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JANUARY 2010 CXC MATHEMATICS - PAPER 2
Section I
1. a. Required to calculate:
Calculation: The arithmetic is not difficult in this question, but we could save valuable time by using the calculator and which is allowable. So,
b. Data: Salesman salary is $3 140 per month together with 2% commission on
sales. (i) Required to calculate: Fixed salary for the year. Calculation: The total fixed salary for the year of 2009 = monthly salary x 12 (ii) Required to calculate: Commission for the year. Calculation: The commission for the year of 2009
(iii) Required to calculate: Total income for the year.
Calculation: The total income for the year of 2009 = The total fixed salary for 2009 + The total earned in commission for 2009
c. Data: cups of milk and 2 cups of pancake mix produces 8 pancakes.
(i) Required to calculate: Number of cups of pancake mix to make 12 pancakes. Calculation: From the given instructions: 8 pancakes are produced from using 2 cups of mix.
27.88.076.2
+
)valueexact(14.79
)calculatortheUsing(69.7545.37.88.076.2 2
=
+=+
121403$ ´=68037$=
000720$of%2=
40014$
000720$1002
=
´=
08052$40014$68037$
=+=
311
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Hence, 1 pancake is produced from cups of mix.
And, 12 pancakes will be produced from cups of mix.
cups of pancake mix.
(ii) Required to calculate: Number of pancakes made from 5 cups of milk. Calculation:
cups of milk produces 8 pancakes.
So, 1 cup of milk will produce pancakes.
Hence, 5 cups of milk will produce pancakes.
pancakes.
pancakes.
2. a. Data: .
Required to calculate:
Calculation: Substituting the values given to the variables to obtain
b. (i) Required to simplify: Solution: Expanding by using the distributive law to get,
82
1282´
3=
311
311
8
5
311
8´
÷øö
çèæ ´´= 5
43
18
30=
8and4,6 =-== cba
bcba
-+2
( ) ( )( )
)termslowestin(322
1282
1232
484364846 22
=
=
=
+-
=
---+
=-+bcba
( ) ( )yxyx 243 ++-
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(ii) Required to simplify:
Solution: Separating the coefficients from the variables to simplify, we use the laws of indices to get
c. Data: . (i) Required to find: x Solution: Using the same principles used to solve equations, we obtain
(ii) Required to find: Smallest integer value of x that satisfies the inequality.
Solution:
The smallest integer that satisfies the inequality is the next integer that is greater than 2 and which is 3.
3. a. Data: , T = {multiples of 3} and E = {even numbers} (i) Required to draw: Venn diagram to represent the information given. Solution:
Hence,
( ) ( )
yxyyxxyxyxyxyx
5738438433243
+=-++=++-=++-
3
42
634xxx ´
3
342
3
42
3
42
22634
634
xx
xxx
xxx
=
=
´´
´=
´
-+
733 -<- xx
22
4224373
733
>÷
>\<
-<+--<-
x
xxxx
xx
{ }3
,...3,2,1,0,1,2,3...,2
min =\---=Î
>
xZZx
andx
{ }12,...,3,2,1=U
{ }{ }12,10,8,6,4,2
12,9,6,3==
ET
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om (ii) (a) Required to list: Members of Solution: The elements that are common to both sets T and E as illustrated on the Venn diagram.
(b) Required to list: Members of Solution: First we list and then deduce the complement
as illustrated on the Venn diagram. b. (i) Required to construct: Triangle ABC where AC = 6 cm, . Solution
ET Ç
{ }12,6=Ç ET
( )¢È ET
{ }( ) { }11,7,5,1
12,10,9,8,6,4,3,2
=¢È
=È
ET
ET
°=°= 60ˆand60ˆ BACBCA
Draw a straight line longer than 6 cm and cut off AC = 6 cm. Construct a 600 angle at A. Construct a 600 angle at C. The line from A meets the line from C at B.
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(ii) Required to complete: Diagram to show the kite ABCD, in which AD = 5 cm.
Solution: Arcs of length 5 cm drawn with centre A and centre C to meet at D on the opposite side of B, to form the kite.
(iii) Required to find:
Solution: (when measured by the protractor).
4. a. Data: Diagram as shown below.
