 Rule. I. Separate the composite number into two or more factors. II. Multiply the multiplicand by one of these factors, and the product by another, and so on until all the factors have been used; the last product will be the product required.  The Complete Arithmetic: Oral and Written - Page 46by Daniel W. Fish - 1874Full view - About this book
 | James S. Russell - Arithmetic - 1847 - 340 pages
...more factors, and multiplying first by one factor, then that product by another factor, and so^on, until all the factors have been used. The last product will be the product required. 30. MODEL OF A RECITATION. If 1 gallon of molasses costs 42 cents, what will be the cost of 1 hogshead... | |
 | Horatio Nelson Robinson - Arithmetic - 1859 - 352 pages
...32 horses, the number bought. 67. Hence we have the following RULE. I. Separate the composite number into two or more factors. II. Multiply the multiplicand by one of these factors, and that product by another, and so on vntil all the factors have been used successively;... | |
 | Horatio Nelson Robinson - Arithmetic - 1860 - 444 pages
...5 times 7, or 35. TTTjT (95, I). Hence we have the following RULE. I. Separate the composite number into two or more factors. II. Multiply the multiplicand by one of these factors, and that product by another, and so on until all the factors have been used successively;... | |
 | Daniel Adams - Arithmetic - 1861 - 452 pages
...72 dollars, cost of IS yards. * When the multiplier exceeds 12, and is a composite number, RULE. 1. Separate the multiplier into two or more factors....II. Multiply the multiplicand by one of the factors, and the product thus obtained by the other factor, and so on if the factors be more than two, until... | |
 | Horatio Nelson Robinson - Arithmetic - 1866 - 200 pages
...factors. II. Multiply the multiplicand l,y one of these factors, and that product by another, and so on until all the factors have been used; the last product will be the product required. EXAMPLES FOR PRACTICE. 1. Multiply 521 by 16=4X4. j Ans. 8336. 2. Multiply 4350 by 25=5X5. Ans. 108750.... | |
 | Charles Davies - Arithmetic - 1866 - 356 pages
...factors. II. Multiply the multiplicand by one factor, and the nroduct by a second factor ; and so on, till all the factors have been used : the last product will be the product required. Examples. 1. Multiply 327 by 12. The factors of 12 are 2 and 6 ; they are also 3 and 4 ; or they are... | |
 | Henry Bartlett Maglathlin - Arithmetic - 1869 - 332 pages
...factors. Multiply the multiplicand by one of these factors, and the product by another, and so on, until all the factors have been used. The last product will be the one required. • 84 PRACTICAL ARITHMETIC. 5. 876 by 28. Ans. 24528. 6. 11350 by 81. 7. 1130 by 54.... | |
 | Henry Bartlett Maglathlin - Arithmetic - 1873 - 362 pages
...factors. Multiply the multiplicand by one of these factors, and the product by another, and so on, until all the factors have been used. The last product will be' the one required. Repent the Knle. Examples. Multiply, using factors, 2. 165 by 24. Ans. 3960. 5. 876 by... | |
 | Daniel W. Fish - 1874 - 320 pages
...3. 6074 by 56 = 7 x 8. 4. 3708 by 60 = 3 x 4 x 5. 5. 36706 by 72. 6. 50047 by 64. 7. 75034 by 108. RULE. — I. Separate the multiplier into two or more...have been used. The last product will be the product sought. 8. What will 75 horses cost, at $197 each ? 9. What will 60 acres of land cost, at $246 an... | |
 | Edward Brooks - Arithmetic - 1874 - 352 pages
...* 1272 RULE. — Multiply the multiplicand by one factor, this product by another factor, and thus continue until all the factors have been used ; the last product will lie the result required. 2. Multiply 85 by 35. 3. Multiply 98 by 16. 4. Multiply 75 by 72. 5. Multiply... | |
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