 | Alexander Ingram - 1883 - 190 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. 5. To the complete divisor add the two lines above it, taking in the last one twice, and to the sum... | |
 | James William Nicholson - Arithmetic - 1885 - 348 pages
...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 the same manner as with the... | |
 | Christian Brothers - Arithmetic - 1888 - 484 pages
...trial divisor for the COMPLETE DIVISOR. V. Multiply the COMPLETE DIVISOR by the last term of the root, subtract 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 used. NOTE. — If a cipher occur in the... | |
 | Edward Brooks - Algebra - 1888 - 344 pages
...their sum will be the COMPLETE DIVISOR. V. Multiply the COMPLETE DIVISOR by the last term of the root; subtract the product from the dividend, and to the...remainder annex the next period for a new dividend. Take 3 time» the square of the root now found, regarded as tens, for a trial divisor, and find the... | |
 | James William Nicholson - Arithmetic - 1889 - 408 pages
...to form the complete divisor. IV. Afultiply 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. V. Proceed with the second, and with each succeeding dividend, in the same manner as with... | |
 | William Frothingham Bradbury, Grenville C. Emery - Algebra - 1889 - 444 pages
...trial divisor, and the SUM will be the TRUE DIVISOR. Multiply the true divisor by the last root ßgure, subtract the product from the dividend, and to the remainder annex the next period for a dividend. Find a new trial divisor, and proceed as before, until all the periods have been employed.... | |
 | William Frothingham Bradbury, Grenville C. Emery - Algebra - 1889 - 428 pages
...for the TRUE DIVISOR. Multiply the tne divisor by this new root fiyure, subtract the product from tlu dividend, and to the remainder annex the next period, for a new dividend. Double the part of the root already found for a trial divisor, and proceed as before, until all tk... | |
 | George Albert Wentworth - Arithmetic - 1889 - 434 pages
...of the root for a complete divisor. Multiply this complete divisor by the last figure of the root, subtract the product from the dividend, and to the remainder annex the next group for a new dividend. Proceed in this manner until all the groups have been thus annexed. The result... | |
 | John Groesbeck - 1891 - 426 pages
...so found for a complete divisor. 5. Multiply the complete divisor by the second figure in the root. Subtract the product from the dividend, and to the...remainder annex the next period for a new dividend. Then proceed as before until all the periods have been used. Note 1. — If the given number contains... | |
 | William James Milne - Arithmetic - 1892 - 440 pages
...complete divisor the Jigure last found, multiply this divisor by the last Jigure of the root found, subtract the product from the dividend, and to the remainder annex the next period for the next dividend. Proceed in this manner until all the periods have been used thus. The result will... | |
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... subtract the product from the dividend, and to the remainder annex the next period for the next dividend. 
