����� ������ ������
��� ������������������
1. The three sides of ∆ABC are extended a distance equal to their length. If the area
For new area:
X
A B
Y
+
x + 90 − x + 180 − x = (11 − 2) 180o
x = 30o
d + 10 x 4 = 25 25 π + 10x
=π
Anew Pnew
h
13 80
2
of ∆ABC is 3 2 , what is the area of ∆XYZ?
���� � o
l arg est = 180 − 30 = 150o 6. A 5 in by by 6 in rectangle piece of paper is folded so that one corner lands on the diagonally opposite corner. What is the length of the crease?
= πd + 2x 2 x = 10 π + 2 x
New changes in the perimeter: Pold = Pnew 20π = 10π + 2x x = 5π
C
So, A new A old
Z ∆ABC = ∆XAB → same ∆XAB = ∆XBY →
base and height
same base and height
(
A = 7∆ABC = 7 3 2
) = 21
2
Answ nswer: 21 2
2. A circular table table is tangent tangent to two two adjacent walls of a room as shown in the figure. Point N is 10 inches from one wall and 5 inches from the other wall. What is the area of the circular table?
=
6−x
4. A sphere with with a 10-cm diameter sits in a cone so that the point of tangency is 12 cm up the cone’s edge from the vertex. How much liquid can be under the sphere if the liquid and sphere just touch one another?
x
6 − 2x
Using Pythagorean theorem:
(6 − x) x=
2
=
x2
+
52
11 12
So, a2 5
a=
5 x
12 h
(r − 5, r − 10 ) r
=
5
θ
r
5 X 10 N
•
5
a
Ans.
5
r
6−x
10 0 π = 0. 0 .75 →
5
25π + 10 ( 5π )
h
( 6 − 2x )
2
+5
2
11 6 − 2 12
2
+ 25
Answer: 6.51 in 12
O
7. There is a wind of 35 mi/h from 320°. Find the airspeed and heading in order that the groundspeed and course be 250 mi/h and 50° 5 0°, respectivel r espectively. y.
Using Pythagorean theorem: 12 2
Z
=
52
+ ( h + 5)
2
h=8 x2
+
y2
=r
2
at (r − 5,r − 10 )
(r − 5)
2
+
(r − 10 )
r 2 − 10r + 25 + r 2 r2
− 30r + 125 =
r1 = 25 and r2
=
2
=
r2
− 20r + 100 =
r2
0 5
So, Area = πr 2
= π ( 25 )
= 62 6 25π →
2
Ans.
3. When a circle of radius 10 cm is squeezed between two parallel lines 10 cm apart, as shown, its area changes but its perimeter remains constant. What fraction of the old area is the new area if the “ends” are semicircles?
Using tangent function: 12 o → θ = 67.38 tan θ = 5 Using ratio and proportion: 5 x 5 x = → = h+5 h 13 8 40 x= 13 Using cosine function: x cos ( 90o − θ ) = r 40 o o cos ( 90 − 67.38 ) = 13 r 10 r= 3 So, V=
1 2 πr h 3
V=
1 10 π (8) 3 3
r = 10
2
Using Pythagorean theorem: airspeed = 352
2
+ 2 50
= 252
mi/h Using tangent function: 250 tan ( 40 − α ) = 35 o α = 42 Answer: 252 mi/h, 42° 8. From a boat sailing due north at 16.5 km/h, a wrecked ship K and an observation tower T are observed in a line due east. One hour later, the wrecked ship and the tower have bearings bear ings S34° S34 °40’ E and S6 5°10’E. 10’ E. Find Fi nd the distance between the wrecked ship and the tower.
Answer: 93.08 cm3
x
10
10
x For the circle (old area): A old = πr 2 = 100π Pold
=
2πr
=
5. Eight of the the angles of of an undecagon undecagon have measures whose sum is 1380°. Of the remaining three angles, two are complementary to each other and two are supplementary to each other. Find the measure of the largest of these three angles.
20π
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Distance AC = 16.5 km Using tangent function: ������������������ ������ ������ ��, ����
����� ������ ������
��� ������������������ them. If only 5 of the 11 people can drive, in how many ways can they be seated?
