each of the sides a, b, and c. But the root, 2, (tens,) already found is the length of one of these sides; we therefore square the root 2 (tens)=20 x 20 = 400, for the square contents of one side, and multiply the product by 3, the number of sides; 400 x 3 1200: or, which is the same, in effect, and more convenient in practice, we may square the 2 (tens) and multiply the product by 300, thus, 2x2 = 4, and 4 X 300 1200, for the divisor, as before.

OPERATION-CONTINUED.

13824(24 Root.
8

The divisor, 1200, is contained in the dividend 4 times, consequently, 4 feet is

Divisor, 1200)5824 Dividend. the thickness of the addition made to

each of the three sides, a, b, c. And 4 X 1200 4800, is the solid feet contained in these additions; but, if we look at Fig. II, we shall perceive that this addition to the 3 sides, does not complete the cube; for there are deficiencies in the 3 corners n, n, n. Now the length of each side, that is, 2 (tens) = 20, and their width and thickness, are each equal to the last quotient figure (4); their contents, therefore, or the number of feet required to fill these deficiencies, will be found by multiplying the square of the last quotient figure, (42)=16, by the length of all the deficiencies, that is, by 3 times the length of each side, which is expressed by the former quotient figure 2 (tens,) 3 times (2 tens) is 6 (tens) = = 60; or, which is the same, in effect, and more convenient in practice, we may multiply the quotient figure 2 (tens) by 30, thus, 2 X 30

60, as before, then 60 X 16=960 contents of the three deficiencies n, n, n.

20

Looking at Fig. III., we perceive there is still a deficiency in the corner where the last blocks meet. This deficiency is a cube, each side of which is equal to the last quotient figure 4. The cube of 4, therefore, (4 x 4 x 464,) will be the solid contents of this corner, which in Fig. IV, is seen filled.

Now the sum of these several additions, viz., 4800+960 + 645824, will make the subtrahend, which, subtracted from the dividend, leaves no remainder, and the work is done.

263. From the foregoing example and illustration we have the following

RULE

FOR EXTRACTING THE CUBE ROOT.

I. POINT OFF the given number, from unit's place, into PERIODS OF THREE FIGURES each.

II. FIND the greatest cube in the left hand period, and put its ROOT in the quotient.

III. SUBTRACT the root thus found from the said period, and to the remainder bring down the next period, and call the sum which it makes the DIVIDEND.

IV. MULTIPLY the square of the quotient by 300, and call it the DIVISOR.

V. SEEK how often the divisor may be had in the dividend, and place the result in the quotient; multiply the divisor by this quotient figure, and write the product under the dividend.

VI. MULTIPLY the square of this quotient figure by the former FIGURE, OF FIGURES, of the root, and this product by 30, and place the product under the last; under ALL write the cube of the last quotient figure, and call their sum the SUBTRAHEND.

VII. SUBTRACT the subtrahend from the dividend, and proceed as before.

NOTE 1. If the divisor cannot be had in the dividend, put a cipher in the root, bring down the next period for a new dividend, and proceed as before directed.

2. If necessary, the root may be carried to decimals, by annexing three ciphers to each subtrahend, and continuing the operation.

What is a cube? What is the ROOT of a cube? What is it to extract the cube root? Why is the square of the quotient multiplied by 300? Why, in finding the subtrahend, do we multiply the square of the last quotient figure by 30 times the former figure or figures of the root? Why do we cube the quotient figure? How may we prove the operation?

EXAMPLES FOR PRACTICE.

1. How many solid feet are there in a cubical box, one side of which measures 2 feet? Ans. 8 feet. 2. How many cubic feet in one 8 times as large? and what would be the length of one side? Ans. 64, and one side would be 4 feet. 3. What is the side of a cubical pile, equal to one 576 feet long, 432 feet wide, and 96 feet high? Ans. 288 feet.

4. There is a cubical box, measuring 1 foot each way; what is the side of a box 8 times as large? 27 times? -64 times?

Ans. 2, 3, and 4 feet. 264. Hence, it appears, that the sides of cubes are as the cube roots of their solid contents, and consequently, their conTENTS are as the CUBES of their SIDES. The same is also true of the SIMILAR SIDES, or of the DIAMETERS of all solid figures of SIMILAR FORMS.*

5. There is a cubical box, one side of which is 5 feet; what would be the side of one containing 27 times as much? 64 times as much? 125 times as much? Ans. 15, 20, and 25. 6. If a ball, weighing 4 pounds, be 3 inches in diameter, what will be the diameter of a ball of the same metal, weighing 32 pounds?

