 | James Bates Thomson - 1858 - 384 Seiten
...3.) 336. From these illustrations we have the following RULE FOR MULTIPLICATION OF DUODECIMALS. I. Place the several terms of the multiplier under the corresponding terms of the multipliiand. II. Multiply each term of thf multiplicand by each term of the multiplier separately,... | |
 | Horatio Nelson Robinson - 1859 - 336 Seiten
...of the indices of those factors ; thus 7' X 8' — 56" ; 4" X 5'" = 20"'". Hence the RULE. I. Write the several terms of the multiplier under the corresponding terms of the multiplicand. II. Multiply each term of the multiplicand by each term of the multiplier, beginning with the lowest... | |
 | James Bates Thomson - 1860 - 422 Seiten
...1 516* From these illustrations we derive the folio wing RULE FOR MULTIPLICATION OF DUODECIMALS. 1. Place the several terms of the multiplier under the corresponding terms of the multiplicand. II. Multiply each term of the multiplicand by each term of the multiplier separately, beginning with... | |
 | Horatio Nelson Robinson - 1860 - 432 Seiten
...Adding the partial products, we have 44 ft. 3' V for the product required. Hence the RULE. I. Write the several terms of the multiplier under the corresponding terms of the multiplicand. II. Multiply each term of the multiplicand oy each term of the multiplier, beginning with the lowest... | |
 | James Bates Thomson - 1862 - 422 Seiten
...(». From these illustrations we derive the following • RULE FOR MULTIPLICATION OF DUODECIMALS. 1. Place- the several terms of the multiplier under the corresponding terms of the multiplicand. II. Multiply each term of the multiplicand by each term of the multiplier separately, beginning with... | |
 | Thomas Percy Hudson - 1866
...above them. The above remarks will sufficiently explain the following Ride for Cross Multiplication. Place the several terms of the multiplier under the...corresponding terms of the multiplicand. Multiply each term of the multiplicand by each term of the multiplier separately, beginning with the lowest... | |
 | Shelton P. Sanford - 1872 - 373 Seiten
...products, we get 15ft. 5' 6" for the answer. Ant. 15ft. 5' 6" Hence, we have the following KULE. 1. Place the several terms of the multiplier under the corresponding terms of the multiplicand, II. Begin at the right, and multiply each term of the multiplicand liy each term of the multiplier... | |
 | Horatio Nelson Robinson - 1875 - 456 Seiten
...Adding the partial products, we have 44 ft. 3' 8" for the product required. Hence the RULE. I. Write the several terms of the multiplier under the corresponding terms of the multiplicand. II. Multiply each term of the multiplicand by each term of the multiplier, beginning with the lowest... | |
 | Montagu H. Foster - 1881
...measured by the rod of Щ ft., and 27 2£ square ft. of brickwork. MULTIPLICATION. RULE. Write the terms of the multiplier under the corresponding terms of the multiplicand. Multiply every term in the multiplicand, beginning at the lowest, by each term in the multiplier beginning at... | |
 | Horatio Nelson Robinson - 1888
...is equal to the sum of the indices of those factors; thus: 7'x8'=56"; 4" x 5"'=2(X"". RULE. I. Write the several terms of the multiplier under the corresponding terms of the multiplicand. II. Multiply each term of the multiplicand by each term of the multiplier, beginning with the lowest... | |
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RULE. I. Write the several terms of the multiplier under the corresponding terms of the multiplicand. II. Multiply each term of the multiplicand by each term of the multiplier, beginning with the lowest term in each, and call the product of any two denominations... 
