To refolve any question of this nature; fay, as the ratio in length is to the ratio in breadth :: fo is the num ber of trees to a fourth number; whofe fquare root is the number in breadth; And as the ratio in breadth : is to the ratio in length: fo is the number of trees: to a fourth, whofe root is the number in length. As 3 2 600 : 400. And in breadth. 400 20 number As 23: 600.900. And 900=30-number in length. As: 7:30-1: 203. And as I 7 :: 20—1 to 133. And 203X133=26999 fquare yards the Anfwer. PROB. VII. Admit a leaden pipe 4 inch diameter will fill a ciftern in 3 hours; I demand the diameter of nother pipe, which will fill the fame ciftern in 1 hour. RULE. As the given time is to the fquare of the given diameter, fo is the required time to the fquare of the required diameter. 75; and ,75 × 75=5625: Then, as 3h:,5625: the: 1,6875 inverfely, and √11,6875=1,3 inch nearly, Answer. PROB. VIII. If a pipe, whofe diameter is 1,5 inch," -fill a ciftern in 5 hours; in what time will a pipe, whose diameter is 3,5 inches fill the fame ? 1,5X1,5 2,25; and 3,5X3,5=12,25: Then, a 2,25: 5: 12,25: 91 hour inverfely, 54 minutes 39 feconds. Anf PROB. IX. If a pipe 6 inches bore, will be 4 hours in running off a certain quantity of water; in what time will 3 pipes, each 4 inches bore, be in difcharging double the quantity? 6×6=36, 4X4—16, and 16×3=48. Then, as 36: 4h. :: 48: 3h. inverfely, and as 1w.: 3h. :: 2w.: 6h, Answer. g of ». PROB. PROB. X. Given the diameter of a circle to make an other 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 diam. eter 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 bethe diameter required. There is a circle, whofe diameter is 4 inches; I demand the diameter of a circle 3 times as large? 4×4=16; and 16X3=48; and ✔ 48=6,928+ inches, Anfwer. PROB. XI. To find the diameter of a circle equal in area to an ellipfis (or oval) whofe tranfverfe and conju gute 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 1728=41,569+the Anf. PROB. XII. Two fhips fail from the fame port; one goes due north 45 leagues, and the other due west 76 leagues: How far are they afunder ? 45X45 2025. 76×76=5776. Then 5776X2025 780, and, 7801-88,32 leagues, the Anf. EXTRACTION * The tranfverfe and conjugate are the longest and shortest diameters of an ellipfis; they pass through the centre, and crofs each other 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 METHOD. 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 greatest 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, call ing it the triple fquare, and the quotient by 30, calling it the triple quotient, and the fun of thefe call the di vifor. 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 last 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 last; under all, fet the cube of the laft quotient figure, and call their fum the fubtrahend. 7. Subtract the fubtrahend from the dividend, and to the remainder bring down the next period for a new div. idend, 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 square root. EXAMPLES. Brought over. 75×75×300 = 1687500 = 2d Triple square. 75×30 2250 2d Triple Quotient, 4. What is the cube root of,008649 ? Anf. 327. Anf. 4,39. Anf.,2052+ 5. What is the cube root of? Anf. 4. To find the true denominator, to be placed under the remain der, 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 |