 | Milton Browning Goff - Arithmetic - 1876 - 462 pages
...exclusive of the right-hand figure. The result we annex to both divisor and root already found, and then multiply the divisor thus increased by the last figure of the root, and subtract the product from the dividend. Result, 895. EXAMPLE 3. — Find the square root of 5852.25.... | |
 | Joseph Ray - Arithmetic - 1877 - 402 pages
...the products to the trial divisor ; the sum is the complete divisor. 5. Multiply the complete divisor by the last figure of the root; subtract the product from the dividend, and to the remainder bring floirn the next period for a new dividend. 6. Find a new trial divisor as before, and continue the... | |
 | Albert Newton Raub - Arithmetic - 1877 - 348 pages
...5. Put this figure of the root in place of the cipher, and then multiply the entire divisor by this last figure of the root; subtract the product from the dividend, and to the difference annex the next period for a new dividend. 6. Double the root already found, with a cipher... | |
 | Horatio Nelson Robinson, Daniel W. Fish - Arithmetic - 1877 - 372 pages
...the result will be the complete divisor. V. Multiply the complete divisor by the trial figure, and subtract the product from the dividend, and to the remainder bring down the next jjeriodfor a new dividend. VI. Add the square of the last figure of the root, the last term in column... | |
 | William Guy Peck - Arithmetic - 1877 - 430 pages
...product to the 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... | |
 | James Bates Thomson - Algebra - 1878 - 322 pages
...place the result in the quotient. II. Multiply the wl&U divisor by the term placed in the quotient ; subtract the product from the dividend, and to the remainder bring down as many terms of the dividend as the case may require. Repeat the operation till all the terms of the... | |
 | Shelton Palmer Sanford - Algebra - 1879 - 348 pages
...trial divisor to form• the complete divisor. III. Multiply the complete divisor by the second term of the root, subtract the product from the dividend, and to the remainder bring down as many terms as may be necessary for a new dividend. IV. Double the terms of the root already found... | |
 | Horatio Nelson Robinson - Algebra - 1879 - 332 pages
...result will be the complete divisor. V. Multiply the complete divisor by the last figure cf the roof, subtract the product from the dividend, and to the remainder bring down another period for a new dividend. VI. Add together the square of the last figure of the root, the... | |
 | George E. Seymour - Arithmetic - 1880 - 332 pages
...annexing the square of the last figure of the root. VII. Multiply the complete divisor thus found, by the last figure of the root, subtract the product from the dividend, and bring down the next period. VIII. Square the root noio found and multiply by three for a trial divisor.... | |
 | Arithmetic - 1882 - 392 pages
...the 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... | |
| |
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. 
