7. What is the cube root of 373248? 8. What is the cube root of 21024576 ? 9. What is the cube root of 84'604519? 10. What is the cube root of '000343 ? 11. What is the cube root of 2? 12. What is the enbe root of ?

Ans. 72.

Ans. 276.

Ans. 4'39.

Ans. '07. Ans. 1'25 +. Ans.

Note. See 105, ex. 10, and T 108, ex. 14.

Ans.

Ans. z

Ans. 125 +.
Ans.

SUPPLEMENT TO THE CUBE ROOT,

1. What is a

QUESTIONS.

cube ? 2. What is understood by the cube root? 3. What is it to extract the cube root? 4. Why is the square of e quotient multiplied by 300 for a divisor? 5. Wny, in finding the subtrahend, do we multiply the square of the last quotient figure by 30 times the former figure of the root? 6. Why do we cube the quotient figure? 7. How do we prove the

operation?

EXERCISES.

1. What is the side of a cubical mound, equal to one 288 feet long, 216 feet broad, and 48 feet high? Ans. 144 feet. 2. There is a cubic box, one side of which is 2 feet; how many solid feet does it contain ? Ans. 8 feet. 3. How many cubic feet in one 8 times as large? and what would be the length of one side?

Ans. 64 solid feet, and one side is 4 feet. 4. There is a cubical box, one side of which is 5 feet; what would be the side of one containing 27 times as much? 64 times as much? 125 times as much?

Ans. 15, 20, and 25 feet.

27

5. There is a cubical box, measuring 1 foot on each side; what is the side of a box 8 times as large? Ans. 2, 3, and 4 feet.

times?

64 times?

111. Hence we see, that the sides of cubes are as the cube roots of their solid contents, and, consequently, their contents are as the cubes of their sides. The same proportion is true of the similar sides, or of the diameters of all solid figures of similar forms.

6. If a ball, weighing 4 pounds, be 3 inches in diameter, what will be the diameter of a ball of the same metal, weighing 32 pounds? 4: 32: 33: 63 Ans. 6 inches. 7. If a ball, 6 inches in diameter, weigh 32 pounds, what will be the weight of a ball 3 inches in diameter? Ans. 4 lbs. 8. If a globe of silver, 1 inch in diameter, be worth $6, what is the value of a globe 1 foot in diameter ?

Ans. 10368. 9. There are two globes; one of them is 1 foot in diameter, and the other 40 feet in diameter; how many of the smaller globes would it take to make 1 of the larger?

Ans. 64000.

10. If the diameter of the sun is 112 times as much as the diameter of the earth, how many globes like the earth would it take to make one as large as the sun? Ans. 1404928. 11. If the planet Saturn is 1000 times as large as the earth, and the earth is 7900 miles in diameter, what is the diameter of Saturn? Ans. 79000 miles.

12. There are two planets of equal density; the diameter of the less is to that of the larger as 2 to 9; what is the ratio of their solidities? Ans. ; or, as 8 to 729,

T*

Note. The roots of most powers may be found by the square and cube root only: thus, the biquadrate, or 4th root, is the square root of the square root; the 6th root is the cube root of the square root; the 8th root is the square root of the 4th root; the 9th root is the cube root of the cube root, &c. Those roots, viz. the 5th, 7th, 11th, &c., which are not resolvable by the square and cube rcots, seldom occur, and, when they do, the work is most easily performed by logarithms; for, if the logarithm of any number be divided by the index of the root, the quotient will be the logarithm of the root itself.

ARITHMETICAL PROGRESSION.

112. Any rank or series of numbers, more than two, increasing or decreasing by a constant difference, is called an Arithmetical Series, or Progression.

When the numbers are formed by a continual addition of the common difference, they form an ascending series; but when they are formed by a continual subtraction of the common difference, they form a descending series.

Thus, { 13, 13; 11; 9; 17, 13, 3, &c. is a descending series.

3, 5, 7, 9, 15, &c. is an ascending series.

(

9, 5,

The numbers which form the series are called the terms of the series. The first and last terms are the extremes, and the other terms are called the means.