(i) Required to calculate: x. Calculation:
CAD ˆ
°= 53ˆCAD
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Applying Pythagoras’ theorem to the right-angled triangle NKM
(ii) Required to calculate: Calculation: Triangle NLK is a right–angled triangle. So,
b. Data: Map with scale 1 : 1 250
(i) Required to find: Distance from S to F on the map. Solution: SF measures 7.8 cm (to 1 decimal place by the ruler).
(ii) Required to calculate: Actual distance SF.
Calculation: The actual distance SF according to the scale. (100 cm = 1 m)
(iii) Data: Time taken to run from S to F = 9.72 seconds.
(a) Required to calculate: Average speed in ms-1
Calculation:
(b) Required to calculate: Average speed in kmh-1
Calculation: 1 km =1 000 m
( ) ( )
cm10100
6436
)Theorem'Pythagoras(86 222
==
+=
+=
x
x
q
1
6 cmsin12 cm
6sin12
3030
q
q
-
° =
æ ö\ ° = ç ÷è ø
\ ° = °=
( )cm25018.7 ´=
placedecimal1tom5.97
m10025018.7
=
´=
)placedecimal1to(ms10ms03.10s72.9m5.97
takentimeTotalcovereddistanceTotalspeedAverage
1
1
-
-
=
=
=
=
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1 h = 3 600 s So
Average speed
5. a. Data: Line passes through T (4, 1) and has a gradient of .
Required to find: Equation of the line. Solution:
To find the equation of the line we require the coordinates of 1 point on the line and the gradient of the line. These we have and applying the formula we get
OR any other equivalent form.
b. (i) Required to draw: Triangle ABC, from the coordinates of the vertices given. Solution: This is shown on the diagram below
(ii) Required to draw: The line . Solution: This is shown on the diagram below
(iii) Required to draw: Image of triangle ABC under a reflection in the line
and label it . Solution: This is shown on the diagram below
h600372.9
km00015.97
=
)figurestsignifican3to(kmh1.36kmh11.36
1
1
-
-
=
=
53
( ) ( )
735123554315
53
41
-=-=--=-
=--
xyxyxy
xy
2=y
2=y CBA ¢¢¢
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(iv) Required to draw: using the coordinates for the vertices given. Solution: This is shown on diagram below.
(v) Required to describe: Single transformation that maps onto .
Solution: The image is congruent to the object and is neither flipped nor re-
oriented. The transformation is a translation. We can deduce the translation vector by observing one object-image set of points.
The point C is mapped onto by a horizontal shift of 9 units to the left
and a vertical shift of 1 unit upwards. This may be represented by the
translation, T, where .
Hence, is mapped onto by a translation, T, where we
define .
CBA ¢¢¢¢¢¢
ABCDCBA ¢¢¢¢¢¢D
CCT
¢¢¾¾ ®¾÷÷ø
öççè
æ -=
19
C ¢¢
÷÷ø
öççè
æ-=
19
T
ABCD CBA ¢¢¢¢¢¢D
÷÷ø
öççè
æ-=
19
T
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6. Data: Records of the distances travelled to school by 26 students of a class. a. Required to complete: Frequency table to represent the data given. Solution: The data is that of a continuous variable. UCL-Upper class limit LCL-Lower class limit LCB-Lower class boundary UCB-Upper class boundary
Distance in km
L.C.L-U.C.L
X L.C.B-U.C.B
Frequency
1 – 5 1 6 – 10 2
11 – 15 4 16 – 20 6 21 – 25 7 26 – 30 3 31 – 35 2 36 – 40 1
b. Required to qraw: Histogram to represent the data given. Solution:
c. Required to calculate: Probability that a randomly chosen student travelled 26
km or more to school.
5.55.0 <£ x5.105.5 <£ x5.155.10 <£ x5.205.15 <£ x5.255.20 <£ x5.305.25 <£ x5.355.30 <£ x5.405.35 <£ x
å = 26f
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Calculation:
d. Required to explain: Whether the mean, mode or median is most appropriate for
estimating the cost of transportation service for the school. Solution: The mean is most appropriate for estimating the cost of transportation. This is
because the mean takes into consideration each student and the actual distance that is covered by each student. If the mean is required, we may use the formula:
, where
= mean mid-class interval or the mid interval value of the classes frequency 7. Data: Graph representing .