CK 16.5 CK = 11.41 km tan34o 40' =
CK + KT 16.5 KT = 16.5 tan 65o10 '− CK tan65o10' =
Answer: 24.2 km 9. Today is Monday, 1 July 2002. What day of the week will be 2 9833 days from now? 2n can be a multiple of 7 23
=
8
8 − 1 = 7 = 7 (1)
26
=
64
64 − 1 = 63 = 7 ( 9)
9
2
=
512 − 1 = 511 = 7 ( 73)
512
23n − 1
= 7k
29833
=
3 3277 ) + 2 2 (
=
(2 (
=
( 7k + 1) 4
=
3 3277)
���� �
=
3 3277 ) 2 ( x4
)4
− 1+ 1
4 ( 7k ) + 4
add 4 days after monday Answer: Friday 10. What is the smallest counting number that is divisible by each of the first fifteen counting numbers? Get the LCM of the first fifteen counting numbers (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15) 23 x32 x5 x7 x11x13 = 360, 360 Answer: 360,360 11. Suppose that the tires on the rear of your brand new Hilux will wear out after 33,600 kilometers, whereas tires on the front will wear out after 46,400 kilometers. Also, suppose 5 identical tires, including the spare, come with the car. What is the maximum distance that you can drive? Rate of changing the tires (tires/km) 2 2 + → thread wear per km 33, 600 46, 400 Total distance (tires/rate) 5 → 48,720 km 2 2 + 33, 600 46, 400
5 ways to choose a driver 10! Ways to arrange the remaining passengers N = 5 (10! ) = 18,144,000 ways 14. Jay-R press carrots by hand to make carrot juice. One-eighth of the juice is extracted from the first pressing. Each subsequent pressing extracts one-eighth of the remaining juice from the carrots. How many times will a bunch of carrots need to press to extract at least threefourths of the juice?
(
a1 1 − rn
S=
)
(
)
15. In celebration of her birthday, Jen threw a big party. Every two persons shared a bowl of rice, every three persons shared a bowl of viand, and every four persons shared a bowl of fruit salad. There were 130 bowls used altogether. How many guests were present? → no.
of bowls used for the rice
→ no.
of bowls used for the viand
→ no.
of bowls used for the fruit salad
x x x + + = 130 2 3 4 x = 120 Answer: 120 16. The sum of the first 4 terms of an arithmetic sequence is 68, and the sum of the first 12 terms of the sequence is 254. Find the first term. 4 ( 2a1 + ( 4 − 1) d) 2 8a1 + 12d = 136 68 =
12 ( 2a1 + (12 − 1) d) 2 24a1 + 132d = 508 254 =
247 16
Answer: 48,720 km
a1 =
12. Forty students took a final exam on which the passing score was 70. The mean score of those who passed was 75, the mean score of those who failed was 63 and the mean of all scores was 72. How many students did not pass the exam?
17. How many possible combinations of dimensions of a rectangle with integer value side lengths in which the numerical value of the area is twice the perimeter?
(Total no. of passers) + (Total no. of nonpassers) = (Total no. of students) 75x + 63 ( 40 − x ) = 72 ( 40)
→
Ans.
xy = 2 (2x + 2y ) 4y y−4
2
P
=
5−2
1 1 5C2 2 2
P = 0.3125 → Ans.
∫
∞
0
xp −1e− x dx = (p − 1)!
