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The interest of 50 dollars for 6 months, is
And, the interest of 1 dol. 50 cts. for 6 months, is

Ans. Rebate, 81 46 2. What is the rebate of 150l. for 7 mouths, at 5 per cent.?

4

nterest of 1501. for 7 months, is nterest of 41. 7s. 6d. for 7 months, is

$ cla

1 50

s. d.

7 6

1

2 61

Ans. £4 4 114 nearly.

By the above Rule, those who use interest tables in their counting-houses, have only to deduct the interest of the interest, and the remainder is the discount.

A concise Rule to reduce the currencies of the different States, where a dollar is an even number of skillings, to Federal Money.

RULE I.

Bring the given sum into a decimal expression by inspection, (as in Problem 1. page 87) then divide the whole by,5 in New-England and by,4 in New-York currency, and the quotient will be dollars, cents, &c.

EXAMPLES.

1. Reduce 541. 88. Sjd. New-England currency, to Federal Money.

,3)54,415 decimally expressed.

Ans. $181,38 cts.

2. Reduce 7s. 114d. New-England currency, to Federai Money.

72 11 d. 0,599 then,,S),599

Ane. 81984,601

Ans. $1,53

3. Reduce 515. 16. 10d. New-York, &c. currency, to Federal Money.

,4)513.842 decimal

4. Reduce 19s. 51d. New-York, &c. currency, to Federal Money.

,4)0,974 decimal of 19s 5fd.

82,431 Ans.

5. Reduce 641. New-England currency, to Federal Money.

,8)64000 decimal expression.

8213,55 Ans.

NOTE. By the foregoing rule you may carry on the decimal to any degree of exactness; but in ordinary practice, the following Contraction may be useful. RULE II.

To the shillings contained in the given sum, annex 8 times the given pence, increasing the product by 2; then divide the whole by the number of shillings contained in a dollar, and the quotient will be cents.

EXAMPLES.

1. Reduce 45s. 6d. Now-England currency, to Federal Money.

6x8+250 to be annexed,
6)45,50 or 6)4*50

8 cts.

87,58 Ans. 758 cents.=7,58 2. Reduce 21. 10s. 9d. New-York, &c. currency, to Federal Money.

9x8+2-74 to be annexed. Or thus, 8)50,74

Then 8)5074

$ cts. Ans. 634 cents.=6 $4

86,34 Ans

N. B. When there are no pence in the given sum, you must annex two cyphers to the shillings; then divide as before, &c.

5. Reduce. Sl. 5s. New-England currency, to Federal money.

St. 58.651. Then 6)6500

Ane. 1083 cents.

SOME USEFUL RULES,

FOR FINDING THE CONTENTS of SUPERFICIES AND

SOLIDS.

SECTION I. OF SUPERFICIES.

The superficies or area of any plane surface, is composed or made up of squares, either greater or less, according to the different measures by which the dimensions of the figure are taken or measured :—and because 12 inches in length make 1 foot of long measure, there fore, 12x12=144, the square inches in a superficial foot, &c.

ART. I. To find the area of a square having equal sides.

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Multiply the side of the square into itself, and the product will be the area, or content.

EXAMPLES.

1. How many square feet of boards are contained in the floor of a room which is 20 feet square ?

20×20=400 feet, the Answer.

2. Suppose a square lot of land measures 26 rods on each side, how many acres doth it contain? NOTE.-160 square rods make an acre. Therefore, 26X26-676 sq. rods, and 676÷160=4a. 36r. the Answer.

ART. 2. To measure a parallelogram, or long square. RULE.

Multiply the length by the breadth, and the product will be the area, or superficial content.

EXAMPLES.

1. A certain garden, in form of a long square, is 96 ft. ong, and 54 wide; how many square feet of ground are contained in it ? Ans. 96×54-5144 square feet. 2. A lot of land, in form of a long square, is 120 rods in length, and 60 rods wide; how many acres are in it ? 120×60 7200 sqr. rods, then 1200=45 acres. Ans. 3. If a board or plank be 21 feet long, and 18 inches road; how many square feet are contained in it?

118 inches 1,5 feet, then 21×1,5-S1,5 Ans.

Or, in measuring boards, you may multiply the length in feet by the breadth in inches, and divide by 12, the quotient will give the answer in square feet, &c.

Thus, in the foregoing example, 21×18÷÷12=-31,5 as before.

4. If a board be 8 inches wide, how much in length will make a square foot?

RULE.-Divide 144 by the breadth, thus, 8)144

5. If a piece of land be 5 rods wide, how length will make an acre?

Ans. 18 in. many rods in

RULE.-Divide 160 by the breadth, and the quotient will be the length required, thus, 5)160

Ans. 32 rods in length.

ART. 3. To measure a Triangle. Definition.-A Triangle is any three cornered figure which is bounded by three right lines.*

RULE.

Multiply the base of the given triangle into half its perpendicular height, or half the base into the whole perpendicular, and the product will be the area.

EXAMPLES.

1. Required the area of a triangle whose base or longest side is 32 inches, and the perpendicular height 14 inches. 32x7=224 square inches, the Answer.

2. There is a triangular or three cornered lot of land whose base or longest side is 51 rods; the perpendicular from the corner opposite the base measures 44 rods; how many acres doth it contain?

51,5×22=1133 square rods,=7 acres, 13 rods.

*A Triangle may be either right angled or oblique ; in either case the teacher can easily give the scholar a right idea of the base and perpendicular, by marking it down on a slate, paper, &c.

TO MEASURE A CIRCLE.

ART. 4. The diameter of a Circle being given, to find the Circumference.

RULE.

As 7 is to 22 : so is the given diameter: to the circumference. Or, more exactly, As 113: is to 355 ; : &c. the diameter is found inversely.

NOTE. The diameter is a right line drawn across the circle through its centre.

EXAMPLES.

1. What is the circumference of a wheel whose diameter is 4 feet As 7: 22 :: 4 : 12,57 the circumfe

rence.

2. What is the circumference of a circle whose diame Ler is 35 P-As 7: 22:: 55: 110 Ans.-and inversely as 22 : 7 : : 110: 35, the diameter, &c.

ART. 5. To find the area of a Circle.
RULE.

Multiply half the diameter by half the circumference, and the product is the area; or if the diameter is given without the circumference, multiply the square of the diameter by 7854 and the product will be the area.

EXAMPLES.

1. Required the area of a circle whose diameter is 12 inches, and circumference 37,7 inches.

18,85 half the circumference.
6 half the diameter.

115,10 area in square inches.

2. Required the area of a circular garden whose diame Ler is 11 rods ?

,7854

By the second method, 11x11 = 121

Ans. 95,0334 rods.

SECTION 2. OF SOLIDS.

Solids are estimated by the solid inch, solid foot, &c 28 of these inches, that is 12x12×12 make 1 cubic solid foot.

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