12167 (23 root.

2X2X2=8

4167

2

2x300x2x\4167

30=1260

1200x3=3600

60X33=540

3×33=27

416.7

but the value of 8 is 8000, and the 2 is 20, that is, 8000 feet of the stone will make a pile measuring 20 feet on each side, and (12167-8000) 4167 feet remain to be added to this pile in such a manner as to continue it in the form of a cube. Now it is obvious that the addition must be made upon 3 sides; and each side being 20 feet square, the surface upon which the additions must be made will be (20 203=22300) 1200 feet, but when these additions are made, there will evidently be three deficiencies along the lines where these additions come together, (20 feet long, or 20 3=230 =) 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 (41671260) 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, (12003 =) 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 (60X33) 540 fest. 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 (3×3×3×3=) 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 process may be summed up in the following

RULE.

271. 1. Having distinguished the given number into periods of three figures each, find the greatest cube in the left hand period, and place its root in the quotient. Subtract the cube

from the left hand period, and bring down the next period for a dividend. Multiply the square of the quotient by 300, calling it the triple square, and the quotient by 30 calling it the triple quotient, and the sum of these call the divisor.

Seek how often the divisor may be had in the dividend, and place the result in the quotient. Multiply the triple square by the last quotient figure, and write the product under the dividend; multiply the triple quotient by the square of the last quotient figure, and place this product under the last; under these write the cube of the last quotient figure, and call their sum the subtrahend. Subtract the subtrahend from the dividend, and to the remainder bring down the next period for a new dividend, with which proceed as before; and so on till the whole is finished.

QUESTIONS FOR PRACTICE.

2. What is the cube root of 1815748?

2?

5. What is the cube root of Ans. 1.25+

The decimals are obtained by

1×1300=300|1815748(122 | annexing ciphers to the remainder,

30 30 1

as in the square root, with this difference, that 3 instead of 2 are an nexed each time.

6. What is the cube root of 27054036008? Ans. 3002.

7. What is the cube root

272. Solids of the same form are in proportion to one another as the cubes of their similar sides or diameters.

1. If a bullet weighing 72 es in diameter ? lbs, be 8 inches in diameter, what is the diameter of a bullet weighing 9 lbs. ?

72: 839: 464 Ans. 4 in.

2. A bullet 3 inches in diameter weighs 4 lbs. what is the weight of a bullet 6 inch

3x3x3=28 & 6×6×6=216 ib. Thus 27: 4 :: 216. Ans. 32 lbs.

3. If a ball of silver 12 inches in diameter be worth $600, what is the worth of another ball, the diameter of which is 15 inches?

Ans. $1171.87-+

EXTRACTION OF ROOTS IN GENERAL.

ANALYSIS.

273. The roots of most of the powers may be found by repeated extractions of the square and cube root. Thus the 4th root is the square root of the square root; the sixth root is the square root of the cube root, the 8th root is the square root of the 4th root, the 9th root is the cube root of the cube root, &c. The roots of high powers are most easily found by logarithms. If the logarithm of a number be divided by the index of its root, the quotient will be the logarithm of the root. The root of any power may likewise be found by the following

RULE.

274. Prepare the given number for extraction by pointing off from the place of units according to the required root. Find the first figure of the root by trial, subtract its power from the first period, and to the remainder bring down the first figure in the next period, and call these the dividend. Involve the root already found to the next inferior power to that which is given, and multiply it by the number denoting the given power for a divisor. Find how many times the divisor may be had in the dividend, and the quotient will be another figure of the root. Involve the whole root to the given power; subtract it from the given number as before, 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 till the whole is finished.

REVIEW.

MISCELLANEOUS RULES.

1. Arithmetical Progression.

275. When numbers increase by a common excess, or decrease by a common difference, they are said to be in Arithmetical Progression. When the numbers increase, as 2, 4, 6, 8, &c. they form an ascending series, and when they decrease, as 8, 6, 4, 2, &c. they form a descending series. The numbers which form the series are called its terms. The first and last term are called the extremes, and the others the means.

276. If I buy 5 lemons giving for the first 3 cents, for the second 5, for the third 7, and so on with a common difference of 2 cents; what do I give for the last lemon?

Here the common difference, 2, is evidently added to the price of the first lemon in order to find the price of the last, as many times less 1, (3+ 2+2+2+2=11 Ans.) as the whole number of lemons, Hence,

I. The first term, the number of terms, and the common difference given to find the last term.

RULE. Multiply the number of terms less I, by the common difference, and to the product add the first term.

2. If I buy 60 yards of cloth, and give for the first yard 5 cents, for the next 8 cents, for the next 11, and so on, increasing by the common difference, 3 cents, to the last, what do I give for the last yard? 59x3=177, and 177+5= 182cts. Ans.

3. If the first term of a series be 8, the number of terms 21, and the common difference 5, what is the least term? 20548 108 Ans.

277. If I buy 5 lemons, whose prices are in arithmetical progression, the first costing 3 cents, and the last, 11 cents, what is the common difference in the prices?

Here 11-3-8, and 5-1-4; 8 then is the amount of 4 equal differences, and 4) 8 (=2, the common difference. Hence,

II. The first term, the last term, and the number of terms 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.

16*