where is p is a positive integer. If p is a non-integer: Γ ( p ) = ( p − 1) Γ ( p − 1) Γ ( p ) = ( p − 1)( p − 2) Γ ( p −
2)
Γ ( p ) = ( p − 1)( p − 2)( p − 3) Γ ( p −
3)
Γ ( p ) = ( p − 1)( p − 2) ... (p − n ) Γ ( p − n ) ∞
∫x
19. Solve
p −1 − x
e dx , where p is a positive
0
integer. A. infinity B. (p – 1)! * C. p! D. p/(p+1) Use reverse engineering. (Assign a value of p (p > 1) and let ∞ → any value >>> zero. 20. Given
Γ (1.5 ) =
0.8862 , determine
Γ ( 3.5 ) . Γ ( p ) = ( p − 1)( p − 2) Γ ( p −
2)
Γ ( 3.5 ) =
2.5Γ ( 2.5 )
Γ ( 3.5 ) =
2.5 (1.5 ) Γ (1.5 )
Γ ( 3.5 ) =
2.5 (1.5 )( 0.8862 ) = 3.32325
21. Given
Γ (1.5 ) =
0.8862 , determine
Γ ( −0.5 ) . Γ ( p ) = ( p − 1)( p − 2 ) Γ ( p −
2)
Γ (1.5 ) =
(1.5 − 1)(1.5 − 2) Γ (1.5 − 2) Γ (1.5) = ( 0.5 )( −0.5 ) Γ ( −0.5 ) 0.8862 = ( 0.5)( −0.5 ) Γ ( −0.5 )
Area = 2 ( Perimeter )
x=
P = nCr pr qn−r
Γ (p) =
1 n 1− (7 / 8) 8 = 7 1− 8 n = 11pressings
x 2 x 3 x 4
In order for Kirby to be back to its initial position after tossing the coin five times, the results of the tossing should be 2 heads and 3 tails. So using binomial distribution or repeated trial,
For #19 – 22 The gamma function of a number is given by
1− r
3 4
backward two squares. What is the probability that he is back in the square where he started?
→
tabulate
Γ ( −0.5 ) = −3.5448 →
Ans.
∞
22. Solve
∫x e
8 −x
dx , where p is a positive
0
integer. A. 100 B. 92 C. 40,320 * D. 20,203
x = 30
Possible dimensions: ( 5, 20) , ( 6,12)( 8,8)
40 − x = 10 → ans
Answer : 3
Use reverse engineering. 23. Find the general solution ux
13. Eleven people plan a trip and hire a mini bus that can exactly accommodate all of
18. Kirby is standing on a sidewalk made of cement squares. He tosses a coin five times. For each head, he moves forward three squares. For each tail, he moves
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= cos y
if
u ( x, y ) is a function of x and y.
������������������ ������ ������ ��, ����
����� ������ ������ ux
=
��� ������������������
cos y
duxy dx
=
cos y if
n x (n ) = − z
= −2
= −2x +
uxy
24. Find the general solution uy
Derivative:
= −2
uxyx
du = cosy dx u = xcos y + f ( y ) → Ans.
dux dy
=
cos y
du dy
=
cosy
=
3 if
+ h(x)
= −2xy +
ux
= −2xy + g ( y ) + h ( x )
x [z] =
du dx
= −2xy + g ( y) + h ( x)
x ( n ) = 3 δ ( n ) + δ ( n − 2 ) + δ ( n + 2)
∑ x (n ) z
−n
n =−∞
so,
∫
h ( x ) dx
∞
∑
x [z] =
∞
n = −∞
3
2
n = −∞
∑ δ (n + 2 ) z
u ( x, 0 ) = 4x + 1
n =−∞
1
n=0
0
n≠0
δ ( n ) =
so, +z
x [ z ] = 3 + z −2
4x + 1 = 3 ( 0 ) + f ( x )
−n
Remember :
x [ z ] = 3z −0
2
x y + 2 =1 2 a b x2 y2 + =1 452 302
u = 3y + f ( x)
−n
∞
29. The arch of a bridge is in the shape of a semi-ellipse having a horizontal span of 90 m and a height of 30 m at its center. How high is the arch 25 m to the right or left of the center?
3
∑ δ (n − 2 ) z
3δ (n ) z− n + +
u ( x, 0 ) = 4x + 1
=
∫ f ( y ) dy
ux
u = − x2 y + xg ( y ) + i ( x ) → Ans.
u ( x, y ) is a function of x and y and
du dy
30. Find the z-transform of each of the following sequences x ( n ) = 3 δ ( n ) + δ ( n − 2 ) + δ ( n + 2)
f (y )
u = − x2 y + xg ( y ) +
25. Find the general solution uy
=
dz
∞
u = sin y + f ( x ) → Ans.
uy
d ( X ( z ))
f (y)
= −2x +
u ( x, y ) is a function of x and y. uy
���� �
−2
+z
+z
2
2
f ( x ) = 4x + 1
at x = 25
31. Find the z-transform, zeroes, poles and the region of convergence of
u = 3y + 4x + 1 → Ans.
