... the right. 2. Find the greatest cube number, in the left-hand period, and place the 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... The accomptant's guide, a new system of arithmetic - Page 174by James Morrison (accountant.) - 1801Full view - About this book
| A. Melrose (Teacher) - Arithmetic - 1795 - 140 pages
...number itfelf from faid left-hand period; and to the remainder annex the next period for a dividend. Find a divifor, by multiplying the fquare of the part...dividend by it ; and put the quotient figure for the next figure of the root Multiply the part of the root formerly found by the laft figure placed in it,... | |
| William Vogdes - Arithmetic - 1847 - 324 pages
...down the next period to the remainder for a new dividend. 3. Find a divisor by multiplying the square of the part of the root found by 300, divide the dividend by it, and place the quotient figure for the second figure in the root. 4. Multiply the part of the root formerly... | |
| James Gray - Arithmetic - 1854 - 120 pages
...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 by it, and put the quotient figure for the next figure of the root. 4. Multiply the part of the root formerly found by the last figure placed... | |
| James Gray - Arithmetic - 1883 - 154 pages
...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 by it, and put the quotient figure for the next figure of the root. 4. Multiply the part of the root formerly found by the last figure placed... | |
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