(i) Required to find: when Solution:
When , or f (0) = 3
( ) No. of student travelling³26kmstudent travelled 26kmTotal no. of students
P ³ =
133266
26123
=
=
=++
=å f
åå=ffx
x
x=x=f
( ) cbxaxxf ++= 2
( )xf 0=x
0=x ( ) 3=xf
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(ii) Required to find: x when Solution:
When , and . These are the points where the curve cuts the
horizontal axis. (iii) Required to find: Coordinates of the maximum point. Solution:
The coordinates of the maximum point are (-1.3, 6.1). (iv) Required to find: Equation of the axis of symmetry. Solution:
Equation is , the vertical which passes through the maximum point. (v) Required to find: x when . Solution: The horizontal line f (x) or y = 5 cuts the curve at 𝑥 = −2 and at 𝑥 = −0.5
( ) 0=xf
( ) 0=xf21
=x 3=x
3.1-=x
( ) 5=xf
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(vi) Required to find: . Solution: We are seeking the range of values of x for which f (x) lies above the line y = 5
8. Data: Table as shown.
No. n, of hexagons in the pattern
1
2
3
4
5
20
n
No. of sticks, S used for the pattern
6
11
16
x
y
z
S
a. (i) Required to find: x Solution: Notice that the value of S = 6, 11, 16, …which are multiples of 5 and exceeded by 1 because we notice that:- 6 = 11 = 16 = Hence, the pattern or relation that connects S and n is (i) When ,
( ) 5>xf
{ }5.02: -<<- xx
( ) 151 +( ) 152 +( ) 153 +
( ) 15 += nS
15 += nS4=n ( ) 145 +=S 21= 21=\ x
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(ii) Required to find: y Solution: When (iii) Required to find: z Solution: When b. Required to find: Expression in terms of n for S. Solution: (as found and shown above)
c. Data: 76 sticks were used to make a pattern of h hexagons. Required to calculate: h Calculation: When
Hence, (data)
The number of hexagons = 15 and .
5=n ( ) 155 +=S 26= 26=\ y
20=n ( ) 1205 +=S 101=101=\ z
15 += nS
76=S
( )
15755
1576
==
+=
nn
n
15=h
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Section II
9. a. Data:
(i) Required to express: v in terms of E and m. Solution:
We cross multiply to make the equation in a linear and more manageable form.
(ii) Required to calculate: v when and . Calculation: When we substitute and we obtain
b. Data:
(i) Required to express: in the form . Solution:
Half the coefficient of x is
2
21 mvE =
1221
21
2
2
2
´=´
=
=
mvE
mvE
mvE
mEv
mEv
Emv
2
22
2
2
=
=
=
45=E 13=m
45=E 13=m( )
places)decimal2to(63.2163.2
139013452
==
=
=v
( ) 283 2 +-= xxxg( )xg ( ) cbxa ++ 2
2383283 22 +÷
øö
çèæ -=+- xxxx
34
38
21
-=÷øö
çèæ-
*343283
22 +÷
øö
çèæ -=+- xxx
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And where the value of * is to be determined.
Hence, and which is of the form
, where
and .
OR We could expand the desired form and equate the coefficients with the given form
Equating the coefficient of we get Equating the coefficient of x we get
Equating the constant we get
which is the required form.
(ii) Required to solve: Solution:
3 x - 43
æèç
öø÷
2
= 3 x2 -83x + 16
9æèç
öø÷
= 3x2 - 8x + 5 13
- 313= *
2
313
343283
22 -÷
øö
çèæ -º+- xxx
( ) cbxa ++ 2
ÂÎ-=ÂÎ=34,3 ba ÂÎ-=
313c
( )( )
cababxaxcbbxxa
cbxaxx
+++=
+++=
++=+-
22
22
22
22
283
2xÂÎ= 3a
( )
( )
ÂÎ-=
-=
-=
34328832
b
b
ÂÎ-=
=+
=+÷øö
çèæ-
313
2315
2343
2
c
c
c
313
343283
22 -÷
øö
çèæ -º+-\ xxx
( ) 0=xg
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Using the quadratic equation formula
OR Using the form obtained when we completed the square
(iii) Required to sketch: The graph of . Solution:
( )( ) ( ) ( )( )
( )
placesdecimal2to39.2or28.0738.2or927.0
64086
24648
3223488
02832
2
==
±=
-±=
--±--=
=+-=
x
xxxg
( )
39.2or28.0738.2or927.0
3104
310
34
rootsquaretheTaking910
34
3313
34
313
343
0313
343
0283
2
2
2
2
2
==
±=
±=÷øö
çèæ -
=÷øö
çèæ -
=÷øö
çèæ -
=÷øö
çèæ -
=-÷øö
çèæ -\
=+-=
x
x
x
x
x
x
xxxg
( )xg
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om (a) Required to find: The y – coordinate of A. Solution: At A, Hence, A = (0, 2) and the y – coordinate of A is 2. (b) Required to find: The x – coordinate of C. Solution:
The curve cuts the x – axis at , that is at the points
and on the horizontal axis. This
approximates to (0.28, 0) and (2.39, 0). C is the point with the greater x value. Therefore, the x –
coordinate of C is .
