 | 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 next figure of the root, and also annex it to the trial divisor, to form the complete divisor. Multiply the complete divisor by the second figure...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
...product to the 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... | |
 | Edward Brooks - Algebra - 1888 - 344 pages
...iht root; 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,... | |
 | James William Nicholson - Arithmetic - 1889 - 408 pages
...figure of the root, and also annex it to the trial divisor, to form the complete divisor. IV. Afultiply the complete divisor by the second figure of the root;...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... | |
 | George Albert Wentworth - Arithmetic - 1889 - 676 pages
...add the second part 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... | |
 | 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.... | |
 | George Albert Wentworth - Arithmetic - 1889 - 434 pages
...second part. This sum will be the complete divisor. Multiply the complete divisor by the second part 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 annexed. The result... | |
 | John Groesbeck - 1891 - 426 pages
...the root, and also at the right of the trial divisor. 4. Multiply the complete divisor by the last figure of the root; subtract the product from the...dividend, and to the remainder annex the next period for a new dividend. 6. Proceed in the same manner with all the periods to the last. The result will be the... | |
 | 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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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... 
