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41. What is the square root of 538 to the nearest hundredth ?

42. What is the value of V71 to the nearest thousandth ?

43. What is the value of 70.002 to the nearest ten-thousandth ?

376. To extract the root of a fraction,

First change it to smallest terms; then, if both terms are perfect squares, take the root of each ; otherwise change the fraction to a decimal, and find the root.

Mixed numbers may be changed either to improper fractions or to mixed decimals.

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How many

57. What is the square root of f to the nearest thousandth ?

58. What is the square root of td to the nearest thousandth ?

59. A general has an army of 226576 men. must he place rank and file to form them into a square

? 60. A gentleman has 3 fields, one containing 3 acres 1 square rod, another 5 acres 69 square rods, and the third 6 acres 91 square rods. He wishes to exchange them for a square field of equal area. Find the side of the

square

field ?

CUBE ROOT.

377. To extract the Cube Root of a number is to find one of the three equal factors which produce it.

378. The cube of a number contains three times as many figures as the root, or three times as many less one or two. Thus,

13 = 1. 103 = 1'000. 1003 = 1'000'000. 43 = 64. 253 = 15'625. 2433 14'348'907. 93 = 729. 993 = 970'299. 9993 = 997/002/999.

379. If a third power is separated into periods of three figures each, beginning with the decimal point, the number of periods will show the number of figures in the root. Thus,

The cube root of 4'080'659.192 contains one decimal and three integral figures.

380. The cube of the highest order of units in the root is found in the highest period of the power. Thus,

9 tens, or 903 = 729 thousands, or 729'000.
9 hundreds, or 9003 = 729 millions, or 729'000'000.

381. Taking any number composed of tens and ones, as 36, separate it into its tens and ones, cube it, keeping the products distinct, and we have 36

30 + 6 36=

30 + 6 216

(30 X 6) + 62 108

(30 X 6) 1296 =

302 + 2 (30 X 6) + 62

30 + 6
7776

(302 X 6) + 2 (30 X 62) + 63
3888 303 + 2 (302 x 6) + (30 X 62)
46656 = 303 + 3 (302 X 6) + 3 (30 x 62) + 63

302 +

36 =

That is, the cube of any number that can be separated into tens and ones equals

382. The cube of the tens, plus three times the product of the square of the tens and the ones, plus three times the product of the tens and the square of the ones, plus the cube of the ones.

This statement may be represented by the following formula in which initial letters are used, and the sign of multiplication is the period.

tö + 3 t2.0 + 30.0% + 03.

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Solution. — This power has two periods ; hence its root has two figures, tens and ones (Art. 379). The cube of the tens is in the highest period (Art. 380). The greatest tens: in 405 thousands is 343000, which, subtracted, and its cube root placed at the right, leaves 62224, which equals “3 ta xo + 30 x 0 + 03." 62224 consists principally of 3 ta x 0; hence, if we divide it by 3 t, or 702 X 3 = 14700, we shall have the ones, which we find to be 4. Finding 3t X o, or 70 X 3 X 4= 840, and oʻ, or 42 = 16, and adding them to 14700, we have 14700 + 840 + 16 = 15556, which equals 3 to + 3t xoto?. Multiplying this by the ones, 4, we have 62224, or 3 ta x 0 + 3 t X 0% + 03. Hence we conclude that 74 is the cube root required.

Geometrical Explanation of Cube Root

383. As the length of a cube is the cube root of its contents, the method of extracting the cube root of a number may be illustrated by the process of ascertaining the length of a cube, its contents being given.

62. It is required to find the length of a cube containing 13824 cu. in.

The length of a cube containing 13824 cu. in. cannot be as much as 30 in., for a 30-inch cube contains 27000 cu. in. It must be more than 20 in., for a 20-inch cube contains only 8000 cu. in. The length of the given cube, then, must be between 20 and 30 inches.

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Removing from the given cube, A, a 20-inch cube, B, containing 8000 cu. in., there remains a solid containing 5824 cu. in. An inspection of this solid shows that it is largely made up of three rectangular solids, C, D, E, having square faces corresponding in area to the faces of

202 the cube removed. It is evident that the thickness of these solids added to the length of the cube removed will be the length of the given cube. Now the thickness of a rectangular solid is found by dividing its contents by the area of its face (Art. 224). We find the area of one of the square faces of

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each of these rectangular solids to be 20', or 400 sq. in., and the area of their combined square faces to be 202 X 3, or 1200 sq. in.

Dividing the approximate contents of these solids, 5824 cu. in., by this area of their square faces, 1200 sq. in., we have as their probable thickness 4 inches.

Removing the three rectangular solids, C, D, E, there remain three smaller rectangular solids, F, G, H, whose length is evidently that of the

cube B removed, and whose width is the thickness of the larger rectangular solids. The area of one face

of one of these solids is 20 X 4, and the area of one face of all of them is 20 X 4 X 3, or 240 square inches.

Removing these three smaller solids, there remains the little cube I, whose face corresponds in area to the end of one of the smaller rectangular solids removed, or 4*, or 16 square inches.

Combining the area of the faces of

the six rectangular solids with that of a face of the little cube, we have a total area of 1200 + 240 + 16, or 1456 sq. in. Multiplying the area of one face of these seven

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solids by their thickness, 4, we have 4 x 1456, or 5824 cu. in., which is the number of cubic inches remaining of the original cube, A, after the removal of the cube B. We therefore conclude that the length of the given cube is 24 inches.

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