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Let the foregoing example be solved hy this Rule. A note for 1000 dols. dated Jan. 4, 1797, at 6 per cent 1st payment February 19, 1798.
8200 2 payment June 29, 1799.
500 Sd payinent November 14, 1799.
260 How much remains due on said note the 24th of De. cember, 1800 ?
Scts. Principal, January 4, 1797,
1000,00 Interest to Feb. 19, 1798, (13) mo.)
Paid February 19, 1798,
Remainder for a new principal,
867,50 Interest to June 29, 1799, (16) mo.) 70,84
Amount, 958,34 Paid June 29, 1799,
500,00 Remains for a new principal,
458,34 Interest to November 14, 1799, (4) mo.) 9,86
November 14, 1799, paid
Remains a new principal,
Balance due on said note, Dec. 24, 1800,
8 cts. The balance by Rule I. 200,579
By Rule II. 200,990
Difference, 0,411 Another Example in Rule II. A bond or note, dated February 1, 1800, was given for 500 dollars, interest at 6 per cent. and there were pay. ments endorsed upon it as follows, viz. 1st payment May 1, 1800,
40,00 ped payment November 14, 1800
3d payment April 1, 1801.
12,00 4th payment May 1, 1801.
30,00 How much remains due on said note the 16th of Septe tember, 1801
8 cts. Principal dated February 1, 1800,
500,00 Interest to May 1, 1800, (s mo.)
Amount, 507,50 Paid May 1, 1900, a sum exceeding the interest, 40,00
New principal, May 1, 1800,
467,50 Interest to May 1, 1801, (1 year.)
Amount, 495,55 Paid Nov. 4, 1800, a sam less than the interest then due,
8,00 Paid April 1, 1801,
12,00 Paid May 1, 1801, a sum greater, 30,00
New principal May 1, 1801,
445,55 Interest to Sept. 16, 1801, (4) mo.)
10,02 Balance due on the note, Sept. 16, 1801, $455,57
The payments being applied according to this Rule, keep down the interest, and no part of the interest ever forms a part of the principal carrying interest.
COMPOUND INTEREST BY DECIMALS.
RULE. MULTIPLY the given principal continually by the amount of one pound, or one dollar, for one year, at the rate per cent. given, until the number of multiplications are equal to the given number of years, and the product will be the amount required.
Or, In Table I. Appendix, find the amount of one dal lár, or one pound, for the given number of years, whicha multiply by the given principal, and it will give amount as before.
EXAMPLES. 1. What will 4001. amount to in 4 years, at 6 per cent per annum, compound interest
400 X 1,06 x 1,06 x 1,06x1,06= 4,504,99+ or
[£ 504 19s. 9d. 2,75grs. t Ans.
Whole amouut=6504,98800 .. Required the amount of 425 dols. 75 cts. for 3 years, at 6 per cent. compound interest. Ans. $307,7 jcts. +
3. What is the compound interest of 555 dols. for 14 years, at 5 per cent. ? By Table I. Ans. 8543,86cts. +
4. What will 50 dollars amount to in 20 years, at 6 per cent, cuinpound interest : Ans. $160 35cts. 6fm.
INVOLUTION. Is the multiplying any number with itself, and that product by the former multiplier; and so on; and the several products which arise are called powers.
Tl.e number denoting the height of the power, is called the index, or exponent
of that power.
What is the square of 17,1 ?
ins. 292,41 Ans. ,007225 Anš. 16387,064
EVOLUTION, OR EXTRACTION OF ROOTS. WHEN the root of any power is required; the business of finding it is called the Extraction of the Root.
The root is that number, which by a continual multiplication into itself, produces the given power.
Although there is no number but what will produce a perfect power by involution, yet there are many numbers of which precise roots can never be determined. But, by the help of decimals, we can approximate towards the rout to any assigned degree of exactness.
The roots which approximate, are called surd roots, and those which are perfectly accurate are called rational roots.
A Table of the Squares and Cubes of the nine digits. Roots.
| 1 | 2 | 3 | 4 | 5 | 61 81 91 Squares. |1|4| 9 | 16
49 64 81 Cubes. 118 | 27 | 64 | 125 | 216 543 5127729
EXTRACTION OF THE SQUARE ROOT. Any nuinber multiplied into itself produces a square.
To extract the square root, is only to find a number, which being multiplied into itself, shall produce the given number.
RULE. 1. Distinguish the given number into periods of two figures each, by putting a point over the place of units, another over the place of hundre's, and so on; and it there arc decimals, point them in the same manner, Irown units lurvards the right handl; which points show tie number of figures the root will consist of.
2. Find the greatest square number in the first, or left band period, place the root of it at the right hand of the given number, (after the manner of a quotient in division) for the first figure of the root, and the square number under the period, and subtract it therefrom, and to the rniainder bring down the next period for a dividend.
3. Place the double of the root, already found, on the left hand of the dividend for a divisor.
4. Place such a figure at the right hand of the divisor, and also the same figure in the root, as when multiplied into the whole (increased divisor) the product shall be equal to, or the next less than thé dividend, and it will be the second figure in the root.
5. Subtract the product from the dividend, and to the remainder join the next period for a new dividend.
6. Double the figures already found in the root, for a new divisor, and from these find the next figure in the root as last directed, and continue the operation in the same manner, till you have brought down all the periods.
Or, to facilitate the foregoing Rule, when you have brought down a period, and formed a dividend, in order to find a new figure in the root, you may divide said divi. dond, (omitting the right hand figure thereof,) by double the root already found, and the quotient will comi.only be the figures sought, or being made less one or two, will generally give the next figure in the quotient.
1. Required the square root of 141225,64. 141225,64(375,8 the root exactly without a remainder; 9
but when the periods belonging to any
given number are exhausted, and still 512 leave a remainder, the operation may 469
be continued at pleasure, by annexing
periods of cyphers, &c. 745)4525