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 Βιβλία Βιβλία 131 - 140 από 160 για Then multiply the second and third terms together, and divide the product by the.... 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. Ruger's Arithmetick, with Questions and Answers: A New System of Arithmetick ... - Σελίδα 242
των William Ruger - 1841 - 263 σελίδες
Πλήρης προβολή - Σχετικά με αυτό το βιβλίο ## Crittenden Commercial Arithmetic ...

John Groesbeck - 1872
...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, and divide the product by the first term: the quotient will be the fourth term, or answer. EXAMPLES. 1. If 25 barrels of flour cost \$165,... ## A progressive arithmetic for elementary schools, ed. by F. Young. 6 pt. [in ...

Francis Young (F.R.G.S.) - 1872
...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. Ex. 1. — If... ## Common School Arithmetic: Combining the Elements of the Science with Their ...

John Homer French - 1872 - 335 σελίδες
...the third term ; and for the second, when it is to be greater. IH. For the fourth or unknown term, multiply the second and third terms together, and divide the product by the first'term. NOTES. — 1. When the terms of the first ratio are of different denominations, they must... ## Elementary arithmetic. [With] Answers, Βιβλίο 1

H T. Sortwell - 1873
...place. Step III. If necessary, reduce the first and second terms to the same denomination. Step IV. Multiply the second and third terms together, and divide the product by the first. It Example (1). If a man can walk 112 miles in 7 days, how far will he walk in 48 days 1 , .. Here,... ## Arithmetic for Schools

Barnard Smith - 1873 - 347 σελίδες
...second terms together for a final second term, and retain the former third term. In this final stating 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 denomination to which the third term was reduced."... ## Advanced arithmetic. [With] Key by W.M. Mansfield

1874
...reduce them to one denomination before proceeding any further. II. RULE FOR SOLVING THE PROPORTION.— Multiply the second and third terms together, and divide the product by the first. Ex. 1. If 6 men earn 42s., how much will 36 men earn ? Here the answer must evidently be in wages,... ## The High School Arithmetic: Containig All the Matter Usually Presented in a ...

Philotus Dean - 1874 - 454 σελίδες
...when the answer should be greater than the third term; otherwise write the less far the second term. Multiply the second and third terms together, and divide the product by the first term. NOTE 1. — If the number sought is made any other term of the proportion than the fourth, each... ## A Treatise on Arithmetic

James Hamblin Smith, Thomas Kirkland, Scott, William, b. 1845 - 1877 - 345 σελίδες
...place; if less, in the first. Then having reduced the first and second terms to the same denomination, multiply the second and third terms together, and divide the product by the first term. The quotient will be the answer required. NOTE. — After the third term has been written down... ## New Elementary Algebra: Designed for Common and High Schools and Academies

Shelton Palmer Sanford - 1879 - 332 σελίδες
...Arithmetic, where we have three terms given to find a fourth. The rule as practised in Arithmetic is to " multiply the second and third terms together, and divide the product by the first ;" which is equivalent to saving that The last extreme is equal to the product of the tiro means, divided... ## Elementary Arithmetic for Canadian Schools

Barnard Smith, Archibald McMurchy - 1879 - 192 σελίδες
...second terms together for a final second term, and retain the former third term. In this final stating 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 denomination to which the third term was reduced....