 | Frederick Emerson - Arithmetic - 1834 - 300 pages
...column. Md the numbers of the lowest denomination together, and divide their sum by that number which is required of this denomination to make a unit of the next higher: write the remainder under the column added, and carry the quotient to the next column. Thus proceed... | |
 | Frederick Emerson - Arithmetic - 1839 - 300 pages
...Jldd the numbers of the lowest denomination together, and divide their sum by that number which is required of this denomination to make a unit of the next higher: write the remainder under the column added, and carry the quotient to the next column. Thiu proceed... | |
 | Charles Davies - 1852 - 344 pages
...denomination, and set down the remainder. III. Proceed in the same way to the required denomination, and the last quotient, with the several remainders annexed, will be the answer. NOTE. — Every remainder will be of the same denomination as its dividend. PROOF. — After a number has... | |
 | Charles Davies - Arithmetic - 1856 - 450 pages
...each succeeding quotient in the same manner, till the unit is reduced to the required denomination: the last quotient with the several remainders annexed, will be the answer. EXAMPLES. 1. Reduce £3 14s. id. to pence. We first multiply the £3 by 20, which gives CO shillings.... | |
 | Charles Davies - Arithmetic - 1861 - 496 pages
...each succeeding quotient in the same manner, till the number is reduced to the required denomination: the last quotient, with the several remainders annexed, will be the answer. Examples. 1. Reduce £3 14s. 4d. to pence. We first multiply the £3 by 20, which gives 60 shillings.... | |
 | Charles Davies - Arithmetic - 1863 - 346 pages
...next scale, and set aside the remainder: proceed in the same way, through all the denominations; and the last quotient, with the several remainders annexed, will be the answer. PROOF. — The proof, in either case, is made by reversing the operation. Examples. 1. Reduce £15... | |
 | James Bates Thomson - 1875 - 392 pages
...farthings there are £23, 73. sd. i far. Hence, the EUEE. — Divide the given denomination by the number required of this denomination to make a unit of the...successive denominations, till the one required is reached. T/ie last quotient, with the several remainders annexed, will be the answer. NOTE. — The remainders,... | |
 | James Bates Thomson - 1876 - 400 pages
...farthings there are £23, 7s. 5d. I far. Hence, the KULE.—Divide the given denomination by the number required of this denomination to make a unit of the...the several remainders annexed, will be the answer. NOTE.—The remainders, it should be observed, are the same denomination as the respective dividends... | |
 | James Bates Thomson - Arithmetic - 1882 - 450 pages
...QA 1 Hence, the " m- 4JaRULE.— Div iile the given denomination by the number required to make one of the next higher. Proceed in this manner with the...annexed, will be the answer. NOTE. — The remainders are the same denomination as the respective dividends from which they arise. 400. PROOF. — Reduction... | |
 | James Bates Thomson - Arithmetic - 1882 - 416 pages
...required of the next lower to make a unit of the higher, and to the product add the lower denomination. Proceed in this manner with the successive denominations, till the one required is reached. 2. In 5 mi. 12 rd. 4 yd. 2 ft. how many feet ? 3. Reduce 143 Ib. 3 oz. 6 pwt. to grains. 4. Reduce... | |
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Div iile the given denomination by the number required to make one of the next higher. Proceed in this manner with the successive denominations, till the one required is reached. The last quotient, with the several remainders annexed, will be the answer.... 
