 | Charles Davies - 1861 - 322 σελίδες
...the right of the divisor. IV. Multiply the divisor, thus augmented, by the lastfyurt of the root, and subtract the product from the dividend, and to the...remainder bring down the next period for a new dividend. But if any of the products should be greater than the dividend, diminish the last figure of the root... | |
 | Daniel Adams - 1861 - 452 σελίδες
...the root (the breadth of the additions), and subtract the contents of the additions thus obtained, from the dividend, and to the remainder bring down the next period for a new dividend. IV. Double the root already found for a new divisor, and continue the operation as before, until all... | |
 | Benjamin Greenleaf - 1861 - 338 σελίδες
...figure last annexed by the figure annexed to the root. and subtract the product from the dividend. To the remainder bring down the next period for a new dividend. Double the root already found for a new divisor, and continue th*. operation as before, till all the... | |
 | Elias Loomis - 1862 - 312 σελίδες
...the root and the divisor. 4. 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. If the product should be greater than the dividend, diminish the last figure of the root. 5. Double... | |
 | John Flint (inspector of schools.) - 1862 - 152 σελίδες
...of the three parts will be the complete divisor, which multiply by the last figure of the root, and subtract the product from the dividend, and to the...remainder bring down the next period for a new dividend. Proceed in same way, until all the periods are brought down. EXAMPLE. Find the cube root of 111215247129.... | |
 | James Bates Thomson - 1862 - 436 σελίδες
...of the partial divisor ; multiply the divisor thus completed by the figure last placed in the root ; subtract the product from the dividend, and to the...remainder bring down the next period for a new dividend. IV. Double the root already found for a new partial divisor, divide, &c., as before, and thus continue... | |
 | James Bates Thomson - 1862 - 428 σελίδες
...of the partial divisor; multiply the divisor thus completed by the figure last placed in the root; subtract the product from, the dividend, and to the remainder bring down t/ie next period for a new dividend. IV. Double the root already found, for a new partial divisor,... | |
 | Horatio Nelson Robinson - 1863 - 432 σελίδες
...trial divisor for a complete divisor ; multiply the complete divisor Ъу the trial figure in the root, subtract the product from the dividend, and to the...remainder bring down the next period for a new dividend. V. Take the last complete divisor, doubling its right-hand figure, for a new trial divisor, with which... | |
 | Charles Davies - 1863 - 346 σελίδες
...annex it to the divisor: IV. 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 : NOTES. — 1. The left-hand period may contain but one figure ; each of the others will contain two.... | |
 | Elias Loomis - 1864 - 386 σελίδες
...square of the second figure. 5. 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. 6. Take three hundred times the square of the whole root now found for a new trial divisor, and continue... | |
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Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. 
