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299. In £13, how many dollars, cents and mills ?
Now, as the pound has different values in different places, the amount in Federal Money will vary according to those values. In England, $1=4s. 6d=4.5s.=£*=£0.225, and there £13–13:0.225=$57.777. In Canada, $1=5s= £0.25, and there £13=13;0.25=$52. In New England, $1=bs.=£$=£0.30, and there, £13=13:0.3=$43.333. In New York, $i=88.
==£0.4, and there, £13=13:0.45 32.50. In Pennsylvania, $1=7s. 6d.=7.53.-£75=£0.375, and there, £13=13:0.375=$34.666. And in Georgia, $1=
4.6+ 43.8d.=4.6 s.=£20.0=£0.2333+, and there, £13–13:0.2333 $55.722.
300. In £16 7s. 8d. 2qr., how many dollars, cents and mills ?
Before dividing the pounds, as above, 7s. 8d. 2qr., must be reduced to a decimal of a pound, and annexed to £16. This may be done by Art. 143, or by inspection, thus, shillings being 20ths of a pound, every 23. will be 1 tenth of a pound: therefore write half the even number of shillings for the tenths £0.3. One shilling being 1 20th=£0.05; hence, for the odd shilling we write £0.05. Fart!ıings are 96Cths of a pound, and if 960ths be increased by their 24th part, they are 1000ths Hence 8d. 2qr.(=34qr.+1=0.035; and 16+0.3+0.05+0.035 =£16.385, which, divided as in the preceding example, give for English currency, $72.822, Can. $65.54, N. Y $40.962, &c. Hence,
301. To change poun:ls, shillings, pence and farthings to Federal Money, and the rererse.
RULE.-Reduce the shillings, &c. to the decimal of a pound ; then, if it is English currency, divide by 0.225; if Canada, by 0.25; if N. E., by 0.3; if N. Y., by 0.4; if Penn., by 0.375, and if Georgia, by 0.23 ;-the quotient will be their value in dollare, cents and mills. And to change Federal Money into the above currencies, multiply it by the preceding deciinals, and the prodluct will be the answer in pounds and decimal parts.
3. In £91, how many dol- 9. Reduce £25 158. N. E., larg ? £91 E.=$404.444. to Federal Money. Can. $364. N. E., $303.333.
Ans. $85.833. N. Y. $227.50, &c. Ans. 4. Reduce £125, N. E. to
10. In £227 17s. 5 d. N. E. Federal Money.
how many dollars, cents and
mills ? Ans. $416.666.
Ans. $759 57ots. 3m. 5. Change $100 to each of the foregoing currencies. 11. In $1.612, now many
$100-£22 10s. Eng.£25 shillings, pence and farthings? Can.-£30 N. E.-£40 N. Y.
9s. 8d. N. E. £37 10s. Penn.
6. In $1111.111, how many pounds, shillings, pence and
12. Reduce £33 13s. N. Y. farthings ?
to Federal Money. £333 63. 82. N. E.
Ans. $84.125. Ans. 444 83. 101d. N. Y.
13. In £1 ls. 10 d. Penn., 7. In £1 ls. 103d. N. E.,
how how many dollars ?
Ans ont Ans.p.040. 8. In £1 ls. 104d. N. Y., 14. In £1 ls. 10fd. Can., how many dollars ?
how many dollars ? Ans. $2.735.
Ans. 3 12. 104d. N. Y.
many dollars ?
302. The following rules, founded on the relative value of the several currencies, may sometimes be of use :
To change Eng. currency to N. E. add }, Ń. E. to N. Y. add $, N. Y. to N. E. subtract 4, N. E. to Penn. add , Penn. to N. E. subtract 5, N. Y. to Penn. subtract 16, Penn. to N. Y. add I's, N. E. to Can. subtract }, Can. to N. E. add $, &.
'15. In $255.406, how many 16. Change £240 158. N. pounds, shillings, pence and E. to the several other curfarthings?
rencies. £76 12s. 5d. N. E.
