International Correspondence Schools.

# The elements of railroad engineering (Volume 5) online

. (page 5 of 15)
Online LibraryInternational Correspondence SchoolsThe elements of railroad engineering (Volume 5) → online text (page 5 of 15)
Font size
72 as the product, which, added to 36 of column (2), gives
108. Annexing two ciphers to 108, we have 10,800 for the
trial divisor. Dividing the dividend by the trial divisor, we
see that it is contained about 8 times, so we write 8 as the
second figure of the root.
Add the first figure of the root
to the first correction, and we obtain 18 as the second cor-
rection. To this annex one cipher, and add the second figure
of the root, and we obtain 188. This, multiplied by the sec-
ond figure of the root, 8, equals 1,504, which, added to the
trial divisor • 10,800, forms the complete divisor 12,304.
Multiplying the complete divisor 12,304 by 8, the second
figure of the root, the result is 98,432. Write 98,432 under
the dividend 111,680; subtract, and there is a remainder
of 13,248. To this remainder annex the next period 000,
thereby obtaining 13,248,000 for the next new dividend.

(3) Adding the second figure of the root, 8, to the num-
ber in column (1), 188, we have 196 for the first new

76 ARITHMETIC.

correction. This, multiplied by the second figure of the root,
8, gives 1,568. Adding this product to the last complete
divisor, and annexing two ciphers, gives 1,387,200 for the
next trial divisor. Adding the second figure of the root, 8,
to the first new correction, 196, we obtain 204 for the new
second correction. Dividing the dividend by the trial divisor
1,387,200, we see that it is contained about 9 times. Write
9 as the third figure of the root. Annex one cipher to the
new second correction, and to this add the third figure of the
root, 9, thereby obtaining 2,049. This, multiplied by 9, the
third figure of the root, equals 18,441, which, added to the
trial divisor, 1,387,200, forms the complete divisor 1,405,641.
Multiplying the complete divisor by the third figure of the
root, 9, and subtracting, we have a remainder of 597,231.
We then find the fourth figure by division, as shown.

{V) 4 16 f 7 4'0 8 8 = 42 Ans.

4 32 64

8 4800 10088

4 244 10088

120

5

125

120 6044
2

122

(^) 4 16 f 9 2'416 = 45.212— Ans.
4 32 64

^ 4800 28416
i_ 6 25 2 712 5

5425 1291000
650 1220408

5 607500 612912)70592.000(.115

130 2704 612912

i_ 610204 930080

^^^^ 2708 612912

2

J352 612912 3171680

2 3064560

1361 107120

ARITHMETIC. 77

(^) 7 49 f .373'248 = .72 Ans.

7 98 343

14 14700 30248

7 424 30248

210 15124
2

212

(133)

1
1

1

2

300
64

364

68

-^2.0000 0000 = 1.259921+ Ans
1

2
1

1000
728

80
2

272000
225125

82
2

43200
1825

46875000
42491979

84
2

45025
1850

4755243 )4383021.000( 9217 or .922-

42797187

360
5

4687500
33831

10330230
9510486

865
6

4721331
33912

8197440
4755243

870
6

4755243

34421970

3750
9

3759
9

3768

This example shows what a great saving of figures is
effected by using the short method. The figures ob-
tained by the division are 9217, thus making the last figures
of the answer 922, according to Art. 272. This is not
correct in this case; the true answer to eight decimal
places being 1.25992104 + ; hence, the first three figures

78 ARITHMETIC.

found by division should be used in this case. The reason
for the apparent failure of the method in this case to give
the seventh figure of the root correctly is because the fifth
figure (the first obtained by division) is 9. Whenever the
first figure obtained by division is 8 or 9, it is better to carry
the root process one place further, before applying Art.
272, if it is desired to obtain absolutely correct results.

'^1'7 5 8.416'7 43 = 12.07 Ans.
1

(134)

1

1

(«)
1
2

2
1

300

64

30
2

364
68

32
2

4320000
25249

34
2

4345249

758
728

30416743
30416743

3600
7

3607

(d) 1 1 i?^l'191'016 = 106 Ana

12 1

2 30000 191016

1 1836 191016

300 31836

6

306

'32 8 ^8 2

(^) 4^ = ^ = 3 ^„^
512 fSia 8

ARITHMETIC. 1Q

(135) 1 1 .^3.00 00 00 00 = 1.442250- Ans.

