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15/ 20.

ANALYSIS.-By Case II we find the proportional quantities to be 35 lb. of 50 ct. tea, and 10 lb. each of 75 and 80 ct. tea. Adding these amounts, we have 55 pounds; but we need 100 pounds: hence we must have 100 times the proportional quantity of each article.

RULE. Find the proportional quantities by Case II, and divide the given quantity by their sum. Multiply each of the proportional quantities by the quotient thus obtained, and the products will be the required quantities.

2. How much water must be mixed with cider worth 20 cents a gallon, in order that 40 gallons of the mixture shall be worth 15 cents a gallon? By CASE II. So, 5

5 x = 10 gal. of water.

15 x 40 = 30 “ cider. 3. A farmer has in his barn oats worth 30 cents a bushel, wheat worth 90 cents a bushel, and rye worth 50 cents a bushel; how many bushels has he of each kind, if the average price of the whole is 54 cents, there being 500 bushels in the barn?

4. Sugars worth 8, 9, 10, 11, and 12 cents a pound, are to be mixed, so that 200 pounds of the mixture shall be worth 10 cents a pound; how much of each sort must be taken?

5. 100 pounds of gold-bearing quartz were bought for $100. There were three kinds, costing $4 a pound, 25 cents a pound, and 50 cents a pound respectively; how many pounds were there of each kind ?

6. How shall I mix gold that is 10 carats fine, with gold 14 carats, and 18 carats fine, respectively, to obtain 6 ounces of a mixture 16 carats fine?

HORNER'S METHOD OF EVOLUTION. 1. Extract the square root of 625.

ANALYSIS.—The greatest square 625(25

contained in the left-hand period is 4

4, the root of which is 2; this we 225 write as the first figure of the required 225

root. We next add the figure thus

found to the cipher at the head of 40 Trial Divisor.

the first left-hand column, and obtain 5

2, which being multiplied by the 2 in 45 Complete Divisor. the root gives 4. We write the 4

under the first left-hand period of the given number, and subtract it, leaving 2 as a remainder. We next add the quotient figure 2 to the 2 in the first column, obtaining 4, to which we annex one cipher for a trial divisor, and, bringing down the next period 25, we have 225 as the new dividend. We find that the trial divisor 40 is contained in this dividend 5 times. We next add the 5 thus found to the last term of the first column, completing the divisor. Multiplying this result by the quotient figure 5 gives 225, which we subtract from the dividend, and, as there is no remainder, 25 is the exact square root of 625.

RULE. Prepare as many columns as there are units in the index of the required root, placing the given number at the top of the first right-hand column, and a cipher at the head of each of the others.

Separate the given number into periods of as many figures each as there are units in the index of the required root, and find by trial the first figure of the root.

Add the figure thus found to the cipher at the head of the first left-hand column, multiply the sum by the root figure, and add the product to the cipher at the head of the second column. Proceed in a similar manner until the last product

falls under the first left-hand period of the given number. Subtract this product from this first left-hand period, and to the remainder annex the next period for a new dividend.

Add the figure of the root just found to the amount in the first column, multiply the sum by the same figure, adding the product to the next column, and so continue till the last product falls in the last column but one.

Proceed in a similar manner, stopping each time one column farther to the left, until the last operation is the addition of the first figure of the root, to the sum in the first left-hand column.

Annex one cipher to the number at the foot of the first column, two ciphers to the number at the foot of the second column, and so on.

Consider the number standing in the column next to the last as the trial divisor, find how often it is contained in the dividend, and write the quotient as the second figure of the root.

Add this figure to the amount in the first column, multiply the sum by the same figure, and proceed as before, till the last product falls under the trial divisor. The sum of the last product and the trial divisor will be the complete divisor. Multiply by the figure of the root just found, subtract the product from the dividend, and proceed as before till all the periods have been brought down.

Note 1.—When any dividend will not contain the trial divisor, place a cipher in the root, bring down the next period, annex one cipher to the number in the first column, two ciphers to the number in the second column, and so on, using the trial divisor with the proper number of ciphers annexed as a new trial divisor.

Note 2.—When the exact root cannot be extracted, periods of ciphers may be annexed, and the work extended to any required number of decinal places.

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3. What is the fourth root of 390625?

390625(25 16

)230625 24

230625 32000 Trial Divisor. 12

14125 2400 46125 Complete Divisor.

425 2825

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4. What is the square root of 2.5 to three decimal places?

5. What is the cube root of 2.7?
6. Find the value of 0.5.
7. What is the fourth root of 4?

8. Find the value of 0 56, or, the cube root of the sixth power of 5.

9. Extract the third root of 3 to three decimal places.

10. Find the difference between the square root of the cube of 3 and the cube root of the square of 3, or V 33 —

32.
11. Extract the third root of .01.
12. What is the value of V.2?

13. Find the square root of the square root of.2, to two decimal places, or V V.2. .

14. What is the value of Y V 6561 ? 15. Extract the cube root of 523606616. 16. What is the value of 50625? 17. What is the value of ý 1048576? 18. What is the fifth root of 1889568? 19. Find the third root of 32.768. 20. What is the fourth root of 1099511627776? 21. What is the sixth root of 5 to one decimal place?

ARITHMETICAL PROGRESSION. An Arithmetical Progression is a series of numbers either increasing or decreasing by a common difference. Thus: 1, 3, 5, 7, 9 is an increasing series.

9, 7, 5, 3, 1 is a decreasing series. The common difference in each series is 2.

The numbers which form the series are called the Terms. The first term and the last term are called the Extremes, and the other terms, the Means.

The following five quantities are considered in Arithmetical Progression: The first term; the last term; the common difference; the number of terms; and the sum of the series.

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