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A. The board has no power to act contrary to the vote of the district in this matter, and its members make themselves liable for an unauthorized expenditure of the district funds.
B. Can a district board, after keeping open the school five months, lawfully continue the school as Swedish school?
A. No school is a lawful public school that is not taught in the English language and by English speaking teachers. If some of the people want a school of a different character, they must support it at private expense. To use the district funds in any way for the support of any other than an English school, taught by a qualified teacher, is entirely illegal.
Q. If a tax of $200 is voted, and by mistake $300 is returned, assessed and collected, must the town treasurer pay it back to the individual tax-payers, or to the district treasurer?
A. It is simply a mistake; it should simply be corrected in the most convenient and effectual way. The money may be returned directly to the individual tax-payers, or the district treasurer, if willing, may be made the agent for paying it back. The money does not belong to the district, as a district, but to the individuals from whom it has been collected by mistake.
Q. Can an appeal be taken to the State Superintendent, from the action of a County Superintendent, annulling a teacher's certificate?
A. An appeal can be taken if the action is deemed unjust. (See School Code, p. 132.)
Q. Must persons, over twenty, residing in the district, pay tuition, if they attend public school?
A. The public schools are free, only to those between four and twenty, residing in the district. (See School Code, p. 50.)
Q. Is there any difference between an ordinary district and a “ union district,” in regard to the right to attend free?
A. There is not. The school is free in such district, only to persons of school age; and these must also come up to the “ standard of qualifications" fixed for admission.
Q. Under section 55, page 84, have parents and guardians a legal right to demand that their children and wards be taught geography, grammar and arithmetic, when the district board are satisfied that the general interests of the district require the teachers' time in giving instruction in spelling, reading and writing? Or in geography and grammar, when only one or two pupils require such instruction, in a district school?
A. The board has the control of this matter, and parents and guardians have no power of dictation. The board must however carry out the provisions of the section, so that opportunity will be given for instruction in all the branches named in the section, at suitable times, and as the wants and ages of the different pupils may require. As a sort of general rule, the wants of the younger pupils most need to be considered in the summer school, and those of the older ones in the winter. An article in this number of the Journal, by the superintendent of Grant county, may be referred to in this connection.
ANSWERS TO QUESTIONS. 59.—The measure of a water pail across the top is twelve inches, across the bottom eight inches, depth twelve inches. How deep is the water when two thirds full? The number of cubic inches in the pail=122(42 +62 +24), 2=3.14159+ % the
3 cubic contents=82(42 +62 +24).
If we let x==the depth of the water, then the radius of the surface=24+x. Hence 8z(70)=172(16+(24,+~)* + 96 + 40); And 608=2(16+(24 + 2)* +96 + 4.2). Expanding, clearing of fractions and uniting, x + 722° +1728x=21888. Solving this equation by approximation, we find x= =8.93099477 in.-LAMBDA.
Solution by the Proposer of the Problem.-If the sides of the pail were extended, it would come to a point in 36 inches, or 24 inches from the bottom. Cube 36= 46,656. Cube 24=13,824. Difference 32,832. Two-thirds of difference is 21,888, to which add the 13,824=35,712. Get cube root of 35,712=32.93+. Subtract 24 =8.93 inches, depth of water.
Explanation.-By supposing a cube with the supposed point for a center, any one can easily see the whole cube will be divided in the right proportion; but this cube may be supposed made up of cones, each with the height of 36 inches, point centering in center of cube and base at surface of cube, then tach cone, no matter of what size or shape, will be divided in same proportion as whole cube.-N. DARROW, Reedsburg.
B. R. A, also sends a solution.
60.-A log (round or square) is 20 feet long; across large end 12 inches, across small end 4 inches, uniform solidity. One man carries small end; where shall two men place their pole underneath so as to carry each 1/2 times as much as the other man.
First find center of log, as in (59). It will point in 30 feet, or 10 feet from small end. Cube each, and subtract, 27,000—1,000=26,000. Take one-half of 26,000= 13,000 and add 1,000=14,000. Get cube root, which is 24.101 + feet. Subtract 10 feet=14.101 + feet, distance of center of log from small end. If one man gets 14.101 + feet from center of the log the two men must get one-third of 14.101 + feet or 4.7 + feet from center, as we can suppose the whole weight to bu in center, which will make it the same as carrying a kettle on a pole, then they would be 1.198+ feet from large end.-N. D.
61.-A tree 100 feet high is broken in the wind; in falling, the broken end rests on the stump, while the top rests on the ground 30 feet from the foot of the tree. Required the height of the stump.-W. H., Loroville.
It is proved in Algebra and Geometry that the sum of two numbers multiplied by their difference is oqual to the difference of their squares, then the difference is equal the difference of squares divided by sum, as (x+y) (2y)=2?—yo and
The 100 feet is the sum of two numbers, and (30)?=900 is the difference between hypothenuse and perpendicular. 900+100=9 is the difference of numbers. (100—9)+2=4512 less (100+9)=2=5412 greater.-N. DARROW, Reedsburg.
Another Solution.—<+y=100 (1); x=100—Y(2). According to rale: a?=y2 +900 (3); and in the given example, x2 + 2xy + y'=10,000 (4).
Substitute the value of x from (3) we have y2 +900+ 2xy + y?=10,000; or 2y + 2xy=9,100; or y2 + xy=4,550. (5)
Adding the value of x from (2) we have yo +100y-yo=4,550; or y=4512.
62.—Three men are to carry a stick of timber 100 feet long, and of uniform size throughout. Each is to carry one-third of the stick, and two of them are to carry together. If the man who carries alone is at the end of the stick, how far from him must the others be that they may carry their share?
One man get 50 feet from center, then the two men must get one-half of 50 feet, or 25 feet from center, or 75 feet from the one man. A long log for three hands to carry.-N. D.
