... in the multiplicand ; and as either factor may be made the multiplier, so, if the decimals had been in the multiplier, the same number of places must have been pointed off for decimals. Hence it follows, we must always point off in the product as... Adam's New Arithmetic - Page 51by Daniel Adams - 1845 - 180 pagesFull view - About this book
| Daniel Adams - Arithmetic - 1810 - 190 pages
...pointed off for decimals. Hence it follows, we must always point off in the product as many places Jar decimals as there are decimal places in both factors....decimals in the pro duct. The reason of pointing off this 375 number may appear still more plain, if We 150 consider the two factors as common or vulgar fractions.... | |
| Daniel Adams - Arithmetic - 1828 - 286 pages
...number of places must have been pointed off for decimals. Hence it follows, we must always point of in the product as many places for decimals as there are...'75 by '25. OPERATION. in this example, we have 4 de|75 cimal places in both factors ; we 2* must therefore point off 4 places 375 for decimals in the... | |
| Daniel Adams - Arithmetic - 1828 - 266 pages
...number of places must have been pointed off for decimals. Hence it follows, we must always point off in the product as many places for decimals as there are...in both factors. 2. Multiply '75 by '25. OPERATION. ln this example, we have 4 de'^5 cimal places in both factors ; we '25 must therefore point off 4 places... | |
| Arithmetic - 1829 - 196 pages
...DECIMALS, we have this JL4-J). RCLB.* Multiply as in whole numhers, and from the product point off so MANY PLACES for decimals, as there are decimal places in BOTH FACTORS ; hut if there are not so many places, supply the deficiency hy prefixing ciphers. / EXAMPLES. 4. What... | |
| Daniel Adams - Arithmetic - 1830 - 294 pages
...number of places must have been pointed off for decimals. Hence it follows, we must always point off in the product as many places for decimals as there are...'75 by '25. OPERATION. In this example, we have 4 de'75 cimal places in both factors ; we '25 must therefore point off 4 places for decimals in the product.... | |
| Roswell Chamberlain Smith - Arithmetic - 1830 - 286 pages
...off two more places for decimals, which, counting both, would make 4 Hence we must alwayspoint off in the product as many places for decimals, as there are 'decimal places in hoth the factors. 2. Multiply ^5 by ,5. In this example, aere being 3 decimal places in both the factors,... | |
| Daniel Adams - Arithmetic - 1831 - 276 pages
...number of places must have been pointed off for decimals. Hence it follows, we must always point of in the product as many places for decimals as there are...in both factors. 2. Multiply '75 by '25. OPERATION. ln this example, we have 4 de'75 cimal places in both factors ; we '25 must therefore point off 4 places... | |
| Roswell Chamberlain Smith - Arithmetic - 1831 - 282 pages
...two more places for decimals, which, counting both, would make 4 Hence we must always point off in the product as many places for decimals, as there are decimal places in both the factors. 2. Multiply ^25 by ,5. i 5 In this example, there being 3 decimal places in both the factors,... | |
| Daniel Adams - Arithmetic - 1833 - 268 pages
...number of places must have been pointed off for decimals. Hence it follows, we must always point of in the product as many places for decimals as there,...'75 by '25. OPERATION. in this example, we have 4 de, V? cimal places in both factors ; we •°" must therefore point off 4 places 375 for decimals... | |
| Roswell Chamberlain Smith - Arithmetic - 1837 - 314 pages
...çomteà oft VNO for decimals, which, counting both, would make 4. Hence, we must always point off in the product as many places for decimals, as there are decimal places in both the factors. 2. Multiply ,25 by .5. • In this example, there being 3 decimal places in both the factors,... | |
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