12 1

2

1_

80

300 3000

136 1744

4 436 256000

34 ^52 241984

_4 58800 14016000

88 1696 12458888

_±_ 60496 6288092) 1557 11 2.000 (.2496 or. 260-

420 1712 12476184

6220800 30949360

8644 24952368

4

424

4

428 6229444 59969920

4 8648 56142828

4320 6238092 3827093

2

4322
2

4324

(13 6) {a) (^)

1 i/1'2 3.2 1 = 11.1 Ans. 1 |/l'14.9 2'10 = 10.72 +

]_ 1 1 1 Ans.

20 23 200 1492
_1 11 7 1449

21 221 207 4310

1 221

220

1

221

7 4284

2140 26
2

2142

(0 ' (^)

7 -/5 0'2 6'8 1 = 709 Ans. 2 4/.0 0'0 4'l 2'0 9 = .0203
7 49 2 Ans.

140 12681 400 04

12681 3 4

1400 403 1209

9 1209

1409

80 ARITHMETIC.

(137) {a)

1 1 f .0 6'5 O'O = .18663 - Ans

1 2 1

2 300 5500

1 304 4832

30 604 668000

8 368 602856

3 8 9 7 200 103788 ) 65 144.00 (.627 or. 63-
8 3276 622728

46 100476 287120

8 3312 207576

640 103788 79544

6

646
6

652

{^)
2 4 ^.0 21'0 0'0 = .2759— Ans.
2 8 8

4 1200 13000

2 469 11683

60 1669 1317000

7 518 1113875

67 218700 226875 ) 203 12 5.0 (.89 or .9-
7 4075 1815000

74 222775 216250
7 4100

810 226875
5

815
5

iio

ARITHMETIC. 81

(^)

2 4 f 8'0 3 6'0 5 4'0 2 7 = 2,003 Ans.

2 8 8

4 12000000 036054027
2 18009 36054027

6000 12018009
3

6003

1 1 -^.0 0'0 4'0 9 6 = .016 Ans.

1 2 000

2 300 004

1 216 1

30 516 3096

6 3096

36

2 4 -i?'17.0 00'00 = 2.5713- Ans.
2 8 8

4 1200 9000
2 325 7625

60 1525 1375000

5 350 1349593

1200

325

1525

350

187500

5299

192799

5348

65 187500 198147)25407.0 0(.128or.l3—
5 5299 198147

70 192799 559230

5 5348 396294

tTo 198147 162936

7

757
7

764

82 ARITHMETIC.

(138^ {a) In this example the index is 4, and equals
2X3. The root indicated is the fourth root, hence the
square root must be extracted twice. Thus, |/ = |/ of
the /~ and 4/656I = WMoi = 4/8I = 9. Ans.

8 -/6 5'61 = 81 4/8I = 9 Ans.

8 64

160 161

1 161

161

(d) In this example the index is 6, and 6 equals 3 X 2 or
2x3. The root indicated is the sixth root; hence, extract
both the square and cube root, it making no particular dif-
ference as to which root is extracted first. Thus,

|/ = |/ of the 4/ , or |/ of the 4/ .
Hence, ^117,649 = Vvil7M^ = V^ = 7. Ans.
3 4/ll'7 6'4 9 = 343 |/343 = 7 Ans.

— L_

60 276

4 256

64 2049

I

680
3

2049

683

{c) 4^.000064 = 1^^.000064^= .2. Ans.

4/. 000064 =.008. K008 = .2. Hence, 4/ . 000064 = . 2.

Ans.

ARITHMETIC. &3

{d)

y I = ? I = .375, since 8)3.000

o o .

.375

7 4 9 v'.3 7 5'0 0'0 = .72112+ Ans.

