EXAMPLES. 1 What is the squared Biquadrate of 48 ? Ans. 281.9230429056. 2 What is the square biquadrate Ans. 96. root of 721389578838336? 3 What is the square biquadrate. Ans. 48. root of 28179280429056 ? 4. What is the square biquadrate Ans. 384. root of 472769874482815188096? S OF THE CUBED.CUBE ROOT. Q. What is a Cubed-Cube ? A. Any number involved nine times is a Cubed: Cube. EXAMPLES. 1 What is the Cubed-Cube-Root Ans. 96. 2. of 692533995824480256 ? 2 What is the Cubed-Cube-Root Ans. 48. 09. of 1352605460594688 ? 3. What is the Cubed-Cube-Root Ans. 384. 5. of 18154.363180141255228864 ? OF THE SQUARE SURSOLID ROOT. Q. What is a Squared Sursolid? A. Any number involved ten times, produces a Squared Sursolid. EXAMPLES. 1. What is the Squared Sursolid Root Ans. 48. of 64925062108545024? 2. What is the Squared Sursolid Root Ans. 96. of 66483263599150104576 ? 3. What is the Squared Sursolid Root Ans. 384. 3. of 697437 54611742420055883776? OF THE THIRD SURSOLID ROOT. Q. WHAT is a Third Sursolid ? A. Any number involved eleven times, produces a third Sursolid. EXAMPLES. 1. What is the third sursolid root Ans. 23. of 952809757913927 ? 2. What is the third sursolid root Ans. 48. of 3116402981210161152 ? 3. What is the third Sursolid root? of 6382393305518410039296 ? of} Ans. 96, OF THE SQUARED SQUARE CUBE ROOT. HAT is a Squared Square Cube ? EXAMPLES. 1. What is the root of this Squared Square? A. 48. Cube 149587343098087735296 ? 2. What is the root of this Squared Square A. 96. Cube 612709757329767363772416? 3. What is the root of this Squared Square? A. 384. Cube 102795639440290,9029176039807836? A GENERAL RULE FOR EXTRACTING THE ROOTS OF ALL POWERS. 1. PREP RE the given number for extraetion, by pointing off from the unity place, as the root required directs. 2. Find the first figure in the root by your own judgment, or by inspection into the table of powers. 3. Subtract it from the given number. 4. Augment the remainder by the next figure in the given number, that is by the first figure in the next point, and call this your dividend. 5. Involve the whole Root, last found, into the next inferior power to that which is given. 6. Multiply it by the index of the given power, and call this your divisor. 7. Find a quotient figure by common Division, and annex it to the root. 8. Involve all the root thus found, into the given power 9. Subtract this power (always) from as many points of the given power as you have brought down, beginning at the lowest place. 10. To the remainder bring dowo:the first figure of the next point for new dividend. 11' find a new divisor as before, and in like manner proceed till the work is ended. 4 x 4 X 3 =48 Divisor 48 X 8 X 48 =110592 Subtrahend 48 X 48 X 3 =6912 Divisor 487 X 487 X 487 =115501303 Subtrahend. ? What is the Biquadrate root of 56249134561 ? 4 X 4 X 40-256 Divisor 48 X 48 X 48 x 48=5303416 Subtrahend 48 X 48 X 48 X 4=442368 Divisor 487 X 487 X 487 X 487=56249134561 Sub. Note. This General Rule I receive from my worthy Friend, William Montaine, E:. F.R. S and teacher of the Mathematics at Shad-Thames, OF SIMPLE INTEREST. What particular Letters are used here? . A. These : P, any Principal. T, the time. A, the Amount. A. It signifies only the simple interest of 11 for one year, at any proposed rate of interest per cent, and is thus found, 100 : 6 :: 1 : 0.06 100 : 5 :: 1 : 0.05 A TABLE OF RATIONS. CASE 2. Q. When P, T and R are given to find A, how is it discovered ? A. Thus ptr +p=A. Note. Any quantity of Letters put together like a word, denote continual Multiplication. EXAMPLES. 1. What sum will 5671 108 amount to in 9 years, at 6 per cent per annum ? Ans. 8731, 198. 2. What will 5081 148. amount to in 1 year at 5 per cent. per annum ? Ans. 5341 2s 8d 1. 6 qrs. 3. What will 600l 14s. amount to in 10 years, at 4 per cent. per annum? Ans. 8711 os. 3d. 2.4.qrs. 4. What will 40001, amount to in 5 years, at 3 per cent. per annum ? Ans. +7001. Note. When the time given does not consist of whole years, then reduce the odd time into Decimal parts of a year. And unless such parts of a year chance to be 1-10 of a year, the best way will be to reduce the odd times into days, and then work with the Decimal parts of a year, that are equivalent to those days. 10 300 70 .75 A Table for the ready finding the Decimal Parts of a year equal to any Number of Days, or Quarters of a Year. Days | Dec. Pts. | Days Dec. Pts. Days Dec. Pts. 1 1.00274 .027397 100 .273973 2 .005479 20 .054794 200 .547945 3 .008219 30 .082192 .821918 010959 40 .109589 360 .000000 5 .013699 50 .136986 6 016438 60 .164383 7 .019 178 .191781 of a Year .25 8 .021918 80 .219178 of a Year .50 9 .024651 90 .246575 1 ) 3 of a Year Note. When the true Number of Days cannot be found at one view in this Table, then both them and their Decimals must be taken out of the Table at twice or thrice, as their Number requires, and added together. So the Decimal Parts of a year=236 days are ihus found, 2005.547945 EXAMPLES. 5 What will 72001 amount to in 6 years, at 5 per cent. per annum? Ans. 95401. 6 What will 1110l 18s amount to in 12 years at 5 per cent. per annum? Ans. 18191 is 11d 2.8qrs. 7 What will 2801 108 amount to in 3 years and 148 days, at 5 per cent. per ann. ? Ans. 3281 58 2d 3.58+qrs. 8 What will 1961 amount to in 189 days, at 4 per cent. ann. ? Ans. 2001 1s 2d 1.23+qrs. CASE 2. Q. When A, T. & R. are given to find P, how is it discovered? a A. Thus:-=P trti EXAMPLES. 1 I demand what principal will amount to 8731 19s in 9 years, at 6 per cent. Ans. 5671 10s. 2 1 demand what principal will amount to 5341 28. 8d 1.6 qr. in one year, at 5 per cent. per ann.? Ans. 5081 148. 3 I demand what principal will amount to 95401 in 63 years, at 5 per cent. per ann. ? Ans. 72001. 4 1 demand what principal will amount to 18191 18 11d 2.8 qrs. in 12 years, at 5 per cent. per ann.? Ans. 1110) 18s. per |