 | Daniel Leach - 1851 - 280 σελίδες
...period for a partial dividend. Multiply the square of the root by 3? and annex two ciphers for the first trial divisor. Find how many times the trial divisor is contained in the partial dividend, and write the result in the quotient. Complete the divisor by adding to the trial... | |
 | Daniel Leach - 1853 - 622 σελίδες
...period for a partial dividend. Multiply the square of the root by 3, and annex two ciphers for the first trial divisor. Find how many times the trial divisor is contained in the partial dividend, and write the result in the quotient. Complete the divisor by adding to the trial... | |
 | George Payn Quackenbos - 1872 - 350 σελίδες
...three times the square of the root already found ; and, annexing two ciphers, place it on the left as a trial divisor. Find how many times the trial divisor is contained in the dividend (making some allowance), and annex the quotient to the root already found. Complete the trial... | |
 | George Payn Quackenbos - 1874 - 444 σελίδες
...three times the square of the root already found; and, annexing two ciphers, place it on tJie left as a trial divisor. Find how many times the trial divisor is contained in the dividend (making some allowance), and annex the quotient to the root already found. 4- Complete the... | |
 | Philotus Dean - 1874 - 472 σελίδες
...second period for a dividend. Square the root found, annex two ciphers, and multiply the result by 3 for a trial divisor. Find how many times the trial divisor is contained in the dividend, and write the quotient as the next figure of the root. To the trial divisor add three times... | |
 | Joseph Ray - 1877 - 350 σελίδες
...remainder bring down the next period for a dividend. 3. Square the root found, and multiply it by 300 for a trial divisor. Find how many times the trial divisor is contained in the dividend, and place the quotient in the root. 4. Multiply the preceding figure, or figures, of the... | |
 | William Estabrook Chancellor - 1902 - 178 σελίδες
...the next period for a new dividend. Multiply the square of the part of the root already found by 300 as a trial divisor. Find how many times the trial divisor is contained in the dividend, and put the figure thus obtained in the root. To the trial divisor add the part of the root... | |
 | Joseph Ray - 1903 - 366 σελίδες
...period for a dividend. EVOLUTION 3. Double the root found, and place it on the left of the dividend for a trial divisor. find how many times the trial divisor is contained in the dividend, exclusive of the right-hand figure ; place the quotient in the root, and also on the right... | |
 | Joseph Archibald Ewart, Wilbur Stanwood Field, Adelbert Harland Morrison - 1912 - 210 σελίδες
...next period, as before. 1'53'76.(12 1 Repeat this process until all the periods have been used, taking twice the part of the answer already found as a trial: divisor. 1'53/76(124. , Ans. 1 22) 53 44 244) 976 976 If at any time the dividend, without the last figure,... | |
 | Earl Adolphus Saliers - 1923 - 1726 σελίδες
[ Λυπούμαστε, το περιεχόμενο αυτής της σελίδας είναι περιορισμένο ] | |
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Subtract its cube from the first period, and to the remainder annex the second period for a dividend. 3. Take three times the square of the root already found ; and, annexing two ciphers, place it on the left as a trial divisor. 