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" Subtract the square of this figure from the left-hand period, and to the remainder annex the next period for a dividend. 3. Double the root already found, for a trial divisor; find how often it is contained in the dividend, exclusive of the righthand... "
Elementary Algebra - Page 255
by George William Myers, George Edward Atwood - 1916 - 338 pages
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Bradbury's Elementary Algebra: Designed for the Use of High Schools and ...

William Frothingham Bradbury - Algebra - 1877 - 280 pages
...Find the greatest cube in the left-hand period, and place its root at the right. Subtract this cube from the left-hand period, and to the remainder annex the next period for a dividend. Square the root figure, annex two ciphers, and multiply this result by three for a TRIAL DIVISOĽ ;...
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The Standard Arithmetic: For Schools of All Grades and for ..., Volume 2

James E. Ryan - Arithmetic - 1877 - 212 pages
...Find the greatest cube in the left-hand period, and write its root as a quotient. SuUract the cube from the left-hand period, and to the remainder annex the next period for a dividend. Consider the root figure already found as tens, and take three times its square for a TRIAL DIVISOR,...
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The Practical Arithmetic on the Inductive Plan: Including Oral and Written ...

William James Milne - Arithmetic - 1877 - 402 pages
...root. Square this root and subtract the result from the left-hand period, and annex to the remainder the next period for a dividend. Double the root already found for a trial divisor, and by it divide the dividend, disregarding the right-hand figure. The quotient or quotient...
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Complete Arithmetic: Theoretical and Practical

William Guy Peck - Arithmetic - 1877 - 428 pages
...9 ; we therefore write 9 for the first figure of the root, subtract its square, 81, from the first period, and to the remainder annex the next period for a dividend. We then double 9 for a trial divisor, and find how many times it is contained in the dividend, exclusive...
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The New American Arithmetic, Part 3

Samuel Mecutchen, George Mornton Sayre - Arithmetic - 1877 - 200 pages
...product falls under the first left-hand period of the given number. Subtract this product from this first left-hand period, and to the remainder annex the next period for a new dividend. Add the figure of the root just found to the amount in the first column, multiply the...
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Bradbury's Eaton's Practical Arithmetic: Combining Oral and Written Exercises

William Frothingham Bradbury - Arithmetic - 1879 - 392 pages
...greatest cube in the left-hand period, write its root at the right of the given number, subtract the cube from the left-hand period, and to the remainder annex the next period for a dividend. 3. Square the root figure, annex two ciphers, and multiply this result by three for a trial divisor...
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Key to the Olney's Science of Arithmetic: In which are Given Full Solutions ...

Edward Olney - Arithmetic - 1879 - 176 pages
...4th power in the left-hand period as the first figure in the root. Subtract this highest 4th power from the left-hand period, and to the remainder annex the next period, forming a dividend. 3. Take 4 times the 3d power of the part of the root already found (regarded as...
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Elementary algebra, with brief notices of its history

Robert Potts - 1879 - 676 pages
...this number will be the first figure of the root. 3. Subtract the cube of this number from the last period, and to the remainder annex the next period for a dividend. 4. Divide this number (omitting the last two figures) by three times the square of the first figure...
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Elementary algebra: with brief notices of its history

Robert Potts - Algebra - 1879 - 668 pages
...this number will be the first figure of the root. 3. Subtract the cube of this number from the last period, and to the remainder annex the next period for a dividend. 4. Divide this number (omitting the last two figures) by three times the square of the first figure...
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Graded Problems in Arithmetic and Mensuration

Samuel Mecutchen - Arithmetic - 1880 - 262 pages
...product falls under the first left-hand period of the given number. Subtract this product from this first left-hand period, and to the remainder annex the next period for a new dividend. Add the figure of the root just found to the amount in the first column, multiply the...
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