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... 4840 - (4840—2x) ? = area of path,

= 1 of 4840 ;
.. (V4840 – 2x)2 = 4235,
and ✓ 4840 – 2x = 4235,

2x = V 4840 - 4235,
x = $(/4840 - 4235) the width of the path.

X. The cube roots are the following 1. Q-1. 2. X+3-1. 3. a? -a-1. 4. a? +1. 6. 2a3 – 36%. 7. a? - 2a +1. 8. 3a: - 4ax +o?. 2

2 10. x?

11.

23a2'

5. ax - by. 9. 2x2 – 3x +1.

a

12. -2-5

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XI. The following are the cube roots of the numbers :1. 64, 81, 125, 145, 165, 173. 2. 25, 52, 256, 652, 2567, 7652. 3. •4, *04, -21, 074, 094, '0301.

4. 1.0033 ... 2:124 ..., 7368 .. •3419 ..., .15874..., 2:38.. 1.508..., 2.070..., 2.083..., 1.241 ..., 2:155 ...

5. 1.01, 9.6, 21:1, 3:43, 8.88, 90-4. 6. 1.259, 1.912, 2.154, 4.641, 1.464. 7. 793..., .956..., 650... 1.5. 1.493 ..., 10.102..., 1}, •8735...

2

2

XII.
2. Shew that x? is greater than (x+1)*
4. (x+1)+= {

(x + 1)(x+2)
+2)}}-{012+}}'
1 xx)

=X3 +30+ 3x +1.
1.2
See note, p. 14, Section X.
5. See p. 9, Art. 9, cote.
6. See V.,

Ex. 7. 7. See V., Ex. 8. 8. The fourth, sixth, and ninth roots of the given numbers are each •13.

XIII. 1. See Art. 11. 2. If a, b, c be any three numbers, then shew that

(a+b+c): - (as +63 +c8)=3(a+b)(b + c)(c+d). 3. (2+1)* + x(x+1)+x* = (2+1) 9 — 23. 5. Let x, 2+1, 2+2 denote any three consecutive numbers, then x(x + 1)(x+2)+(x+1)=(x+1); also (x - 1)(x - 2)(x – 3) + (x - 2)=(x - 2)".

6. If a, 6 be the numbers, shew that a: +63 +1 is greater then 3ab. See Ex. 18, p. 45, Section IV.

7. The number expressed by its prime factors is 2.32.58.7, and the required multiplier is 22.3.72.

8. Here 33 +48 +58 = 6s; multiply these equals by xs ; then (3x)* +(4x) * + (5x) *=(6x)*, in which x may be taken equal to 1, 2, 3, 4, &c. Also, since 68 +(20)* + (36)&=(38)*; then (6.x) * + (20x)* +(36x) * = (38x)*, and a may be fractional as well as integral.

3

XIV. 1. 96 square inches in its surface, 216 cubic inches in its volume.

2. The edge of the cube is 10 inches, its surface is 600 square inches, the diagonal of one of the faces 10/2, the diagonal of the cube 10/3 inches.

4. The content is }(V3+1) cubic inches.

7. The cube root of the number which expresses the content of the parallelopiped will be the edge of the equivalent cube.

11. Find first the content of the three cubes, and the cube root of their sum will be the edge of the cube required.

16x1728 } +

XV. 1. The dimensions are 50, 100, 200 inches respectively.

2. The cube root of the content gives the length of an edge of the cube, and 6 times the square of this number will give the number of square yards.

4. The length, breadth, and height are respectively 32, 16, and 8 feet.

5. The height is 9 feet, and the length and breadth are respectively 32 and 24 feet. 6. The length of a side is {

13613701

feet. 7. The length, breadth, and depth are respectively (17920). feet.

10 8. Difference between external and internal surfaces is 24m(atm) square inches, and content of the boards of which the chest is composed is 2m(3a? – 6am + 4in?), cubic inches.

9. The length of the edge is 8 feet. 10. The quotient arising from the division of £8,584 18s. 7d. by £4 3s. 4d. gives the number of cubic inches of the mass. The cube root of this number gives the lineal inches in the edge of the cube.

12. The balls are supposed to be in layers so that the diameters of the balls where they touch one another are in vertical and horizontal straight lines. And the num. ber of balls which can be packed in the box is equal to the number of cubes which also could be packed in the box, if the edge of each cube be equal to the diameter of each ball. The portion of vacant space in the box is the difference between the content of the cubes and the content of the balls.

The content of a sphere is two-thirds of its circumscribing cylinder, or if r. denote the radius of sphere, then the content of the sphere

4r3
x (273) or
*

where 7=314159...

3 If the balls be each m inches in diameter the vacant space will be expressed by

40 m: as {mo

In the question, suppose the edge of the box 25 inches, and find the vacant spaces in the three cases, when the balls are one inch, half an inch, and one-third of an inch in diameter.

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XVIII. These products differ from the exercises in Section IV. in this circumstance—that the indices of the quantities here are fractional and not integral.

