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the length of one side into itself will show the surface of one side, and this multiplied by 3, the number of sides, gives the contents of the surface of the three sides. Thus 30 X30=900, which multiplied by 3=2700 feet.

Now as we have 5768 solid feet to be distributed upon a surface of 2700 feet, there will be as many feet in the thickness of the addition, as there are twenty-seven hun. dreds in 5768. 2700 is contained in 5768 twice; there. fore 2 feet is the thickness of the addition made to each of the three sides.

By multiplying this thickness, by the extent of surface (2700 X2) we find that there are 5400 solid feet contained in these additions.

[blocks in formation]

But if we examine Fig. 2, we shall find that these additions do not com. plete the cube, for the three corners a a a need to be filled by blocks of the same length as the sides (30 feet) and of the same breadth and thickness as the previous additions (viz. 2 feet).

Now to find the solid contents of these blocks, or the number of feet required to fiil these corners, we multi. ply the length, breadth, and thickness of one block together, and then multiply

this product by 3, the num. ber of blocks. Thus, the breadth and thickness of each block has been shown to be 2 feet; 2X2=4, and

this multiplied by 30 (the 30 length=120, which is the

solid contents of one block. But in three, there will be

three times as many solid 30

feet, or 360, which is the number required to fill the deficiencies.

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30

1

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Fig. 3.

In other words, we square 30

the last quotient figure (2) 30

22 multiply the product by the

first figure of the quotient (3 tens) and then multiply the last product by 3, the 'number of deficiencies.

But by examining Fig. 3, 30

it appears that the figure is

not yet complete, but that a 30

small cube is still wanting, 30

where the blocks last added meet. The sides of this small cube, it will be seen, are each equal to the width of these blocks, that is, 2 feet. If each side is 2 feet long, the whole cube must contain 8 Fig. 4.

solid feet (because 2x2x2 32 feet.

=8), and it will be seen by Fig. 4, that this just fills the vacant corner, and

completes the cube.

32 We have thus found, 32

that the additions to be made around the large cube (Fig. 1) are

as fol. 32

lows.

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5400 solid feet upon three sides, (Fig. 2).
360

to fill the corners a a a.
8 “ to fill the deficiency in Fig. 3.
Now if these be added together, their sum will be 5768
solid feet, which subtracted from the dividend leave no
remainder and the work is done. 32 feet is therefore the
length of one side of the given cube.

The proof may be seen by involving the side now found to the third power, thus ; 32X32X32=32768; or it may be proved by adding together the contents of the several parts, thus,

27000 feet=contents of Fig. 1.
5400 66 =addition to the three sides.

360" =addition to fill the corners a a a.

8 " =addition to fill the corner in Fig. 3.

32768 Proof. From these illustrations we see the reasons for the fol. lowing rule.

RULE FOR EXTRACTING THE CUBE ROOT.

1. Point off the given number, into periods of three figures each, beginning at the right.

2. Find the greatest cube in the left hand period, and subtract it from that period. Place the root in the quotient, and to the remainder bring down the next period, for a divi. dend.

3. Square the root already found (understanding a cipher at the right) and multiply it by 3 for a divisor.

Divide the dividend by the divisor, and place the quotient for the next figure of the root.

4. Multiply the divisor by this quotient figure. Multiply the square of this quotient figure by the former figure or fi. gures of the root, and this product by three. Finally cube this quotient figure, and add these three results together for a subtrahend.

5. Subtract the subtrahend from the dividend. To the remainder bring down the next period, for a new dividend, and proceed as before.

If it happens in any case, that the divisor is not con. tained in the dividend, or if there is a remainder after the last period is brought down, the same directions may

be observed, that were given respecting the square root. (See page 251.)

Repeat the process of illustration. What is the rule for extracting the cube root ?

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EXAMPLES.

What is the cube root of 373248 ?

373249(72
343

70% X3=14700)30248(First Dividend.

29400 22 X 70 X3= 840

235 8

30248 Subtrahend.

0000 Find the cube root of 941,192,000. A. 980. Of 958,585,256. A. 986. Of 478,211,763. A. 782. Of 494,913,671. A. 791. Of 445,943,744. A. 764. Of 196,122,941. A. 581. Of 204,336,469, A, 589. Of 57,512,456. A. 386. Of 6,751,269. A. 189.

Of 39,651,821. A. 341. Of 42,508,549. A. 349. Of 510,082,399. A. 799. Of 469,097,433. A. 777.

Find the cube root of 7. A. 1.912933. Of 41. A. 3.448217. Of 49. A. 3.659306. Of 94. A. 4.546836. Of 97. A. 4.610436. Of 199. A. 5.838272. Of 179, A. 5.635741. Of 389. A. 7.299893. Of 364. A. 7.140037. Of 499. A. 7.931710. Of 699. A. 8.874809. Of 686. A. 8.819447. Of 886. A. 9.604569. Of 981. A. 9.936261.

The cube root of a fraction, is obtained by extracting the root of numerator and denominator, but if this cannot be done, it may be changed to a decimal, and the root extracted.

Find the cube root of 137: A. Of 244. A. 11. Of 199923. A. 77. Of 2014. A. 1. Of 19348417. Ans. 273.

Find the cube root of. A. .8549879. Of is. A. .5593445. A. .4578857.

A. .4562903. Of 131. A. .9973262.

How is the cube root of a fraction obtained ?

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2 57

ARITHMETICAL PROGRESSION, Any rank, or series of numbers, consisting of more than two terms, which increases or decreases by a common difference, is called an Arithmetical series, or progression.

When the series increases, that is, when it is formed by the constant addition of the common difference, it is called an ascending series, thus,

1, 3, 5, 7, 9, 11, &c. Here it will be seen that the series is formed by a continual addition of 2 to each succeeding figure.

When the series decreases, that is, when it is formed by the constant subtraction of the common difference, it is called a descending series, thus,

14, 12, 10, 8, 6, 4, &c. Here the series is formed by a continual subtraction of 2, from each preceding figure.

The figures that make up the series are called the terms of the series. The first and last terms are called the extremes, and the other terms, the means.

From the above, it may be seen, that any term in a se. ries may be found by continued addition or subtraction, but in a long series this process would be tedious. A much more expeditious method may be found.

1. The ages of six persons are in arithmetical progres. sion.

The youngest is 8 years old, and the common dif, ference is 3, what is the age of the eldest ? In other words, what is the last term of an arithmetical series, whose first term is 8, the number of terms 6, and the common differ. ence 3 ?

8, 11, 14, 17, 20, 23. Examining this series, we find that the common difference, 3, is added 5 times, that is one less than the number of terms, and the last term, 23, is larger than the first term, by five times the addition of the common difference, three; Hence the age of the elder person is 8+3 X5=23.

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

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

When are numbers said to be in arithmetical progression ?

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