252 y2 + =1 452 302 y = 24.94 m
x ( n ) = 2n u ( n )
Answer: 24.94 m
x [z] =
x ( n ) = 2n u ( n) .
∞
26. Find the general solution uxx
3 if
=
u ( x, y ) is a function of x and y.
uxx
dux =3 dx ux = 3x + f ( y ) du dx
u = 1.5x
2
+
Z − Transform Formulas
1 u ( n) = 0
x (n)
so,
↔
X (z)
n α u ( n) ↔
27. Find the general solution uxy
3
=
8xy if
nαnu ( n) ↔
u ( x, y ) is a function of x and y. =
uxy dux dy ux du dx
−nα
8xy3
=
8xy
=
2xy4
=
4
2xy
u = x2 y 4 u = x2 y 4
x [z] =
1
1 − αz
αz
−1
∫
×5
−n
1 1 − 2z −1
1 = 1.667 1 − 2 × 5−1 Zero (numerator = 0 ) → z = 0
−1
2
Pole ( denominator = 0 )
1 − ( cos ωo ) z −1 1 − 2 ( cos ωo ) z
−1
+
→ 1 − 2z
z
−2
1 − ( sin ωo ) z −1 1 − 2 ( cos ωo ) z −1 + z−2 X
(
α
−1
z
)
Time shifting: x ( n − no ) ↔ z −no X ( z ) = −2 if
u ( x, y ) is a function of x and y.
= 1.667
n =0
(1 − αz 1)
n α x (n ) ↔
Ans.
28. Find the general solution uxyx
−n
−1
=
0→z=2
Region of convergence
( answer ≠ ∞ ) → 2z
−1
<
0→ z
Multiplication by exponential:
f ( x ) dx + h ( y )
+ g( x ) + h ( y ) →
n
2
αz
n
f (x )
∑2
Answer :
−
u ( −n − 1) ↔
x [z] =
−1
(1 − αz 1)
cos ( nωo ) u ( n) ↔
+
n
15
1
sin ( nωo ) u ( n) ↔
∑2 z
Reverse engineering: let z = 5
1 − αz −1
3
+ f (x )
n<0
n =0
−
+
n≥0
∞
n −α u ( −n − 1) ↔
xf (y ) + g ( y ) → Ans.
−n
n =−∞
δ (n) ↔ 1
3x + f ( y )
=
n
Remember :
For #30 - 35
=3
∑ 2 u (n ) z
Time reversal:
( )
x ( −n ) ↔ X z−1
Answer:
X (z ) = z
>2
1 1 − 2z −1
, zero : 0, pole : 2
>2
32. Find the z-transform of n
n
1 1 x ( n ) = 2δ ( n ) + 3 u ( n ) − u ( n) 2 4
Linearity : ax ( n ) + by ( n ) ↔ aX ( z) + bY ( z) Convolution: x (n) ∗ h (n) ↔ X ( z ) H( z )
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������������������ ������ ������ ��, ����
����� ������ ������
��� ������������������
n
1 1 x ( n ) = 2 δ ( n ) + 3 u ( n ) − u ( n) 2 4
1 x ( n ) = u ( n + 2 ) + 3n u ( −n − 1) 2
n
∞
1 1 2 u (n) → 1 1 − z−1
−n
1
∞
∑ 4
−
n
n
1 3 z −n n =0 2 ∞
+
n +2
1 2
∑
n
1 −n ∑ z n= 0 4 ∞
−
n
x [z] = 2 +
1
15
∑ 3 2
1
∞
×5
−n
∑ 4
−
n=0
x [z] =
n
×5
−n
S
z2 1 1 − z− 1 2
MODE → CMPLX
=
SHIFT → CMPLX → r∠θ → Magnitude:146 lb
4z2 1 − −1 1 1 − z −1 1 − 3z 2 1 < z <3 ROC : 2 X( z) =
Y ( z) =
3 3 − 1 −1 1 −1 1− z 1− z 2 4
Answer: 1 < z 2
33. Find the z-transform of x ( n ) = u (n ) − u ( n − 10 )
=
145.77 ∠ − 30.96
2
3 1 − Answer : x [z ] = 2 + 1 −1 1 −1 1− z 1− z 2 4 3 1 244 − = x [z] = 2 + 1 1 −1 −1 57 1 − ( 5) 1 − (5) 2 4
(ref is North for angle )
125∠0 + 75 ∠ − 90
4z2 1 2 u (n + 2) → 1 1 − z−1
244 57
Answer: X ( z ) = 2 +
75
n
n =0
E
W
z2 u (n + 2) → 1 1 − z −1 2
1 1 u (n + 2) → 4 2
Reverse engineering: let z = 5 n
125
2
u ( n ) z −n
n =−∞
x [ z ] = 2z −0
N
n
1 x [ z ] = ∑ 2δ (n ) z + ∑ 3 u (n ) z −n n = −∞ n = −∞ 2 ∞
���� �
n
n
Course: N 31o W
4z 2 1 − 1 −1 1 − 3z −1 1− z 2
39. A tower 27.5 m tall makes an angle of 126°with the inclined road on w hich it is located. Determine the length from the top of the tower to a point down the road 35 m from its foot.