Second Solution.-Let the log be poised on a fulcrum at 50 feet from each end, and let them bear down instead of lifting, then the man at the end bears one pound as often as the others bear down two; hence their lever must be one-half of 50, or 25 feet from end of log.-B. R. A.
65.-Suppose an opening to be made directly through the center of the earth, and a cannon ball to be dropped into the abyss, where would the ball come to a state of rest?-L. C., Door Creek.
My impression is that the operation of the law of gravity would be the same as in rolling a ball down the side of a hill, there being another hill on the other side of the valley, that it would vacillate to and fro until it came to a state of rest in the center of the earth.-B. R. A.
Another Solution.-'The ball would come to a state of rest at the center of the earth, where the attraction ceases, as the attraction of the earth to its center is the only power that makes the ball or anything else fall toward the center of the earth. -J. BR., New Holstein.
66.—What causes the explosion when a mixture of oxygen and hydrogen is ignited ?-W. F. BUNDY, Sauk City.
We only know that oxygen and hydrogen have a strong affinity for each other, and the principle is the same as when oxygen and carbon are united, except that they are more violent.-B. R. A.
67.—Find two mixed numbers whose sum and product are equal.-P. T.
Any two numbers less than two will have a greater sum than their product, and any two greater than two will have a less sum than their product, so one of the
numbers must be less and the other greater than two. Let x and y=numbers.
y Then x+y=xy; transpose, ay-x=y; factor, x(y-1)=y; divide by y-1, x=77. If we take y for the less number it must be more than one and less than two, for if it be one or less, y-1 is less than nothing, and the operation is impossible; then we can form as many couplets to suit the question as there can be fractions between one and two which cannot be numbered. If the numerator of the fraction part of y is 1, the x is a whole number; thus, (y=112,x=3), (y=143, x=4), (y= 114, =5), etc.
If the fraction part of y has 2 for a numerator, as 13, 13, 14, 15, 2= 23, 33, 43, 54, etc. If y
has 3 for its fractional numerator, as 13, 14, 11, 12, 13, X=21, 23, 33, 33, 41, etc.
To find the value of x and y. + x= 7 B.R. A., as requested, has promptly sent a solution to this problem, as corrected, as have also W. B. MINAGHAN, of Chilton, and C. L. POWERS, of Mukroonago, but we withhold them till next month, and give place to the following.-EDS.
Messrs. Editors-Will the following solution (?) do to allay the spirit of the restless problem (No. 68) that is constantly arising to trouble young algebraists?
75.—If one cow and one ox cost $57, and twenty cows and thirty oxen cost $1,500, what is the cost of each per head?
If one cow and one ox cost $57, twenty cows and twenty oxen will cost twenty times $57, which are $1,140; but ten oxen more, or twenty cows and thirty osen cost $1,500 hence ten oxen will cost $1,500-$1,140, which are $360; and if 10 oxen cost $360, one ox will cost one-tenth of $360, $36. If one ox and one cow cost $57 one cow will cost $57-$36, or $21.-C.S. FULLER, Windsor.
R. W. sends two solutions, the second of which we give.-EDS.
(1) Twenty cows and thirty oxen cost $1,500.
(2) Multiplying (1) by twenty, twenty cows and twenty oxen cost $1,140. (3)
Subtracting (3) from (2), ten oxen cost $360.
(4) Hence, one ox costs $360--10=$36; and ope cow costs $57-$36=$21.-R. W..
H. P., Alma, B. R. A., Kilbourn City, W. BRIER, Baraboo, J. C. DOUGLASS, Dunkirk, and C. L. POWERS, Mukuonago, have also sent solutions to 75, for which we cannot make room.-EDS.
76.-Wanted to know the names and respective terms of office of the several Chief Justices of the United States.
See Townsend's Civil Government, page 296.-R. W.
This may be regarded as two operations in division, in each of whi:h the true value of the quotient is to be proportionately increased. The true quotient arising from the division of 6 by 3=2; but the supposed quotient is 3, which is 192 times the true quotient. In the second couplet, 20:4=5, and multiplying this by 142, as in the first case, gives 742 for the answer. Or, by Proportion–4:3:40 : a.
3 X 20 Then,
3 x 20 x 3 Reducing,
=712, for the fourth term.--R. W. Another Solution.-12 of 6=2. 4 of 20=5. Then if 3=2, what equals 5. If 3=2, then 142=1; 5 will then equal 142 ×5=742; or, what equals 14 of 20.
By Proportion–3:2::(?):5=712; or, 2 : 3::5:(?)=712.-J. C. DOUGLAS, Dunkirk.
83.—The centers of two given spheres are at the extremities of a given right line; required the locus of a point from which the greatest portion of spherical surface is visible.
Let AB represent the line joining the centers of the spheres, P the point whose locus it is required to find, and C a point where the locus intersects the line AB. Place AB=2a, AC=m, BC=n, then it is evident that AP: BP::m:n.
Let x and y be the co-ordinates of the point P, the origin being at the center of the line AB.
Then AP= v(a + x)" + 1" BP=v(a−2)2 + yo
[ m2 + na Consequently x2 + y2 –
m2-n2 This is the equation of a circle whose circumference is the locus of P.-L. CAMPBELL.
84.-Alice is one-third as old as her mother; five years since she was but onefifth as old; in how many years will she be one-half as old ?-B. R. G.
Since Alice five years since was one-fifth her mother's age, four-fifths or twelvefifteenths her mother's age at that time, must be the difference in their ages. Alice is now one-third her mother's age; the difference in their ages now is twothirds or ten-fifteenths the mother's present age. And as the difference in their ages must always be the same, twelve-fifteenths the mother's first age must equal