17. Reduce 37 poles to the decimal of a mile. Ans. 115625. 18. Reduce 3 hours, 30 minutes to the decimal of a day. Ans. 14583'. 19. Reduce 15 minutes, 30 seconds to the decimal of an hour. Ans. 2583'. 20. Reduce 3cwt. 1qr. 7lbs. to the decimal of a ton. Ans. 165625' 21. If the diameter of a circle be 1, the circumference is 34 nearly, or 3 more nearly. Express each of these decimally.* Ans. 3.142857', and 3.1415929, &c. 11 22. If the circumference of a circle be 1, and the diameter is nearly, or more nearly. Express each of these decimally. 3'18', and 308309859, &c. 23. An American acre is, 12 the equivalent decimal. of an Irish acre required Ans. 61734693877551, &c. 24. The pound troy is 14 of a pound avoirdupois : required the equivalent decimal. Ans. 822/85714'. PROBLEM II. To find the value of a given decimal in the parts contained in the integer. 125. RULE.-Multiply the given decimal by the numbers which, if it were an integer, would reduce it to the lower denominations contained in it, and, after each multiplication, point off for decimals as many figures toward the right hand as there were figures in the given decimal. The figures remaining on the left will express the required value. Example 5. Required the value of 3945 of a day in hours, &c. * The circumference true to 20 places, is 3.14159265358979323S46. 10000 Here the given decimal is multiplied by 24, the number of hours in a day, and four figures being cut off towards the right hand, it appears that 3945 day is equal to 96% or 96 hours. The decimal 4680 is then multiplied by 60, and four figures being cut off, there results 28.0800 or 28.08 minutes. By continuing the process, the value of the given decimal is found to be 9 hours, 28 minutes, 4 seconds, 48 thirds.* .3945 day. 15780 7890 9.4680 hours. 60 28.0800 min. 4.8000 sec. 48.0000 thirds. 6. Required the value of 805' of a yard in long mea sure. •805' of a yard 3 2.4166 12 4.9999 inches. This, and similar exercises, may be wrought either by converting the proposed decimal into a common fraction, in the way shown in page 150, or more easily by employing an ap proximate process, as in the margin. Here we carry 1 to the product of 3 and 5, because had the decimal been continued farther, it is evident 1 must have been carried from the preceding product. For a similar reason, 7 is carried to the product of 12 and 6. The result is found to be 2 feet, 4.9 inches. (see page 149.) Another 5 was added to the given decimal, that the result might be more distinct and certain. Exercises. Required the values of the following decimale. 1. Required the value of 0675 of a cwt. Ans. 724lbs. 4625 of a ton. 2. 3. £.5937. Ans. 9cat. 1qr. Ans. 11s. 10rd. 10000 *With respect to the reason of this process, it is only necessary to observe, that it is exactly the same as finding the value of 35 day by problem VII. page 132, the pointing off of decimals serving the pur pose of dividing by the denominator. 4. Required the value of 8845 of an acre. Ans. 3 roods, 211p. ⚫00213 of a day. Ans. 3 min. 4125 sec. 2'85714' of a cwt. •4' foot, long measure. Ans. 5 inches, 4 lines. 13. What is the value of 67 of a league. Ans. 2m. 3p. 1yd. 3žin. 14. What is the value of 61 of a tun of wine. Ans 2hhd. 27gal. 2qt. 1pt. Addition of Decimal Fractions. 126. RULE.—Arrange the given numbers so that the separating points may be all in the same column. Find the sum as in addition of whole numbers. Point off as many of the decimals, as there are in the given number which contains the most. If any of the given numbers contain periodical decimals, let these be carried out to as many places as there are in the longest line of the finite decimals; or, if greater accuracy be required, let them be carried as far as may be judged necessary. * * In the application of decimals to practical purposes, it is generally known from the nature of the case under consideration, to how many places it is necessary, that the result may be true. When a result is thus required to be true to an assigned number of places of decimals, it is proposed to carry the decimals, which consist of more places, to at least one place beyond the assigned number, and to reject the last figure. In this case, it is proper to observe, that when a decimal is not carried out to its full length, the last figure of the part retained, should be increased by a unit, if the succeeding figure be 5 or greater than 5. 81.4632 9.75 47.388 Example 1. Add together 81.4632, 9.75, and 47.388. Here, the numbers are arranged as in the margin, and added as in addition of whole numbers. The reason of the arrangement and operation is quite manifest, those figures being added together which are of the same local value. 138.6012, sum. 3.73737 -873 51-77778 108.2 2. Add together 3-7'3', -873, 51-7', 108-2, and 73.463128; so that the result may have four places of decimals true. In this example, the first, third, and fifth numbers are carried to five places each, and the last figure of the third is made 8, because the next figure would be 7. In like manner the fifth figure of the last line is made 3, because the succeeding figure is 8.* 73 46313 238.0513, sum. Exercises.-1. Add together 1·83, 5·674, 3125, 18.3, 100, 38-62, 4.3957, and ⚫5. Ans. 169.6222. 2. Required the sum of 93-617843, 7·836, 12.25, 71375, 4.391, 7-839, 3.7674285, and ⚫8693. Ans. 131.2843215. 3. Required the sum of 7354, 7354′, 735'4', '7'354', 07354, and ⚫0735'4'. Ans. 3.088857'991'. 4. Required the sum, true to five places, of the numbers given in exercises 5th, 6th, and 13th of Addition of Fractions, the several fractions being previously reduced to de cimals. Ans. 6-0078125, 3907'14285', and 1.561011'90476, Subtraction of Decimal Fractions. 127. RULE.-Set the less number so that each figure in it may stand below a figure of the same local value in the greater. Then find the difference as in subtraction of whole numbers, and place the separating point as in addition of decimals. * The reason of this is evident, since 30 is nearer 28 than 20 is, and 30, by the rejection of the last figures, becomes 3. In the addition, the sum of the last column is 18, from which 2 is carried, because 18 is nearer 20 than 10. The correct sum, found by carrying the decimals farther, is 239-0512795'1', which by retaining only four figures of the decimal, and increasing the last of them by a unit, because it is fol lowed by 79, &c. becomes 238-0513, the same as before. Example 1. From 3.5/4' take 1.34265. Here, the greater number is extended, and the remainder is found to be 2.202804'5'. 3.5454545 1.34265 2.202804'5' diff. From 8.6 Rem. 5.8222 2. Required the difference of 8.6 and 2.7 Here, the less number is carried to four places, that the true answer may be discovered with greater certainty. In the subtraction, ciphers are conceived to be annexed to the greater number, and 1 is carried to the repeating figure first used, because this must have been done, had the less number been carried one place farther. The answer is found to be 5.82'. Exercises.-1. Required the difference of 3.468 and Ans. 2.2089. 1.2591. 2. Required the difference of 6.45 and 1.34'5'. Ans. 5.104'5'. 3. Required the difference of 13.6′ and 4.345. Ans. 9.3216'. 4. Required the difference of 682 and 09647. Ans. 58553. 5. Required the difference of 5.83 and 4.1'7'. 128. RULE.-Multiply the factors as in multiplication of whole numbers, and point off in the product as many of the decimals as there are in both factors, sup * In this exercise and the next two, the given fractions are to be reduced to decimals, and the difference taken according to the rule. |