Multiply the complete divisor by the second figure of the root, subtract the product from the dividend, and to the remainder annex the next period for a new dividend. New Practical Arithmetic - Page 363by Eugene L. Dubbs - 1901 - 440 pagesFull view - About this book
| John Davidson, Robert Scott (writing master) - Arithmetic - 1818 - 172 pages
...the divisor. The tum of these three parts will be the complete divisor, which multiply by the last **figure of the root, subtract the product from the dividend, and to the remainder** bring down the next part for a new dividend. Proceed in the same manner as before to find the divisor... | |
| John Radford Young - Algebra - 1832 - 352 pages
...divisor's place, and the divisor will be completed. Multiply the complete divisor by the last term **of the root, subtract the product from the dividend, and to the remainder** connect the three next terms, and proceed as before. For (by Art. 37), the cube of a + b is a3 + 3a2¿... | |
| Samuel Young (of Manchester.) - 1833
...both in the root and on the right of the Disisor; also by it multiply the Divisor thus completed, and **subtract the Product from the Dividend, and to the Remainder annex the next period for** a new Dividend. To the completed Divisor add the figure last put in the root ; the sum is a new Divisor,... | |
| Benjamin Peirce - Algebra - 1837 - 288 pages
...is also to be placed at the right of the divisor. Multiply the divisor, thus augmented, by the last **figure of the root, subtract the product from the dividend, and to the remainder** bring down the next period for a new dividend. Double the root now found for a new divisor and continue... | |
| Benjamin Peirce - Algebra - 1837 - 288 pages
...is also to be placed at the right of the divisor. Multiply the divisor, thus augmented, by the last **figure of the root, subtract the product from the dividend, and to the remainder** bring down the next period for a new dividend. Double the root now found for a new divisor and continue... | |
| John Radford Young - 1839
...divisor's place, and the divisor will be completed. Multiply the complete divisor by the last term **of the root, subtract the product from the dividend, and to the remainder** connect the three next terms, and proceed as before. For (by Art. 37,) the cube of a+b is a»+ 3a2¿>... | |
| John Husband (math. master, Berwick.) - 1841
...right ; add together these two lines for the complete divisor; multiply the sum by the second figure in **the root ; subtract the product from the dividend, and to the remainder annex the** third period for a new dividend. Place the square of the second figure of the root under the complete... | |
| George Roberts Perkings - Arithmetic - 1841 - 252 pages
...the result will be the TRUE DIVISOR. Multiply the true divisor by this second figure of the root, and **subtract the product from the dividend, and to the remainder annex the next** period,for a SECCUD DIVIDEND. . ft IV. To the last TRUE DIVISOR, add the Jastfgure of the root, for... | |
| Arithmetic - 1843 - 129 pages
...product write also the square of the trial-figure, then multiply the sum of these by the trial-figure, **subtract the product from the dividend, and to the remainder annex the next period for** a new dividend. EVOLUTION.ciphers for a new divisor, with which find anolner trialfigure, and proceed... | |
| George Roberts Perkins - Arithmetic - 1846 - 258 pages
...the result will be the TRUE DIVISOR. Multiply the true divisor by this second figurt of the root, and **subtract the product from the dividend, and to the remainder annex the next period, for** a SECOND DIVIDEND. IV. To the last TRUE DIVISOR, add the last figure of the not, for a new TRIAL DIVISOR,... | |
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