 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...  A system of arithmetic - Page 70by John Husband (math. master, Berwick.) - 1841Full view - About this book
 | 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... | |
 | John Rose - Arithmetic - 1835 - 192 pages
...the result* to the quotient and also to the divisor. 5. Then multiply and subtract as in Division, and to the remainder, a.nnex the third period for a new dividend. 6. To the (last) divisor, add the last figure of the root for a new divisor, find another quotient... | |
 | 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... | |
 | 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... | |
 | 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¿>... | |
 | 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,... | |
 | 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,... | |
 | 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... | |
 | James Bates Thomson - Arithmetic - 1846 - 354 pages
...also on the right of the partial divisor ; multiply the divisor thus completed 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, as before. IV. Double the root already found for a new... | |
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