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.MATHEMATICS-Answers

Use This No1 pls is more acurate

1)

tabulate

Number-

5932

*6141

num

3679

*3113

den

Log

3.7732

3.7882

7.5612

3.4931

7.0588

anti log

7.5612

7.0588/0.5024/3=0.1675

anti log

=1471*10

================================

3a)

P=100I/RT=100*30,000/4*3

P=N250,000

================================

5)

tabulate

Mark- 21-25,26-30,31-35,36-40,41-45,46-50

F- 6,8,12,4,6,4=40

X- 23,28,33,38,43,48

FX- 138,224,396,152,258,192=1360

Class Boundaries- 20.5-20.5, 25.5-30.5,35.5-40.50,40.5-45.5

5i)

MEAN

(X)=EFX/EF=1360/40=34

MODE

L1+[Fm-Fa/2fm-fa-fb]c

=30.5+[12-8/2(12)-8-4]5

=30.5+[4/12]5

=30.5+1.67

=32.17

6i)

7+8+x = 47

15+ 2x =47

2x = 47- 15

2x/2 = 32/2

divide both side by 2

x=16

6ii)

tran only;:

7+8 + 3 + y=30

18 +y =30

y=30 -18 =12

:. Tran only =12 iii) at least two: 7+ 16+ 3 +8 =34

iv)16 + 7 +12 +16 +8 + 3 + Q=95

62+ Q =95

Q= 95 -62

Q=33 :. Car only = 33

(1)

TABULATE

No|Log|

5932|3.7732|

*6141|3.7882|

Num|7.5612|7.5612

3679|3.5657|

*3113|3.4931|

Den|7.0588|7.0588

| |0.5024/3=0.1675

Antilog:-1.471*10^0

=1.471

==================================

(2)

a = 3, b = 20, C = -7

Sum is

(A/1 + 1/B + B/1 + 1/ A) Note A means Alpha and B means beta

A^2 B + B^2 + A + B / AB

= -7/3 (-20/3) + (-20/3) = 140/9 – 20/3 / -7/3

= 80/9 x -3/7 = -80/21

Product:

-7/3 + 2 + 1/-7/3 = -7/3 + 2/1 – 3/7

= -49+42-9/21

= -16/21

x^2 – (sum)x + product= 0

x^2 – 80/21x + (-16/21) = 0

Final Answer : 21x^2 – 80x – 16 = 0

================================

(3a)

P=100I/RT=100*30,000/4*3

P=N250,000

(3b)

#7000 to Francs

#1 = 8 francs

.:.#7000 = ? france

CROSS MULTIPLY

8/1 * 7000 = 56,000 franc

Spending 49,400 franc

56, 000- 49,400 = 6600

Converting 6 ,600 franc to # AT 10 franc TO 1#

10 franc —– 1#

6600 FRANCS —? #

= 6,600/10 X 1

=#660

===================================

(4i )

difference in latitude = 36 degree + 36 degree = 72degree

distance travelled =@/ 360* 2pier

= 72 / 360* 2* 22 / 77* 64

= 20275200 / 2520

= 8045 . 71

= 8050 km ( 3 s.f )

(4ii)

given that speed = 800 km / hr

distance travelled = 804571 km

time = distance travelled / speed

= 8045 . 71/ 800

= 10 .057

= 10 hrs ( to the nearest hour )

===================================

(5)

Tabulate

Mark – 21- 25, 26- 30, 31- 35,36 -40 ,41 -45 ,46- 50

F – 6 ,8, 12,4 ,6, 4= 40

X – 23,28 ,33 ,38, 43, 48

FX – 138, 224, 396, 152, 258, 192= 1360

Class Boundaries- 20 .5 -20 .5, 25 .5 -30 .5, 35. 5- 40. 50,40 .5 -45 .5

(5i)

MEAN

( X )= EFX / EF = 1360/ 40 = 34

MODE

L 1+ [Fm – Fa / 2fm – fa -fb ] c

= 30 .5+ [12 -8/ 2 ( 12 ) -8- 4] 5

= 30 .5+ [4/ 12 ]5

= 30 .5+ 1.67

= 32 .17

===================================

(6)

(i)

7+8+x = 47

15+ 2x =47

2x = 47- 15

2x/2 = 32/2

Divide both side by 2

x=16

(ii)

Train only;:

7+8 + 3 + y=30

18 +y =30

y=30 -18 =12

.:. Train only =12

(iii)

at least two: 7+ 16+ 3 +8 =34

(iv)

16 + 7 +12 +16 +8 + 3 + Q=95

62+ Q =95

Q= 95 -62

Q=33 :. Car only = 33

=======================================

(7a)

a+2d=11—eq1

-(a+8d)=29

-6d/-6=-18/-6

d=3

Sub for d=3 in eq1

a+2(3)=11

a+6=11

a=11-6

a=5

(7b)

L = ar^n-1

729/3 = 3/3 x 3^n-1

243 = 3^n-1

3^5 = 3^-1

5 = n-1

N = 5+1

= 6

Sn = a(r^-1)/r-1 = 3(3^6 – 1)/3-1

Sn = 3( 729 -1)/2

S6 = 3 x 728/2

S6 = 1092

(7c)

3x^3/3 + x | 2,1

(2^3 + 2) – (1^3+1)

(8+2) – (1+1)

= 8

=================================

(8ai)

y=(4x+9)^3

Let u=4x+9

y=u^3

du/dx=4,dy/du=3u^2

dy/dx=dy/du*dy/dx

=4*3u^2

=12u^2

Recall u=4x+9

therefore dy/dx=12(4x+9)^2

(8aii)

y=(3x-2)^3(x^2+4)^2

u=(3x-2)^3, v=(x^2+4)^2

du/dx=3.3(3x-2)^3-1

dv/dx=2x.2(x^2+4)^2-1

du/dx=9(3x-2)^2

dv/dx=4x(x^2+4)

dy/dx=Vdu/dx +Udv/dx

=(x^2+4)^2[9(3x-2)^2+(3x-2)^3][4x(x^2+4)]

=(x^2+4)(3x-2)^2(9x^2+36+12x^2-8x)

=(x^2+4)(3x-2)^2(21x-8x+36)

(8b)

y-y1=m(x-x1)

y-6=2(x-2)

y-6=2x-4

y=2x-4+6

y=2x+2

===============================

(9a)

y proportional x proportional 1/x

y=kx/1 +k/x

4=2k/1+k/2

4=(4k+k)/2

8=5k

k=8/5

Relationship:

y=kx/1+k/x

y=8x/5+8/5x

(9b)

4(3^x+1)-3^2x=27

4(3^x *3^1)-362x=3^3

let 3^x=p

4(3p)-p^2=27

12p-p2=27

p^2-12p+27=0

p^2-9p-3p+27=0

p(p-9)-3(p-9)=0

(p-9)(p-3)=0

p=9 or p=3

Recall 3^x=p

when p=9

3^x=3^3

x=3

when p=3

3^x=3^1

x=1

therefore x=2 or 1

==================================

(11a)

Tabulate

X- |-1| |0| |1| |2| |3| |4|

6 |6| |6| |6| |6| |6| |6|

+X |-1| |0| |1| |2| |3| |4|

-X^2 |-1| |0| |-1| |-4| |-9| |16|

Y- |4| |6| |6| |4| |0| |-6|

(11b)

Draw the graph

(11ci)

gradient at x/x

dy/dx =B-A/B-C

=7-4/2-1=3/1=3

(11cii)

roots of equation from the graph

x=-2.2 and x=3

(11ciii)

minimum value of y

=6.4

================

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