This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 185? Excerpt: ...+barrels, the Answer; that is, 6 bar2850 rels and 526 thousandths of another-----barrel. By annexing a cipher to the first 23'5 remainder, thereby reducing it to 1250 l0ths, and continuing the division, 950 we obtain from it '5, and a still fur ther remainder of 125, which, by an 3000 nexing another cipher, is reduced ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 185? Excerpt: ...+barrels, the Answer; that is, 6 bar2850 rels and 526 thousandths of another-----barrel. By annexing a cipher to the first 23'5 remainder, thereby reducing it to 1250 l0ths, and continuing the division, 950 we obtain from it '5, and a still fur ther remainder of 125, which, by an 3000 nexing another cipher, is reduced to 2850 l00ths, and so on. The last remainder, 150, is f of 0 a thousandth part of a barrel, which is of so trifling a value, as not to merit notice. If now we count the decimals in the dividend, (for every cipher annexed to the remainder is evidently to be counted a decimal of the dividend, ) we shall find them to be five, which corresponds with the number of decimal places in the divisor and quotient counted together. 3. Under IT 71, ex. 3, it was required to multiply '125 by '03; the product was '00375. Taking this product for a dividend, let it be required to divide '00375 by '125. One operation will prove the other. Knowing that the number of decimal places in the quotient and divisor, counted together, will be equal to the decimal places in the dividend, we may divide as in whole numbers, being careful to retain the decimal points in their proper places. Thus, OPERATION. The divisor, 125, in 375 goes 3 '125) '00375 ('03 times, and no remainder. We have only to place the decimal point in Qqq the quotient, and the work is done. There are rive decimal places in the dividend; consequently here must be five in t'e divisor and quotient counted together; and, as there are three in the divisor, there must be two in the quotient; and, since we have but one figure in the quotient, the deficiency must be supplied by prefixing a cipher. The operation by vulgar fractions will bring us to the same result. Thus, '125 is and '00375 is n&y...
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