(c) Required to find: The x and y coordinates of B.
Solution:
(Assuming that B is the minimum point of .)
The axis of symmetry of is
The vertical at
( ) 283 2 +-= xxxg0=x ( ) ( ) ( ) 208030 2 +-=g
2=
( )xg3104 ±
÷÷ø
öççè
æ - 0,3104
÷÷ø
öççè
æ + 0,3104
39.23104
»+
( ) 283 2 +-= xxxg( )xg
( )xg ( )( )328--
=x
34
=x
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The axis of symmetry passes through the minimum point of .
Therefore, the x – coordinate of the minimum point is .
To find the corresponding the y coordinate we get
The y – coordinate of B is and so .
OR
( )xg
34
=x
313- ÷
øö
çèæ -=
313,
311B
( )
( )
311
34
034
034
0343
at313
3130
0343
313
343
283
2
2
min
2
2
2
=
=
=-
=÷øö
çèæ -
=÷øö
çèæ -
-=
-=\
"³÷øö
çèæ -
-÷øö
çèæ -=
+-=
x
x
x
x
xg
xx
x
xxxg
313
2332
316
2348
343
34 2
-=
+-=
+÷øö
çèæ-÷
øö
çèæ=÷
øö
çèæg
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The x – coordinate of B = and the y – coordinate of B =
and B has coordinates .
OR
Since is a quadratic graph and its shape is a parabola, then
the x – coordinate of B is halfway between and ,
that is
And
Therefore, x coordinate of B is and y coordinate of B is
and the coordinates of B is .
311
313-
÷øö
çèæ -
313,
311
( )xg
3104 -
3104 +
34
2310
34
310
34
23104
3104
=
++-=
++
-
=x
313
34
-=÷øö
çèæg
34
313-
÷øö
çèæ -
313,
311
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10. Data: Manager wishes to make x small pizzas and y large pizzas and the oven can hold no more than 20 pizzas. a. (i) Required to find: Inequality to represent the information given. Solution: Number of small pizzas = x Number of large pizzas = y Total must not be more than 20, Hence …(1) (ii) Data: Ingredient for each small pizza cost $15 and ingredients for each
large pizza cost $30. Manager plans to spend no more than $450 on ingredients.
Required to find: Inequality to represent the information given Solution: Cost of x small pizzas at $15 each Cost of y large pizzas at $30 each Total must be no more than $450. So must be $450 We to simplify the inequation to obtain …(2) b. (i)-(ii) Required to draw: Graphs of the lines associated with the inequalities and shade the feasible region. Solution: We draw the line by obtaining 2 points on the line .
x y 0 20 20 0
We draw a horizontal to intersect the line. The side with the smaller (acute) angle will represent the region.
The region which satisfies is
yx + 20£20£+\ yx
( )x´= 15$( )y´= 30$
( )yx 3015$ + £15÷302 £+ yx
20=+ yx
£
20£+ yx
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Obtaining 2 points so as to sketch the line .
x y 0 15 30 0
A horizontal is drawn through the line. The side with the smaller angle will represent the region.
The region which satisfies is
The feasible region is the area in which both shaded regions overlap.
302 =+ yx
£
302 £+ yx
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(iii) Required to state: Coordinates of the shaded region. Solution: The shaded region OABC represents the feasible region where , , and .
c. (i) Required to find: Expression in terms of x and y for the total profit made on the sale of pizzas. Solution: Let the total profit be P. The profit on x small pizzas at $8 each The profit on y large pizzas at $20 each
(ii) Required to calculate: Maximum profit made. Calculation:
The vertices used for identifying the maximum profit are A (0, 15), B (10, 10) and C (20, 0). It makes no sense to test x = 0 and y = 0.
