| Daniel Adams - Arithmetic - 1807 - 248 pages
...Therefore, 1. If the Multiplier be a Composite number ; Multiply first by one of the component parts, and that product by the other ; the last product will be the answer sought. EXAMPLES. 1. Multiply б 7 by J 5. OPERATION. 6 7 5 one of the component parts. 335 3 the other... | |
| Osgood Carleton - Arithmetic - 1810 - 264 pages
...or of thousands, and multiply the value of one hundred, or of one thousand, by one of those numbers, and that product by the other ; the last product will be the value of the whole. JVors. When the price is very" small, proceed as in note to Case IV. EXAMPLES.... | |
| Nathan Daboll - Arithmetic - 1815 - 250 pages
...take any two such numbers as when multiplied together, will exactly produce the given quantity, and multiply first by one of those figures, and that product by the other ; and the last product will be the answer, ' EXAMPLE8, What cost 28 yards of.cloth, at 6s 10d. per... | |
| Nathan Daboll - Arithmetic - 1818 - 246 pages
...composite number, tiiat is, •when it is produced by multiplying any two numbers in the table together ; multiply first by one of those figures and that product by the other ; and the last product wijft be the total required. EXAMPLES. Multiply 41364 by 35. 7x5=55. f " 289548... | |
| Thomas Tucker Smiley - Arithmetic - 1825 - 224 pages
...the divisor exceeds 12, but is the exact product of two figures in the multiplication table. Divide first by one of those figures and that product by the other. Examples. 4 $ с<*. $ els. s. Divide 366.18! by 3 Ans. 122.06A. 4. — 384.871 by 6 _ 64.14^+2. 5.... | |
| Nicolas Pike, Dudley Leavitt - Arithmetic - 1826 - 214 pages
...composite number, (or exactly equal to the product of any two figures in the multiplication table ; — RULE. — Multiply first by one of those figures,...by the other, the last product will be the answer. f EXAMPLES. 1. 2. 3. Multiply 59375 by 35 39187 by 48 91632 by 56 7 415625 5 Product, 207813* * This... | |
| Nicolas Pike, Dudley Leavitt - Arithmetic - 1826 - 222 pages
...composite number, (or exactly equal to the product of any two figures in the multiplication table) — RULE — Multiply first by one of those figures, and...by the other, the last product will be the answer.! • EX.'.Mi'LES. I. 2. 3. Multiply 59575 by 35 39187 by 48 91132 by 56 7 7X5=35 415625 5 Prod. 2078125... | |
| Nathan Daboll - Arithmetic - 1829 - 252 pages
...take any two such numbers as when multiplied together, will exactly produce the given quantity, and multiply first by one of those figures, and that product by the other ; and the last product will be tiie answer. EXAMPLES. What cost 28 yards of cloth, at 6s. 10'! . per... | |
| Nathan Daboll - Arithmetic - 1829 - 268 pages
...composite number, that is, when it is produced by multiplying any two numbers in the table together ; multiply first by one of those figures and that product by the otfier ; and the last product will be the total required. EXAMPLES. Multiply 41364 by 35. 7X5~35. 7... | |
| Thomas Tucker Smiley - Arithmetic - 1830 - 192 pages
...the multiplier is exactly equal to the product of any two figures in the multiplication table, tl<e operation may be performed by the following Rule....Multiply first by one of those figures, and that product byfhe other; the iant product will be the answer. Question. Repeat the rule for performing the operation... | |
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