 | Webster Wells - 1885 - 368 σελίδες
...subtract its square from the number, and to the result bring down the next period. Divide this remainder, omitting the last figure, by twice the part of the root already found, and annex the quotient to the root and also to the divisor. Multiply the complete divisor by the figure of the root... | |
 | Webster Wells - 1885 - 370 σελίδες
...subtract its square from the number, and to the result bring down the next 2/eriod. Divide this remainder, omitting the last figure, by twice the part of the root already found, and annex the quotient to the root and also to the divisor. Multiply the complete divisor by the figure of the root... | |
 | Elias Loomis - 1886 - 396 σελίδες
...square is contained in the left-hand period ; this is the first figure of the required root. &ibtract its square from the first period, and to the remainder bring down the second period for a dividend. 3d. Double the root already found for a divisor, and find how many times... | |
 | Edward Albert Bowser - 1888 - 868 σελίδες
...square is contained in the left period, and place it on the right ; this is the first figure of the root ; subtract its square from the first period, and to the remainder bring doivn the next period for a dividend. Double the root already found for a trial divisor, and see how... | |
 | Webster Wells - 1890 - 560 σελίδες
...-figure from the left-hand period, and to the result annex the next period. Divide this remainder, omitting the last figure, by twice the part of the root already found, and annex the quotient to the root, and also to the trial-divisor. Multiply the complete divisor by the root-figure... | |
 | Webster Wells - 1893 - 382 σελίδες
...root-figure from the first period, and to the result annex the next period. Divide this remainder, omitting the last figure, by twice the part of the root already obtained, and annex the quotient to the root, and also to the trial-divisor. Multiply the complete... | |
 | Webster Wells - 1897 - 422 σελίδες
...root-figure from the left-hand period, and to the result annex the next period. 181 Divide this remainder, omitting the last figure, by twice the part of the root already found, and annex the quotient to the root, and also to the trial-divisor. Multiply the complete divisor by the root-figure... | |
 | Webster Wells - 1897 - 434 σελίδες
...root-figure from the left-hand period, and to the result annex the next period. 181 Divide this remainder, omitting the last figure, by twice the part of the root already found, and annex the quotient to the root, and also to the trial-divisor. Multiply the complete divisor by the root-figure... | |
 | Webster Wells - 1897 - 426 σελίδες
...root-figure from the left-hand period, and to the result annex the next period. 181 Divide this remainder, omitting the last figure, by twice the part of the root already found, and annex the quotient to the root, and also to the trial-divisor. Multiply the compílete divisor by the root-figure... | |
 | Henry G. Abbott - 1898 - 392 σελίδες
...the first period and to the remainder bring down the second period. Divide the number thus formed, omitting the last figure, by twice the part of the root already obtained, and "annex the quotient to the root and also to the divisor. Then multiply the divisor, as... | |
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Divide this quantity, omitting the last figure, by twice the part of the root already found, and annex the result to the root and also to the divisor, then multiply the divisor as it now stands by the part of the root last obtained for the subtrahend. 
