SOLUTION.-We find the first term of the root, the first trial divisor, and the first true divisor, as in the ordinary method.

To find the second trial divisor, we take the sum of the square of 2, the true divisor, and the previous correction, and we have 8112. We find the next true divisor by adding the usual corrections to the trial divisor, and have 820596.

We find the third trial divisor by taking the sum of the square of 6, the previous true divisor, and the corrections, and have 830028. We find the next true divisor as before, etc.

7. 277167808.

Ans. 652. 12. 4.080659192. Ans. 1.598 8. 1878080904. Ans. 1234. 13. 16503.467336. Ans. 25.46.

APPLICATIONS OF CUBE ROOT.

649. The Applications of cube root to problems involving geometrical volumes, such as cubes, parallelopipedons, spheres, etc., are extensive.

650. The Edge of a cube is equal to the cube root of its contents.

EXAMPLES FOR PRACTICE.

1. What are the dimensions of a cubical chest which shai, contain 64000 cubic feet? Ans. 40 ft.

2. Required the number of square feet in one face of a cubical block whose contents are 405224 cu. ft. Ans. 5476. 3. What is the entire surface of a cube whose cubical contents are 91125 cubic feet? Ans. 12150 sq. ft.

4. What is the edge of a cube which shall contain as much as a solid 20 ft. 6 in. long, 10 ft. 8 in. high?

wide, and 6 ft. 9 in.

Ans. 11.4 ft.-.

5. What is the depth of a cubical cistern which shall contain 200 gal. (231 cu. in.) of water?

6. A farmer had a cubical bin which els of grain; what was its depth?

7. What would it cost to plaster the & cubical reservoir which contains 100 6 cents a square foot?

Ans. 35.9 in.-. contained 50 bush.

Ans. 3.962 ft.+. bottom and sides of barrels of water, at Ans. $16.85.

SIMILAR VOLUMES.

651. Similar Volumes are such as have the same shape, but differ in size; as, cubes, spheres, etc.

652. A Dimension of a volume is a length, breadth, height, diameter, radius, circumference, etc.

653. The Principles of similar volumes are derived from geometry.

PRINCIPLES.

1. Similar volumes are to each other as the cubes of their like dimensions.

2. Like dimensions of similar volumes are to each other as the cube roots of those volumes.

1. A man has two balls, one 6 in. in diameter, the other 2 in.; the first is how many times as large as the second?

OPERATION. 1st 2d 63: 23 1st=2dx()=2dx33 1st = 2dX27

SOLUTION.-By the principle above we have the proportion 1st: 2d::63:23; and since the first term equals the 2d term multiplied by the ratio of the 3d to the 4th, we have 1st term =2d term multiplied by the ratio of 63 to 23, which is 2d term X (), or 2d x33, or 2d× 27. Hence the first is 27 times as large as the second.

EXAMPLES FOR PRACTICE.

2. Required the relation of two cubes whose dimensions are 3 in. and 15 in respectively. Ans. 2d is 125 times 1st. 3. If a ball 3 in. in diameter weigh 9 pounds, how much will a ball 4 in in diameter weigh? Ans 214 1h

4. If a cubical box 4 ft. long, hold 51.43 bu. of grain, how much will a cubical box 6 ft. long hold? Ans. 173.58 bu. 5. If a haystack 12 feet in diameter contain 15 tons, what is the diameter of a similar stack of 120 tons? Ans. 24 ft. 6. If a man 5 ft. high weigh 150 lb., what is the weight of a man of similar build whose height is 6 ft? Ans. 2591lb.

7. The sun is 885680 miles in diameter, and the earth 1912 miles; the sun is how many times as large as the earth? Ans. About 1123.

8. There are two balls whose diameters are respectively 3 in. and 4 in.; what is the diameter of a ball whose contents are equal to them both? Ans. 4.5 in. nearly.

SUG.-Cube 3 and 4, take their sum, and then compare this with either of the given balls.

654. Any root whose index contains only the factors 2 or 3 can be extracted by means of the square and cube root, according to the following principle:

PRINCIPLE.

A root of a number equals a root of a root of the number, in which the product of the indices of the two latter roots equals the index of the former.

Since the square of the cube of a number equals the sixth power, the sixth root of a number equals the square root of the cube root of the number, and the same is true in any other case.

1. Extract the sixth root of 4096.

SOLUTION. To find the sixth root of 4096 we first extract the square root, which we find to be 64, and then find the cube root of 64, which is 4. Hence the sixth root of 4096 is 4.

NOTE. For a general rule of Evolution see Brooks's Higher Arithmetic

SECTION XI.

SUPPLEMENT.

The Supplement contains additional matter for advanced classes only. It should generally be omitted, or, if deemed necessary, such parts of it may be need as the interests of the pupils require.

THE PROGRESSIONS.

655. A Series is a succession of numbers, each derived from the preceding by some fixed law.

656. The Terms of a series are the numbers which compose it. The Extremes are the first and last terms; the Means are the terms between the extremes.

657. An Ascending Series is one in which the terms increase from left to right; a Descending Series is one in which the terms decrease from left to right.

ARITHMETICAL PROGRESSION..

658. An Arithmetical Progression is a series of numbers which vary by a common difference; as 3, 5, 7, 9, etc. 659. The Common Difference is the difference between any two consecutive terms; thus, in the above series the common difference is 2.

660. The Quantities considered are five, any three of which being given, the others may be found.

QUANTITIES CONSIDERED.

1 The first term.

The last term.

3. The common difference.
4. The number of terms.

5. The sum of all the terms.

CASE I.

661. Given, the first term, the common difference, and the number of terms, to find the last term.

1. The first term is 3, the common difference 2, and num. ber of terms 10; required the last term.

SOLUTION.-The first term is 3, the second term equals 3 plus once the common difference, the third term equals 3 plus twice the common difference, etc.; hence the tenth term equals the first term plus nine times the common difference, which equals 3+2x9=21.

OPERATION.

2d=3+2 3d=3+2x2=7 4th 3+2x3=9 hence 10th=3+2x9=21

Rule.—The last term equals the first term increased by the common difference multiplied by the number of terms less one.

NOTE. In a descending series we must subtract instead of adding.

2. Given the first term 4, the common difference 3, to find the 12th term. Ans. 37. 3. The first term is 3, the common difference 4; what is the 22d term? Ans. 87. 4. Required the 76th term of a descending series, the 1st term being 80 and common difference. Ans. 30.

5. A man bought 50 yards of muslin at cent for the first yard, 1 cent for the second, 1 for the third, and so on; what did the last yard cost? Ans. 25 cents.

6. The amount of $100 at 5 per cent. for years, is respectively $105, $110, $115, etc.; amount for 25 years?

1, 2, 3, etc., what is the Ans. $225.

CASE II.

662. Given, the last term, the common difference, and the number of terms, to find the first term.

1. Required the first term, the last term being 41, the number of terms 20, and the common difference 2.

SOLUTION. From the rule in Case I., we have 41 1st term +19 times 2, hence we find first term =41-19 × 2=3.

OPERATION.

411st+2 X19
1st -41-2 X19 =3.

Rule. The first term equals the last term, diminished by the common difference multiplied by the number of terms less one.

2. Required the first term, the last term being 95, common difference 5, and number of terms 18. Ans. 10.

3. A woman bought 25 yards of muslin at the rate of 25 sents for the last yard, 24 for the next to the last, and on; what did the first yard cost? Ans. 13 cents.