<3
d
27.5 m 126o 35
36. An airplane flies 100 km in the direction S38°10’E. How far south and how far east of the starting point is it?
x ( n ) = u ( n) − u ( n − 10 )
road
∞
X ( z) =
∑ x (n ) z
−n
d2
n =−∞ 9
=
E
∑ (1) z
−n
=
z
−0
+z
−1
+z
−2
+ ... + z
−9
100 38 10'
(
a1 1 − rn
From G.P. S =
(126o )
d = 55.8 m
1− z
−1
Answer: X ( z ) =
S
=
1 − z−10 1− z
South : 100 cos ( 38o10 ') = 78.6 km
−1
1 − z −10 1 − z− 1
n
1 transform of x ( n ) = u ( −n) 3
240 kph
1 x ( n ) = u ( −n ) 3
E
S
u (n) 1 1 − 3z −1
( )
But with time reversal: x ( −n ) = X z
−1
<
1 3
35. Find the region of convergence of ztransform of
+
x2
= 12
v 2 + x2
=
u2
= 10
+
y2
PI = 5
1 x ( n ) = u ( n + 2 ) + 3n u ( −n − 1) 2 ����� ��� ��������, ���� ���., ���� ���� ���. ������� � �����������
2
72 2
v 2 + y2 2
+7
=
2
(u2 + y2 ) + ( v 2 + x2 ) − ( u2 + x2 )
− 12
2
=
38. A body is acted upon by a force of 75 lb, due west, and a force of 125 lb, due north. Find the magnitude and direction of the resultant force.
n
I
v 2 + y2
=
Course : S80.54 o E <
=
= 10
Groundspeed: 243 kph
1 , z 1 − 3z
10
PI2
243.31∠99.46
Answer: Y ( z ) =
7
N
240∠90 + 40∠180
o
1 3
y
PI2
H
P
=
SHIFT → CMPLX → r∠θ →
1 1 − 3z ROC : 3z < 1 X ( z) =
12
PI2
MODE → CMPLX
v
x
u2
40kph
−n
x ( −n ) = ( 3 ) u ( n ) →
u
C
37. A plane is headed due east with airspeed 240 km/h. If a wind is blowing from the north at 40 km/h, find the groundspeed and course.
n
n
Solution:
East : 100sin ( 38o10 ') = 61.8 km
34. Find the region of convergence of z-
1 x ( −n ) = 3
40. Let CHIN be a rectangle and let P be a point inside the rectangle. IF PC = 12, PH = 7, and PN = 10, then PI =?
10
( )
)
1− r
1 1 − z−1
z
2
o
n =0
X ( z) =
2
= 35 + 27.5 − 2 ( 35 )( 27.5 ) cos
41. Identify the directrix of the parabola 2y 2 – x – 5y + 9 = 0.
2y 2
−
2 y2
x − 5y + 9 = 0 2
−2.5 2
− 2.5y +
2 ( y − 1.25 )
( y − 1.25 )
2
2
=
2 −2.5 = x − 9 + 2 2
x − 5.875
= 0.5 ( x − 5.875 )
Vertex :(5.875, 1.25)
������������������ ������ ������ ��, ����
����� ������ ������
��� ������������������
From Eq. 1,length of the latus rectum is the coefficient of x. So, 4a = 0.5
45. Write in rectangular coordinates form r = 6 tan θ sec θ r
a = 0.125 The parabola is opening to the right: y a
1.25
���� �
=
y2 = 16x y
6tan θ sec θ
y
tan θ cos θ r cos θ = 6 tan θ r
=
6
x=6 x2
=
y 2
y x
0
6y
x x=4
dx
5.875 46. Find the volume of the solid generated when the region bounded by the y-axis,
Directrix : x
=
y = e x and y = 2 is rotated round the yaxis.
