HORNER'S METHOD. Ex. 1. Required the cube root of 9295677. OPERATION. 0 16 64 16 28959 32 27125 4800 1834677 625 1834677 120 5425 5 650 125 5 130 5 1350 607500 0 Ans. 453. The greatest cube 92959677 453 contained in the lefthand period we find to be 64, whose root, 4, we write as the first figure of the required root. This figure, 4, we write under the cipher of the first column; and adding it to the cipher obtain 4, which sum multiplied by the gives 16; and the result, 16, we write under the cipher of the second column, and by addition obtain 16, which sum, multiplied by the 4, gives 64; and the result, 64, we write in the last column under the left-hand period, 92, of the given number. The 64 subtracted from the 92 above it gives for a remainder 28. We next add the 4 to the last term, 4, of the first column, obtaining 8; and the result, 8, multiplied by the 4, gives 32, which we write under the last term, 16, of the second column, and, adding the same together, obtain We next add the 4 to the last term, 8, of the first column, obtaining 12; and, annexing one cipher to the last term, 12, of the first column, obtaining 120, two ciphers to the last term of the second column, obtaining 4800, and to the remainder in the last column bringing down the next period, 959, obtaining 28959, we complete the work preparatory to the finding of the second root figure. 3 1353 48. To determine that root figure, we take the last term, 4800, of the second column, for a trial divisor, and the last term, 28959, of the last column, for a dividend; and, dividing, 6 would appear to be the second figure of the root. This, on trial, however, is found to be too large; we therefore take 5, which answers. This 5 we add to the last term, 120, of the first column, obtaining 125; which sum, 125, multiplied by the 5, gives 625, and that product, added to the last term, 4800, of the second column, gives 5425; and this result, 5425, multiplied by the 5, gives 27125, which, written in the last column and subtracted from the figures above it, gives a remainder 1834. Then we add the 5 to the last term, 125, of the first column, obtaining 130; and the result, 130, multiplied by the 5, gives 650, which we write under the last term, 5425, of the second column, and by addition obtain 6075. We next add the 5 to the last term, 130, of the first column, obtaining 135; and, annexing one cipher to the last term, 135, of the first column, obtaining 1350, two ciphers to the last term, 6075, of the second column, obtaining 607500, and to the remainder in the last column bringing down the next period, 677, obtaining 1834677, the work is completed preparatory to finding the third root figure. To determine that figure, as before, we divide the last term of the third column by the last term of the second. We thus obtain 3, which, added to the last term of the first column, gives 1353, which sum, multiplied by the 3, gives 4059; and that product, being added to the last term of the second column, gives 611559; and that sum, multiplied by the 3, gives 1834677, which being exactly as large as the last term of the third column, on being written under it and subtracted there is no remainder. The given number is therefore a perfect power, and the cube root sought is 453. In practice, the work may be performed with less figures, by writing down in the several columns only the results. RULE.- Commence as many columns as there are units in the exponent of the root to be extracted, by writing the given number as the head of the right-hand column, and a cipher as the head of each of the others. Separate the given number into as many periods as possible of as many figures each as the exponent of the root requires; and having found the nearest root of the left-hand period, write it as the first figure of the required root. Write this figure in the first column, and, having added it to what stands above it, multiply the sum by the same figure, and write_the_product in the second column; add, in like manner, in the second column, and multiply the sum by the same figure, writing the product in the third column; and so proceed, writing the last product in the last column, and subtracting it from what stands above it. Then add the same figure to the last term of the first column, multiply the sum by the same figure, and add the product to the last term of the second column; and so on, writing the last product in the last column but one. Repeat the process, stopping each time with one column farther to the left, till the last product shall fall in the second column. Add the figure found for the root to the last term of the first column; annex one cipher to the last number in the first column, two ciphers to the last number in the second column, and so on; and to the last number in the last column bring down the next period for a dividend. Take the last term of the column next to the last for a trial divisor, and see how often it is contained in the dividend, and write the result as the next figure of the root. Add this figure to the last term of the first column, multiplying the sum by the same figure, add the product to the second column, and so on; proceed as before, till all the periods have been brought down, or an answer sufficiently exact has been obtained. NOTE 1.