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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 X 30=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.
32768(32 But if we examine Fig. 2, we shall 27
find that these additions do not com.
plete the cube, for the three corners 2700)5768 a a a need to be filled by blocks of the
same length as the sides (30 feet) and 5400
of the same breadth and thickness as 360
the previous additions (viz. 2 feet). 8
Now to find the solid contents of
these blocks, or the number of feet 5768 required to fill these corners, we multi
ply the length, breadth, and thickness 0000 of one block together, and then multiply Fig. 2.
this product by 3, the num30
ber of blocks. Thus, the 30
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 30
solid contents of one block. But in three, there will be
three times as many solid 30
feet, or 360, which is the 530
number required to fill the deficiencies.
In other words, we square 30
the last quotient figure (2) 30
12 multiply the product by the 2
first figure of the quotient (3 tens) and then multiply the last product by 3, the 30 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.
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, thus32X32X32=32768; or it may be proved by adding together the contents of the several parts, thus,
27000 feet=contents of Fig. 1.
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 of 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 figures 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 contained 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 ?
What is the cube root of 373248 ?
503 X3=14700)30248(First Dividend.
29400 22 X 70 X3= 840
23 = 8
0000 Find the cube root of 941,192,000. A. 980.
Of 958,585,256. A. 986. Of 478,211,768. 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.
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 ex. tracted.
Find the cube root of itt: A. t. Of ***+. A. ft. Of 96143. A. 71. Of 7991384 A. 14 246441. Ans. 37.
Find the cube root of s. A. .8549879. Of. A. .5393445. Of A. .4578857.
A. .4562903. Of 197. A. .9973262. How is the cube root of a fraction obtained ?
28 730 895
ARITHMETICAL PROGRESSION. Any rank, or series of numbers, consisting of more than two terms, which increases or decreases by a common dif. ference, 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 con. tinual 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
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.
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 ?