_7 98 343

14 14700 32000

7 424 30248

210 15124 1752000

2 428 1557361

212 1555200 1659523 ) 1 9 4 6 3 9.0 (.124 or .12 +
'^ 2161 1559523

214 1557361 3868670

2 2162 3119046

2160 1559523 749624

1

2161
1

2162

Hence, y| = . 72112 +. Ans.

o

(l<5y) W r 5476-y^476- 3 9

60 325

6 325

66

„ ./1225 35 . ,

Hence, |/g^=^. Ans. 7 4/5476 = 74

7 49

140 676
4 576

144 '

84 ARITHMETIC.

(^)

(0

5

i/.3 3'6 4 =

.58 3

5

25

Ans. 3

LOO

864

60

8

864

1

lOB

61
1

4/.10'0 0'0O'OO=.31623^

_9 Ans.

100
61

3900
3756

620 632) 14 4.0 (.227 or .23

6 126 4

626 1760

6 12 64

632 4 96

(^) 25.o| = 25.075.

5 4/2 5.0 7'5 0'0 0'0 0'00 = 5.00749 +

5 2 5 Ana

10000 075000

7 70049

10007 495100

7 400576

100140 9452400

4 9 013401

100144 438999

4_

1001480

9

1001489

{e) .000^ = .0004444444+.

2 |/.0 0'0 4'4 4'4 4'4 4 = .0210 8-1-

3_ 00 Ans

40 04

1 4

41 44

1 41

4200 34444

8 33664

4208 780

ARITHMETIC. 85

(140) (^) 1^ = /72.
1 4/2.0 O'OO'OO'OO = 1. 41421356 -f

1_ 1

20 100
4 96

24 400

4 281

280 11900
1 11296

281 60400

1 56564

2820 28284 ) 3836.0000 ( .13562 or .1356 +

4 28284

2824 100760

4 84852

28280 159080

2 141420

28 2 82 176600

2 169704

28284 6896

1 i/1.41'42'13'5 6 =1.1892+ Ans.

L 1

20 — n

_i 21

21 2042

1 1824

220 21813
8 213 21

22 8 4 9 2 5 6 It is required in this prob-
8 4 7 5 6 4 1^"^ to extract the fourth

2 3 6 1692 root of 2 to four decimal

9 places; hence, we must ex-

2 3 6 9 tract the square root twice,

9 since ^ = |/ of the |/ .

2 3 7 8 In the first operation we carry the root to 8

2 decimal places, in order to carry the root in the

23 7 8 2 second operation to 4 decimal places.

86 ARITHMETIC.

{b) f 6 = VTa

2 4/G.O O'O O'O O'O O'O = 2.4494897428 +

2 4

40 200
4 176

44 2400
4 1936

480 46400
4 44001

484 239900

4 195936

4880 4396400

9 3919104

48 8 9 489896)47 7 29 6.00000 (.974280 or .97428 +
9 4409064

4 8980 ■ 3638960
4 3429272

48984 2096880

4 1959584

489880 1372960

8 979792

489888 3931680

8 3919168

489896 12512

It is required in this problem to find the sixth root of 6 ;
hence it is necessary to extract both the square and cube
roots in succession, since the index, 6, equals 2x3 or
3x2. It makes no particular difference as to which root
we extract first, but it will be more convenient to extract
the square root first. The result has been carried to 10
decimal places; since the answer requires but 5 decimal
places, the remaining decimals will not affect the cube root
in the fifth decimal place, as the student can see for himself
if he will continue the operation.

ARITHMETIC. 87

1 1 -^^2.4 4 9'4 8 9'7 42'8 00= 1.34801 -

12 1 Ans.

2 300 1449

1 9 9 119 7

30 399 252489

3 108 209104

33 50700 43385742

3 1576 43352192

3 6 522 76 • 5451312 ) 3 3 5 5 0.000 ( .006 or .01 —

3 1592 32707872

394 5419024

4 32288

398 5451312

4

4020
8

4028
8

4036

(141) (a) 1

1

390 5386800 842128

4 32224

20

7

27
7

340

7

347

7

354 204

i/3.14'16 =

1

2 14
1 89

1.7725-
(.245+ or

Ans.

