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Multiply the second and third terms together, and divide their product by the first term , the quotient will be the answer...
The Modern House-carpenter's Companion and Builder's Guide: Being a Hand ... - Seite 164
von William Allen Sylvester - 1883 - 210 Seiten
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## Daboll's Schoolmaster's Assistant: Improved and Enlarged : Being a Plain and ...

Nathan Daboll - 1813 - 240 Seiten
...third terms to the same denomination, and reduce the second term to the lowest name mentioned in it. 3. Multiply the second and third terms together, and divide their product by the first term; and the quotient will be the answer to the question, in the same denomination you left the second term...
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## Daboll's Schoolmaster's Assistant: Improved and Enlarged, Being a Plain ...

Nathan Daboll - 1815 - 240 Seiten
...terms to the same denomination, aiid reduce the second term to the lowest name ir.antioned in it. 3. Multiply the second and third terms together, and...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...
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## Daboll's Schoolmaster's Assistant: Improved and Enlarged. Being a Plain ...

Nathan Daboll - 1818 - 240 Seiten
...terms to the same denomination, and reduce the second term to the lowest naiae mentioned in it. 3. Multiply the second and third terms together, and...product by the first term, the quotient will be the answer to the question, in fue same denomination you left the second term in, which may be brought...
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## Arithmetic Modernised: Or, A Complete System of Arithmetic, Adapted to ...

John Davidson, Robert Scott (writing master) - 1818 - 172 Seiten
...write the greater as thefirst, and the other as the second proportional. The numbers being thus stated, multiply the second and third terms together, and divide their product by thefirst, the quotient will be the answer, in ihe same name in which the third term is9 when you arc...
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## A New and Concise System of Arithmetick: Containing Vulgar, Decimal, and ...

Beriah Stevens - 1822 - 423 Seiten
...extreme is the divisor. 3. Place the divisor on the left hand, and the other extreme on the right ; then multiply the second and third terms together, and divide their product by the first, ami the quotient gives the answer ; which is always of the same name with the middle term. When the...
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## Daboll's Schoolmaster's Assistant: Improved and Enlarged, Being a Plain ...

Nathan Daboll - 1824 - 240 Seiten
...third terms to the same denomination, and reduce the second term to the lowest name mentioned in it. 3. Multiply the second and third terms together, and...their product by the first term ; the quotient will b« the answer to the question, in the same denomination you left the second term in, which may be...
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## Elements of Arithmetic

Etienne Bézout - 1824 - 219 Seiten
...consequently, for performing the Simple Rule of Three Direct, a« explained in article (179), is as follows ; Multiply the second and third terms together and divide their product by the first ; the quotient will be the answer, or fourth term sought. -I he following examples will explain the...
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## Daboll's Schoolmaster's Assistant, Improved and Enlarged

Nathan Daboll - 1825 - 240 Seiten
...second term to the lowest name mentioned in it. 3. Multiply the second and third terms together, and j 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...
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## The tutor's expeditious assistant; being a system of practical arithmetic ...

John White - 1826 - 108 Seiten
...first and third terms into the same denomination, and the second into the lowest name mentioned. 2. Multiply the second and third terms together, and divide their product by the first, and their quotient will be the answer, in the same denomination as that in which the second was left....
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## The Juvenile Arithmetick, and Scholar's Guide: Wherein Theory and Practice ...

Martin Ruter - 1828 - 166 Seiten
...it. Reduce, likewise, the first and second terms to the lowest denomination that either of them has. Then multiply the second and third terms together,...divide their product by the first term. The quotient thus obtained will be the answer. It will not be necessary to distinguish between direct and inverse...
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