 | William Guy Peck - 1875 - 348 σελίδες
...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... | |
 | Horatio Nelson Robinson - 1875 - 468 σελίδες
...to the trial divisor for a complete divisor ; multiply the complete divisor by the trial figure in the root, subtract the product from the dividend, and to the remainder briny down the next period for a new dividend. V. Multiply the last figure of the last complete divisor... | |
 | Milton Browning Goff - 1876 - 462 σελίδες
...figure of the root ; 2d. The square of the last figure of the root. Multiply the divisor thus increased 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. Form, in the same manner, successive divisors, and find corresponding... | |
 | Thomas Henderson - 1876 - 114 σελίδες
...30, and also the square of the last figure. Multiply the complete divisor by the last root figure, subtract the product from the dividend, and to the remainder bring down the next period, if any, or add three cyphers, for a new dividend. To the last complete divisor add the two numbers... | |
 | Edward Brooks - 1876 - 588 σελίδες
...the 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 times the square of the root now found, regarded as... | |
 | George Augustus Walton - 1876 - 358 σελίδες
...true divisor thus obtained by the last term of the root, and subtract this product from the dividend ; to the remainder bring down the next period for a new dividend. Double the terms of the root already found for a new trialdivisor, and proceed as before. NOTE I. —... | |
 | Joseph Ray - 1877 - 402 σελίδες
...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... | |
 | William Guy Peck - 1877 - 430 σελίδες
...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... | |
 | Albert Newton Raub - 1877 - 348 σελίδες
...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 - 1877 - 372 σελίδες
...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... | |
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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. 