251 6
242 9

354)8 7.0
708

25-

1620
1416

88 ARITHMETIC.

8
8

-/.7 8'5 4'0 = .8862 + A:
64

160

8

188

1454
1344

8

110

1760

1059 6

6
1766

1772 ) 404.0 (.22 or.2 +
3544

'6

496

1772

(142) (a)

1 1 >(^3.141'6 00'000= 1.4646 -

12 1 Ans.

420
6

426
6

432
6

4380
4

4384
4

2 300 2141

1 136 , 1 744

30 436 397600

4 152 368136

34 58800 29464000

4 2556 25649344

38 61356 6429888 ) 3 814656.0 (.59 or. 6-

4 2592 32149440

6394800 6998120

17536

6412336
17552

6429888

4388

ARITHMETIC. 89

(^)

8 6 4 v'.5 2 3'G0 0'0 0=.80599 + or.8060-

8 128 512 Ans.

16 1920000 11600000
8 12025 •9660125

2 400 193 2 025 1944075)1 9 3 9 8 7 5.00 ( .99
5 12050 17496675

2405 1944075 1902075
5

2410

(143) 11.7 : 13::20 : ;ir. The product of the means

11.7^=13x20 equals the product of the

11.7^ = 260 extremes.

_ 260
'^~ 11.7) 260.00 (22.22+ Ans.
234

260
234

260
234

260
234

26

(144) («) 20 + 7 : 10+8::3:;r.

27 : 18::3 : x

27;r= 18 X 3

27.r = 54

54 „ .
X — — —% Ans.

{b) 12' : 100':: 4 : x.
144 : 10,000:: 4 : x
144.r= 10,000 X 4
144 .r= 40,000

90 ARITHMETIC.

40,000

144)40000.0(277.7+ Ans.

288

1120

1008

1120

1008

1120

1008

112

is

equivalent

to 4 :

; X'.

:7 ;

:21

4 7
(145) {a) - = 21' ^^ equivalent to 4 : ;r::7 : 21. The

product of the means equals the product of the extremes.
Hence,

7;r = 4x21

7;r = 84

x= — ox 12. Ans.

{b) In like manner,

X 8

r-r = t:; is equivalent to ;»- : 24::8 : 16.

»4 16

16;r=24x8
16;r=192

X = —- = 12. Ans.
Id

2 X
(r) — = — - is equivalent to 2 : 10 : : ;r : 100.

10^ = 2 X 100

10;r = 200

200 ^^ .
X = — — = 20. Ans.

(rtf ) — - = — is equivalent to (e) — — = — — is equivalent to
' 45 -r ^ ^ ' 150 600

15 : 45 :: 60 : x. 10 : 150 :: x : 600.

15^ = 45X60 150;r = 10X600

15 4r= 2,700 150.r = 6,000

2.700 ,„^ 6,000 .. .

X = '^ = 18a X = ■ ' - = 40. Ans.
15 . 150

Ans.

ARITHMETIC. 91

(146) -r : 5 :: 27 : 12.5. (147) 45 : GO :: x : 24
5 GO.i' = 45X24

12.5 ) 1 3 5.0 ( loj Ans. ^^'^= ^'^^^

125 ' ^ = L080 = i8.Ans.

100 4

60

125 5

(148) ;r : 35 :: 4 : 7. (149) 9 : x :: 6 : 24.

7.r=35x4 6.ar=9x24

7;r=140 6;r=216

140 „^ . 216 ^^ .

X = -y- = 20. Ans. X = -— = 36. Ans.

(ISO)

-^^1,000 : fl,331 :: 27 : ;r. -f^T^OOO = 10.

10: 11:: 27 : jr. V^i;33T=ll.

10-t:=297. 1 1 1'331(11

297
10

297 ^^ „ 1 2 1

;r = — - = 29.7. _ _

Ans.

2 300 331

1 31 331

30 331

1

31

(151) 64 : 81 = 21' : .r'.

Extracting the square root of each term of any proportion
does not change its value, so we find that -/el : -/Sl =
|/2p : 4/^' is the same as

8 : 9 = 21 :.r

8 ^ = 189
;tr= 23.625. Ans.

