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MATH REFRESHER –ENGR. FUENTES

Cebu: JRT Bldg., Imus Avenue, Cebu City Tel. 2685989 – 90 | 09173239235 Manila: 3rd

 & 4th

 Fl. CMFFI Bldg. R. Papa St. Sampaloc Tel. 7365291

( ) ( )

( ) ( )

A ab 3 2 6

V 2 RA 2 3 6

V 355.31

= π = π = π

= π = π π

=

( )

( )

( )( ) ( )( )

( )( ) ( )( )

2

22

22

111

2

2

2

1

h3r h

V 3hV

3r h3

223 15 22

V 3 4.7V 8

3 15 83

π−

=π

−

π−

= =

π−

26.

( )

( )

2 1

2 2

2

1 2

42

0

Using shell method:

V 2 RdA 2 y x x dyx y 25

x 25 y ; x 5

44V 2 y 5 25 y dy

3

= π = π −

+ =

= − =

π= π − − =

∫ ∫

∫

 

27.

( )2

3 122

1

3 12 3 2 2

3S 1 y ' dx 1 x dx 4.66

2

3y x y x y ' x

2

= + = + =

= → = → =

∫ ∫

 28.

29.

Equate 1 and 2. Then solve for x. x = 2cm

30.2 2

2 2

y x1

9 4

a 9 b 4

a 3 b 2

+ =

= =

= =

 

31.

32.

( )33 0

3

2r 180V 1.25 rad

270 270

V 6.67 m

π   π θ= = ×

π

=

 

33.

( )

( )

( ) ( )

2

2

2

4y x 2x 1

x 2x 4y 1 0

2h 1

2Subst. h for x and solve for

y which is k:

1 2 1 4y 1 0

y k 0

V : 1, 0

Length of latus rectum:coef. of y

L 4a 4

a 1

The parabola is opening upward.

2 2 8A bh 4 1 sq. units

3 3 3

= − +

− − + =

−= =

−

− − + =

= =

= =

=

= = =

 

34.

( )

1 1 1 1

2 2 2 2

2 2

Ax By C x 2y 8d

A B 1 2

5 2 2 84.02

1 2

+ + + += =

± + + +

+ − += =

+ +

 

35.The minute hand will pass the number 8 onthe clock.

( )t 8 5.4545 43.636 min

60s43 min .636min

1min

t 43min 38.2s

Time : 2 : 43: 38.2

= =

= +  

= +

 

36.

( )

3 5

1 3

1

1

2 2 1r GeometricPr ogression

2 2 4

a 2 2Sum

1 r 1 1 4 3

− −

− −

−

= = = →

= = =− −

 

37. Reverse Engineering:

( ) ( )

( )

( )( )   ( )( )

If there are choices :

1g x x 2 correct

3

f x 3x 2

1

g f x 3x 2 2 x satisfied3

= − →

= +

= + − = →

 

38.H.P.: 1, 1/4,1/x, 1/10. The reciprocal should

form arithmetic progression:

A.P.: 1, 4, x ,10. Therefore x is 7.

 

39.

40.

( )

2 2

22

a 25 b 9

a 5 b 3

2 32bL 3.6

a 5

= =

= =

= = =

 

41.

( ) ( ) ( )

( ) ( ) ( )

2 2 2

2 1 2 1 2 1

2 2 2

r d x x y y z z

r 8 0 1 0 6 0 10.05

= = − + − + −

= − + − + − =

 

42.

The problem can be solve by reverse sincethe curve will pass through (0.5, 0) which isa center of one the circles. Then, (0.5, 0)will satisfy the equation of the locus.

43. Use Reverse Engineering.44.

3

S S

L S S LL L

3

L

L

L

V dV V 1260 V 1260 V

V d

1260 V 2

V 3

V 972

= + = → = −

−   =

=

 

45.

( )

( ) ( ) ( )

( )

( )

( ) ( )

( )

( ) ( )

2 2

2 22 2

22

1 1 1 1

2 2 22

22

x y 4x 6y 12 0

4 6h 2; k 3 C : 2,3

2 2

r h k f 2 3 12 5

A r 5 25

Ax By C 2x 3y 12d

A B 2 3

2 2 3 3 126.93

2 3

Using 2nd proposition of Pappus:

V 2 RA 2 6.93 25 3419.82

+ + − − =

−= = − = = −

− −

= + − = − + − − =

= π = π = π

+ + − −= =

± +   − + −

− − −= =

− + −

= π = π π =

 

46. To ensure that there will be three balls of thesame color, consider the worst casesituation. The situation is 10 balls are taken

but no three balls of the same color: 2 white,2 green, 2 black, 2 red, 2 yellow. Then the11th ball will definitely produce 3 balls of thesame color. The answer is 11.

47.

( ) ( ) ( )100 0.04 x 0 100 x 0.05

x 20kg

+ = −

=

 

48.

