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A figure in any place, is ten times as great as the same figure, in the next right hand place.
If a figure be removed, then, from the units' to the tens' place, from the tens' to the hundreds', or from any place to the next higher, it will become ten times as great as before ; that is, it will be multiplied by 10.
Annexing a cypher to a number, removes every figure of that number a place higher.
It therefore makes every figure ten times greater, and multiplies the whole number
Annexing two cyphers, removes every figure two places higher, and consequently multiplies the whole number by 100.
Annexing three cyphers, multiplies by 1,000, four cyphers, by 10,000, and so on.
Wher we use the word ANNEX, we mean PLACE AT THE RIGHT; when we use the word PREFIX, we mean PLACE AT THE LEFT.
It will be seen that the number of cyphers, annexed, is just equal to the number of cyphers in the multiplier. Hence, WHEN THE MOLTIPLIER IS 1, WITH CYPHERS ANNEXED, MULTIPLICA
ANNEXING THE SAME NUMBER OF CYPHERS TO THE MULTIPLICAND.
EXAMPLES FOR PRACTICE. 2. At 10 dollars a barrel, what cost 357 bbls. of flour ? Ans. 3,570 dolls.
3. If I were to give 75 men 100 dollars apiece, what should I give the whole ? Ans. 7,500 dolls.
4. What would 275 pounds of beef come to, at 10 cts. pr. pound. Ans. 27 dolls. 50 cts.
The money of the UNITED STATES, is called FEDERAL MONEY. It is reckoned in EAGLES, DOLLARS, DIMES, CENTS and MILLS. These are called its denominations.
di. 10 dimes
1 dollar 10 dollars
E. NOTE. The mill is merely a nominal coin, since there is no real coin in use, of that value.
IS PERFORMED BY
1. How many cents in 2di ? in 3di ? in 5di ? in 7di ? in 12di?
2. How many dimes in 2 dolls ? in 3 dolls ? in 4 dolls? in 7 dolls ? in 9 dolls ?
3. How many mills in 2 cts? in 3 cts ? in 4 cts ? in 5 cts ? in 9 cts ? in 10 cts?
4. How many mills in 1 di.? in 2 di.? in 3 di.? in 4 di.? in 5 d.? in 7 d.? in 10 d.?
5. How many mills in 1 doll? in 2 dolls ? in 3 dolls ? in 4 dolls ? in 10 dolls ?
6. How many mills in 1 E.? in 2 E.? in 3 E.? in 7 E.? in 9 E.? in 10 E.?
Then, it seems that a dollar is 10 dimes, or 100 cents, or 1,000 mills. Hence,
DOLLARS ARE MADE DIMES BY ANNEXING ONE CYPHER ; CENTS BY AN. NEXING TWO CYPHERS; AND MILLS BY ANNEXING 3 CYPHERS.
If a sum be cents all the even hundreds will be dolls., of course ; if it be dimes all the even tens will be dolls.; if it be mills, all the even thousands will he dolls. Hence,
TO CHANGE A SUM FROM DIMES TO DOLLARS, POINT OFF ONE FIGURÉ ÓX THE RIGHT; FROM CENTS TO DOLLARS POINT OFF TWO FIGURES ; FROM MILLS TO DOLLARS, POINT OFF TURPB PIGURES, ALL ON THE LEFT OF THE POINT WILL BE DOLLARS.
For this purpose a period (.) is the most convenient point. It is called the SEPARATRIX, because it is used to separate dollars from lower denominations. With the exception of this point the numbers, follow one after ano. ther like common numbers. If there be but one figure after the point, the number may be read, dollars and dimes, or the whole may be read as dimes. If two figures follow the point, the number may be read dollars dimes and cents, or dollars and cents, or all together as cents. If three, dollars, dimes, cents and mills, or, dolls. cents and mills, or all together as mills. Many more variations in reading these numbers may be made, but it is of no use to repeat them, since they are not used in practice. The teacher, if he sees fit, may require them of the pupil, who will readily discover them for himself.
Accounts are kept in dollars and cents, the other denominations being disregarded. Thus instead of writing 3 E. 5 dolls. 4 di. 6 cts, we write 35 dolls. 46 cts. or, which more common $35.46. When the character $ is placed before a number, the names of the other denominations are omitted, and dollars are to be understood for the figures before the separatrix, cents for the first two after it, and mills for the third, if there be any. Hence, to read Federal Money,
CALL ALL THE FIGURES BEFORE THE SEPARATRIX, DOLLARS AND ALL AFTER IT CENTS, OR CENTS AND MILLS.
To write Federal Money. WRITE EAGLES AND DOLLARS, TOGETHER AS DOLLARS ; AND PLACE THE SEPARATRIX : THEN WRITE DIMES AND CENTS TOGETHER AS CENTS, AND MILLS, IF THERE BE ANY, NEXT.