4 : 32 :: 33 : 216, and 3/216 = 6 in. Ans.

7. If a ball, 6 inches in diameter, weigh 32 pounds, what will be the weight of a ball 3 inches in diameter ? Ans. 4lbs. 8. If a cube of silver, whose side is 4 inches, be worth $125, what will be the side of one worth 4 times as much? Ans. 6.349in.+

9. If a globe of silver, of 3 inches diameter, be worth $100, what will be the value of another globe 1 foot in diameter ?

Ans. $6400.

10. There are two globes; one of them is I foot in diameter, and the other 30 feet: how many of the smaller globes will it take to make one of the larger? Ans. 27000.

Ans. 343in.

11. There is a cubical vessel, whose side is 2 feet; I demand the side of a vessel which shall contain 3 times as much. 12. If the diameter of the sun be 112 times that of the earth, how many globes like the earth would it take, to make one as large as the sun?

Ans. 1404928. 13. If the planet saturn is 1000 times larger than the earth, and the earth is 7900 miles in diameter, what is the diameter of saturn?

Ans. 79000 miles. 14. If the British national debt be 13 hundred millions of pounds sterling, what would be the diameter of a globe of gold of equal value, allowing a cubic inch of the metal to be worth £53 2s. 8d.? +

Ans. 360.199944492 inches.

ARITHMETICAL PROGRESSION. 265. ARITHMETICAL PROGRESSION is any rank or series of numbers, more than two, INCREASING OF DECREASING by a constant difference.

266. If the numbers increase, they are said to form an ASCENDING SERIES; but if they decrease, they form a DESCEND

ING SERIES.

Thus,

3, 5, 7, 9, 11, 13, &c. is an ascending series. 13, 11, 9, 7, 5, 3, &c. a descending series.

...

267. The numbers which form the series are called the

*How are the sides of cubes proportional?

their CONTENTS?

* The solidity of a globe divided by .5236 gives the cube of the diameter.

TERMS of the series; of which the first and last, are called the EXTREMES, and the other terms the MEANS.

268. There are FIVE THINGS in arithmetical progression, any three of which being given, the other two may be found. 1st. The FIRST term.

2nd. The LAST term.

3d. The NUMBER of terms.

4th.

The COMMON DIFFERENCE.

5th. The sUM OF ALL THE TERMS.

1. Bought 10yds. of cloth, giving 3 cents for the first yard, 5 for the second, 7 for the third, and so on; what did the last yard cost me ? The common difference, 2, being added to every term except

1st. term 2d. 3d.

4th. 5th.

7th.

8th.

9th.

10th.

6th. 3+2=5+2=7+2=9+2=11+2=13+2=15+2=17+2=19+2=21.

the last, it is plain, the last yard will be 9 X 2 18 cents more than the first. Therefore 3+18 21 cents, Ans.

Hence, when the first term, the common difference, and the number of terms, are given, to find the last term,

269. RULE. Multiply the number of terms LESS ONE, by the common difference, and to the product add the first term. 2. A person travelling into the country, went 3 miles the first day, 8. the second, and so on, for 12 days; how far did he travel the last day? Ans. 58 miles. 3. A man had 8 sons, whose ages differed alike, the youngest was 4 years old, and the common difference 4; what was the age of the eldest? Ans. 32 years.

4. What will be the amount of $1 for 50 years, at 6 per cent.? 270. It is plain, the yearly amounts, at simple interest, form an arithmetical progression, of which the principal is the first term, the last amount the last term, the yearly interest the common difference, and the number of years 1 less than the number of terms. Ans. $4.

5. A gentleman bought 100yds. of Osnaburgh, and by agreement was to give what the last yard would come to, reckoning 2s. for the first, 5 for the second, 8 for the third, and so on; what did it cost him?

Ans. £14 19s.

6. Bought 10yds. of cloth in arithmetical progression, giving for the first yard 3 cents, and for the last 21; what was the common difference? This being the reverse of example 1st, we proceed with it inversely; thus, 21-3= 18, and 1892, Ans.

Hence, when the extremes and number of terms are given, to find the common difference,

271. RULE. Divide the difference of the extremes by the number of terms LESS ONE, and the quotient will be the answer. 7. If the extremes be 7 and 315, and number of terms 78; what is the common difference? Ans. 4. 8. A man had 8 sons in arithmetical progression, the youngest was 4 years old, and the eldest 32; what was the common difference of their ages? Ans. 4 years. 9. If A puts out $1, at simple interest, for 50 years, and receives at the end of the time $3; what is the common difference, or rate per cent.? Ans. $.06.