( ) ( ) ( ) ( )( )

2 2 2

2 2 2

o

Solving for θ using

Cosine law:

c a b 2abcos

130 190 180 2 190 180 cos

41.02

= + − θ

= + − θ

θ =

 

( ) ( ) ( ) ( )   ( )

2 2 2

2 22 o

Solving for L using

Cosine law:

c a b 2abcos

L 95 180 2 95 180 cos 41.02

L 125

= + − θ

= + −

=

 

o

o

base

2 2 2

2 2 2

s r

36 150180

s 30

s C 2 r

30 2 r

r 15

L h r

36 h 15

h 32.73

= θ

π =

= π

= = π

= π

=

= +

= +

=

i

( ) ( )

2

2

1V r h

3

115 32.73

3

V 7711

= π

= π

=

( )

( )

22 2

22

5 R 5 x

R 5 5 x Eq.1

= + −

= − − →

10 10 x

5 R

10 xR Eq.2

2

−

=

−= →

( )3,0,0

( )3,3,0

( )0,0,4

( )0,3,4

W 3=

L 5=   ( )( )2

A LW

5 3

A 15 m

=

=

=

( ) ( )

( ) ( )

2 2

2 2 2

2 2 2

2

x 1 x 2 y 0

x 1 x 2 y

x 2x 1 x 4x 4 y

y 6x 3 0

+ = − + −

+ = − +

+ + = − + +

− + =

( )h,k

2x 3y 12 0− − =

d R=

180130

95 95

θ

L

( )x,yx

x

1

( )2,0

( )0.5 ,0

1x

y 4=

x 5=

2xdy

y

x axis−

0150

36

s r= θ

L 36=

r

h

5

x 3=

R

5

105

x

RR

5

R

5 x−

R

10

x

10 x−

2h 22=

1h 8=

15

8/10/2019 Math R1 Soln26-99

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MATH REFRESHER –ENGR. FUENTES

Cebu: JRT Bldg., Imus Avenue, Cebu City Tel. 2685989 – 90 | 09173239235 Manila: 3rd

 & 4th

 Fl. CMFFI Bldg. R. Papa St. Sampaloc Tel. 7365291

( )

( )( )

22

213

1

Using Circular disc method:

V r dx 1 y dx

1 x x 1 dx

16V

105

−

= π = π −

= π − − +

π=

∫ ∫

∫

( )2, 9−

2

9−

f

a 1/12=

a 1/12=

107y

12= −

1092,

12

−

x

y

83.

( )( )

( )

( )

x yi 1 2i 7 4i

x 2xi yi 2y 1 7 4i

x 2y 2x y i 7 4i

x 2y 7 Eq. 1

2x y 4 Eq. 2

Solve the two equations:

x 3; y 2

+ − = −

− + − − = −

+ + − + = −

+ = →

− + = − →

= =

 

84.

( )   ( )1.18 j 1.18jln 5 12i ln 13e ln13 lne

2.565 1.18j

+ = = +

= +

 

85.

( ) ( )

( )

( ) ( )

( )

( ) ( )

2 2

2 2

2 2

2 2

2

2

2 x 4x a 3 2x bx 1

5 cx 5x 6 23x 17x 5

2x 8x 2a 6x 3bx 3

5cx 25x 30 23x 17x 5

2 6 5c x 8 3b 25 x

2a 3 30 23x 17x 5

2 6 5c 23; 8 3b 25 17;

2a 3 30 5

c 3; b 0; a 19

a 3b 2c 19 3 0 2 3 13

⇒   − + − − + −

+ + + = + −

⇒   − + + − +

+ + + = + −

⇒   + + + − − +

+ + + = + −

⇒   + + = − − + =

+ + = −

= = = −

+ + = − + + = −

 

86. Reverse Engineering 

87. For vertical asymptote, equate thedenominator to zero:

x 2 0

x 2 ans

+ =

= − → 

88. .

1C k

N

10.0125 k 250

k 3.125

=

=

=

 

89.

1 1x k x 36

y y

1 19 k x 36 2

4 18

k 36

= = −

− = = − =−

= −

 

90 and 91:

( )

( ) ( )

2

2

2

3x 12x y 21 0

Divide the equation by 3:

1x 4x y 7 0 Eq.1

3

Vertex : h, k

4h 2

2

To solve for k, substitute h to x in the

Eq. 1 and solve for y. y k.

12 4 2 y 7 0

3

y 9

k 9

Vertex :(2, 9)

From Eq. 1,length of the

− + + =

− + + = →

−= =

−

=

− + + =

= −

= −

−

 latus rectum is the

coefficient of y. So,

14a

3

1a

12

The parabola is opening downward:

=

=

 

1 109focus : 2, 9 2,

12 12

1 107Directrix : y 9

12 12

− − → −

= − + = −

 

92 and 93

( ) ( )

( )

2 2x 4y 2x 32y 27 0

2 32h 1; k 4

2 1 2 4

C : 1,4 ans

− + + − =

= = − = =− − −

− →

 

( ) ( )

( ) ( )

( ) ( )

2 2

2 2

2 2

2 2

2 2

x 4y 2x 32y 27 0x 2x 1 4 y 8y 16 27 64 1

x 1 4 y 4 36

y 4 x 11

9 36

a 9 b 36

a 3 b 6

The transverse axis is parallel with y - axis.

− + + − =

+ + − − + = − +

+ − − = −

− +− =

= =

= =

 

( )( )   ( )

( )( )   ( )

vertices : 1, 4 3 1, 7

1, 4 3 1, 1

− + → −

− − → −

 

94-98. Reverse Engineering

99.

3

3

3

Intersections:

y x x 1 Eq.1y 1 Eq.2

1 x x 1

x x 0

x 0; x 1 ; x 1

= − + →

= →

= − +

− =

= = = −

 

y 1=

3y x x 1= − +

y

dx

1− 1

r

( )1,4−

( )1,1−

( )1,7−