In adding or subtracting Federal Money, the pupil must be careful to add or subtract the same denominations to or from each other ; that is dollars, to or from dollars ; cents, to or from cents, &c.
EXAMPLES FOR PRACTICE, CONTINUED. 5. How many cents are there in 700 dollars ? Ans. 70,000. 6. How many mills are there in *965.13? Ans. 965,130.
What would 642 gallons of brandy cost at $1.00 pr. gal ? 8. What would 987 barrels of flour cost at $10.00 pr. barrel ? Give the several answers in dolls. cents and mills.
9. What cost 954 pounds of cheeso at $0.10 pr. pound ? 10. Multiply 6,839 by 10 16. 1,349,864
10 9,894 100 17. 2,957,863
100 12. 76,943 1000 18. 2,234,967
1,000 13. 194,972
100 19. 72,777,963 100,000 14.
279,864 10,000 20. 331,000,221 1,000,000 15, 1,943,867 5 100,000 | 21, 17,236 1,000,000,000
XVI. 1. At 20 cents a pound, what will 78 pounds of sugar cost.
Here 20 is the multiplier.
But 20 is a composite number, because it is 2 tens or 2 times 1 ten.
Its factors, then, are 2 and 10.
By the last rule, multiply by 10, annexing thus, a cypher,
2. On one page of a book are 47 lines, how many on 60 ?
Here 60=6X10 is the multiplier. Ans. 2,820 3. At 48 dollars a hogshead, what cost 500 hogsheads of molasses ?
Here 500=5X100 is the multiplier. Ans. 24,000 In all these examples, we multiply by the significant figures of the multiplier, and annex the cyphers to the product.
It is plain that if there are cyphers at the right of the multiplicand, these will, likewise, be found at the right of the product. For example, multiply 30 by 3 ;
4500 Ans. Here we have the same cyphers at the right of the products, which were at the right of the multiplicands. Hence,
WHEN THERE ARE CYPHERS AT THE RIGHT OF EITHER OR BOTH FACTORS, MULTIPLY BY THE OTHER FIGURES, AND ANNEX ALL THE CYPHERS TO THE PRODUCT.
Note. If the significant figures of the multlplier form a composite number, we may maltiply by them according to the rule of composite nambers, and then annex cyphers. In the same manner we inay take advantge of any of the preceding rules.
EXAMPLES FOR PRACTICE. 4. If 60 men can do a piece of work in 79 days, how many days will it take 4 men to do it ?
Ano. 4,740 days. 5. If 27 men can do a piece of work in 300 days, how many men will it take to do it in one day ?
Ans. 8,100 men. 6. If 500 labourers on a canal be paid 75 dollars a piece, how ma. ny cents is that for the whole ?
Ans. 3,750,000 cents. 7. If there are 275 pages in a book, how many pages are there in ap edition of 7,000 books?
Ans. 1,925,000. 8. If 742 pounds of provision will serve a garrison 1 day, how many will serve 300 days ?
Ans. 222,600. 9. If $250.00 will gain $15.00 interest in one year, what will it gain in 70 years ? Give the answer in mills. Ans. 1 050,000 mills.
10. Ifa sum of money will gain $25.00 interest in one year, how much will it gain in 50 years? Give the answer in dimes.
Ano. 12,500 dimes. 11. If $900.00 gain a certain interest in a year, what sum will gain 320 times as much ? Give the answer in cents.
Ans. 28,800,000 cts. 12. If 600 barrels of rum cost 50 dollars each, what cost the whole ? Give the answer in mills. Ans. 30,000,000 mills. 13. Mul. 765,000 by 270 | 16.
300,200 19 14. 500,700 18 17.
561,800 20 15. 200,000 17 | 18
♡ XVII. We will now give à process, adapted to all cases in Multiplication.
1. A 'carpenter received 1,565 dollars for building a house. How much ought he to receive for building 17 houses at that rate ?
Here 17 is the multiplier. First find how much he ought to receive for 7 houses ; thus,
Then find how much for 10 houses, by
annexing a cypher to 1,565 ; thus,
Ans. 26605 2. In a garrison, if the allowance of one man is 23 ounces of provision a day, how many ounces a day ought 634 men to have ?
Here 23 is the multiplier.
First, find what it would amount to at 3 ounces per day: thus,
1902 Thien, at 20 ounces ; thus,
But, 20 and 3 are 23. Therefore,
add the two products; thus,
14582 Ans. In these examples we found the answer from multiplyiug by the units in the multiplier, (according to 31,) then, by the tens, (according to J xvi,) and, finally, adding the two products together.
We may do the same thing a little more concisely. thus :