multiplied by the length of each, (20 feet,) or, as there are 3, by 3 x 20=80, or, which is easier in practice, we may multiply the 2 (tens), in the root, 20, by 30, making 60, and this product by 42 = 16, the square contents = 960 solid feet.

We do the same in the operation, by multiplying the 2 in 20 by 30=60 x 4 X 4 = 960 solid feet, as before; this 960 we write under the 4800, for we must add the several products together by and by, to know if our cube will contain all the required feet.

By turning over the block, with all the additions of the blocks marked B and C, which are now made to A, we shall spy a little square space, which prevents the figure from becoming a complete cube. The little block for this corner is marked D, which the pupil will find, by fitting it in, to exactly fill up this space. This block D is exactly square, and its length, breadth, and thickness are alike, and, of course, equal to the thickness and width of the Cs, that is, 4 feet, the last quotient figure; hence, 4 ft. X 4 ft. x 4 ft. 64 solid feet in the block D; or, in other words, the cube of 4, (the quotient figure,) wlrich in the same as 43 = 64, as in the operation. We now write the 64 under the 960, that this may be reckoned in with the other additions.

We next proceed to add the solid contents of the Bs, Cs, and D together, thus, 4500 +960 + 645824, precisely the number of solid feet which wo bad remaining after we deducted 8000 feet, the solid contents of the cube A. If, in the operation, we subtract the amount, 5824, from the remainder of dividend, 5824, we shall see that our additions have taken all that remained after the first cube was deducted, there being no remainder.

The last little block, when fitted in, as you saw, rendered the cube complete, each side of which we have now found to be 20+4=24 feet long. which is the cube root of 13824 (solid feet); but let us see if our cube containe the required number of solid feet.

In Operation 2d, we see, by neglecting the ciphers at the rigir. of 8, the 8 ta still 8000, by its standing under 3 (thousands); hence, we may point off three figures by placing a dot over the units, and another over the thousands, and

so on.

From the preceding example and illustrations we derive the following

RULE.

Divide the given number into periods of three figures each, by placing a point over the unit figure, and every third figure from the place of units to the left, in whole numbers, and to the right in decimals.

Find the greatest cube in the left-hand period, and place its root in the quotient.

Subtract the cube thus found from the said period, and to the remainder bring down the next period, and call this the dividend.

Multiply the square of the quotient by 300, calling it the divisor.

Find how many times the divisor is contained in the dividend, and place the result in the root (quotient); then multiply the divisor by this quotient figure, placing the product under the dividend.

Multiply the former quotient figure, or figures, by 30, and this product by the square of the last quotient figure, and place the product under the last; under these two products place the cube of the last quotient figure, and call their amount 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, until the whole is finished.

Note 1. When the subtrahend happens to be larger than the dividend, the quotient figure must be made one less, and we must find a new subtrahend. The reason why the quotient figure will be sometimes too large, is, because this quotient figure inerely shows the width of the three first additions to the original cube; consequently, when the subsequent additions are made, the width (queuent figure) may make the solid contents of all the additions more than the cubic feet in the dividend, which remain after the solid contents of the original sube are deductel.

2. When we have a remainder, after all the periods are brought down, we may continue the operation by annexing periods of ciphers, as in the square

3. When it happens that the divisor is not contained in the dividend, a cipher must be written in the quotient (root), and a new dividend formed by bringing dewn the next period in the given sum.

9. What is the cube root of 17576? 30. What is the cube root of 571787? 11. What in the cube root of 970299? 12. What is the cube root of 2000376? 13. What is the cube rout of 3796416? A. 156. 14. What is the cube root of 94818816? A. 456. 15. What is the cube root of 175616000?

A. 198

4. 560.

16. What is the cube root of 7-8613312? A. 908.

17. What is the cube root of 731189187729? A. 9009.

18. What is the cube root of 27? 4. 2.

19. What is the cube root of § 4. §.

Ithe root be a surd, reduce it to a decimal before its root is extracted, as ir the square root.

23. What is the length of one side of a cubical block, which contains 172 solid or cubic inches? A. 12.

24. What will be the length of one side of a cubical block, whose contents shall be equal to another block 32 feet long, 16 feet wide, and 8 feet thick?

3√32 x 16 x 8 = 16 feet, Ans.

25. There is a cellar dug, which is 16 feet long, 12 feet wide, and 12 feet deep; and another, 63 feet long, 8 feet wide, and 7 feet deep; how many solid or cubic feet of earth were thrown out? and what will be the length of one side of a cubical mound which may be formed from said earth? A. 5832. 18.

26. How many solid inches in a cubical block which measures 1 inch on each side? How many in one measuring 2 inches on each side? 3 inches on each side? 4 inches on each side? 6 inches on each side? 10 inches on each side? 20 inches on each side? A. 1. 8. 27. 64. 216. 1000, 8000. 27. What is the length of one side of a cubical block, which contains I solid of eubic inch? 8 solid inches? 27 solid inches? 64 solid inches? 125 solid aches? 216 solid inches? 1000 solid inches? 8000 solid inches ?

A. 1. 2. 3. 4. 5. 6. 10. 20.

By the two preceding examples, we see that the sides of the cube are an the eube roots of their solid contents, and their solid contents as the cubes of their sides. It is likewise true, that the sold contents of all similar figures are in proportion to each other as the cubes of their several sides or diameters.

Note. The relative length of the sides of cubes, when compared with their Bolid contents, will be best illustrated by reference to the cubical blocks socompanying this work.

