The First Part of the United States Arithmetic Designed for Schools

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E.C. & J. Biddle, 1857 - 112 pages

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Page 61 - TABLE. 4 nails, (na.) or 9 inches, make 1 quarter, marked qr. 4 quarters, or 36 inches, - 1 yard, - - - - yd. 3 quarters, ------ 1 ell Flemish, - - E. Fl 5 quarters, ------ 1 ell English, - - EE 6 quarters, ------ 1 ell French, - - E. Fr 27.
Page 47 - Multiply the last remainder by the preceding divisor, or last but one, and to the product add the preceding remainder ; multiply this sum by the next preceding divisor, and to the product add the next preceding remainder ; and so on, till you have gone backward through all the divisors and remainders to the first.
Page 15 - ... any number divided by 9, will leave the same remainder as the sum of its figures, or digits, divided by 9, which may be thus demonstrated.
Page 62 - LIQUID MEASURE 4 gills (gi.) = 1 pint (pt.) 2 pints = 1 quart (qt...
Page 54 - Scale: 4 farthings (far.) = 1 penny (d.); 12 pence = 1 shilling (s.) ; 20 shillings — 1 pound (). 156.
Page 34 - The reason of this method is obvious ; for any number multiplied by the component parts of another, must give the same product as if it were multiplied by that number...
Page 63 - ... ..1 gallon, gal 36 gallons " .1 barrel, bar. 54 gallons
Page 40 - The logarithm of the product of two or more numbers is equal to the sum of the logarithms of those numbers. Let a denote the base of the system ; also, let m and n be any two numbers, and x and y their logarithms. Then, by the definition of logarithms, we have ax=m, (1.) a?in.
Page 91 - Carry the integers, thus found, to the product of the next higher denomination, with which proceed as before ; and so on, through all the denominations -to the highest; then this product, together with the several remainders, taken as one number, will be the whole amount required.
Page 49 - Divide the number by any prime number which will divide it without any remainder ; then divide the quotient in the same way, and so continue until a quotient is obtained which is a prime. Then will the successive divisors, together with the last quotient, be the prime factors required.

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