« ForrigeFortsett »
14. In a garden is a square bowling-green, a side of which is 30 yards, and near to it is a rectangular grass plot. The number of square yards in the area of the grass plot is a mean proportional be. tween and the number of square yards contained in the grass plot and bowling-green together. Also the number of square yards contained in the square described on the diameter of the grass plot, is a mean proportional between 10 and the number of square yards contained in the aforesaid square, increased by the number contained in the bowling-green. Required the area and sides of the grass plot?
Ans. 48 square yards. 15. After A had won four shillings of B, he had only half as many shillings as B had left. But had B won six shillings of A, then he would have had three times as many as A would have had left. How many had each ?
Ans. A had 36, and B 84. 16. A and B playing at Bowls, says A to B, If you will give me a guinea, I will bet you half a crown to eighteen pence on each game, and will play 36 games together. B won his guinea back again, and i 17s. besides. How many games did each win?
Ans. A won 8, and B 28 games. 17. A Vintner sold at one time, 20 dozen of port wine, and 30 of Sherry, and for the whole received £120; and at another time sold 30 dozen of port and 25 of sherry, at the same prices as before, and for the whole received £140. What was the price of a dozen of each sort of wine ?
Ans. The prices of port and sherry per dozen, were 3 and 2 pound respectively.
18. A Person engaged to reap a field of 35 acres, consisting partly of wheat, and partly of rye. For every acre of rye he received five shillings; and what he received for an acre of wheat, augmented by one shilling, is to what he received for an acre of rye as 7 to 3. For his whole labour he received £ 13. Required the number of acres of each sort? Ans. 15 acres of wheat, and 20 of rye.
19. Two pieces of cloth of equal goodness but of different lengths were bought, the one for £5, the other for £6 10. Now if the lengths of both pieces were increased by 10, the numbers resulting, would be in the proportion of 5 to 6. How long was each piece, and how much did they cost a yard?
Ans. Price 5s. and the lengths are 20 and 26 yards. 20. A gentleman had some of his horses at grass at 3 shillings each week, and the rest at livery stables at 10 shillings each a week. The horses in the stables cost him twice as much a week as the horses at grass. But he finds that if he had sent three horses to grass out of the stable, the expence of the stables would have been only 6 shillings a week more than the grass. How many horses had he?
Ans. 15 at grass, and 2 in the stable.
witle water; then water is conveyed out of B into A in the following manner. First, a number of gallons is taken, which is less by two than the square root of the number of gallons in A, then a quanity less than the former by two gallons, and so on. Now when B is in this manner exactly emptied, A is exactly full : and it is known that 8 gallons were taken out of B at one time, after which the quantity ieft in B was 12 gallons. Required the number of gallons of wine in A?
Ans. 256 gallons. 22. There is a certain number, of which 9 times its cube, seven times its square, and 5 times the number itself, will make 547: What is that number?
Ans. 3.64481861. 23. What is the distance between two towns, of which it is known that the cube of that distance, minus 22 times the said distance is 24 English miles ?
Ans. 5:162277. 24. A person being asked how old he was, answered, that the
qua• druple of the cube, the treple of the square, and the double of his age, amounted to 638,712 years. What was his age ?
Ans. 54 years. 25. Required the root of the cubic equation, x3-3r? + 2r=27?
Ans. 4'111069. 26. Required the root of the Biquadratic equation x* +5r9 +4.ro + 313105 ?
BND OF THE ALGEBRA
Definition. An Algebraic product, of which the difference between cvery two adjacent factors is equal to the same given number, is called a factorial.
Notation. In a factorial are to be considered the number of factors, otherwise called the exponent, the first factor and the common difference whether + or
Let m be the first factor, n the number of factors, and c the common difference; then every factorial may be thus indicated malo let n=4, and c=1, then will mak=m4i=m (m+1) (m+2) (m+3). Again, if n=5, and c=-1. Then will makc =msii m (m-1) (m—2) (m—-3) (m—4).
Proposition. Any two factorials in which the base of the one is equal to the sum formed by adding the product of the exponent and common difference of the other to its exponent, may be reduced to one.
