 | William James Milne - Arithmetic - 1906 - 364 pages
...complete divisor the figure last found, multiply this divisor by the figure of the root last 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... | |
 | William James Milne - Arithmetic - 1906 - 364 pages
...complete divisor the figure last found, multiply this divisor by the figure of the root last found, subtract the product from the dividend, and to the remainder annex the next period for the next dividend. 1. When the number is not a perfect square, annex periods of decimal ciphers and... | |
 | Charles Wilbur Leigh - 1906 - 72 pages
...trial divisor, and multiply the new divisor, their sum, by the last figure in the root. Subtract this product from the dividend, and to the remainder annex the next period for the new dividend. After annexing a cipher to that part of the root already found, and doubling the... | |
 | William James Milne - Arithmetic - 1907 - 616 pages
...complete divisor the figure last found, multiply this divisor by the figure of the root last found, subtract the product from the dividend, and to the remainder annex the next period for the next dividend. 22 245 l'.56'25 |1.25 1 56 44 1225 1225 1. When the number is not a perfect square,... | |
 | William James Milne - Algebra - 1908 - 476 pages
...divisor the figure last found, multiply this complete divisor by the figure of the root last 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... | |
 | George Wentworth, David Eugene Smith - Arithmetic - 1909 - 276 pages
...sum will be the complete divisor. Multiply the complete divisor l,y the second figure of the root, subtract the product from the dividend, and to the...remainder annex the next period for a new dividend. Proceed in this manner until all the periods have been annexed. The result will be the cube root required.... | |
 | William James Milne - 1911 - 360 pages
...divisor the figure last found, multiply this complete divisor by the figure of the root last 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... | |
 | Glenn Moody Hobbs - 1912 - 100 pages
...the trial divisor, and multiply the completed divisor by the last figure in the root. Subtract this product from the dividend, and to the remainder annex the next period for the new dividend. Continue the operation until all the periods are used. (d) To place the decimal point... | |
 | George Wentworth, David Eugene Smith - Algebra - 1913 - 478 pages
...of the root for a complete divisor. Multiply the complete divisor by this next figure of the root, subtract the product from the dividend, and to the...remainder annex the next period for a new dividend. Proceed in this manner until all the periods have been thus annexed. The result will be the square... | |
 | George Albert Wentworth, David Eugene Smith - Algebra - 1913 - 310 pages
...of the root for a complete divisor. Multiply the complete divisor by this next figure of the root, subtract the product from the dividend, and to the...remainder annex the next period for a new dividend. Proceed in this manner until all the periods have been thus annexed. The result will be the square... | |
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... subtract the product from the dividend, and to the remainder annex the next period for the next dividend. 
