 | Samuel Read Hall - 1832 - 294 sider
...highest figure in the root required. Subtract the square of the root already found from the left hand period, and to the remainder bring down the next period for a dividend. Double the root already found for a divisor ; seek how many times the divisor is contained in the dividend,... | |
 | Zadock Thompson - 1832 - 186 sider
...highest figure in the root required. Subtract the square of the root already found from the left hand period, and to the remainder bring down the next period for a dividend. Double the root already found for a divisor; seek how many times the divisor is contained in the dividend... | |
 | Daniel Adams - 1833 - 268 sider
...period, and write its root as a quotient in division. Subtract the square number.from the left hand period, and to the remainder bring down the next period for a dividend. . III. Double the root already found for a divisor ; seek how many times the divisor is contained in... | |
 | William Templeton (engineer.) - 1833 - 224 sider
...period ; then from the first period subtract the greatest cube it contains ; put the root as a quotient, and to the remainder bring down the next period for a dividend. Find a divisor by multiplying the square of the root by 300 ; see how often it is contained in the... | |
 | Catharine Esther Beecher - 1833 - 296 sider
...greatest cube in the left hand period, and subtract it from that period. Place the root in the quotient, and to the remainder bring down the next period, for a dividend. 3. Square the root already found (understanding a cipher at the right) and multiply it by 3 for a divisor.... | |
 | Frederick Emerson - 1834 - 300 sider
...hand period, and write its root in the quotient. Subtract the square of this root from the left hand period, and to the remainder bring- down the next period for a dividend. Thirdly — Double the root already found, for a divisor. Ascertain how many times the divisor is contained... | |
 | Benjamin Snowden - 1835 - 108 sider
...and put its root for the first figure of the root sought. 2. — Subtract this cube from the first period, and to the remainder bring down the next period for a dividend. 3. — Find a divisor by multiplying the square of the root by 300. 4. — Divide the dividend by this... | |
 | Stephen Pike - 1835 - 210 sider
...period, and set its root on the right of the given number: subtract said square from the left hand period, and to the remainder bring down the next period for a dividual. 3. Double the root for a divisor, and try hp\v often this divisor (with the figure used in... | |
 | James L. Connolly (mathematician.) - 1835 - 264 sider
...not too great, of the first period, for the first figure of your root, subtract its cube from said period, and to the remainder bring down the next period for a resolvend. Take three times the square of the root for a defective divisor, and seek how often it is... | |
 | Charles Potts - 1835 - 202 sider
...hand period, and set its root on the right of the given number ; subtract said cube from the left hand period, and to the remainder bring down the next period for a dividual. 3. Square the root, and multiply the square by 3 for a defective divisor. 4. Reserve, mentally,... | |
| |
... and to the remainder bring down the next period for a dividend. 3. Place the double of the root already found, on the left hand of the dividend for a divisor. 4. Seek how often the divisor is contained... 
