Module 07 - Sol

Republic of the Philippines GILLESANIA Engineering Review and Training Center

Cebu

BOARD OF CIVIL ENGINEERING MATHEMATICS, SURVEYING & TRANSPO. ENG’G. Friday, October 27, 2017 SOLUTIONS  1

 7

JUAN  ASAWA NI JUAN JUAN

NAKARAAN

KASALUKUYAN

x y = 12

24 = 2y x

dy

m

dy

=

dx dt

Ang asawa ni Juan ay 18 na taong gulang.

dx

dx

√ 

|A| = 32 + 62 + 22 |A| = 7

dy

−1

=

2

m =

−1

dt

=

2

= −2

 8

d π     3  2 

Cos θx = 3/7 θx = cos-1(3/7) θx = 64.6°  3

dt

=

dt

24 – x = x – 12 x = 18

 2

SET A

Module 7 solutions

4

3

2

=

12 π     × 6   2  

3

d = 6 6 = 1 0 .9 .9 cm cm  9

√  √ 

H = 1.5  (ft) H = 1.5 225 225 ft × 1 m/3.28 ft H = 6.86 m

  1145.916 R  = = 176.29 6.5 2

θ

 4

( 1 + i) i=

12

12

e

=

0.09

e

L

=

2RL s

0.09

20

θ= i=

4

2

θ 3

80 = 20

180°

2 × 176. 176.29 29 × 80

− 1  = 0.75282%

  8000 = 1062671.67 P = i

1

L =

π

  = 0.8125°

= 0.27°

 10  5

3

  L   3 L− t   4   t =

40 11

=

=3

L−

L 5

1 + xQ

=

4

2

xQ = 5

t

 11

 6

y

Let x be the money he had at the start

=

2

4 − 3x 3x − x

Bet

Balance nce

Balance Balance afte r result

y' = -3 - 2x

x

x−x

2x

30

2x − 30

3 ( 2 x − 3 0)

m = −3 − 2( 2) m = −7

54

3( 2 x − 30 3 0) − 5 4

4[ 3 ( 2 x − 3 0) − 5 4 ]

72

4[ 3( 2x − 3 0) − 5 4 ] − 7 2

48

4 [ 3 ( 2 x − 3 0) − 5 4 ] − 7 2

=

y − y1 y+ 6 y

48

=

Result

Bet

Amount before Betting

4th 3rd   2nd  1st 

48 120 84 58

72 54 30 x

72 + 48 = 120 120/4 + 54 = 84 84/3 + 30 = 58 58/2 = x = $29

=

m( x − x1)

−7( x − 2)

− 7x + 8

x = 29 Round

=

 12

5 + yQ

yQ = 3

7 11

=

2

3

x

x =

det( x) = x

2744

=

3

 2 −3

2 7 44 44 = 1 4

 V =

2

D =

x ( 3) = 24 .2 5 4

 V =

3

4  

 1 2 −1  = −7  3 −1 2  

 V = 7

3

D π     = 7465.6

 19

 2 

a  A 

B

 13

a (24, h)

30°

θ

(8, 20)

a cos 30°

20 m 16 m

(0, 0)

D

(48, 0)

 θ   2 

t an

θ = 

C

0.5a

=

2atan

a − acos acos( 30° 30°) 0.5     os( 3 0° 0°)   1 − c os

θ =  150°  20

 14

 t  

S 

=

400

v 

v

+

400 40 0 v + 10

=

14 +

2 3

  50   v =

−

 

60

v = 5 0 mp h

11  

47

25

 15

i' =

50 1000

0.05

=

= 0.05

32

i + 0 . 0 2 + 0.02i

50 1000

a = 32

= 0.05

β = 

i=

i = 2.94 % R 

=

2

2

acos 1

 16

c = 25

 a + c − b   =  110.49° 2a c   2ac   2

  0.05 − 0.02 0.02

1 + 0.02 i = 0.02 0.0294 941 1

b = 47

1

a

2

cos ( θ)

=

2

c

cos ( β − θ)

Given a

=

cos ( θ)

c cos ( β − θ)

θ = 

Find( θ) = 5 0 . 3 7 ° a x = t an( θ) = 1 9 . 3 2 2

1

 Alternat  Alt ernat e Solut S olution ion 2

s= 3

 A t = 2

2

( 1 + 2) + 3 = 1 18  17 12³ + 1³ = 10³ + 9³ =  1729

 18

R  =

a+ b + c 2

= 52

s( s − a) ( s − b ) ( s − c ) = 37 3 7 4 .7 ab c 4A t

= 25.09 2

 a   = 19.32 x = R  −  2  2

 21

FN

  Heal_Length

π(

 V w =

=

Heal_Spread   3025 =9 FN = 336.11

4

1

 V t =

 22

2

3.6

) × 10  = 101.79

2

( 3 . 6 − 1) × 1 2 × 5.5  = 85.8

 V R  = V w − V t  = 15.99  V R    5 .5 .5 × 1 2( h − 2 .6 .6) =

θ

15

1

h =

 V R + 2.6  = 2.842 66

 27

30 m

R( x) = 12 0 0 x

15 θ =  acos    = 60 °

 30 

0.10( 10000) ( 2 0 0 )

