 | Daniel Adams - Arithmetic - 1838 - 276 pages
...the two remaining numbers for the second term, and the greater for the fin,t ; and, in either case, multiply the second and third terms together, »;•« divide the product by the first for the sns'.vei, which will always be of the same denomination as (be third term. Note 1. If the first... | |
 | Thomas Holliday - Surveying - 1838 - 404 pages
...the question according to the proper rule or case, multiply the second and third terms together and divide the product by the first, and the quotient will be the fourth term required for the natural number. But in working by logarithms, the second and third terms... | |
 | Arithmetic - 1838 - 218 pages
...till they are ; and if the third term cons of several denominations, reduce it to the lowest. Then multiply the second and third terms together, divide the product by the ßrst term, the quotient be the answer or fourth term. ГОТЕ. — The answer will be of that denomination... | |
 | John Darby (teacher of mathematics.) - 1843 - 236 pages
...third into the lowest terms mentioned. Multiply the second and third terms, thus reduced, together, and divide the product by the first, and the quotient will be the answer to the question in the same denomination you left the third term in ; which may be brought into the... | |
 | James Bates Thomson - Arithmetic - 1846 - 354 pages
...the other two numbers for the second term, and t/ce other for the first. III. Finally, multiplying the second and third terms together, divide the product...by the first, and the quotient will be the answer in the same denomination as the third term. PROOF. — Multiply the first term and the answer or fourth... | |
 | James Bates Thomson - Arithmetic - 1846 - 402 pages
...of the other two numbers for the second term, and the other for the first. III. Finally, multiplying the second and third terms together, divide the product...by the first, and the quotient will be the answer in the same denomination as the third term. PROOF. — Multiply the first term and the answer or fourth... | |
 | James Bates Thomson - Arithmetic - 1846 - 362 pages
...second term, and the other for the first. IIL Finally, multiplying the second and third terms togethf-r, divide the product by the first, and the quotient will be the. answer in the same denomination as the third term. PROOF. — Multiply the first term ami the ansmr together,... | |
 | James Bates Thomson - Arithmetic - 1847 - 426 pages
...place the less of the other two numbers for the second term, and the other for the first. III. finally, multiply the second and third terms together, divide...by the first, and the quotient will be the answer in the same denomination as the third term. PROOF. — Multiply the first term and the answer together,... | |
 | James Bates Thomson - Arithmetic - 1847 - 432 pages
...place the less of the other two numbers for the second term, and the other for the first. III. finally, multiply the second and third terms together, divide...by the first, and the quotient will be the answer in the same denomination as the third term. PROOF. — Multiply the first term and the answer together,... | |
 | James Bates Thomson - Arithmetic - 1848 - 434 pages
...the less of tlie other two numbers for the second term, and the other for the first. III. Finally, multiply the second and third terms together, divide...by the first, and the quotient will be the answer in the tame denomination as the third term. PROOF. — Multiply the first term and the answer together,... | |
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... for the second term, and the other for the first. III. finally, multiply the second and third terms together, divide the product by the first, and the quotient will be the answer in the same denomination as the third term. 
