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" Then multiply the second and third terms together, and divide the product by the first term: the quotient will be the fourth term, or answer. "
Mathematical Dictionary and Cyclopedia of Mathematical Science Comprising ... - Page 80
by Charles Davies, William Guy Peck - 1859 - 592 pages
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Mathematics for Practical Men: Being a Common-place Book of Principles ...

Olinthus Gregory - 1863 - 482 pages
...same denomination ; and if the second term be a compound one, reduce it to the lowest name mentioned. 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 to which you reduced the second term....
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The civil service arithmetic. [With] A key, Book 1

Robert Johnston (F.R.G.S.) - 1863 - 254 pages
...20 159610 ing to the Rule, the first and second 7 terms to shillings, and then multiply 2,0)111727,0 the second and third terms together, and divide the product by the first 12)558631g( = i) term 20, which gives 55863| the 2,0)465,5 3| ""s• in Pence- £232 15s. 3±d. Ans....
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Elements of Geometry, and Plane and Spherical Trigonometry: With Numerous ...

Horatio Nelson Robinson - Conic sections - 1865 - 474 pages
...D = B x C byJi, , * B x C we have D = — ^ — . That is, to find the fourth term of a proportion, multiply the second and third terms together and divide the product by the first term. This is the Kule of Three of Arithmetic. This equation shows that any one of the four terms can be...
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University Arithmetic: Embracing the Science of Numbers, and General Rules ...

Charles Davies - Arithmetic - 1865 - 468 pages
...first place, and when less, write the greater there, and the remaining term in the second place : III. Then multiply the second and third terms together, and divide the product by the first. This rule gives, when quantity and cost are considered ; quantity : quantity : : cost : cost. When...
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An Elementary Arithmetic Serving as an Introduction to the Higher Arithmetic

George Roberts Perkins - Arithmetic - 1865 - 360 pages
...Having written the three terms of the proportion, or, ax usually expressed, having stated the question, then multiply the second and third terms together, and divide the product by the fast term. NOTE. — Since there is a ratio between the first and second lerms, they must be reduced...
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Practical Arithmetic ...

John Purdue Bidlake - Arithmetic - 1866 - 232 pages
...terms to the same name, and the third, if necessary, to the lowest denomination mentioned in it. 5th. Multiply the second and third terms together and divide the product by the first ; the quotient will be the answer in the same name that the third term was reduced to. The workiny...
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The advanced lesson book, by E.T. Stevens and C. Hole

Edward Thomas Stevens - 1866 - 434 pages
...the remaining term in the first place. Reduce the first and second terms to the same denomination. Multiply the second and third terms together, and divide the product by the first. The quotient will be the answer required. COMPOUND PROPORTION. — Find the quantity which is of the...
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The Crittenden Commercial Arithmetic and Business Manual: Designed for the ...

John Groesbeck - Arithmetic - 1868 - 358 pages
...denomination, and if the third term is a compound number, reduce it to the lowest term mentioned in it. Then multiply the second and third terms together,...product by the first term: the quotient will be the fourth term, or answer. EXAMPLES. 1. If 25 barrels of flour cost $165, what will 35 barrels cost? 25...
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The Crittenden Commercial Arithmetic and Business Manual: Designed for the ...

John Groesbeck - Arithmetic - 1868 - 350 pages
...denomination, and if the third term is a compound number, reduce it to the lowest term mentioned in it. Then multiply the second and third terms together,...product by the first term: the quotient will be the fourth term, or answer. EXAMPLES. 1. If 25 barrels of flour cost $165, what will 35 barrels cost? 25...
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The New Federal Calculator, Or Scholar's Assistant: Containing the Most ...

Thomas Tucker Smiley - 1868 - 238 pages
...second terms to the same denomination, and to the lowest denomination mentioned in 3ither of them. 3. Multiply the second and third terms together, and divide the product by the first term ; the result will be he fourth term, or answer, in the same denomination to vhich the third term was reduced....
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