(162) 7 + 8 : 7 = 30 : -r is equivalent to
15 : 7 = 30 : X.
15 ;ir = 7 X 30
15;ir=210

X = -—— = 14. Ans.
15

92 ARITHMETIC.

(163) 2 ft. 5 in. = 29 in. ; 2 ft. 7 in. — 31 in. Stating
as a direct proportion, 29 : 31 = 2,480 : x. Now, it is easy
to see that x will be greater than 2,480. But r should be
less than 2,480, since, when a man lengthens his steps, the
number of steps required for the same distance is less;
hence, the proportion is an inverse one, and

29 : 31 = ;ir : 2,480,
or, 31:r = 71,920;
whence, x = 71,920 -r- 31 = 2,320 steps. Ans.

(154) This is evidently a direct proportion, 1 hr.
36 min. = 96 min. ; 15 hr. = 900 min. Hence,

96 : 900 = 12 : ;r,
or, 96;r= 10,800;

whence, x = 10,800 ^ 96 = 112.5 mi. Ans.

(155) This is also a direct proportion ; hence,

27.63 : 29.4 = .76 : x,
or, 27. 63 ;ir = 29. 4 X . 76 = 22. 344 ;

whence, x = 22. 344 -^ 27. 63 = . 808 + lb. Ans.

(156) 2 gal. 3 qt. 1 pt. = 23 pt. ; 5 gal. 3 qt. = 46 pt.

Hence,

23 : 46 = 5 : 4r,

or, 23.*-= 46 X 5 = 230;

whence, ;r = 230 -^ 23 = 10 days. Ans.

(157) Stating as a direct proportion, and squaring the
distances, as directed by the statement of the example,
6' : 12" = 24 : ;r. Inverting the second couplet, since this
is an inverse proportion,

6' : 12' = X : 24.

Dividing both terms of the first couplet (see Art. 310)
by 6

1» : 2* = 4.' : 24; or 1 : 4 = ^r : 24;
whence, 4 .r = 24, or x = 6 degrees. Ans.

ARITHMETIC.

93

(158) Taking the dimensions as the causes,

;^

15

^

5

2

%

^

f>

= ;^

X, whence, 2 -r = 75, or, x = \$37. 50.

Ans.

(159) 2 hr. = 120 min. ; 14 hr. 28 min. = 8G8 min.
Hence, 120 : 868 = 100 : x,

or, 120 ;r = 86,800;

whence, x = 723J gal. Ans.

Taking the dimensions as the causes,

(160)

2

^

^p

- m

17

57

^

X, whence, 2x = 17 X 57 = 969,
or, X = 484| bbl. Ans.

(161)

or

8 hr. 40 min. = 520 min. Hence,
444 : 1,060 = 520 : x,
130
, , ^ 1,060X m ^ 13M00 ^ 1,241.44 + min. = 20 hr.
^ ^^^ 41.44+ min. Ans.

(162) 1 min. = 60 sec. Hence,
5^ : 60 = 6,160 : x,
60 X 6,160

or, X =

5.5

67,200 ft. Ans.

(163) Writing the statement as a direct proportion,
8 : 10 = 5 : -r, it is easy to see that x will be greater than 6 ;
but, it should be smaller, since by working longer hours,
fewer men will be required to do the same work. Hence,
the proportion is inverse. Inverting the second couplet,

8:10 = ;ir: 5,
4

or, X — ^ ^ — 4 men.
I

Ans.

94

ARITHMETIC.

(164) Taking the times as the causes.

n

'%^

14

^

n

=m

m

;p

X %1l

2

3

^p ; whence, 3;r = 2 X 14 = 28, or ;ir = 9^ hr.

Ans.

(165) Taking the horsepowers as the effects, we have
for the known causes in example 4, Art. 349, 14', 500, and
48, and for the known effect 112 horsepower. Hence,

9

;^

m

m

22

14'

30'.

f>

m

500

660=112

X, or ^0p

m = m

48

42

^

3 ^

^

i^

whence, ;ir=9x22X3 = 594 horsepower. Ans.