C( x) = 80 0 x + 20 0 0 0 +

P = 2 ( 3 6 0 ° − 2 θ) ( 3 0 ) = 2 5 1 . 3 3

12

Given

 23 x

x

R( x) x

=

C( x)

x = Find( x) = 9 1. 67 16 cm

round( x) = 9 2  28 8x – 5(26 – x) = 0 X = 10

√ 2 + 2 = 16  = 4.69 √ 2 = √ 2(4.69) √ 2 = 6.63

 29 D

B

 24

4 3

3

π

3

r + ( 2r) + ( 3r) 3

R  =

3

=

3

π R 3

E 5

2

 A

 A s

=

 A s

=

1 2 3 a 4 2 2

=

173.2

  173.2

= 10

3

2

2

3 −x +

DE =

2

2

4 −x

(

2

)

2

4 −x

2

+

5 −

−(

3

2

(

−x ) 2

2

3

3 45°

6 2

α

3

x

4

= 117.85

5 β θ

D

x

B

)

2

3  − x

C 5

6 2

5

2

 A 

a

10

x+

=

2

Alternate Solution

3

 26

C

x

2

x = Find( x) = 1 . 7 2

2

3a

2

3a

 V =

 52   32

 3

Given

 25

=

3

       �

3

2

 V

x

36r = 3.302r

 A s = 4 π R  = 137.01r

a=

 4

4

4

4

2

2

= 5.657

 42 + ( 3

α = 

acos



2

2) − 5

2 × 43 2



2

 35



L

α =  74.62° 2

( 20m )

0 . 0 0 8 − ( − 0 . 0 0 4) 0.002

( 20m )

L = 1 20 m

x = 6 .07

   sin( 45° + α) 3  =  25.44° x    

β = 

asin

θ=

90° − β   = 64.56° 2

 36

d

( h 1 + 2 Σh i + hn ) 2 15  A  = [ 5 .5 .5 9 + 2( 3 .3 .3 8 + 2 .3 .3   0 + 3.96) + 4.80] 2  A = 222 .53  A 

2

6 . 0 7 + 4 − 2 × 6 . 07 07 × 4 cos ( θ)

DE =

∆g

L =

2

3 + 4 − 2 × 3 × 4 cos ( α + 45° )

x =

g1 − g2

=

DE = 5 .6 .6 56 56

=

 37

 30

12 !

N =

60

1

= 7983360

2

x S ec ec on on d G am ame

T hi hird G am ame

x

x−y−z

2 ( x − y − z)

2 [ 2 ( x − y − z) ]

y

2y 2y

2y − ( x − y − z) − 2z

2 [ 2y − ( x − y − z) − 2z]

z

2z

2 × 2z

2 ×2   z − 2 ( x − y − z) − [ 2 y − ( x − y − z) − 2z]

=

2

x −5

=

3

 38

24

2 [ 2y − ( x − y − z) − 2z]

=

24

2 ×2   z − 2 ( x − y − z) − [ 2y − ( x − y − z) − 2z]

=

24

 39

 39   21   12 

 

t = Pangapat 24 24 24

Pangatlo Pangalawa 12 + 12 + 24 = 48 48/2 = 24 24/2 = 12 6 + 24 + 12 = 42 24/2 = 12 12/2 = 6

Una 24/2 = 12 42/2 = 21 12+21 + 6 = 39

 32

120mi

( 2 5 + 15 ) mp h

Say the last loser was A, the seconds was B and fir st was C. Manglalaro  A B C

2x

1

−

  3  

First G am am e

2 [ 2 ( x − y − z) ]

1

=

 − 5  

let x, y and z be the amount of money of the three who lose in sequence

Find( x , y , z) =

x

x −5

 31

Given

1

−

= 3 hr

S = 100m 100mph ph( 3hr 3hr) = 30 0 mi  40

let x be number of plants in each of the y rows

is the number of vehicles moving in a specified direction on a given lane or roadway that pass a given point during specified unit time and is usually expressed as vehicles per hour or vehicles per day.

