It is stated in the Abeille Medicata that the seed of the pompion is & sure remedy against tape-worm. Aboat an ounce of these seeds, pounded in a mortar with a sufficient quantity of sugar, is administered at a time, and repeated for three days, which is generally sufficient to effect a complete care. LETTER TO THE LITTLE FOLKS, In my last I promised to write you about excuses for Absence and Tardiness. Some little boys think, perhaps, that almost any excuse will answer for Tardiness. Now listen one minute. When I was a little boy, my Mother allowed me to have a party. Of course I was very happy, and I invited all the little boys of my acquaintance, to come at two o'clock and spend the afternoon. When two o'clock came, all the boys were there. One had a bad cold, but he was as ready for play as any one. He had to be a little careful, but that did not prevent his coming. Another had a sore finger, but he came. Another came without his dinner, because, he said, it was two o'clock before it was ready. Another had work to do, but he arose earlier in the morning and kept busy, so that all was done before two o'clock. Another said his Mother wanted him to go to the store, and he ran all the way for fear it would be two o'clock before he shonld get back, All had extra work to do, bat all were there in time. We had a nice play. A person older than I, invited the same little boys to come and see him a day or two afterward. He wished them to come at 9 o'clock in the morning. Several of the boys did not get there till after 9. What do you think they gave as excuses for not coming at the time appointed ? I will tell you some of them. One little boy said he had a bad cold; another had a sore finger; another had to wait for his breakfast ; another had work to do; another had to go to the store for his Mother; all had soncething to do that kept them until after 9 o'clock, Now can you tell me what made the difference? Some little fellow says, one was in the afternoon, and the other in the forenoon." That was not the difference; for the same person invited them again in the afternoon, and the same boys came late with the same excuses. I will tell you. One was a party, and the other was a school. Now tell me,—which is of more value to you, a party or a school ? Parties are good in their place, but schools are of more value, Never give an excuse for being late at or absent from school, that you would not give for being late at or absent from, a party. Then you will have a good time at school, and enjoy yourselves all the time. Then you will be respected by others, and respect yourselves. A great many boys are too sick to be at school, who can go out sliding or skating. Who loves them 8 A Happy New Year to all punctual boys and girls. PLATTEVILLE, FEB. 1859. ONE WHO LOVES THEM Mathematical Department. 1 Solution of Problem 9.-Given x3 + -2 Nx to find a. Let wx=y then 1 x=y and x'=ye. The equation now becomes yo +2y. Clearing of ya fractions and transposing gives yo-2y+1=0. Resolving this equation into factors, we find y-2y+1=(y-1)(y"+y+y+y+y-y-y-1)= 0. Placing y-1=0, we find y=1. Hence Næ=1, and «=l=1. Ans. L. CAMPBELL. 2 2 Solution of Problem No. 12.-ED. JOURNAL:-Allow me to offer the following as a general solution for all problems of the same class as problem 12. It is required to find the value of x in the equation om — 2014 —U, - (1.) when m is less than n, and u a maximum remainder. Let å be increased by the variable quantity y. Then (x+y)a=2* +m x --by + m{m,-1) 200-23 n(n-1) y' + etc.; and (x+y)•=** +n x sady + Lacrambya +, etc.; and by sub m (m-1) tracting and putting o = the remainder, we get 2h + many+ n(n-1) 22-"ya + etc. -*— n 22-dy. =o. (2.) m (m—1) Subtracting (1) from (2), and dividing by y, m 2-1 + y+ etc., n(n-1) (3.) y Since y is variable, we are at liberty to assign to it any value we please. Suppose y=0. This supposition will cause o=u in (2), and reduce (3) to -n 2 mm muro. =0. Hence, mac =n 23; or mx=n x"; 2 2 In problem 12, m=1, and n=3; hence, (4) becomes x=V 1 3 Or thus: Differentiating (1), and observing that because u is a maximum, it is constant, we get mam-ldx-n22ldx=0; and dividing by dx, m m 2-l- n21-1=0, which, as above, reduces to x==m m/ n A. W. WHITOOM. Another solution of Problem No. 12.-Let = the required fraction, and d an infinitely small number. Then x+d-(x+d)——. Hence, x+d -*—3 x ? d—3x d’-d3=24**. Dividing by d+transposing 1=3 x? +3 xd+d'. Since d is infinitely small, l=32*+x=1}. O. E. If 3} acres, with its growth for four weeks, keep 12 sheep for 4 weeks, 1 then will 3 acre with its growth for 4 weeks keep 37* 14,4 sheep 1 week, This result is shown by (3). In the same way we find from (2), that 1 acre, with its growth for 9 weeks, will keep lo xix41 18,9 sheep 1 week; expressed by (4). Subtracting (3) from (4) we find that the growth of 1 acre for 5 weeks will keep 4,5 sheep; expressed by (5). Hence, the growth of 1 acre for 1 week will keep } x 4,5=0,9 sheep for 1 week, expressed by (6), ank the growth for 4 weeks will keep 4x0,9 =3,6 sheep, expressed by (7). Subtracting (7) from (3) we find that 1 acre without the growth will keep 10,8 sheep 1 week, expressed by (8). Consequently 24 acres, without the growth, will keep 24 x x 10,8=14,4 sheep 18 weeks—expressed by (9)—and the growth of 24 acres for 18 weeks will keep 24 x V x 11 x 0,9=21,6 sheep for 18 weeks—expressed by (10). Taking the sum of (9) and (10) we find 24 acres will keep 35 sheep 18 weeks. A. W. WHITOOM. Solution of Problem No. 14.-Given * + 13x? 3 - 39x=81. By trans –892 + 81. Adding (1 13.x 39.2 169x2 posing ** + to 3 1 81 6 36 13x8 169x2 169.x both numbers, *+ + 39x + 81; or extracting the 3 36 36 13x 132 square root x? + +9; therefore, 2-9. Hence, C-+3. 6 + Problem No. 17.-There are fonr numbers ic geometrical proportion, the second of which is less than the fourth by 24; and the sum of the ex• tremes is to the sum of the means as 7 to 3. What are the numbers. Problem No. 18.-Find two numbers such that their product shall be equal to the difference of their squares, and the sum of their squares shall be equal to the difference of their cubes. Problem No. 19.-If three equal circles touch each other externally, and thus inclose one acre of land; required the diameter of each of the three circles the sides of an equilateral triangle described about them, and the diameter of a circle circumscribing the triangle? A solution by numbers is requested. [The principles involved in the above problem were developed in a solation in the October Number of the Journal.Ed.] Problem No. 20.-A globe 40 inches in diameter is to be divided into three equal segments by two parallel planes; required the height of each segment. P. BRONSON. W YOOENA, Wis. Problem No. 21.-Required the diameter of the greatest cylinder that can be cut from a given globe, and the relation between its altitude and the diameter of the globe. A. B. O. [Mr. J. M. Ingalls writes us that A. W. Whitcom's solution of Problem No. 6, in the January Number, is erroneous. Mr. I. sends a correct solution, which we will give in the next issue.-Ed.] Lord Macauley has announced that he will confine himself in future to his closet as a historian, and take no further part in public life. |