"(166) First find the volume of the cylinder in cubic
inches, as in the example, Art. 345. The volume, multi-
plied by the weight of one cubic inch (.261 lb.), will evidently
be the weight of the cylinder. Hence,

10'
20

12'
60

= 1,570.8

100

X, or

^P

144

= 1.570.8

^P

x;

, 144X3X1,570.8 .^o«o«^ • o^u r
whence, x = ttut^ = 6,785.856 cu. m. Therefore,

100
weight of cylinder = 6,785.856 X .261 = 1,771.11 lb.

(167) Referring to the example in Art. 348,

i
5

Ans.

15

20'
10

40

18' = 187
12

100
X, or ^pp

;p

^p

324 = 187
4

whence, x = ^'^^ ^^^ ^^'^ = 484. 7 lb. Ans.

ALGEBRA.

(QUESTIONS 168-217.)

(168) - '"|^7S = ^l"/\rM - Ans. (Art. 482.)

(169) [a) Factoring each expression (Art. 457), we
have 9jr' + 12^y+ 4/ = (3^'+ 2/) (3;r'+ 2/) = (3;r' + 2/)'.

{b) 49^* - 154«'(^' + \^W = (7rt' - 11<^') {7a' - IW) =
{7a' - lUy. Ans.

{c) 64-ry + Q4xy + 16 = 16(2;r>' + 1)'. Ans.

(170) (<2) Arrange the dividend according to the
decreasing powers of x and divide. Thus,

3x—l)9x' + 3x' + x — l{3x*-{-2x-\-l Ans.
9x* — Sx'

T^t,l (Art. 444.)

dx-1

Sx-l

(6) a-b)a' - 2ab' + b' {a' -\- ab - b' Ans.

a' - a'b

a'b - "lab' ,. ^ ....

,, ,, (Art. 444.)

a b — ab ^ '

- ab'-^-b"

- ab'-^b'

{c) Arranging the terms of the dividend according to
the decreasing powers of x, we have

7x-.d) 7x* - Ux' + 58.tr -2l{x'-'dx+7 Ans.
7x* - S.r'

— 21;r' + b%x

— %lx'-{- 9x

AQx - 21
49.r - 21

96 ALGEBRA.

(171) See Arts. 352 and 353.

(172) («) In the expression 4:xy — 12;try -|- Sxy"", it is
evident that each term contains the common factor ixjf.
Dividing the expression by 4-ry, we obtain x'' — 3xy -f 2^
for a quotient. The two factors, therefore, are 4a-j and
;r' - Sxy + 2f\ Hence, by Art. 452,

4;ir'y - 12;iry + 8;i'j>^' = ixyix" - 34r> + 2/). Ans.

(^) The expression {x* —y*) when factored, equals {x'' -j-
/)(-^' -/). (Art. 463.) But, according to Art. 463, r^
—y may be further resolved into the factors (x -\-j^){x — /).

Hence, {x* — y) = (■^' +y )(•*'+/)(■*' —j)- Ans.

(^) Sx' — 27/. See Art. 466. The cube root of the
first term is 2x, and of the second term is 3/, the sign of the
second term being — . Hence, the first factor of Sx^ — 27j'
is 2x — Sf. The second factor we find to be 4;jr* + ^-^f + 9j'»
by division. Hence, the factors are 2x — 3/ and
4;ir' + 6xy + 9^".

(173)- Arranging the terms according to the decreasing
powers of m.

3m' + lOm^n + lO^nn^ + dn'
dm*n — 5m'n'' + dfw'n' — mn^

^nfn + 307«"«" + 30;«';«=' + %m'n*

— Um'n^ — 50m'n' — 50m'n* — 15m'n*

4- 15m^n' + oOm*n* + 507;/'«' + 15m'n*
— dm*n* — 10 ///' ;?^ — IQjM'fi' — 3mn'

%m'n + ISw'w" — 5/«*«* 4- 6»2'«* + 25;«'«' + 5w'«' — 3w«'

Ans.

(174) {2a'bcy = lQa'b*c'\ Ans.