Traffic volume

xy

=

60 0

( x − 5) ( y + 6) 6 0 0  600 + 6  600 ( x − 5)   x   =

=

 33

COURSE  AB BC

DEP

DMD

117.56

y

x

426.20

x =

 −20    25  

-20 is absurd

the re are 25 plants in in a row

DEPAB = 117.56 DMDBC = DMD AB + DEP AB + DEPBC 426.20 = 117.56 + 117.56 + x x = 191.08 m

 41

 34

θ

12 11

x

1

10

=

2

9

3

8

4 7

5 6

x =

15 + 180 11

x 12

= 16

4 11

T Lc

θ T

Rt an an

=

=

=

 θ   2 

Given a+ b a+ c

=

12

180° 180° ( 2 π R − Lc )

b+ c

=

14

 a 

 4 

b =  c 

 180° ( 2 π R − Lc)  Rt an an 2π R   

Given 5280 5280

=

Rtan Rtan

28 0)   180° (2 π R − 5 28 2 π R 





Find( R ) =  1437.44  42

N

=

52

1+

∑ x

+

51

52

N =

10

π R( 360° − θ) π R 

=

=

=1

52 x

52 50

+

52 49

+ ... +

52

Find( a , b , c) =

6  8 

π ( 8 2) =  201.06

 A  =  46

x² = -12y x² = ±4ay LR = 4a = 12  47

1

= 235.98

round( N) = 2 3 6  43 Σd

= 2

R=1

t=4

t=3

t=2 t=5

C = π(1 + 1 + 1 + 2) C = 5π

ST = |f(4) – f(2)| + |f( 5) – f(4)| ST = |-31 - -15| + |-24 - -31| ST = 16 + 7 ST = 23

 44

 48

x

g1 = 4% dθ /dt

 = tan−1 20   =      20 miles

g 2 = − 6%

 A = g1 − g2

L = 1 4 0m

Given L

 A S

2

=

h1

2 ( h1 + =

h2)

2

h2

   89.29  Find( h1 , h 2) = cm  89.29   49

 45

A = ½ d₁ d₂ sin θ d₁ = |A| = 32 + 12 + 22  = 14 14 2 2 2 26 d₂ = |B| = 1  + 3  + 4  = 26

√  √ 



cos  =

√  √ 

  14 √ 26 26 √ 14

3(1) + 1( 3) + ( 2)(4)

 = 114.79° A = ½�√ 14 14√ 26 26 sin 114.79° A = 8.66 square units

S = 10 0 m

 50 3x2 – hx + x – 7 h = 0 3(3)² – h(3) + (3) – 7 h = 0 h=3

 56

S2

g2L

=

48

g 1 − g2

=

 51

 

0.04L

0.0 6 − − 0.0 4

L = 1 2 0.0

r = 5 + 12 = 13 x = tan-1 12/5 x = 67.38° x = 1.176 2

2

 57

  1    6000 

rexi = 13e1.176i 

2   550cm =

2

  1m    100cm 

2

 A t 2

 A t =  1980000m  A t =  198hectare

 52

  =  + 2

 A

 52 +  106 =   = 160 2   160 + 76 ∠ = 160 2 ∠ = 118

76°

 58

B 106° E

52° C

D

 53

 59

E = e(tan θ₁ - tan θ₂) E = 5’[tan 42° - tan (-38)°] E = 8.41’  60 log 64 – log 128 + log 32 = log 2⁶ - log 2⁷ + log 2⁵ log 64 – log 128 + log 32  = 6 log 2 - 7 log 2 + 5 log 2 log 64 – log 128 + log 32  = 4 log 2  61

96 – 2x = 32x 32(6 – 2x) = 32x 2(6 – 2x) = 2x 12 – 4x = 2x 6x = 12 x=2

 54

d = 9 − 7. 7 . 5 = 1.5 an

=

a1 + ( n − 1) d

34.5 7.5 + ( n − 1 ) ( 1.5 ) n = 19 n S ( a1 + an) 2 =

=

19

S =

2

( 7 .5 .5 + 3 4. 4. 5)

 62 A = A₀e0.4t  2A₀ = A₀e0.4t  t = 1.733  63 R = P(-3/2) = 4(-3/2) 3 + 18(-3/2) 2 + 8(-3/2) – 4 R = 11  64

S = 3 99

t

=

L

=

 55

K

=

FC +

K = 30 +

OM i

+

2 0.08

K = 60.98 M

RC

+

1.08

15

−1

gt

g 2

4π

n

( 1 + i) − 1 13

L

2π

L =

2

9 .8 .8 1 × 1 2

2

= 0.24849m

4π L = 2 4. 4. 85 85 c m  65

 are intended to protect the land from Sea wall and Revetment  are wave erosion.