For let male and [m+nc]ble, be the two factorials, the base of the latter being formed as announced in the proposition; then because c is the common difference, and ma is the first factor of the factorial mnt, the second factor will be mtc, the third m +2c, the fourth m+3c, and so on; therefore in n+ 1 factors, the (n + 1)th factor from the first, will be the first factor, together with n times the common difference c; therefore if the factorial (m + nc) olc be annexed to the factorial mnie, as two factors, the product will be the factorial in* + ble
Problem. To resolve a given factorial into two factorial factors in which the factors of each shall have the same common difference as the factors of the given factorial, and the one a given exponent less than that of the given factorial.
Rule. 1. Take the less of the two given exponents from the greater, and the remainder will be the exponent of the factorial factor, which is not given.
2. To the base of the given factorial, apply cither of the exponents of the two factorial factors, and the common difference, and the quantity thus formed will be one of the factorial factors.
3. To the saine base add the product of the exponent and common difference of the factorial factor thus completed, and to the sum as a base apply the remaining exponent of the two factorial factors, and the common difference, then the quantity thus formed will be the other factorial factor.
Examples. 1. Resolve muc into two factorial factors, so that one of them may have the given exponent r.
By rule n— will be the exponent of the other, now if male be the one factorial, (m+rc) x-re will be the other.
Or if mais be the one factorial [m+(n-r)c]rs will be the other.
2. Resolve male into two factorial factors, so that one of them may have the given exponent 1.
By rule n-1 will be the exponent of the other ; now, therefore, if miks =m, be the one factorial factor, then (m-cn-li will be the other, or if ma-slc be the one factorial, then will [m+(n1c]" m+(n-1)c be the other.
3. Resolve mall into two factorial factors, so that one of them may have the given exponent 1.
By rulen-1 will be the exponent of the other; if therefore mili be the one factorial factor, then will (m+1)=-111 be the other, or if the one factorial factor be mawili then will the other factorial factor be (m+7–1)ili=m+1-1.
4. Resolve male into two factorial factors, so that one of them may bave the given exponentt.
By rule n— will be the exponent of the other ; therefore, if małe be the one factorial factor, then will (m—rc)-=rī be the other.
Or if mark be the one factorial factor, then will [m-11-r)cm be the other.
5. Resolve mai into two factorial factors, so that one cf them may have the given exponent 1.
By rule n-1 is the exponent of the other ; therefore if m=m de the one factorial factor, the other will be (m-1)=-11.
Or, if the one factorial factor be moun the other will be (mon+1)=m-n+1, which is the last term of the factorial mais?
Definitions. 1. The different positions in which the same things can be placed, are called their permutations.
2. The different collections that can be formed out of a given number of things, are called combinations.
3. Each collection of things combined out of a given number of things, is called a parcel.
4. The number of things to be combined, are called the base or first order, and the things combined are said to be raised from that base.
5. The whole number of parcels that can be formed by taking a given number of things, out of a greater given number of things into each parcel, is called an order of combination.
6. The number of an order of combination, is the same as the number of the letters forming each parcel.
7. Combinations which have no two letters alike in any parcel, are called combinations without repetitions.
8. Combinations which have parcels containing two or more things alike, are called combinations with repetitions.
Theorem 1. (1.) The number of permutations of m things taken n at a time is mali
As any one of the m quantities may stand first, there will be m different ways of filling up the first place.
Now whichever of the m things stand first, there will remain (m-1) things to make choice of to fill up the second place ; let, therefore, each of the m things be supposed to be placed and to stand first when combined in separate parcels ; then since any one of them may be adjoined with the remaining m-1 things m-
-1 different ways, let this be done, and each one of the m things will be multiplied m 1 times when placed in two's; therefore the whole number of the m things will be multiplied m-1 times; and therefore the number of parcels in two's will be m(m-1)=mii which is equal to the number of their permutations.
Now, since each parcel contains two things, there will remain m-2 things; therefore each parcel in two's may be adjoined m-2 times with each of the remaining things ; let this be done, and each parcel in two's will be multiplied m-2 times; and consequently the whole number of parcels in two's will be multiplied m-2 times, in order to produce the parcels in three's. But the number of parcels in two's is m(m-1); therefore the number of parcels in three's is mm-1) (m-2)=mstí