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north, the other 100 miles east. How far were they then apart?

25. A kite accidentally lodged in the top of a tree, but the line breaking, I measure its length, which is 210 feet. What is the height of the tree, the base being 189 feet from my standing place?

26. Desiring to know the height of a precipice, I drop a stone from the summit, and observe by my watch that it strikes the ground in 3 seconds. What is the height!

27. A bag of sand is dropped from a balloon 14 miles above the surface of the earth. How long will it be in falling?

28. In what time will a stone fall to the bottom of a shast, that is sunk 870ft. below the surface of the ground?

When one number bears the same ratio to a second as the second does to a third, the second number is called a mean proportional between the other two. Thus, in the proportion 3:6:: 6:12, 6 is a mean proportional between 3 and 12.

The mean proportional between any two numbers is equal to the square root of their product.

29. Find a mean proportional between 7 and 252. 30. Find a mean proportional between .75 and 12. 31. Find a mean proportional between and ads. 32. Find a mean proportional between zõ and .875.

33. Find mean proportionals between it and 16; 5 and 6; 25 and 13; Ž and g. We

may often discover, by simple inspection, that any given number is not a perfect square. The following are some of the principal properties of squares:

(1.) Every even square is divisible by 4. Therefore, no even number, which is not divisible by 4, can be a perfect square.

(2.) If a square number contains a prime factor, it also contains the square of that factor. Therefore, no number divisible by a prime factor, and not divisible by its square, can be a square number.

(3.) All squares terminate in 0, 1, 4, 5, 6, or 9. Therefore, no number terminated by 2, 3, 7, or 8, can be a square number.

(4.) Every square number terminated by 5, is also terminated by 25. Therefore, no number ending in 5 can be a square, unless the 5 be preceded by a 2. We may remark moreover, that the 2 must be preceded by 0, 2, or 6.

(5.) The zeroes terminating any perfect square, must be of an even number. Therefore, no number terminating in an odd number of zerves, can be a square number. And if the zeroes be even, unless they are preceded by a square number, the number itself is not a square. Thus 2500 is a square number, but 1500 is not.

EXTRACTION OF THE CUBE ROOT. The cube root of a number is the number which, when raised to the third power, will produce the given number.

In the following table are the numbers from 1 to 10 inclusive, and beneath them are their cubes, therefore the numbers of the second line have for their cube roots the numbers of the first.

Roots 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Cubes 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.

Thus we see that there are only nine integral cubes between 1 and 1000. All the other intermediate integers are imperfect cubes, and their roots can only be obtained approximately.

All perfect cubes from 1 to 1000, evidently have but one integral figure in their cube root. All numbers between 1000 or 103, and 1000000 or 1003, will have two figures in their root. And generally, if we divide a cube into periods of three figures each, by placing a point over units, and one over every third figure from units, the number of points will show the number of figures in the root.

EXAMPLES FOR THE BOARD.

In order properly to understand the principles of the cube root, the student should be provided with the following blocks :

1. A cubical block, each side measuring 3 inches, to represent the cube OF THE TENS.

2. Three blocks, each 3 inches square and jy of an inch thick, to represent THE SQUARE OF THE TENS MULTIPLIED BY THE UNITS.

3. Three blocks, each of an inch square and 3 inches long, to represent THE SQUARE OF THE UNITS MULTIPLIED BY THE TENS.

4. A cubical block, each side measuring jy of an inch, to represent THE CUBE OF THE UNITS.

If block No. 1 be placed upon a table, the three blocks No. 2, laid against three of its sides, the three blocks No. 3, in the deficiencies left by No. 2, and No. 4 in the corner still unfilled, we shall have a new cube, which will represent THE CUBE OF THE TENS, plus THREE TIMES THE SQUARE OF THE TENS MULTIPLIED BY THE UNITS, plus THREE TIMES THE SQUARE OF THE UNITS MULTIPLIED BY THE tens, plus THE CUBE OF THE TENS un

This set of blocks is readily applicable in showing the difference between 303 and 373. But it will be easily perceived that the same result would have been obtained, if, instead of 3 inches, and io of an inch, we had employed 4 inches, and is of an inch, -11 inches, and of an inch,-or any other numbers to represent tens and units. Therefore, the cube of any number whateveris equal to the cube of the tens + three times the square of the tens X the units, + tñree times the tens x the square of the units, + the cube of the units.