( - 3«'^V)^= - 243a"'<^'V*. Ans.
( - Im'nxyy = 497n'n\vy\ Ans.

(1 75) (a) 4a' - ^' factored = {2a + /^)(2« - d). Ans.

(<^) IB;!;" - 1 factored •^. {Ax' + l){4x' - 1). (Art. 463.)

Ans.

ALGEBRA. y7

{c) 16x* — Sx*y* + xy*, when factored =

(4.r» - xy'){Ax' - xy^). (Art. 467, Rule.)

But, (4a-' - ;r/) = ^'(2^ + j)(2;r - j). (Arts. 462 and
463.)

Hence, 16^* - 8;try + .ry = .r" (2^ + j) (2;r + j) {2x — y)
{}lx — y). Ans.

(176) 4a' - 12^^ +5«*.r»+6a';r*+aV(2a='-3aV-a4r*
4a* Ans.

4a* — 3a\r

4a'— 6a'jr— a^

— 12a'^ + ha*x

- 12a'x + QaV"

— 4a'jr* + 6a'.r' + a';ir*

— 4a*;tr' + 6a';ir' + a V

(177) (a) 6a*/^^ + a*^'-7a»/^» + 2a(5^ + 8.

{d) 3 + 2a/^c + a'd' - 7a'd' + 6a^^\

(c) 1 + a^ + a" + 2 a'. Written like this, the a in the
second term is understood as having 1 for an exponent;
hence, if we represent the first term by a", in value it will
be equal to 1, since a" = 1. (Art. 439.) Therefore, 1
should be written as the first term when arranged according
to the increasing powers of a.

(178) i^lGa^W? = ± 2a'dc\ Ans. (Art. 621.)
V-32a"= - 2a'. Ans.
f/-l,728aV"jry = - 12aV*Ay. Ans.

(179) (a) {a-2x-^4y)-{32+2d-c). Ans. (Art. 408.)

{d) —^b — A:C-{-d— (2/— Ze) becomes
_ [3^ _[- 4^ _ ^-|- (2/— 3f)J when placed in brackets pre-
ceded by a minus sign. Ans. (Art. 408.)

[c) The subtraction of one expression or quantity from
another, when none of the terms are alike, can be repre-
sented only by combining the subtrahend with the minuend
by means of the sign — .

98 ALGEBRA.

In this case, where we are to subtract
2^— {'Sc -]-2d) —a from x, the result will be indicated by
x-[2d- {3c + 2d)-a.'] Ans. (Art. 408.)

(180) {a) 2x' -]- 2x' -{- 2x - 2
X -1

2x* + 2x' + 2x' - 2x

- 2x' - 2;ir' - 2^ + 2

2x* — 4x-\-2 Ans.

(d) x^ — Aax -\- c

2x -{- a

2x^ — Sax'' + 2cx

ax' — Aa'x + ac

2x' — lax' + 2^-*" — 4«'-«" + ac Ans.

(c) - a*-irZa'b-2b*
6a' + ^ab

- ha" + 16a' b - IQa'b'

- %a'b + 21a'b' - ISab*

- ba" + Qa'k - lOa'b' + 21a'b' - ISab*
Arranging the terms according to the decreasing powers

of a, we have - ba" + %a'b + 21 a' b' - 10a' b' - lSab\ Ans.

(181) (a) ^xyz The sum of the coefficients

— Zxys of the positive terms we find to

- 6xyz be + 13, since (+ 3) + (+ 6) +
(Sxyz (+4) = (+13).

— %xyz When no sign is given before
"dxyz a quantity the -f sign must al-

— 4,xvz Ans ways be understood. The sum

of the coefficients of the nega-
tive terms we find to be — 17 since (— 9) + (— 5) + (— 3) =
(— 17). Subtracting the lesser sum from the greater, and
prefixing the sign of the greater sum ( — ) (Art. 390, rule
II), we have (+ 13) + (— 17) = — 4. Since the terms are
all alike, we have only to annex the common symbols xyz to
— 4, thereby obtaining — A:xyz for the result or sum.

