 | James Bates Thomson - Arithmetic - 1849 - 438 pages
...cost of 100, or 1000 articles, pounds, &c., is given, the price of one is found by simply removing the decimal point in the given cost or dividend, as many...1000 feet of hemlock boards cost $6.40, what will one /oot cost ? .- 30. Bought 42 cwt. of tobacco for $565.82 : what is that per cwt. ; and what per pound... | |
 | George Roberts Perkins - Arithmetic - 1849 - 346 pages
...•57". We may, obviously, divide any decimal by 10, 100, 1000, &c., by removing the decimal point as many places to the left as there are ciphers in the divisor : when there are not so many figures at the left of the decimal point, we may prefix ciphers. 10 100... | |
 | George Roberts Perkins - Arithmetic - 1850 - 364 pages
...2-223+. 57. We may, obviously, divide any decimal by 10, 100, 1000, &c., by removing the decimal point as many places to the left as there are ciphers in the divisor ; when there are not so many figures at the left of the decimal point, we may prefix ciphers. 1-212.... | |
 | George Roberts Perkins - Arithmetic - 1851 - 356 pages
...2-223+. Sf. We may, obviously, divide any decimal by 10, 100, 1000, &c., by removing the decimal point as many places to the left as there are ciphers in the divisor; when there are not so many figures at the left of the decimal point, we may prefix ciphers. 10 100... | |
 | Calvin Tracy - 1851 - 214 pages
...(See 4th sum.) § 90, — To divide a decimal by 10, 100, 1000, &c. RULE. Remove the decimal point as many places to the left as there are ciphers in the divisor. 1. Divide 30515.50 by 100. Ans. 305.1550. 2. Divide 36.5 by 10. Ans. 3.65. 3. Divide 36.10 by 100.... | |
 | John Hunter - Arithmetic - 1852 - 184 pages
...annexed, is used as a divisor, the quotient may be represented by merely shifting the decimal point of the dividend as many places to the left as there are ciphers in the divisor. Thus, * 860-=- 100 = 860-0 -r- 100 = 8'6; 45 -4- 1000 = 45-0 -4- 1000 = -045; 23-47 -h 10000 = -002347.... | |
 | David Henry Cruttenden - Arithmetic - 1853 - 330 pages
...remainders as if added to the dividend. If the divisor be 10 or 100, etc., simply removing the point as many places to the left, as there are ciphers in the divisor, completes the division. (See Case V., Remark.) 17. Divide .192816 by .312. Ans. .618. 18. Divide 1.28... | |
 | William Frederick Greenfield - 1853 - 228 pages
...the point, then (Cor. 1 Prop. 37) ciphers must be supplied. Hence the Rule : remove the decimal point as many places to the left as there are ciphers in the denominator, supplying ciphers if necessary. Prop 41. — To prove the Rule for Multiplication of Decimals.... | |
 | George Roberts Perkins - Arithmetic - 1855 - 388 pages
...2-223+. 57. We may, obviously, divide any decimal by 10, 100, 1000, &c., by removing the decimal point as many •places to the left as there are ciphers in the divisor ; when there are not so many figures at the left of the decimal point, we may prefix ciphers. 10 100... | |
 | John Fair Stoddard - Arithmetic - 1856 - 312 pages
...2448 by '012 ART. 152. A decimal may be divided by 10, 100, 1000, &c., by removing the decimal point as many places to the left as there are ciphers in the divisor. If necessary, prefix ciphers to the dividend. f4'36 1 Divide ;3o:42; \ by w. [ 431-2 J T 146-34 T Divide... | |
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... dividend, as many places to the left as there are ciphers in the divisor. 
