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PROB. X. Given the diameter of a circle to make another circle, which shall be 2, 3, 4, &c. times greater or lefs than the given circle.

RULE.-Square the given diameter, and if the required circle be greater, multiply the fquare of the diameter by the given proportion, and the root of the product will be the required diameter;-But if the required circle be lefs; divide the fquare of the diameter by the given proportion, and the root of the quotient will be the diameter required.

There is a circle, whofe diameter is 4 inches; I demand the diameter of a circle 3 times as large?

4X4 16; and 16 × 3=48; and √ 486,928+ inches, Anfwer.

PROB. XI. To find the diameter of a circle, equal in area to an ellipfis (or oval) whofe tranfverfe and conjugate diameters are given.*

RULE.-Multiply the two diameters of the ellipfis together and the Square Root of that product will be the diameter of a circle equal to the ellipfis.

Let the tranfverfe diameter of an ellipfis be 48, and the conjugate 36; What is the diameter of an equal circle?

48x36 1728, and 172841,569 + the Anfwer.

PROB. XII. Two fhips fail from the fame port; one goes due north 45 leagues, and the other due weft 76 leagues: How far are they afunder ?

45 452025. 76x76=5776. Then, 5776 x 2025 780, and,✓ 7801-88,32 leagues, the Aufwer.

EXTRACTION

* The tranfverfe and conjugate are the longest and shortest diameters of an ellipfis; they pafs through the centre, and cross each ether at right angles.

EXTRACTION of the CUBE ROOT.

A Cube, is any number multiplied by its fquare. To extract the cube root, is to find a number which being multiplied into its fquare, fhall produce the given number.

FIRST METHO D.

RULE 1. Separate the given number into periods of three figures each, by putting a point over the unit figure, and every third figure beyond the place of units.

2. Find the greateft cube in the left hand period, and put its root in the quotient.

3. Subtract the cube, thus found, from the faid period, and to the remainder bring down the next period, and call this the dividend.

4. Multiply the fquare of the quotient by 300, calling it the triple fquare, and the quotient by 30, calling it the triple quotient, and the fum of thefe call the divijar.

5. Seek how often the divifor may be had in the dividend, and place the refult in the quotient.

6. Multiply the triple fquare by the laft quotient figure, and write the product under this dividend; multiply the fquare of the laft quotient figure by the triple quotient, and place this product under the laft; under all, fet the cube of the laft quotient figure, and call their fum the fubtrahend.

7. Subtract the subtrahend from the dividend, and to the remainder bring down the next period for a new dividend, with which proceed as before, and fo on till the whole be finished.

Note. The fame rule must be obferved for continuing the operation, and pointing for decimals, as in the fquare root.

EXAMPLES.

EXAMPLES.

Required the cube root of 436036824287?

436026824287(7583 the rcot.

343

1st Divifor 14910)93036=1st Dividend.

73500
5250

125

78875-1ft Subtrahend.

2d Divifor 1687750)14161824 = 2d Dividend.

13500000
144000

512

13644512 = 2d Subtrahend.

3d Divifor=172391940)517312287 = 3d Dividend.

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To find the true denominator, to be placed under the remainder after the operation is finished.

In the extraction of the cube root, the quotient is faid to be fquared and tripled for a new divifor : But is not really fo, till the triple number of the quotient be added to it; therefore, when the operation is finished, it is but fquaring the quotient, or root, then multiplying it by 3, and to that number adding the triple number of the root, when it will

become

become the divifor, or true denominator to its own frac tion, which fraction must be annexed to the quotient, to complete the root.

Suppofe the root to be 12, when fquared it will be 144, and multiplied by 3, it makes 432, to which add 36, the triple number of the root, and it produces 468 for a denominator.

SECOND METHOD.

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RULE 1. Having pointed the given number into pe riods of three figures each, find the greateft cube in the left hand period, fubtracting it therefrom, and placing its root in the quotient; to the remainder bring down the next period, and call it the dividend.

2. Under this dividend write the triple fquare of the root, fo that units in the latter may ftand under the place of hundreds in the former; and under the faid triple fquare, write the triple root, removed one place to the right hand, and call the fum of thefe the divifor.

3. Seek how often the divifor may be had in the dividend, exclufive of the place of units, and write the refult in the quotient.

4. Under the divifor write the product of the triple fquare of the root by the laft quotient figure, fetting down the unit's place of this line, under the place of tens in the divifor; under this line, write the product of the triple root by the fquare of the laft quotient figure, fo as to be removed one place beyond the right hand figure of the former; and, under this line, removed one place forward to the right hand, write down the cube of the laft quotient figure, and call their fum the fabtrabend.

5. Subtract the fubtrahiend from the dividend, and to the remainder bring down the next pood for a new dividend,

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