Multiply the divisor, thus increased, 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. 5. Double the whole root already found for a new divisor, and continue the... American Common-school Arithmetic ... - Page 160by Rufus Putnam - 1849Full view - About this book
| A. Melrose (Teacher) - Arithmetic - 1795 - 140 pages
...divifor, thus increafed, by the laft. figure placed iu the root : Subtract the produft from the drwdend, and to the remainder annex the next period for a new dividend ; with which proceed in the fame manner, and'fo on,. till all the periods are ufed. I Required the... | |
| John Davidson, Robert Scott (writing master) - Arithmetic - 1818 - 190 pages
...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... | |
| Samuel YOUNG (of Manchester.) - 1833 - 272 pages
...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, with which... | |
| Benjamin Peirce - Algebra - 1837 - 300 pages
...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 - 302 pages
...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 - 332 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 a»+ 3a2¿>... | |
| John Husband (math. master, Berwick.) - 1841 - 126 pages
...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 Perkins - Arithmetic - 1841 - 274 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 - 142 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 as before. 1.... | |
| George Roberts Perkins - Arithmetic - 1846 - 266 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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