解析 To write expression as a multiplication problem, we count the amount of the same numbers that are being added. In this case, there are four numbers of -15 in the problem(-15)+(-15)+(-15)+(-15).Hence, the answer is (-15)+(-15)+(-15)+(-15)=4(-15)....
百度试题 结果1 题目3. Use the multiplication table to answer the division equation:75 35 相关知识点: 试题来源: 解析 7 反馈 收藏
There are 13 storybooks on the shelf in a bookstore. Each book is sold for 32 dollars. (1) 3 2 × 1 3 (2) The total price of 3 books is dollars. (3) The total price of 10 books is dollars. (4) The total price of 13 books is dollars.相关...
The students should say if the answer was right or wrong. Rather than counting, or writing on the paper, the students pulled the answers from m ___. It’s as if(就像)the answers to basic addition, subtraction, and multiplication problems are kept in a long-term storage(储存)area in...
The answer in a multiplication equation is called the product. A multiplication sign (x) is written between the two factors. Dividend is the number that’s being divided. Divisor is the number that tells how many times a dividend should be divided. The answer we get in a division equation...
and made correspondin and made no answer and make a new name and make people laugh and make relevant rec and make sure everyth and make sure you get and make systemic che and many other advant and many other progra and many trees still and maps and pictures and mary said my soul and ...
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1 Look at the diagram and then use multiplication to calculate the answer.How many * are there in total?关关关Think this way:I use multiplication to find it Split 13 into 10 and 3 first and out.13 ×6=?multiply each by 6. Then add the two products.First calculate:()× ()=(),(...
The basic parts of a multiplication problem consist of at least two factors that are multiplied together to result in one product.More than two factors can be involved in a multiplication problem, but the answer always consists of only one product. ...
The student quickly picks up that this can be easily accomplished by adding a certain number of zeros to themultiplicand(the first number in a multiplication problem, because these problemsalwayslook like “55 x 1,000” and never “1,000 x 55”) and then moving on to the next problem....