- When any dividend will not contain its corresponding trial divisor, write a cipher in the root, bring down to the dividend another period, annex an additional cipher to the last term of the first column, two additional ciphers to the last term of the second column, and so on; and use the same trial divisor as before, increased however by the additional ciphers. NOTE 2. When the given number does not have an exact root, periods of ciphers may be annexed. NOTE 3. When the root is required to many places of decimals, the work may be contracted by rejecting one figure at the right from the number in the column next to the last, two from the number in the column next farther to the left, and so on, and otherwise proceeding as directed in the rule, except that the new figure of the root is not added to the first column. As soon as all the figures are rejected in the number of the first column, the remainder of the work may be performed as in contracted division of decimals (Art. 276). EXAMPLES. 2. Required the fourth root of 1.016397 to eight places of 4. What is the cube root of 43614208? 5. What is the cube root of 1.05 to six places of decimals? 6. What is the fifth root of 184528125? 7. Required the fourth root of 100 to mals? Ans. 1.016397. Ans. 45. six places of deciAns. 3.162278. 8. Required the fifth root of the fourth power of 9 to seven places of decimals? Ans. 5.7995466. APPLICATIONS OF POWERS AND ROOTS. 534. A TRIANGLE is a figure having three sides and three angles. When one of the sides of a triangle is perpendicular to another side, the opening between them is called a right angle, and the triangle is called a right-angled triangle. The lowest side, A B, is called the base of the triangle AB C, the side BC the perpendicular, the longest side, A C, the hypothenuse, and the angle at B is a right angle. Also, the line BC, being perpendicular to the base, is the altitude. A Hypothenuse. Base. Perp Perpendic. A 535. The square described upon the hypothenuse of a rightangled triangle is equivalent to the sum of the squares described upon the other two sides. 536. To find the hypothenuse, the base and perpendicular being given. Add together the square of the base and the square of the perpendicular, and extract the square root of the sum. Thus, if the base be 4 and the perpendicular 3, the hypothenuse will equal √42 +32 = 25 = 5. 537. To find the perpendicular, the base and hypothenuse being given. Subtract the square of the base from the square of the hypothenuse, and extract the square root of the remainder. Thus, if the base be 4 and the hypothenuse 5, the perpendicular will equal No 52 42 = 93. 538. To find the base, the hypothenuse and perpendicular being given. Subtract the square of the perpendicular from the square of the hypothenuse, and extract the square root of the remainder. Thus, if the perpendicular be 3 and the hypothenuse 5, the base will equal 523 16 4. = = 539. All triangles having the same base are to each other as their altitudes. All similar triangles, and other similar rectilineal figures, are to each other as the squares of their homolo gous or corresponding sides. Thus, the triangles ACE and A C D, having the same base, A C, are to each other as the altitude E C of the one is to the altitude D C of the other. A B E D Also, the triangles A CE and B C D, having their corresponding angles the same, and their sides in direct proportion, are said to be similar, and are to each other as the squares of their corresponding sides, or as (A E)2 is to (B D), (A C) is to (B C)2, and (C E)2 is to (C D). Likewise the larger square, of which A C is one of the equal sides, is to the smaller square, of which B C is one of the equal sides, as (A C) is to (B C). 540. All circles (Art. 143) are to each other as the of their diameters, semidiameters, or circumferences. The circumference of a circle is the line which bounds it; and the diameter is a line drawn through the centre, and terminated by the circumference; as A B and C D. Then, the larger circle, of which A B is the diameter, is to the smaller, of which C D is the diameter, as (A B)2 is to (CD)2, &c. A squares B 541. To find the side, diameter, or circumference of any surface, which is similar to a given surface. State the question as in Proportion, and square the given sides, diameters, or circumferences, and the square root of the fourth term of the proportion will be the answer required. Thus, if 12 feet be the length of a side of a triangle whose area is 72 square feet, the length of the corresponding side of a |