ALGEBRA. 99

(6) So* -\- 2alf + 4(^' When adding polynomi-

5a* — Sad -\- b^ als, always place like terms

— tf * + 5rt^ — b* under each other. (Art.

18a' - 20^^ - 19^' 393.)

14a' — ^^ab + I^V The coefficient of a* in

39a''-24a^+ 5^' Ans. ^^^ result will be 39, since

(+14) + (+18) + (-l) +
(+ 6) + (+ 3) = 39. When the coefficient of a term is not
written, 1 is always understood to be its coefficient. (Art.
359.) The coefficient of ab will be — 24, since (- 3) +
(-20) + (+ 5) + (- 8) + (+ 2) = - 24. The coefficient of b*
will be ( + 20) + (-19) + (-l) + (+l) + (+4) = +5.
Hence, the result or sum is 39a' — 24a^ + hb*.

{c) ^mn + 3a^ — ^c

+ Imn - ^ab + 3^ + 3w' — 4/

%mn — ab — 4:C -{- 3x -{- dm^ — 4/ Ans.

(182) Thereciprocalof 3.1416is-— i— =.3183+. Ans.

o.141d

Reciprocal of . 7854 = -=^ = 1. 273 + . Ans.
^r 1 1 .. 64.32 „, „^ .

°f 602 =-T- = ^ X -T- = '^'^- ,t"\ . .„, .

(183) (a) —^ \- ^ ~ ^ . If the denominator of the

^ ' ^ ' X — y y — X

second fraction were written x — y^ instead oi y — x^ then

X — y would be the common denominator.

By Art. 482, the signs of the denominator and the sign

X 1/ X y

before the fraction may be changed, giving —.

We now have

_^c__x_-j!^x-x^y^_j^ Ans.
X — y X — y X — y x — y

X* X X

(V) —5 - I r— 7 — z . If we write the denominator

^ ' x^ — \ ^ X -\-\ 1 — X

of the third fraction x — 1 instead oil — x^ x"* — 1 will then

be the common denominator.

100 ALGEBRA.

By Art. 482, the signs of the denominator and the sign

X

before the fraction may be changed, thereby giving -.

We now have

x" X X _ X* + x{x — 1) + x{x + 1)

x' -1 ' x-\-l ' x-1 x'-l

X* + x' — X + x'' -\- X Sx*

x'-l

Ans.

, . Sa — U 2a — d -\- c , Ida — Ac

{c) r; q — \- ■ T^ > when reduced to

a common denominator

_ 12{3a - U) - 2S{2a - 3 + ^) + 7(13^ - 4^)
"" 84

Expanding the terms and removing the parentheses, we
have

S6a - 48^ - 5Ga + 283 - 28^ + 91^ - ^Bf
84
Combining hke terms in the numerator, we have as the

result,

71a - 20b - 56<r .
84 • ^^^•

(184) {a) 45jry-90^y-360;ry=
A5xy{xy-2x-%y). Ans. (Art. 452.)

{b) a^b^ + 2abcd-\- c'd'' = {ab + cd)\ Ans. (Art. 457.)

{c) {a-\-bY -{c — dY=.{a^b-^c-d){a-\-b-c-\- d).
Ans. (Art. 463.)

(185) {a) If a man builds 20 rods of stone wall, and
we consider this work as positive, or -|- , the work which he
does in tearing it down maybe considered as negative, or — .
If he tore down 10 rods, we could say that he built — 10
rods.

{b) See Arts. 388 and 398.

(186) (.)^^l±^-^_£-=£(^^±4-)x^^^.
^ ' ^ ' a^ — X* a — X a — x^ x

(Art. 502.)

ALGEBRA. 101

Canceling common factors, the result equals -j— — — 5.

Ans.
a — x) a^ — x"" {^a^ ■\- ax ■\- X*
a* — a*x

a'x — X'
a^x — ax'*

ax^ — jr*
ax"^ — x'

{b) Inverting the divisor and factoring, we have

Online LibraryInternational Correspondence SchoolsThe elements of railroad engineering (Volume 5) → online text (page 5 of 15)