SECOND METHOD BY APPROXIMATION. RULE 1. Divide the refolvend by three times the affumed root, and referve the quotient. 2. Subtract one twelfth part of the square of the affumed root from the quotient. 3. Extract the square root of the remainder. 4. To this root add one half of the affumed root, and the fum will be the true root, or an approximation to its -Take this approximation as the assumed root, and, by repeating the process, a root farther approximated will be found, which operation may be farther repeated as often as neceffary, and the root discovered to any affign. ed exactness. NOTE. In order to find the value of the fire afsumed root, in this or any other power, divide the refolvend into periods, by beginning at the place of units, and including in each period, fo many figures as there are units in the exponent of the root; viz, 3 figures in the cube root; ---4 for the biquadrate, and so n : Then by a table of powers, or otherwise, find a figure, which, (being involved to the power whose exponent is the fame with that of the required root) is the nearest to the value of the first period of the refolverd at the left hand; and to Dat annex fo many cyphers as there are peri periods remaining in the integral part of the refolvend; this figure, with the cypbers annexed, will be the affumed root: And it is of no importance whether the figure thus chofen, be, when in volved, greater or less than the left hand period. 1. i What is the cube root of 436036824287? 7000==affumed root... 8 : 21,000)436036824287(20763658,2994 Subtr.7000 × 7000-12-4083333-3333* 16680324,9661-4084,15 Add the affumed root,=3500 And it gives the approximated root=7584,15. こ For the fecond operation, use the approximated root as the affumed one, and proceed as above. THIRD METHOD BY APPROXIMATION. f. Assume the root in the usual way, then multiply the square of the affumed root by 3, and divide the refolvend by this product; to this quotient add of the af fumed root, and the sum will be the true root, or an ap. proximation to it. : 2: For each fucceeding operation let the last approx. imated root be the assumed root, and proceeding in this manner, the root may be extracted to any assigned ex. actness. : What is the cube root of 7?" Let the affumed root be 2. the divisor. Then 2X2X3=121 12)7,0(,583, to this add of 2=1,333, &c. that is, 583+1,333 1,916 approximated root. Now affume. 1,916 for the root, then, by the second 7 process, the root is-2+1,916=1,9126, &c. 3×1,916 APPLICATION AND USE OF THE CUBE ROOT. 1. To find two mean proportionals between any two given numbers. RULE I. Divide the greater by the less, and extract the cube root of the quotient. 2. Multiply the root, so found, by the least of the given numbers, and the product will be the leaft. 3. Multiply this product by the same root, and it will. give the greatest.. Uz EXAMPLES 1 : 750÷6=125, and 125=5. Then 5×6=30=leafst, and 30×5=150=greateft. Anf. 30 and 150. Proof. As 6:30 :: 150: 750. 2. What are the two mean proportionals between 56 and 19208? Ans. 392 and 2744 NOTE. The folid contents of fimilar figures are in proportion to each other, as the cubes of their similar fides or diameters. 3. If a bullet 6 inches diameter weigh 32lb.; what will a bullet of the fame metal weigh, whose diameter is 3 inches? 6X6×6=216. 3×3×3=27. As 216: 32lb. :: 27: 4lb. Ans. 4. If a globe of filver of 3 inches diameter, be worth £45, what is the value of another globe of a foot di ameter? 3×3×3=27. 12X12X121728. As 27:45 1728: £2880 Ans. The fide of a cube being given, to find the fide of that cube which shall be double, triple, &c. in quantity to the given cube. RULE. Cube your given fide, and multiply it by the given proportion between the given and required cube, and the cube root of the product will be the side sought. If a cube of filver, whose side is 4 inches, be worth 150, I demand the fide of a cube of the like filver, whose value shall be 4 times as much? 5. 3 4X4X4-64, and 64X4=256.256 6,349 Xinches, Anfwer. 6. There is a cubical vessel, whose side is 2 feet; I demand the fide of a vessel, which shall contain three times as much? X2X2=8, and 8×3=24. inches, Answer. 5 13 24=2,8842 feet, 7. The diameter of a bushel measure being 18 inches, and the height 8 inches; I demand the side of a cubick box, which shall contain that quantity? Ans. 12,908 inches. 8. Suppofe a ship of 500 tons has 89 feet keel, 36 feet beam, and is fixteen feet deep in the hold; what Mare the dimenfions of a fhip of 200 tons, of the same mould and shape? 89×89×89-704969=cubed keel. As 500: 200 :: 704969: 281987,6 cube of the required keel. 3 281987,665,57 feet, the required keel. As 86: 65,57:: 36: 26,522=260 feet beam, nearly. As 89: 65,57 :: 16 : 11,7 feet, depth of the holde EXTRACTION OF THE BIQUADRATE ROOT. RULE. Extract the square root of the refolvend, and then, the square root of that root, and you will have the biquadrate root. What is the biquadrate root of 20736? 1. Divide the refolvend by fix times the square of the affumed root, and from the quotient fubtract i's part of the square of the affumed root. 2. Extract the square root of the remainder. 3. Add of the assumed root to the square root, and the sum will be the true root, or an approximation to it. 4. For every succeeding operation (either in this or in the following, method) proceed in the farne manner as in the first, each time using the last approximated root for the assumed root.. The biquadrate root of 20736 is required. Here, 10 is the affamed Root. 10X10X6=600) 20736(34,56 Subtract, 10X10+18= 5,5555 1 √29,0044-5,38 Add of 10=6,66 Approximated root 12,C4, to be madi the affumed root for the next operation. Divide the refolvend by four times the cube of the af firmed root: To the quotient add three fourths of the af. fumed root, and the sung will be the true root, or an approximation to it. Let the biquadrate of 20736 be required, as before. 10 X 10 X 10X4=4000) 20736(5,184 Add of 107,5 Approximated root 12,684, to be made the affumed root for the next operation. EXTRACTION OF THE SURSOLID ROOT, I. BY APPROXIMATION. A Particular RULE. Divide the refolvend by five times the affumed root, and to the quotient add one twentieth part of the fourth power of the same root.. 2. From |