| Charles Vyse - Arithmetic - 1815 - 340 pages
...name mentioned. 3. Multiply the second ^nd third terms together, and divide that product by the «rst. The quotient will be the answer to the question, in the same denomination you left your second term in. 4. If there happen to be a remainder after the division, reduce it into the next... | |
| Daniel Staniford - Arithmetic - 1818 - 332 pages
...same name with the answer required, in the second or middle place. GENERAL RULE TO WORK THE QUESTION. Multiply the second and third numbers together, and divide the product by the first, the quotient will be the answer in the same denomination with the second or middle term. PROOF. Invert... | |
| Robert Patterson - Arithmetic - 1819 - 174 pages
...= be, and this divided by a, will give d = —. In words — multiply the a second and third terms together, and divide the product by the first, and the quotient will be the fourth, or term required. II. IN ALLIGATION ALTERNATE. Let a, A, = the respective rates of two ingredients... | |
| William Jillard Hort - 1822 - 308 pages
...first and third terms into the same denomination, and bring the second into the lowest term mentioned. Multiply the second and third numbers together, and divide the product by the first. The quotient will be the answer, of the same denomination as the second term. VOL. I. K Example. —... | |
| Jacob Willetts - Arithmetic - 1822 - 200 pages
...several denominations reduce it to the vest one mentioned! 6. Then multiply the second and third term together and divide the product by the first, and the quotient will be the fourth term or answer ; which will be in the same denomination as the second, or as that to which the... | |
| Thomas Keith - Arithmetic - 1822 - 354 pages
...the second into the lowest denomination mentioned. Multiply the second and third numbers together, divide the product by the first, and the quotient will be the answer in the same denomination as the second number.. If there be a remainder after division, it is always... | |
| Nathan Daboll - Arithmetic - 1823 - 262 pages
...it. 3. Multiply the second and third terms together, and divide their product by the first term .• the quotient will be the answer to the question, in the same denomination you left the second term in, which may be brought into any other denomination required. fhe method of proof is by inverting... | |
| John Penrose (teacher of arithmetic.) - Arithmetic - 1824 - 320 pages
...quantities, bring them to the same denomination ; then altogether disregarding their denomination, multiply the second and third numbers together, and divide the product by the first, taking the advantage of any abreviations the numbers admit of. JFfrrft «Frample. If 81bs of beef can... | |
| Poplar House Academy - 1826 - 100 pages
...and the third to the lowest denomination mentioned in it. Then, multiply the second and third terms together, and divide the product by the first, and the quotient will be the answer, in the same denomination that the third term was reduced to ; which must be brought again, if necessary,... | |
| Nicolas Pike, Dudley Leavitt - Arithmetic - 1826 - 214 pages
...and the other on the left for the fast term. 3. Multiply the second and third terms together, divide by the first, and the quotient will be the answer to the question, which, 231. What is the nature ofthete useleti distinctions ? 232. What is the general (as also the... | |
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