 | Clement Mackrow - 1879 - 552 pages
...left-hand period, and put the root of it in the quotient : subtract this square fn m the left-hand period, and to the remainder bring down the next period for a dividend, and to the left hand of it write double the quotient for a divisor; then consider what figure if annexed... | |
 | Webster Wells - Algebra - 1879 - 468 pages
...left-hand period, and place its root on the right ; subtract the cube of this root from the lefthand period, and to the remainder bring down the next period for a div1dend. Divide this dividend, omitting the last two figures, by three times the square of the root... | |
 | Joseph Ray - Arithmetic - 1880 - 420 pages
...first period on the left; place its root on the right, like a quotient in division; subtract the square from the period, and to the remainder bring down the next period for a dividend. 3. Double the root already found, as if it were units, and write it on the left for a trial divisor ;... | |
 | Horatio Nelson Robinson - 1875 - 472 pages
...and write its root for the first figure in the required root ; subtract the cube from the left hand period, and to the remainder bring down the next period for a dividend. III. At the left of the dividend write three times the square of the first figure of the root, and... | |
 | James Morton - Circle-squaring - 1881 - 236 pages
...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... | |
 | Montagu H. Foster - 1881 - 182 pages
...as in division. Square the root thus found, and place that square under the first period. Subtract, and to the remainder bring down the next period for a dividend. Double the root already found, and place the product for a divisor. Divide the dividend exclusive of... | |
 | James Bates Thomson - Arithmetic - 1882 - 416 pages
...left, we find the greatest cube in 15 is 8, the root of which is 2. Placing the 2 on the right, we subtract its cube from the period, and to the remainder bring down the next period for a dividend. This shows that we have 7625 solid feet to be added. OPERATION. 15625 (25 8 1200 300 7625 7625 3. We... | |
 | James Bates Thomson - Arithmetic - 1882 - 416 pages
...15 is 8, the root of which is 2. Placing the 2 on the right, we subtract its cube from the pariod, and to the remainder bring down the next period for a dividend. This shows that we have 7625 solid feet to be added. 3. We square the root already found, which in... | |
 | James Gray - Arithmetic - 1883 - 154 pages
...root of that number as the first figure of the root sought : subtract the number itself from the said period, and to the remainder bring down the next period for a dividend. 3. Find a divisor by multiplying the square of the part of the root found by 300, divide the dividend... | |
 | William Allen Sylvester - Building - 1883 - 220 pages
...and write its root for the first figure in the required root ; subtract the cube from the left-hand period, and to the remainder bring down the next period for a dividend. 3. At the left of the dividend write three times the square of the first figure of the root, and annex... | |
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... 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... 
