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. A System of Arithmetic - Page 159by Samuel Webber - 1812 - 248 pagesFull view - About this book
| Elias Loomis - Algebra - 1858 - 359 pages
...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... | |
| Thomas William Silloway - Carpentry - 1858 - 180 pages
...likewise on the right hand of the divisor. Multiply the divisor by the last quotient-figure, subtract the **product from the dividend, and to the remainder bring down the next period for a new dividend,** 4th, Double the quotient already found for a partial divisor, and from these find the next figure in... | |
| Benjamin Greenleaf - Arithmetic - 1858 - 444 pages
...figure, for a true divisor. Multiply the true divisor 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.** Multiply the square of (lie root figures already found by 3, and to the product annex two ciphers for... | |
| Benjamin Greenleaf - Arithmetic - 1858 - 444 pages
...the same for a true divisor. Multiply the true divisor by the last figure of the root : subtract the **product from the dividend, and to the remainder bring down the next period** far a new dividend. To the last true divisor and the number immediately over it, add the square of... | |
| Charles Haynes Haswell - Measurement - 1858 - 322 pages
...period, and place its root in the quotient ; subtract the square number from the left-hand period, **and to the remainder bring down the 'next period for a new dividend.** Double the root already found for a divisor ; find how many times this incomplete divisor is contained... | |
| Benjamin Greenleaf - Arithmetic - 1858 - 324 pages
...represent? What is the rule for extracting the cube root ? Subtract the subtrahend from the dicidend, **and to the remainder bring down the next period for a new** dicidend, with which proceed at before; and so on, till the whole is completed. NOTE 1. — In separating... | |
| Horatio Nelson Robinson - 1859 - 336 pages
...complete divisor ; multiply the complete divisor by the trial figure in the root, and subtract the **product from the dividend, and to the remainder bring down the next period for a new dividend.** V. Multiply the last figure of the last complete divisor ly 2, and annex one cipher for a new trial... | |
| Charles Guilford Burnham - 1859
...write the cube of the last quotient figure, and call their sum the subtrahend. Subtract the subtrahend **from the dividend, and to the remainder bring down the next period, for a new dividend,** and proceed as before, till the work is finished. EXAMPLES. 2. What is the cube root of 1906624 ? ,... | |
| James Bates Thomson - Arithmetic - 1860 - 422 pages
...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... | |
| Horatio Nelson Robinson - Arithmetic - 1860 - 432 pages
...will be the complete divisor. V. Multiply the complete divisor by the trial figure ; subtract the. **product from the dividend, and to the remainder bring down the next period for a new dividend.** VI. Add the square of the last figure of the root, the last term in column II, and the complete divisor... | |
| |