evidently be three deficiencies along the lines where these additions come together, (20 feet long, or 20×3=2 × 30=) 60 feet, which must be filled in order to continue the pile in a cubic form. Thus the points upon which the additions are to be made, are (1200+60=) 1260 feet and 4167 feet, the quantity to be added divided by 1260, the quotient is (4167 1260) 3, which is the thickness of the additions, or the other figure of the root. Now if we multiply the surface of the three sides by the thickness of the additions, the product, (1200 X 3 ) 3600 feet, is the quantity of stone required for those additions. Then to find how much it takes to fill the deficiencies along the line where these additions come together, since the thickness of the additions upon the sides is 3 feet, the additions here will be 3 feet square, and 60 feet, and the quantity of stone added will be (60×3×3=) 540 feet. But after these additions there will be a deficiency of a cubical form, at the corner, between the ends of the last mentioned additions, the three dimensions of which will be just equal to the thickness of the other additions, or 3 feet, and cubing 3 feet we find (3×3×3=) 27 feet of stone required to fill this corner, and the pile is now in a cubic form, measuring 23 feet on every side, and adding the quantities of the additions upon the sides, the edges, and at the corner together, we find them to amount to (3600+-540+27=) 4167 feet, just equal to the quantity remaining of the 12167 after taking out 8000. To illustrate the foregoing operation, make a cubic block of a convenient size to represent the greatest cube in the left hand period. Make 3 other square blocks each equal to the side of the cube, and of an indefinite thickness, to represent the additions upon the three sides, then 3 other blocks, each equal in length to the sides of the cube, and their other dimensions equal to the thickness of the square blocks, to represent the additions along the edges of the cube, and a small cubic block with its dimensions each equal to the thickness of the square blocks, to fill the space at the corner. These placed together in the manner described in the above operation, will render the reason of each step in the process perfectly clear. The learner may adopt the following general RULE. 1. Distinguish the given sum into periods of 3 figures each, beginning at the right hand. 2. Find the greatest cube in the left hand period, and write down its root in the place of the quotient in division. 3. Subtract the cube from the period, and bring down the next period at the right hand of the remainder; call this a dividend. 4. Square the quotient and multiply that product by 300. This is the triple square. Multiply the quotient by 30. This is the triple quotient. Add these products together for a divisor. 5. Ascertain how many times the divisor is contained in the dividend for the 2d figure of the root. 6. Multiply the triple square by the last quotient figure: then multiply the triple quotient by the square of the last quotient, and cube the last quotient, and add these products together and take the amount from the dividend. and to the remainder bring down the next period. Proceed as before, till the whole is finished. 1. What is the cube root of 15625? 15625(25 root 8 7625 dividend. 6000 The greatest cube in 15, the left hand period, is 8, of which the root is 2; and 2×2=4, and 4×300=1200, the triple square : 2×30=60, the triple quotient; their sum is 1260 the divisor, which is contained in the 7625, the dividend, 5 times. 1200×5=6000; the square of 5 is 25 and 60×25=1500, and the cube of 5 is 125; the sum of 7625=dividend. these products is 7625=the dividend. The work is therefore done. The cube root is 25. 1500 125 2. What is the cube root of 2? Ans. 1.25+ In this example ciphers must be added as decimals to the 2. II. Solids of the same form are to one another as the cubes of their diameters, or similar sides. 1. If a ball which weighs 72 lbs. is 8 inches in diameter, what is the diameter of one which weighs 9 lbs. ? 8X8X8=512. 72 : 9 :: 512 to the cube of the small64, the cube root of which is 4, the Answer. er ball 2. A ball 4 inches in diameter weighs 6 lbs.; what is the weight of one 8 inches in diameter ? 4X4X4 64 and 8X8X8-512. 64 : 6 :: 512 : 48 lbs., the Ans. To find two mean proportionals, between any two given numbers, Divide the greater by the less, and extract the cube root of the quotient. Multiply the least given number by the root for the lesser, and this product by the same root for the greater of the two numbers sought. 1. What are the two mean proportionals between 2 and 16? 162-88-2; and 2×2-4 the lesser, and 4X2=8, the greater. Proof. 2: 4 :: 8: 16. 2. What are the two mean proportionals between 6 and 162? Ans. 18 and 54. [The learner may derive important assistance, in understanding the extraction of the square root, by using pieces of pasteboard, so cut as to enable him to make the additions to the sides of a square which are required, to continue it in that shape. By the help of "the blocks," he will be better able to understand the principles involved in extracting the cube root, than he can, by any written demonstration.] EXTRACTION OF ROOTS IN GENERAL. It is not very often required in business transactions, to find the roots of other powers. As it may be useful, however, to be acquainted with an expeditious mode of extracting any large root, the learner may attend to the following RULES. 1. Point off according to the root specified if the fourth root is required, the period must have four figures; if the fifth, it must have five, &c. 2. Find the greatest root in the left hand period by trial; subtract its power from that period, and at the right hand of the remainder place the first figure of the next period. This is a dividend. 3. Involve the root already found to the power next less than that of which the root is required, multiply this by the number expressing the given power, and use this as the divisor of the dividend. 4. Ascertain how many times the dividend contains the divisor, and put the quotient at the right hand of the other figure of the root. 5. Involve the root already found to the given power, and subtract this from the given numbers. Bring down the first figure of the next period to the right hand of this remainder, for a new dividend, and proceed as before, till the work is finished. SECTION XXVII, ARITHMETICAL PROGRESSION. If any three numbers are in arithmetical progression, the sum of the two extremes must be equal to double the mean; and when four numbers are in progression, the sum of the two means must be equal to the sum of the two extremes. If a larger number are in progression, then the sum of the extremes must be equal to the sum of any two means equally distant from the extremes. For the reasons of which, see illustrations in the Rule of Three. Arithmetical progression is a series of numbers either increasing or decreasing by a common difference. Five things are to be considered, viz. the first and last terms, the number, common difference, and sum of all the terms. If any three of these are given, the others can be readily found. If I buy 6 apples, giving 2 cents for the first, 4 for the second, 6 for the third, and so on, increasing by a common difference of 2, what must I give for the last?-There are several ways in which I can obtain the answer. As the first costs 2 cents, the second 4, &c. I can add the price of the first to itself to make that of the second; the price of the first, or 2 cents, added to the price of the second, will make that of the third, and so on. But in large sums, this would be inconvenient. As we have the first term, the number of terms, and the common difference given, we can in this, and all similar cases, find the answer by the following RULE. Multiply the number of terms less 1 by the common difference, and to this product add the first term. The sum will be the answer. 2. If the first term be 8, common difference 4, and number of terms 10, what is the last term? RULE 2. Ans. 44. When the first term, the last term and number of terms are given, to find the common difference, Divide the difference between the first and last by the number of terms less 1, and the quotient must be the common difference. Example. I bought 12 books. The cheapest cost 2 dollars; the dearest 35. What is the common difference in the price? 35 11)33(3 Ans. RULE 3. If the first and last terms and common difference are given, to find the number of terms,— Divide the difference of the extremes (that is, the first and last terms) by the common difference, and the quotient increased by 1 will be the answer. Example. If the first term be 5, the last 35, and the common difference 3, what is the number of terms? 35 3)30(10 1 11 Ans. RULE 4. When the first, last and number of terms are given, to find the sum of all the terms or series, Multiply half the sum of the extremes by the number of terms, and the product will be the answer. |