28. If a ball, 3 inches in diameter, weigh 4 pounds, what will a ball of the same metal weigh, whose diameter is 6 inches?

33 : 63 :: 4 : 32: Ratio, 23 x 4 = 32 lbs., Ans. 99. If a globe of silver, 3 inches in diameter, be worth $160, what is the value of one 6 inches in diameter? 33: 63 :: $160 : $1280, Ans.

30. There are two little globes; one of them is 1 inch in diameter, and the other 2 inches; how many of the smaller globes will make one of the larger?

A. 8.

31. If the diameter of the planet Jupiter is 12 times as much as the diameter of the earth, how many globes of the earth would it take to make one as large as Jupiter? A. 1728.

32. I the sun is 1000000 times as large as the earth, and the earth is 9008 smiles in diameter, what is the diameter of the sun? A. 800000 miles.

Note. The roots of most powers may be found by the aquare and cube roots aly; thus the square root of the square root is the biquadrace, or fourth rest, the sixth root is the cube of this square root.

ARITHMETICAL PROGRESSION.

¶ LXXXVIII. Any rank or series of numbers mor than 2, increasing by a constant addition, or decreasing by a constant subtraction of some given number, is called an Arithmetical Series, or Progression

The number which is added or subtracted continually is called the common difference.

When the series is formed by a continual addition of the * common difference, it is called an ascending series; thus, 2, 4, 6, 8, 10, &c., is an ascending arithmetical series; but 10, 8, 6, 4, 2, &c., is called a descending arithmetical series, because it is formed by a continual subtraction of the common difference, 2.

The numbers which form the series are called the terms of the series or progression. The first and last terms are

called the extremes, and the other terms the means.

In Arithmetical Progression there are reckoned 5 terms, any three of which being given, the remaining two may be found, viz.

1. The first term.

2. The last term.

3. The number of terms.
4. The common difference.
5. The sum of all the terms.

The First Term, the Last Term, and the Number of Terms, being given, to find the Common Difference ;—

1. A man had 6 sons, whose several ages differed alike; the youngest was? years old, and the oldest 28; what was the common difference of their ages! The difference between the youngest son and the eldest evidently shows the Increase of the 3 years by all the subsequent additions, till we come to 28 years; and, as the number of these additions are, of course, I less tha.. the number of man (5), it follows, that, if we divide the whole difference (28 -3 =), 25, by the number of additions (5), we shall have the difference between each one separately, that is, the common difference.

Thus, 28

-

325; then, 25+5=5 years, the common difference.
A. 5 years.

Hence, To find the Common Difference ;

Divide the difference of the extremes by the number of terms, less 1, and the quotient will be the common difference.

8. Ifthe extremes be 3 and 23, and the number of terms 11, whai ki ( mea difference? 4. 9.

3. A man is to travel from Boston to a certain place in 6 days, and to go only miles the first day, increasing the distance travelled each day by an equal excess, so that the last day's journey may be 45 miles; what is the daily inBrease, that is, the common difference? A. 8 miles.

4. If the amount of $1 for 20 years, at simple interest, be $2,20, what is the te per cent.?

In this example, we see the amount of the first year is $1,06, and the last year $2,20; consequently, the extremes are 106 and 220, and the number of term 90. 4. $,066 per cent.

5. A man hought 60 yards of cloth, giving 5 cents for the first yard, 7 for the second, 9 for the third, and so on to the last; what did the last cost?

Since, in this example, we have the common difference given, it will be easy to find the price of the last yard; for, as there are as many additions as there yards, less 1, that is, 59 additions of 2 cents, to be made to the first yard, follows, that the last yard will cost 2 x 59 118 cents more than the first, and the whole cost of the last, reckoning the cost of the first yard, will be 118+5= $1,23. A. $1,23.

Hence, When the Common Difference, the First Term, and the Number of Terms, are given, to find the Last Term ;—

Multiply the common difference by the number of terms, less 1, and add the first term to the product.

6. If the first term be 3, the common difference 2, and the number of terms 11, what is the last term? A. 23.

7. A man, in travelling from Boston to a certain place in 6 days, travelled the first day 5 miles, the second 8 miles, travelling each successive day 3 miles farther than the former; what was the distance travelled the last day? A. 20.

8. What will $1, at 6 per cent., amount to, in 20 years, at simple interest? The common difference is the 6 per cent.; for the amount of $1, for 1 year, in $1,06, and $1,06+$,06 $1,12, the second year, and so on. A. $2,20.

9. A man bought 10 yards of cloth, in arithmetical progression; for the first yard he gave 6 cents, and for the last yard he gave 24 cents; what was the amount of the whole?

In this example, it is plain that half the cost of the first and last yards will be the average price of the whole number of yards; thus, 6 cts. +24 cts. =30+ 315 cts., average price; then, 10 yds. X 15150 cts. $1,50, whole cost. A. $1,50.

Hence, When the Extremes, and the Number of Terms, given, to find the Sum of all the Terms ;

Multiply half the sum of the extremes by the number of terms, and the product will be the answer.

10. If the extremes be 3 and 273, and the number of terms 40, what is the suMD of all the terms? A. 5520.

11. How many times does a regular clock strike in 12 hours? A. 78.

12. A butcher bought 100 oxen, and gave for the first ox $1, for the second 22, for the third $3, and so on to the last; how much did they come to at task sate? A. $5050.

13. What is the sum of the first 1000 numbers, beginning with their natumi erder, 1, 2, 3, &c.? A. 500500.

14. If a board, 18 feet long, be 2 feet wide at one end, and come to a point at the other, what are the square contents of the board? 4. 18 feet.