XIX. The same remark applies to these exercises.

XX. These exercises require no remarks.

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XXI. 1. X-2*+1. 2. 2:2 - 1+1.

! 3. a-2all) - 6. 4.

US
ai i 1
5.

6. 1
at, axi

at 1 y
+
43-2

7. +2W-1. 8.
2 3

y! 211 9. ata +bx+cixt.

22 23
10. 1+

x 3.22 323 324
+
+ &c.
11. 1+
+

+, &c. 16

8 16

128 6 362 368 12. atot

+, &c. 2 8a

16a?

1+ + 3*

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XXIII. The verification of these expressions involves the same operations as in p. 33, Section IV., the only difference being that in these examples the indices are fractional and not integral.

XXIV. 1. 5. 2. 1}. 3. x(x - 1)2 +18=0. 4. 2. 5. 5+22.

-1 6.

22 1+2 1-2 2(3-8) 1+(1+)*1+(1 x)*** x

3 8. 7/3

(3-1)at 9. 3.

a’ +1+(1+2a® - 15a“)} 10.

11. 6

sa?

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7. When z=V%, (2+v3)1=1+V3, and

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XXV. These exercises are reduced in the same manner as those in pp. 13, 14 in Bection V.

1.

5.

1-25

2+1 2.

4.

23 – 2x - 1
2} +500*
22 +2-1

23 – 2x2 - 6x-3 - 2x(x + 1)(x + 1)
(x+1)(x2 - x - 1)

22 - 2-1
(x+1)(– 3x-3) - 2x(x+1)(x+1)!x2 – 3x – 3 – 2x(x + 1)]
(003 - 3x – 2) + (x2 - 1)(x2 – 4)*_(x2 - 1)(x-2)+(x2 - 1)(x2 - 4)
(23 – 3x+2)+(c2 - 1)(~2 - 4)+ (x2 - 1)(x + 2) + (x2 - 1)(x2 - 4)*
_(at2 - 1)(x - 2)}. {(x - 2)+(x+2)"}=+2S

S
(23-1/+2). {(x+2)++ (x - 2)"}
– 2.cłyt +2actyl - y_(act - yt)(a +akys – atyk+y)

ic - audyt + y (-y)(x1 - kytactyl - y)
(a +x)+(a – 2)a +(a? – x2)

9. a?" +Q2"-'+1. (a+x) -(a - 2)

a

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XXVI. The verification of these equivalents is effected by exactly the same processes as those in p. 24., Section V., the only difference being in the fractional indices. In Example 13, one side of the equivalent is omitted. It should be

2ab - {(1 - a*)(1 - 64)}"

1+62 +a?(1 – 62)

XXVII. 1. See Sect. IV., Art. 3, and notes pp. 2, 3; and in this Section VI., Art. 3, and notes pp. 3, 4.

2. a? + b = (a+W - 1)(a-bv-1).
3.

2+3V-1_2+3-14-5v-1-23 + 2V -1.
4+5/-1~4+57-1*4-5-1-+-41
a+bV-1_a+bV-1,6-dv-1_ac + bd , (bc-ad)v -1

+
c+dy -1c+dy -1c-dy-1*c+da

c+02
4. See Art. 3, pp. 2, 3.
5. See Art. 18, and note pp. 17, 18.
6. First (-a)tax (-6)*n=lax - 1)** X16X – 1)**

(ox -
atr. (-1) **.837 – 1) in

1
=(ab).(-)***

1

1

1

1

1

1

1

=

=a

1

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=
=(atá. (-ujka

1

1

Hence the product involves ( - 1)?“, an even root of -1, which is an impossible expression. Next (-ajintax (=bj*nta =(ax – 1 )n+ex (8x – 1)++?

=latta. ( - 1 panta x binta. (-1jinta
= (at)*+z. (-1;&nta

+(ab)uta.(-1***l- (a} Hence the product involving an odd root of -1 which is always possible and equal to - 1.

1

9

1

2

10. The third power of -1!V - 3 is +1; hence this expression exhibits the two symbolic cube roots of +1. And the third power of

+1EV -3

is -1, and the expression gives the two symbolic cube roots of -1.

-1+1 -3 -1 +1 -3.
And

2
-1-V-3
-x

-1-1-3-1-7 -3°

2

-1-V-5

2

2

XXVIII. The first eight examples offer no difficulty beyond the exact performance of the operations indicated.

9. This may be shown in two ways: by extracting the square roots of each quantity and taking the sum of the results, and by squaring each side of the ex. pression, and showing that the result when reduced is 32+68=100, an obvious identity.

XXIX.
The first five examples require no remark.
6. The expression may be put into the form

y-
2-1 , - 2xx -1, 2x YV -1, (x – YV –1)(y+-1

✓ (x2 + y2)V-1
y - XW-1 (2x YN - 1)(y + x - 1)

✓(x2 + y2) V-1
(y2 + x2) (2x YN - 1) -(y2 + x2).
- (2x - y - 1)(x2 + y)

.(x2 + y)).
V(x2 + yo)

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y + 2x - 1

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