We will now cube 37, both by our rule and by direct multiplication. The cube of the tens,

=27 thous.

37 3x the square of the tens x the

37 units,

=189

hund. 3x the tens x the square of the

372= 1369 units, = 441 tens,

37 The cube of the units,

343 units,

373=50653 The cube of 37,

=50653 Let it now be required to extract the cube root of 50653.

50653(37 Dividing the number into pe27

riods, we find there will be two

figures in the root. 27 is the 3x32 27,49 23653 greatest cube contained in the 3x30x7= 6 30 23653 first period, and its root 3, is ev

idently the tens' figure of the comp. div. 3379

root sought. Subtracting 27 from 50, and annexing the next period to the remainder, we have 23653, which must evidently contain 3 x the square of the tens x the units,+3x the tens X the square of the units+the cube of the units, which is the same as, the units X (3x the square of the tens +3x the tens X the units+the square of the units). To discover the units' figure, we will employ 3x the square of the tens, or 27 hundreds, for a trial divisor. 27 hundreds is contained in 236 hundreds 8 times. But if we complete our divisor with 8, it will be found too large; we therefore try 7. 3x the tens X the units =63 tens or 630. The square of the units is 49 units, which is annexed to the 27 hundreds. Adding these numbers, we obtain 3379 for our complete divisor, which being multiplied by 7, gives

tr. div.

23653,—the remaining part of the root. Hence we derive the following

RULE. Separate the number into periods of three figures, by placing a point over the units' figure, and one over each third figure to the left, (and also to the right, if decimals are desired in the root). Write in the quotient the root of the greatest cube contained in the left hand period, and subtract the cube from the period, annexing the second period to the remainder.

Take three times the square of the root already found, for a trial divisor, and find how many times this divisor is contained in the hundreds of the dividend. Place the result in the quotient, and its square at the right of the trial divisor (supplying the place of tens with a zero, if the square is less than ten).

Multiply 30 times the root already found, by the last root figure ; this product added to the divisor, will give the complete divisor.

Multiply the complete divisor by the last root figure, subtract the product from the dividend, annex the next period to the remainder, and proceed as before until the whole root is extracted.

If any complete divisor is not contained in its dividend, a zero must be annexed to the root, and two to the trial divisor, and the next period brought down for a new dividend.

If any figure obtained for the root is too great, dimin. ish it by 1 or more, and repeat the work.

[It may be well to illustrate to the pupil the algebraical formation of the cube, by raising 30+7 to the third power.

In forming the complete divisor, we say, “Multiply 30 times the root already formed, by the last root figure,” because 3x the tens x the units would give tens, and a zero must be annexed for units.

The square of the units is written at the right of the trial divisor, to render the work more compact. This square being units, and the trial divisor hundreds, the two may very properly be united.]

What is the cube root of 226802438.843904 ?

226802438.843904(609.84

63=216
tr. div.
3x 62=108,01

108,02 30 x 6x1= 1 80

10981 c. div. which is not contained in the dividend.

tr, div. 30 x 602=10800,81 3 x 60 x 9= 16200

108024,38

9866529

1096281 com. div.

tr. div. 3X 6092= 1112643,64 30 X 609 X 8= 146160

9359098,43

111410524 com.div. 891284192

tr, div. 3X 60982= 111556812,16 446256519,04 30 x 6098 X4= 731760

44625651904 11156412976 com. div. The trial divisors, after the first, may be more conve. niently found, by adding to the last complete divisor the last number used to complete it, and twice the square of the last root figure. Thus in the foregoing example, our third trial divisor=

1096281

16200 2x925 162

trial divisor, 1112643 1. What is the cube root of 328509? 2. What is the cube root of 1986091 ? 3. What is the cube root of 9129329? 4. What is the cube root of 345.357380096 ? 5. What is the cube root of .062570773? 6. What is the cube root of 24; 25; 1000

512? 7. What is the cube root of 5,13; 1134; 2745 ? 8. What is the cube root of 6; 1 ; 21; 95; 8.71?

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