23 4
y
x = 5.875 − 0.125 x=
23 4
→
y2
− 2x +
x
=e
+ 2y =
( x − 1)
2
so, r 2
=
+
( y + 1)
2→r
C = 2πr
=
2
2
y = ex
=
2π 2
∫
V = π x 2dy
= 8.88 =π
∫
x2
− 4y + 2x + 32y − 27 =
2
x2
+ 2x + 1 − 4 y − 8y +
0
−8 2
2
2
2
(ln y ) dy A
47. Find the area bounded by the p arabola x2 = − y + 2 and the line x = y.
= 27 + 1 −8 2
2
x=y
y2 1
2
+ ( −4 )
( x + 1)
2
( y − 4)
− 4 (y −
2
−
4)
( x + 1)
2
c2 c
2
dx
=9;
b2
= 36
2
x2
= −y + 2
x2
= − ( y − 2)
( −1, 4 − 3 5 ) and (−1,4 + 3 5 ) Ans.
tan2θ =
3xy + y 2
−2
3y = 0.
A −B
vertex : (0, 2 )
x
2
2θ = 120 60o
(− 3 )
o
→
=
x2
+y
2
z2
=
62
+8
2
z = 10 d 2 ( z = x2 + y2 ) dt dz dx dy = 2x + 2y 2z dt dt dt dz = 2 ( 6 )( 2 in/s ) + 2 (8 )( -3 in/s ) 2 (10 ) dt dz = −1.2 in/s dt
20m
= −x + 2 +
x−2=0
x = 1 ; x = −2
500 kg
y = 1 ; y = −2 Point of intersection: (1,1) & ( -2,-2) 1
1
−2
−2
2 ∫ ( y2 − y1 ) dx = ∫ ( 2 − x − x ) dx =
9 2
−2
3 3 −1
z2
= −y + 2
A=
C
2θ = tan−1
θ=
+ 2x + 2
C
x=y x2
44. Find an appropriate angle through which to rotate axes and sketch a graph of the equation
y
50. A 60 m cable that weighs 4 kg/m has a 500 kg weight attached at the end. How much work is done in winding up the last 20 m of the cable?
Points of intersection: x2
tan2θ =
1
=1
5
3x 2
y1
2
= a + b = 9 + 36
=3
−2
= −36
9 36 Center : ( −1, 4 ) ; a2
z
V = 0.592
2
43. Given: x − 4y +2x + 32y − 27 = 0 . Find the foci of the hyperbola.
∫
4
x
x
1
2
2π
B
→ ln y = 2
=
49. If one leg AB of a right triangle increases at the rate of 2 in/s, while the other leg AC decreases 3 in/s, find how fast is the hypotenuse is changing when AB = 6 and AC = 8 feet.
x
=2
5
∫
0
( x2 − 2x + 1) + ( y 2 + 2y + 1) = 1 + 1
+
y 2 + 5 ydx 0 4 16x + 5 16xdx = 1072.33 V = 2π 0 2
1
0
y 2
y = −5
∫
x
dy
=
−5
V = 2π R dA
y=2
Ans
42. The equation of a circle in x-y planes is given by x 2 − 2x + y 2 + 2y = 0 . Find the circumference of the circle. x2
R
48. The area in the first quadrant bounded by the parabola y2 = 16x, the line x = 4, and the x – axis is revolve about the line y = 5. Find the volume.
W
= Fxd
W
=
W
= 10, 800
20m ( 4kg/m )( 20m ) + 500kg ( 20m) 2 kg-m
Note: The total weight of cable is assumed be at its center.
Ans
Note :angle measure from x-axis and counterclockwise
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