31. What is the cube root of 33? 3/33 = 3/27=&= 1}, Ans. 32. What is the cube root of 417? 64 ? 74088 336. APPLICATION OF THE CUBE ROOT. 1. Solids which are of the same form and have their like lines proportional are similar; thus, if one of two rectangular solids has its length 4 feet, its breadth 2 feet and its thickness 1 foot, and the other solid has its length 12 feet, its breadth 6 feet and its thickness 3 feet, those two solids are similar. 2. The like lines or parts of two similar solids or figures are called homologous lines or parts. 3. By Geometry it is easily proved that the solidities, i. e., the solid contents of all similar solids are to each other as the cubes of their homologous lines; thus, the solidities of two spheres are to each other as the cubes of their radii, as the cubes of their diameters, or as the cubes of their circumferences, etc., etc.; the solidities of cubes are to each other as the cubes of their edges; etc., etc. Ex. 1. How many lead balls Ans. 64. of an inch in diameter will be required to make a ball 1 inch in diameter? 2. If a man dig a cubical cellar whose edge is 5 feet in one day, how long will it take him to dig a similar cellar whose edge is 25 feet? Ans. 125 days. 3. Suppose the diameter of the sun is 886144 miles and that of the earth 7912 miles, how many bodies like the earth will make one as large as the sun? Ans. 1404928. 4. If an iron ball 5 inches in diameter weighs 16lb., what is the weight of an iron ball 20 inches in diameter? Ans. 1024lb. 5. If a globe of gold 1 inch in diameter is worth $100, what is the diameter of a globe worth $2700? Ans. 3 inches. 6. A, B, C and D own a conical sugar loaf which is 16 inches high and weighs 161b.; what part of the height shall each take off in the order A, B, C and D, so that each shall take 4lb.? Ans. A, 10.079 in.; B, 2.620in.; C, 1.837in.; D, 1.464in. 7. A half-peck measure is 9 inches in diameter and 4 inches deep; what are the dimensions of a similar measure that will hold a bushel? Ans. 18 by 8 inches. 8. A rectangular bin, containing 327680 cubic inches, has its width, height and length in the ratio of 1, 2 and 5; what are its dimensions? Ans. 32in. wide; 64in. high; 160in. long. 9. What is the edge of a cubical box whose solidity is equal to that of a bin whose length, breadth and height are respectively 144, 36 and 9 inches? Ans. 36 inches. 10. Suppose 1000 bodies like the earth are required to make 1 like Saturn and that the diameter of Saturn is 79000 miles; what is the diameter of the earth? Ans. 7900 miles. 11. Four spheres have their solidities to each other in the ratio of the numbers 1, 2, 3 and 4; the diameter of the largest sphere is 5 inches. What is the radius of the smallest and what the successive increase of the radii of the 2d, 3d and 4th? Ans. § 40. TO EXTRACT A ROOT OF ANY DEGREE. 337. RULE.—1. Point off the given number into periods of as many figures each as there are units in the index of the required root, by placing a dot over units, etc. 2. Find by trial, or by the table of powers (312), the greatest power of the same name as the root in the left hand period, and place its root as the first figure of the required root. 3. Subtract the power from the first period and to the remainder annex the first figure of the next period, for a dividend. 4. For a trial divisor, involve the part of the root already found to a power whose index is one less than that of the required root and multiply this power by the index of the root. 5. Divide, and the quotient will be the second figure of the root or something greater. 6. Involve the part of the root found, to a power of the same name as the root, subtract the power from the first two periods, and to the remainder annex the first figure of the next period, for a new dividend. 7. Find a new trial divisor and proceed in a similar manner until the entire root is obtained. NOTE 1. The left hand period may be incomplete. If, in pointing a decimal, the right hand period is incomplete, annex one or more ciphers. - Sections 4 and 5 of the rule aid in finding the successive root figures; still each must be found by trial. NOTE 2. NOTE 3. The last involution in solving a question is, at the same time, a proof of the work. NOTE 4. This rule is founded in Algebra, and cannot be easily explained to pupils unacquainted with that science. Ex. 1. What is the 4th root of 390625 ? Trial Divisor 23 × 4=32 ) 230 Dividend. 254390625 Subtrahend. 0 2. What is the 5th root of 1282388557824? 1282388557824 (264, Ans. 25= 32 1st Trial Divisor 24 x 5=80) 962, 1st Dividend. = 265—11881376, 2d Subtrahend. 2dTrialDiv. 26*X 5-2284880) 9425095, 2d Dividend. 26451282388557824, 3d Subtrahend. == √√625 = Ans. 24810. = NOTE 5.- Such roots as the 4th, 6th, etc., i. e., those roots whose indices are composite numbers may be more easily found by taking a root of a root; thus, the */625 √25 = 5. Again, the 6/64 = 3√√√64 : 3/8=2; or thus, °√64 = √√√√√/64√42; but roots whose exponents are prime numbers, as the 5th, 7th, 11th, etc., cannot be extracted in this way. = § 41. ARITHMETICAL PROGRESSION. 338. Any series of numbers increasing or decreasing by a common difference is said to be in ARITHMETICAL PROGRESSION; thus, 2, 5, 8, 11, etc., is an ascending series, and 30, 25, 20, 15, etc., is a descending series. 339. The several numbers forming a series are called terms; the first and last terms, extremes; the others, means. The difference between any two successive terms is the common differ ence. 340. In Arithmetical Progression, five particulars claim special attention :· 1st. The first term. 2d. The last term. 3d. The common difference. 4th. The number of terms. 5th. The sum of all the terms. 341. These particulars are so related to each other that if any three of them are given the other two can be found. 342. Twenty cases may arise in Arithmetical Progression, but it will be sufficient to notice a few of the more important ones. 343. In an ascending series, let 3 be the first term and 4 the common difference; Then, 3+ 31st term. 7 2d term. 3+4+43 + 2 × 4 = 11 = 3d term. 3+4+4+4 = 3 + 3 × 415 3+4 3+4 +4 +4 +4 = 3 + 4 X 419 4th term. 5th term. 3+4+4+4+4+4=3+ 5 × 4 = 23 = 6th term. Again, in a descending series, let 30 be the first term and 3 the common difference; Thus we see that, in an ascending series, the second term is found by adding the common difference once to the first term ; the third term, by adding the common difference twice to the first term; and, generally, any term is found by adding the common difference as many times to the first term as there are terms preceding the one sought. A similar explanation may be given when the series is descending. Hence, |