 Hence it follows, we must always point off in the product as many places for decimals as there are decimal places in both factors. 2. Multiply '75 by '25. OPERATION. In this example, we have 4 de'75 cimal places in both factors ; we '25 must therefore...  Putnam's Arithmetic - Σελίδα 29των Rufus Putnam - 1849 - 264 σελίδεςΠλήρης προβολή - Σχετικά με αυτό το βιβλίο
 | William Tinwell - 1805 - 212 σελίδες
...-78623? I 4. -620378, and -3? MULTIPLICATION of DECIMALS. CASE i. To multiply finite decimals. RULE. — Multiply as in whole numbers, and point off in- the product as many decimals as there are in both multiplicand and multiplier ; but if the produfl does not contain as... | |
 | 1811 - 210 σελίδες
...take 1.146 7. From 146.265 take 45.3278 ,8. From 4560. take .720 MULTIPLICATION OF DECIMALS. RULE. Multiply as in whole numbers, and point off in the product as many decimal places as tliere are in both factors. If there are not as many places in the product as there... | |
 | 1817 - 214 σελίδες
...100.17 take 1.146 7. From 146.265 take 45.3278 8. From 4560. take .720 MULTIPLICATION OF DECIMALS. RULE. Multiply as in whole numbers, and point off in the product as many decimal places as there are in both factors. If there are not as many places in the product as there... | |
 | Jacob Willetts - 1822 - 200 σελίδες
...an unit, or 1, subtract the millionth part of itself. Am. .999999. MULTIPLICATION OF DECIMALS. RULE. Multiply as in whole numbers, and point off in the product as many decimal places as there are in both factors. If there are not as many places in the product as there... | |
 | Daniel Adams - 1828 - 266 σελίδες
...the same 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 decimal places in both factors. 2. Multiply '75 by '25. OPERATION. ln this example, we have 4 de'^5 cimal places in both factors... | |
 | Daniel Adams - 1828 - 286 σελίδες
...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 decimal places in both factors. 2. Multiply '75 by '25. OPERATION. in this example, we have 4 de|75 cimal places in both factors... | |
 | Daniel Adams - 1830 - 280 σελίδες
...the same 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 decimal places in both factors. 2. Multiply '75 by '25. OPERATION. '75 '25 375 150 '1875 Product. MIn this example, we have... | |
 | Daniel Adams - 1830 - 294 σελίδες
...the same 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 decimal places in both factors. 2. Multiply '75 by '25. OPERATION. In this example, we have 4 de'75 cimal places in both factors... | |
 | Roswell Chamberlain Smith - 1830 - 286 σελίδες
...pointed 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,... | |
 | Roswell Chamberlain Smith - 1831 - 282 σελίδες
...have pointed off 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, we point... | |
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