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" ... multiply the second and third terms together, and divide the product by the first for the answer, which will always be of the same denomination as the third term. "
Adams's New Arithmetic - Página 182
por Daniel Adams - 1830 - 264 páginas
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The Teachers' Assistant: Or a System of Practical Arithmetic

Stephen Pike - 1835 - 210 páginas
...and if the third term consist of several denominations, reduce it to its lowest denomination; then, Multiply the second and third terms together, and divide the product by the first term: the quotient will je the answer. Note. — The product of the second and third termsis of he...
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Taplin's improved edition of Walkingame's Tutor's assistant. To which is ...

Francis Walkingame - 1835 - 270 páginas
...proportion, if necessary, to the same name, 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; the quotient will be the answer to the question in the same denomination the third term was reduced...
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A Practical and Theoretical System of Arithmetic...: Also, a Treatise on ...

George Willson - 1836 - 202 páginas
...mentioned in it.* * It is often better to reduce the lower denominations to the decimal of the highest. 3. Multiply the second and third terms together, and divide the product by the first, and the quotient will be the answer, in that denomination which the third term was left in. In arranging...
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Arithmetic made easy

A. Turnbull - 1836 - 368 páginas
...then reduce the third term to the least denomination contained in it. The three terms thus reduced, we multiply the second and third terms together, and divide the product by the first, and the quotient will be the fourth term in the same denomination, to which the third term has been...
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The Common School Arithmetic ...: To which is Added a Dictionary of ...

Peirpont Edward Bates Botham - 1837 - 252 páginas
...question. The first and third terms must be of one name. The second term of -divers denominations. Multiply the second and third terms together, and divide the product by the first term ; the quotient thence arising will be the Answer. OBS. This rule is founded on the obvious principle,...
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A System of Geometry and Trigonometry, with a Treatise on Surveying: In ...

Abel Flint - 1837 - 338 páginas
...is calculated accordingly. GENERAL ROLE. 1. State the question in every case, as already taught : 2. Multiply the second and third terms together, and divide the product by the first. The manner of taking natural sines and tangents from the tables, is the same as for logarithmic sines...
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A complete treatise on practical land-surveying

Thomas Holliday - 1838 - 404 páginas
...answer. 3.—By arithmetical computation. Having stated 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,...
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A Practical and Theoretical System of Arithmetic: Containing a New System of ...

George Willson - 1838 - 194 páginas
...mentioned in it.* * It is often better to reduce the lower denominations to tha daeimil «f the highest 3. Multiply the second and third terms together, and divide the product by the first, and the quotient will be the answer, in that denomination which the third term was bft in. In arranging...
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The parent's guide to a liberal ... education; containing a selection of ...

Robert Simson (master of Colebrooke house acad, Islington.) - 1838 - 206 páginas
...When the terms are stated and reduced, how do you proceed in order to find a fourth proportional? I multiply the second and third terms together, and divide the product by the first, the quotient is the answer. In what name are the product of the second and third terms, the quotient,...
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The Private Instructor, Or Mathematics Simplified: Comprising Every Thing ...

Jason M. Mahan - 1839 - 312 páginas
...denomination, reduce both to the lowest in either, and the third to its lowest denomination mentioned. Multiply the second and third terms together, and divide the product by the first : the quotient will be the answer to the question, in the same denomination you left the third term...
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