There are several. I would go for:
66 * 94 = (60 + 6)*(100 - 6)
= 60*100 - 60*6 + 6*100 - 6*6
= 6000 - 360 + 600 - 36 = 6204
You could go for (60 + 6)*(90 + 4) instead, but other than the fact that all components are positive, I see no advantage. The multiplications are much simpler if (100 - 6) is used.
Very little. An algorithm is a method that has been expressed in a detailed, unambiguous form.
The partial products method is a method for performing multiplication problems. An actual multiplication problem is necessary to demonstrate. See related link.
YES (i think)
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No, it could be a partial sum.
Very little. An algorithm is a method that has been expressed in a detailed, unambiguous form.
The partial products method is a method for performing multiplication problems. An actual multiplication problem is necessary to demonstrate. See related link.
YES (i think)
In division, the partial quotient method involves breaking down the dividend into smaller, more manageable parts to simplify the division process. The partial product method, on the other hand, is commonly used in multiplication and involves multiplying each digit in one number by each digit in the other number and then adding the results. To perform the partial product method, you would multiply each digit of the multiplicand by each digit of the multiplier, starting from the rightmost digit and moving leftwards, and then summing up the products to get the final result.
(40 x 5) + (2 x 5)
the partial products for 84 and 78 6000,500,50,and 2 :)
what is the meaning for partial sums
The answer will depend on the order in which you do partial products. It is quite common in the UK for the first partial product to be the two digits in the tens' place and so that is often the largest. This ties in with the method for multiplying two binomials when they move on to algebra.
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136-79 with ballpark estimate and partial differences method = 57
No, it could be a partial sum.
241 * 31 = (240 + 1) * (30 + 1) = 240*30 + 240*1 + 1*30 + 1*1