When
( )0,0=O ( )15,0=A ( )10,10=B ( )0,20=C
( )x´= 8$( )y´= 20$
( )yxPyxP
208$20$8$
+=+=
15and0 == yx
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When
When
Hence, the maximum profit, P, is $300 as illustrated above and occurs when the shop sells 15 large pizzas.
11. a. Data: Diagram as shown below.
(i) Required to prove: Proof:
Q.E.D. (ii) Required to calculate: Distance PQ. Calculation: In triangle PQR, we have two sides and an included angle. So we can apply the cosine law
( ) ( )300$
152008=
+=P
10and10 == yx( ) ( )280$
1020108=
+=P
0and20 == yx( ) ( )160$
020208=
+=P
°= 126ˆQRP
°=°-°=
°=°+
126116242ˆ
242116ˆ
QRP
QRP
( ) ( ) ( )( )
m.nearest the tom128m0.1285.40416
)(cos126cos10238210238 222
===
°-+=
PQ
lawPQ
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b. Data: Vertical pole, 10 m standing on a horizontal plane with points, K, L and M.
(i) Required to complete: Diagram showing the information given. Solution:
(ii) (a) Required To Calculate: Length of KL. Calculation: Consider the right-angled triangle SKL
(b) Required to calculate: Length of LM.
Calculation: Consider the right-angled triangle SKM
The length of LM = The length of KM – The length of KL
tan21° = 10KL
KL = 10tan21°
= 26.05= 26.1 m( to one decimal place).
tan14° = 10KM
\KM =10
tan14°= 40.11m
= 40.11- 26.05= 14.06= 14.1 m (to one decimal place)
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12. a. Data: Diagram shown below with and .
(i) Required to calculate: Calculation:
(The angle subtended by a chord (GH) at the center of a circle, (angle GCH) is twice the angle it subtends at the circumference, (angle GFH) standing on the same arc).
(ii) Required to calculate: Calculation:
(The opposite angles of a cyclic quadrilateral (DEHG) are supplementary.) (iii) Required to calculate: Calculation:
(Angles at a point on a straight line total 1800).
(The sum of the angles in a triangle = 180°).
(Angles at a point on a straight line totals 1800).
°= 88ˆHCG °= 126ˆEHG
HFG ˆ
( )
°=
°=
44
8821ˆHFG
EDG ˆ
°=°-°=
54126180ˆEDG
FED ˆ
°=°-°=
54126180ˆFHG
( )°=
°+°-°=82
4454180ˆFGH
°=°-°=
9882180ˆHGD
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(The opposite angles of a cyclic quadrilateral are supplementary).
b. Data: GC = 4 cm (i) Required to calculate: Area of Calculation:
We know two sides and the included angle.
So, area of
(ii) Required to calculate: Area of the minor sector GCH. Calculation:
The area of the minor sector
(iii) Required to calculate: Area of the shaded segment. Calculation:
°=°-°=
8298180ˆFED
GCHD
( )( ) °=D 88sin4421GCH
2
2
cm00.8cm995.7
=
=
( )2414.336088
´°°
=GCA
2
2
cm28.12cm028.12
=
=
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om The area of the shaded segment = The area of sector GCH – The area of
13. a. Data: Vector diagram as shown below.
(i) (a) Required to express: in the form .
Solution: If,
This is of the form , where and .
(b) Required to express: in the form .
Solution:
GCHD
2
2
cm29.4cm528.4
995.7280.12
=
=
-=
OB ÷÷ø
öççè
æba
( )
÷÷ø
öççè
æ -=
-=
414,1
OB
B
÷÷ø
öççè
æba
1-=a 4=b
OBOA + ÷÷ø
öççè
æba
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Applying the vector triangle law
which is of the form , where and .
(ii) Required to calculate: Midpoint of AB. Calculation: Using the midpoint formula
is of the form , where and . b. Data: Diagram with OE, EF and MF are straight lines. EF = 3EH, MF = 5MG
and M is the midpoint of OE. and .
(i) (a) Required to express: in terms of u and v. Solution:
(b) Required to express: in terms of u and v.