There are five things in arithmetical progression, any three of which being given, the other two may be found :

1st. The first term.

2d. The last term.

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

1. A man bought 100 yards of cloth, giving 4 cents for the first yard, 7 cents for the second, 10 cents for the third, and so on, with a common difference of 3 cents; what was the cost of the last yard?

As the common difference, 3, is added to every yard except the last, it is plain the last yard must be 99 X 3, cents, more than the first yard.

= 297 Ans. 301 cents.

Hence, when the first term, the common difference, and the number of terms, are given, to find the last term,-Multiply the number of terms, less 1, by the common difference, and add the first term to the product for the last term.

2. If the first term be 4, the common difference 3, and the number of terms 100, what is the last term? Ans. 301. 3. There are, in a certain triangular field, 41 rows of corn; the first row, in 1 corner, is a single hill, the second contains 3 hills, and so on, with a common difference of 2; what is the number of hills in the last row? Ans, 81 hills,

4. A man puts out $1, at 6 per cent. simple interest, which, in 1 year, amounts to $1'06, in 2 years to $1'12, and so on, in arithmetical progression, with a common difference of $ '06; what would be the amount in 40 years? Ans. $3'40.

Hence we see, that the yearly amounts of any sum, at simple interest, form an arithmetical series, of which the principal is the first term, the last amount is the last term, the yearly interest is the common difference, and the number of years is 1 less than the number of terms.

5. A man bought 100 yards of cloth in arithmetical progression; for the first yard he gave 4 cents, and for the last 301 cents; what was the common increase of the price on each succeeding yard?

This question is the reverse of example 1; therefore, 3014297, and 297 ÷ 99 3, common difference.

Hence, when the extremes and number of terms are given, 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.

6. If the extremes be 5 and 605, and the number of terms 151, what is the common difference? Ans. 4.

7. If a man puts out $1, at simple interest, for 40 years, and receives, at the end of the time, $3'40, what is the rate ?

If the extremes be 1 and 3'40, and the number of terms 41, what is the common difference?

Ans. '06. 8. A man had 8 sons, whose ages differed alike; the youngest was 10 years old, and the eldest 45; what was the common difference of their ages? Ans. 5 years.

9. A man bought 100 yards of cloth in arithmetical series; he gave 4 cents for the first yard, and 301 cents for the last yard; what was the average price per yard, and what was the amount of the whole?

Since the price of each succeeding yard increases by a constant excess, it is plain, the average price is as much less than the price of the last yard, as it is greater than the price of the first yard; therefore, one half the sum of the first and last price is the average price.

152

cts.

=

average 15250 cts.=

Ans.

One half of 4 cts. +301 cts. price; and the price, 152 cts. X 100 $152'50, whole cost.

Hence, when the extremes and the number of terms are given, to find the sum of all the terms,-Multiply the sum of the extremes by the number of terms, and the product will be

the answer.

10. If the extremes be 5 and 605, and the number of Ans. 46055. terms 151, what is the sum of the series?

11. What is the sum of the first 100 numbers, in their natural order, that is, 1, 2, 3, 4, &c.? Ans. 5050. 12. How many times does a common clock strike in 12 hours?

Ans. 78.

13. A man rents a house for $50, annually, to be paid at the close of each year; what will the rent amount to in 20 years, allowing 6 per cent., simple interest, for the use of the money?

The last year's rent will evidently be $50 without interest, the the last but one will be the amount of $50 for 1 year, last but two the amount of $50 for 2 years, and so on, in arithmetical series, to the first, which will be the amount of $50 for 19 years = $107.

If the first term be 50, the last term 107, and the number of terms 20, what is the sum of the series? Ans. $1570. 14. What is the amount of an annual pension of $100, being in arrears, that is, remaining unpaid, for 40 years, Ans. $7900. allowing 5 per cent. simple interest?

15. There are, in a certain triangular field, 41 rows of corn; the first row, being in 1 corner, is a single hill, and the last row, on the side opposite, contains 81 hills; how Ans. 1681 hills. many hills of corn in the field?