to fit within one memory location .
由此可见，加数和被加数都应足够小，以便能够合适地装入一个存储单元。 - Switch the two addends and the sum remains the same
生：交换两个加数的位置，和不变。 - Stability of multiplier - addend model and the macro - control policies
加速数模型的稳定性与宏观调控政策 - Switch the two addends and the sum remains the same
7生：两个数相加，交换加数的位置，他们的和不变。 - It's difficult to find addend in a sentence. 用addend造句挺难的
- In addition , switch the two addends and the sum remains the same
生：两个数相加，交换加数的位置，他们的和不变。 - 137 and 357 are the addends . the answer 494 is the sum
生： 137和357在加法算式中叫做加数。他们相加的得数494叫做和。 - The two numbers being added are the addends . the answer is the sum
3电脑出示结论：相加的两个数叫做加数，加的数叫做和。 - Multiplication is the easier way to find the sum of same addends
根据学生的回答总结得出：求几个相同加数的和的简便运算，叫做乘法。 - In addition , add the addends in any order and the sum remains the same
师：这就是加法结合律。利用加法结合律，可以进行简便运算。请做下列题。 - Subtract one addend from the sum . if you can get the other addend , you ' re right
加法的验算。从算出的和里减去一个加数，看是不是等于另一个加数。 - One addend adds one number . the other addend subtracts the same number . the sums are the same
小结：一个加数增加几，另一个加数减少几，增加和减少的数相等，它们的和仍是10 。 - This demonstrates the operation addition performed on two whole numbers , two and three ( known as addends , ) which result in a sum ( total ) of five circles
以上演示了两个整数的加法运算，两个整数分别为2和3 （是作为加数） ，这样就得到和是5个球。 - The way to find an unknown addend from one addend and the sum is called subtraction . this is the essence of subtraction . please read after me
归纳减法的意义：已知两个加数的和与其中的一个加数，求另一个加数的运算，叫做减法。 （让学生读读。 ） - First , with the coding idea of thomas h . labean " logical computation using dna molecules " , we developed a dna - based addition model of two n - system addend
首先，运用labean解决累加异或问题的编码思想，实现n进制的两个加数的加法进位问题。
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