| 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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