new resolvend; to which form a new divisor from the whole root now found, and thence another figure of the root as before, &c. 2. What is the cube root of 1092727 ? Ans. 103. 1. The statute bushel contains 2150,4197+ cubic or solid inches; I demand the side of a cubic box which shall contain just that quantity? 2150,419724-12,907+inch. Ans. NOTE.-The solid contents of similar figures, are in proportion to each other, as the cubes of their similar sides or diameters. TO EXTRACT THE ROOTS OF powers in GENERAL. 179 2. If a bullet 4 inches diameter weigh 9ft., what will a bullet of the same metal weigh, whose diameter is 8 inches? 4x4x4=64 8x8x8-512. As 64: 9 :: 512: 72t. Ans. 3. If a solid globe of silver, of 3 inches diameter, be worth $150, what is the value of another globe of silver, whose diameter is eight inches? 3x3x3=27. 8x8x8-512. As 27: $150 :: 512: $28441 Ans. The side of a cube being given, to find the side of that cube which shall be double, triple, &c. in quantity to the given cube. RULE.-Cube the given side, and multiply it by the given proportion between the given and required cube, and the cube root of the product will be the side sought. 4. If a cube of silver, whose side is 2 inches, be worth $20, what should the side of a cube of like silver be, whose value would be 8 times as much? 2x2x2=8, and 8x8-64. 64-4 inches. Ans. 5. There is a cubical vessel whose side is 4 feet; I demand the side of another cubical vessel, which shall contain 4 times as much? 4x4x4=64. & 64×4=256. 3/256=6,349+feet. Ans. 6. A cooper having a cask 40 inches long, and 32 inches at the bung diameter, is ordered to make another cask of the same shape, but which shall hold just twice as much; what will be the bung diameter, and length of the new cask? 40×40×40×2=128000; then 128000=50,3+ in ches length. Ans. 65536-40,3+ inches bung diam. Ans. TO EXTRACT THE ROOTS OF POWERS IN GENERAL. RULE.-1. Prepare the given number for extraction by pointing off from the place of units as the root required directs; 4th root put a dot over every 4th figure &c. from the place of units; 5th root, over every 5th, &c. from units' place, &c. 2. Find the first figure of the root by trial, and subtract its power from the given number. 3. To the remainder bring down the first figure in the next period, and call it the dividend. 4. Involve the root to the next inferiour power to that which is given, and multiply it by the number denoting the given power, for a divisor. 5. Find how many times the divisor may be had in the dividend, and the quotient will be another figure of the root. 6. Involve the whole root to the given power, and subtract it from the given number as before. 7. Bring down the first figure of the next period to the remainder for a new dividend, to which find a new divisor, and so on, until the whole is finished. EXAMPLES. 1. What is the cube root of 53157376? 53157376(376 root. 32x3=27)261 dividend. 50653=373 372x3=4107)25043 second dividend. 53157376=3763 2. What is the biquadrate root of 34827998976 ? Ans. 431,9.+ 3. What is the sursolid root of 281950621875 ? Ans. 195. 4. What is the square cubed, or sixth root of 16196, 005304479729 ? Ans. 5,03. 5. Find the seventh root of 34487717467,30751. Ans. 32,01.+ NOTE. The roots of most powers may be found by the square and cube roots only; therefore, when any even power is given, the better way will be, especially in very high powers, to extract the square root of it, which reduces it to half the given power, then the square root of that power; and so on till it comes to a square or cube. For example, suppose a 12th power be given; the square root of that reduces it to a sixth power; and the square root of a sixth power to a cube. 6. Extract the eighth root of 7213895789838336. Ans. 96. 7. What is the biquadrate root of 5308416? Ans. 48. ARITHMETICAL PROGRESSION. ANY rank of numbers more than two, increasing by a common excess, or decreasing by a common difference, is said to be in Arithmetical Progression: such are the numbers 1, 2, 3, 4, &c. 7, 5, 3, 1; and ‚8, ‚6, ‚4, ‚2. When the numbers increase, they form an ascending series; but when they decrease, they form a descending series. The numbers which form the series, are called the terms of the progression. Any three of the five following terms being given, the other two may readily be found. 1st. The first term, 2d. The last term, commonly called the extremes. 3d. The number of terms. 4th. The common difference. 5th. The sum of all the terms. PROBLEM 1. The first term, the last term, and the number of terms being given, to find the sum of all the terms. RULE.-Multiply the sum of the extremes by the number of terms, and half the product will be the answer. EXAMPLES. 1. The first term of an arithmetical progression is 1, the last term 21, the number of terms 11; required the sum of the series. 21 2. How many strokes does a Venice clock strike in the compass of a day, going to 24 o'clock ? Ans. 300. 3. If 100 stones be placed in a right line, a yard distant from each other, and the first a yard from a basket; what distance will that man go who gathers them up singly, returning with them one by one to the basket? Ans. 5 miles and 1300 yards. 4. A draper sold 100 yards of cloth at 5cts. for the first yard, 10cts. for the second, 15 for the third, &c., increasing 5cts. for every yard; what did the whole amount to, and what did it average per yard? Ans. Amount was $2524, and the average price was $2,52cts. 5m. per yard. PROBLEM II. The extremes and the number of terms being given, to find the common difference. RULE.-Divide the difference of the extremes by the number of terms less 1, and the quotient will be the common difference. EXAMPLES. 1. The extremes are 3 and 19, and the number of terms is 9; required the common difference, and the sum of the whole series. 2. A man is to travel from Boston to a certain place in 12 days, and to go but three miles the first day, increasing every day by an equal excess, so that the last day's journey may be 58 miles; required the daily increase, and the distance of the place from Boston. Ans. Daily increase 5, distance 366 miles. 2. A man had 12 sons whose several ages differed alike; the eldest was 49, the youngest 5 years old; what was the common difference of their ages? Ans. 4 years. |