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... Elementary Algebra - Page 255by George William Myers, George Edward Atwood - 1916 - 338 pagesFull view - About this book
| George Albert Wentworth - Arithmetic - 1897 - 440 pages
...This sum will be the complete divisor. POWKRS AND ROOTS. Multiply the complete divisor by the second **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 annexed. The result... | |
| George Albert Wentworth - Arithmetic - 1898 - 424 pages
...root. To this partial divisor add the last figure of the root for a complete divisor. Multiply this **complete divisor by the last figure of the root, subtract...the dividend, and to the remainder annex the next** group for a new dividend. Proceed in this manner until all the groujjs have been thus annexed. The... | |
| George Albert Wentworth - Arithmetic - 1898 - 424 pages
...root. To this partial divisor add the last figure of the root for a- complete divisor. Multiply this **complete divisor by the last figure of the root, subtract...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... | |
| George Albert Wentworth - 1898 - 424 pages
...root. To this partial divisor add the last figure of the root for a complete divisor. Multiply this **complete divisor by the last figure of the root, subtract...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... | |
| George Albert Wentworth - Arithmetic - 1898 - 424 pages
...the second figure. This sum will be 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** group for a new dividend. Proceed in this manner until all the groups have been annexed. The result... | |
| George Edward Atwood - Arithmetic - 1899 - 388 pages
...quotient to the root and also to the divisor. Multiply the completed divisor by the last figure in **the root, subtract the product from the dividend, and to the remainder annex the next period for** the next dividend. Proceed in the same manner until all the periods have been used. NOTE 1. If the... | |
| Eugene L. Dubbs - Arithmetic - 1901 - 462 pages
...annex it to the trial divisor for the complete divisor. 4. 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** another dividend. 5. Double the part of the root already found for another trial divisor, and proceed... | |
| William James Milne - Algebra - 1901 - 456 pages
...trial divisor the .figure last found, multiply this complete divisor by the figure 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. The result will be... | |
| John Appley Ferrell, B. F Sisk - Arithmetic - 1901 - 436 pages
...annexed, and the square of the last figure of the root, 3x3x50 = 450; 3 2 =9. 7500+450+9 = 7959, complete **divisor. Multiply the complete divisor by the last figure of the root, subtract the product,** and bring down the next period. (4) Proceed as in (3). 1^8489664 =( ) ? Process: Explanation : If after... | |
| Joseph Benjamin Rider - Engineering - 1901 - 546 pages
...and the other figure or figures of the root, and the square of the last figure to form the complete **divisor. Multiply the complete divisor by the last figure of the root,** and subtract the product from the dividend. To the remainder annex the next period, and proceed as... | |
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