 | George Albert Wentworth - 1898 - 424 pages
...of the root for a complete divisor. Multiply this complete divisor by the last 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 thus annexed. The result... | |
 | George Edward Atwood - Arithmetic - 1899 - 392 pages
...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... | |
 | Edward Brooks - Algebra - 1901 - 248 pages
...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 tens, for a trial divisor, 'and find the... | |
 | William James Milne - Algebra - 1901 - 476 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... | |
 | Eugene L. Dubbs - Arithmetic - 1901 - 462 pages
...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... | |
 | Edward Gideon - 1902 - 272 pages
...sum will be the complete divisor. V. Multiply the complete divisor 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. VI. Proceed in the same manner with all the periods to the last. The result will be the cube root required.... | |
 | George Edward Atwood - Arithmetic - 1902 - 168 pages
...to the root and also to the divisor. Multiply the completed divisor 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, continuing the process until all the periods have been used. NOTE. — If any partial divisor is not... | |
 | William James Milne - Algebra - 1902 - 620 pages
...trial divinar the figure last found, multiply this complete divisor b>/ 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 tJte periods have been used. TJie result icill... | |
 | Jacob Henry Minick, Clement Carrington Gaines - Business mathematics - 1904 - 412 pages
...sum will be the complete divisor. Multiply the complete divisor by the second figure of the root and subtract the product from, the dividend, and to the remainder annex the next period for the next dividend. Take three times the square of the root already found for a second trial divisor,... | |
 | Benjamin Franklin Sisk - Arithmetic - 1905 - 224 pages
...sum add the square of the trial term. 7. Multiply the complete divisor by the trial term in the root, subtract the product from the dividend, and to the...remainder annex the next period for a new dividend. STEPS: POWERS AND ROOTS EXAMPLES 1. What is the cube root of 13144256? OPERATION. 1st trial divisor... | |
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... subtract the product from the dividend, and to the remainder annex the next period for the next dividend. 