£321 Os. Od. N Y. Ans. £102 33. 3d. N. Y.
£300 18s. 9d. Penn. $95 158. 67d. Penn.
£200 12s. 6d. Can. £63 178. Ofa, Can.
8802.50 Fed. Mon
303. Of the most common gold and silver coins, containing their weighs Bineness, and intrinsic value in Federal Money.
Country. | Names of coins. | Weight. | Fieness. | Valuc.
416. 208. 104.
41.6 464.50 232.25
92.90 451.62 386.18 418.47 418.47 450.90 225.45 432.93 216.46 265.68 504.20 162.70 443.80 301.90
1.000 0.500 0.250 0.100 1.111 0.555 0.22% 1.06 0.898 0.991 0.972 1.037 0.519 0.926 0.463 0.615 1.222 0.375 1.009 0.602
Not The current values of several of the above coins differ somewbies from ther wu..usic value, as expressed in the table
1. Mensuration of Superficies. 304. The area of a figure is the space contained within the boonds of He surface, without any regard to thiekness, and is estimated by the namoer of squares contained in the same; the side of those squares being either an inch, a foot, a yard, a rod, &c. Hence the area is said to be so many square inches, square feet, square yards, or square rods, &c.
305. To find the area of a parallelogram (65), whether it be a square, a rectangle, a rhombus, or a rhomboid.
RULE.-Multiply the length by the breadth, or perpendicular height, and the product will be the area.
1. What is the area of a 3. What is the area of a square whose side is 5 feet ?
rhombus, whose length is 12
rods, and perpendicular height 5
Ans. 48 rocks. 5
4. What is the area of ADA. 25 ft.
rhomboid 24 inches long, and 8 wide ?
Ans. 192 inches. 5
5. How many acres in 2. What is the area of a rectangular piece of ground, rectangle, whose length is 9, 56 rods long, and 26 wide ? and breadth 4ft.? Ans. 36ft. 56X26 160—9o. Ana
306. To find the area of a triangle. (64) Rule 1.-Multiply the base by half the perpendicular height, and the product will be the areas
Role 2.-If the three sides only are given, add these togethez and take half the sum ; from the half sum subtract each side separately; multiply the half and the three remainden continually together, and the square root of the last product will be the area of the triangle.
MENSURATION OF SUPERFICIES.
1. How many square feet 3. What is the area of a in a triangle, whose base is 40 triangle, whose three sides are feet, and height 30 fect? 13, 14 and 15 feet? 40 base.
13+14+15=42 155perpend. height
and 42%22l-half sum
21 21 21 200
13 14 15 and 21 X6X7X
ren. 8 7 6 600 feet. Ans.
Then 705612=84 feet, Ana. 2. The base of a triangle is 4. The three sides of a tri6.25 chains, and its height 5.20 angle are 16, 11 and 10 rode ; chains; what is its area? what is the area ? Ans. 16.25 square chains.
Ans. 54.299 roda.
307. To find the area of a trapezoid. (65) RULE.-Multiply half the sum of the two parallel sides by the perpendicular distance between them, and the product will be the area.
1. One of the two parallel 2. How many equare feet sides of a trapezoid is 7.5 in a plank 12 feet 6 inches chairs, and the other 12.25, long, and at one end, I foot and the perpendicular distance and 3 inches, and, at the othen between them is 154 chains; 11 inches wide ? what is the area
Ans. 1347 feet 12.25 75
3. What is the area of a
piece of land 30 rods long, 2)16.75
and 20 rods wide at one end
and 18 rods at the other ? 9.875
Ans. 570 rodei 15.4
4. What is the area of a 39500
hall 32 feet long, and 22 feet 49375
wide at one end, and 20 at the 9875
other? Ans, 672 feet.
152.0750 sf. chains. Ans.
308. To find the area of a trapezium, or an irregular polygon
RULE.--Divide it into triangles, and then find the area of these triangles by Art. 306, and add them together.