EXAMPLES.

I. What are the two mean proportionals between 6 and 750?

750÷6=125, and 125=5. Then 5×6=30-least, and 30×5150-greateft. Anf. 30 and 150.

Proof. As 6: 30: 150 750.

2. What are the two mean proportionals between 56 and 19208? Anf. 392 and 2744.

NOTE. The folid contents of fimilar figures are in proportion to each other, as the cubes of their fimilar fides or diameters. 3. If a bullet 6 inches diameter weigh 32lb.; what will a bullet of the fame metal weigh, whofe diameter is 3 inches?

6X6×6=216. 3×3×3=27. As 216: 32lb. :: 27: 4ib. Anf.

4. If a globe of filver of 3 inches diameter, be worth £45, what is the value of another globe of a foot diameter ?

3X3X3=27 1728: £2880 Anf

12X12X12=1728. As 27: 45 **

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 fide fought. 5. If a cube of filver, whofe fide is 4 inches, be worth £50, I demand the fide of a cube of the like filver, whofe value fhall be 4 times as much?

3

4X4X464, and 64X4=256./256—6,349 Xinches, Anfwer.

6. There is a cubical veffel, whofe fide is 2 feet; I demand the fide of a veffel, which fhall contain three times as much?

X2 X2=8, and 8×3=24. e inches, Anfwer.

:3

242,884—2 feet,

7. The diameter of a bufhel measure being 181/ inches, and the height 8 inches; I demand the fide of a cubick box, which fhall contain that quantity?

Anf. 12,908 inches.

8. Suppofe a fhip of 500 tons has 89 feet keel, 36 feet beam, and is fixteen feet deep in the hold; what "are the dimenfions of a flip of 200 tons, of the fame mould and fhape?

89X89X89=704969-cubed keel.

As 500: 200 :: 704969: 281987,6 cube of the required keel.

281987,6=65,57 feet, the required keel.

As 86 65,57 36 26,522-26 feet beam, nearly.

As 8965,57 :: 16: 11,7 feet, depth of the hold * EXTRACTION OF THE BIQUADRATE ROOT.

RULE. Extract the fquare root of the refolvend, and then, the fquare 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 fquare of the affumed root, and from the quotient, fubtract is part of the fquare of the affumed root.

2.

Extract the fquare root of the remainder.

3. Add of the affumed root to the fquare root, and the fum will be the true root,, or an approximation to it.

4. For every fucceeding operation (either in this or in the following, method) proceed in the farme manner as in the first, each time ufing the last approximated root. for the affumed root..

The biquadrate root of 20736 is required. Here, to is the affumed Root: 10X10X6=600)20736(34,56

Subtract, 10X10+18=

5,5555

29,0044-5,38`

Add 3 of 10=6,66

Approximated root 12,04, to be, mada

the affumed root for the next operation.

METHOD 2:

Divide the refolvend by four times the cube of the affumed root, To the quotient add three fourths of the affumed root, and the fum will be the true root, or an ap proximation to it.

Let the biquadrate of 20736 be required, as before. The affumed root is 10.

10X 10X 10X4=4000)20736(5,184!

Add of 10=7,5

Approximated root 12,684, to be made.the

affumed root for the next operation

EXTRACTION OF THE SURSOLID ROOT, BY APPROXIMATION.

I.

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 fame root..

2. From

2.

From the fquare root of this fum fubtract one fourth part of the fquare of the affumed root.

3. To the fquare root of the remainder add one half of the affumed root, and the fum is the root required, or an approximation to it.

NOTE This Rule will give the root true to five places, at the leaft, (and generally to eight or nine places) at the first process.

Required the Surfolid Root of 281950621875

200 affumed root.

1900) 281950621,875 quotient,

Add 200X200X200X200÷20 S 80000000

✔ 361950621,87519025nearly. Subtract, 200 X 2,00-4—10000

A GENERAL RULE FOR EXTRACTING THE ROOTS OF ALL POWERS.

RULE* I. Prepare the given number for extraction, by pointing off from the unit's place, as the required root directs.

2.. Find

*The extracting of roots of very high powers will, by this rule, be a tedious operation--The following method, when practicable, will be much more convenient.

When the index of the power, whofe root is to be extracted, is a compofite number, take any two or more indices, whofe product is equal to the given index, and extract out of the given number a root anfwering to another of the indices, and fo on to the laft,

Thus, the fourth rootfquare root of the fquare root; the fixth root-fquare root of the cube root; the eighth rootfquare root of the fourth root; the ninth root the cube root of the cube root; the tenth root-square root of the fifth root; the twelfth root-cube root of the fourth, &c.

2. Find the first figure of the root by trial, or by inSpection into the Table of Powers, and fubtract its pow er from the left hand period.

3. To the remainder bring down the first figure in the next period, and call it the dividend.

4. Involve the root to the next inferiour power to that which is given, and multiply it by the number, denoting the given power, for a divisor.

5.

Find how many times the divifor may be had in the dividend, and the quotient will be another figure of

the root.

6: Involve the whole root to the given power, fubtract it from the given number as before.

and

7. Bring down the firft figure of the next period to the remainder for a new dividend, to which find a new divifor, as before, and, in like manner, proceed till the

whole be finished..

EXAMPLES.

⚫. What is the cube root of 20346417?

19683 2d Subtrahend.

272 X 3 2187)6634-2d Dividend.

2733=20346417=3d Subtrahend.

The Method of Operation

X2X28 root of the first period, or 1st subtrahend.

2X24 (next inferiour power) and,