 | Edward Brooks - Arithmetic - 1877 - 444 pages
...CORRECTION ; add the CORRKO riON to the TRIAL DIVISOR, and the result will be the COMPLETE DIVISOR; multiply the COMPLETE DIVISOR by the last term of...root, subtract the product from the dividend, and annex the next period to the result for a new dividend. IV. Square the last term of the root, and take... | |
 | William Guy Peck - Arithmetic - 1877 - 430 pages
...trial divisor for a complete divisor. IV. Multiply the divisor thus completed by the trial -figure of the root, subtract the product from the dividend, and to the remainder annex the following period for a new dividend. V. Proceed as before, continuing the operation till... | |
 | Joseph Ray - Arithmetic - 1877 - 402 pages
...and also on the right of the trial divisor. 306 4. Multiply the complete divisor 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 found, for a new trial divisor,... | |
 | Horatio Nelson Robinson - Algebra - 1878 - 428 pages
...correction, and the result will be the complete divisor. IV. Multiply the complete divisor by ¿he last term of the root, subtract the product from the dividend, and arrange the remainder for a new dividend. V. Add together the last complete divisor, the last correction,... | |
 | Shelton Palmer Sanford - Algebra - 1879 - 348 pages
...times the prod' uct of the first term of the root by the second, and also the square of the second. IV. Multiply the complete divisor by the last term of...root, subtract the product from the. dividend, and bring down as many terms as may be necessary to continue the operation. Y. Square the whole root already... | |
 | Edward Brooks - Arithmetic - 1880 - 584 pages
...a CORRECTION; add the CORRECTION to the TRIAL DIVISOR, and the result will be the COMPLETE DIVISOR; multiply the COMPLETE DIVISOR by the last term of the root, subtract t/ie product from the dividend, and annex the next period to the result for a new dividend. IV. Square... | |
 | Arithmetic - 1882 - 392 pages
...trial divisor, thus forming the complete divisor. 2. 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 dividend. IV. To find the succeeding figures of the root:— Proceed with... | |
 | James Gray - Arithmetic - 1883 - 154 pages
...and annex it also to the divisor. 4. Multiply the divisor thus completed by the last figure placed in the root : subtract the product from the dividend, and to the remainder bring down the third period for a new dividend ; proceed in the same manner till all the periods are... | |
 | James William Nicholson - Arithmetic - 1885 - 348 pages
...the trial divisor, to form the complete divisor. 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 dividend. Proceed with the second, and with each succeeding dividend, in... | |
 | Christian Brothers - Arithmetic - 1888 - 484 pages
...term of the root, and add the product to the trial divisor for the COMPLETE DIVISOR. CUBE ROOT. V. Multiply the COMPLETE DIVISOR by the last term of...the product from the dividend, and to the remainder annex the next period for a new dividend. VI. Follow the same method until all the periods have been... | |
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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... 