Solution: (Since M is the midpoint of OE, then ). Using the vector triangle law, we get
5 10 4
44
OA OB-æ ö æ ö
+ = +ç ÷ ç ÷è ø è øæ ö
= ç ÷è ø
÷÷ø
öççè
æba
4=a 4=b
( )2,2204,
251
=
÷øö
çèæ ++-
=M
( )yx, 2=x 2=y
vOM = uEH =
HF
uuuHF
uEF
EHEF
23
3
3
=-=
=
=
MF
vOMME ==
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aths.c
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(c) Required to express: in terms of u and v. Solution:
(ii) Required to prove:
Proof: Using the vector triangle law
If , then
Q.E.D. (iii) Required to prove: O, G and H lie on a straight line. Solution:
uvEFMEMF
3+=+=
OH
uvEHOEOH
vvvOE
+=+=
=+=
2
2
( )uvOG += 253
MGv
MGOMOG
+=
+=
MGMF 5= MFMG51
=
( )uv 351
+=
( )
( )vu
vu
uvv
uvvOG
253
56
53
53
51
351
+=
+=
++=
++=\
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Since is a scalar multiple, which is of , then and are
parallel. Since O is a common point on both lines then, G lies on (as shown on the diagram). Therefore, O, G and H all lie on the same straight line and are collinear. Q.E.D.
14. a. Data: and
Required to calculate: Calculation: We will multiply N by N to obtain N 2 and subtract the product from L
( )
OG
uvOH
uvOG
352
253
=
+=
+=
OG 53
OH OG OH
OH
÷÷ø
öççè
æ=
4123
L ÷÷ø
öççè
æ -=
2031
N
2NL -
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b. Data: is a singular matrix.
Required to calculate: x Calculation:
If M is singular, then the . Hence,
c. Data: (2, 1) is mapped onto (-1, 2) by .
Required to calculate: The value of p and of q. Calculation:
( ) ( ) ( ) ( )( ) ( ) ( ) ( )
÷÷ø
öççè
æ -=
÷÷ø
öççè
æ----
=
÷÷ø
öççè
æ-÷÷ø
öççè
æ=-
÷÷ø
öççè
æ=
÷÷ø
öççè
æ´+´´+-´´+´-´+-´-
=÷÷ø
öççè
æ
÷÷ø
öççè
æ=
÷÷ø
öççè
æ-÷÷ø
öççè
æ-=
0112
44013213
4031
4123
4031
2230021023310311
2031
2031
2
2221
1211
2221
1211
2
NL
eeee
eeee
N
÷÷ø
öççè
æ=
xx
M312
÷÷ø
öççè
æ=
xx
M312
0det =M
( ) ( )
6360360123
2
2
±==
=-
=´-´
xx
xxx
÷÷ø
öççè
æ=
00q
pR
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Both are 2 x 1 matrices and are equal. So, equating corresponding entries, we obtain
and
d. Data: (5, 3) is mapped onto (1,1) by .
Required to calculate: The value of r and of s. Calculation:
Both are 2 x 1 matrices and are equal. So, equating corresponding entries, we obtain
and
e. Data: The point (8, 5) undergoes the transformations R followed by T.
Required to calculate: The image of (8, 5). Calculation: Let
Firstly we transform under R.
( ) ( )( ) ( )
÷÷ø
öççè
æ -=÷÷
ø
öççè
æ
÷÷ø
öççè
æ=
÷÷ø
öççè
æ´+´´+´
=÷÷ø
öççè
æ÷÷ø
öççè
æ
÷÷ø
öççè
æ -¾¾ ®¾÷÷
ø
öççè
æ ÷÷ø
öççè
æ
21
2
2
102120
12
00
21
12 0
0
qp
qp
qp
qp
qp
1-=p 22 =q1=q
÷÷ø
öççè
æ=sr
T
÷÷ø
öççè
æ=÷÷
ø
öççè
æ++
\
÷÷ø
öççè
æ¾¾ ®¾÷÷
ø
öççè
æ ÷÷ø
öççè
æ=
11
35
11
35
sr
sr
T
415-=
=+rr
213-=
=+ss
÷÷ø
öççè
æ¾®¾÷÷
ø
öççè
æyxR
58
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Secondly we transform under T
Therefore, the image of (8, 5) after the combined transformations R followed by T is (-9, 6).
( ) ( )( ) ( )
÷÷ø
öççè
æ-=
÷÷ø
öççè
æ´+´´-+´
=÷÷ø
öççè
æ÷÷ø
öççè
æ -
85
50815180
58
0110
( ) ( )8,5, -=\ yx
( )( ) ÷÷
ø
öççè
æ-=÷÷
ø
öççè
æ-+-+-
¾¾¾ ®¾÷÷ø
öççè
æ- ÷÷ø
öççè
æ--
=
69
2845
85 2
4T
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