The answer is B

Starting zeros are not significant and so the number of significant digits starts from 3597. The 3 digits lands on 9 but the digit next to it is more than 5 (7) and so 1 is added to the 9 to make it 10 and so the 0 is written down and the 1 is added to the 5 next to it which makes it 6.

NOTE
Zeros between other digits are significant. Only starting zeros are insignificant. For instance 3.052306, all the zeros are significant because they are not starting zeros after a decimal point.

Change the decimal number to a standard form with a complete whole number
∴ 0.064 = 64 x 10^-3
⇒ (0.064)^-1/3 = (64 x 10^-3)^-1/3
Applying, (a x b)ⁿ = aⁿ x bⁿ ⇒ (64 x 10^-3)^-1/3 = 64^-1/3 x 10^{(-3 x -1/3)}
64^-1/3 x 10^{(-3 x -1/3)} = 64^-1/3 x 10¹
Note:-3 x -1/3, negatives cancels each other likewise the 3s
Change the 64 to an indices form in order to cancel the powers
64 = 2 x 2 x 2 x 2 x 2 x 2 = 2⁶
⇒ 64^-1/3 x 10¹ = (2⁶)^-1/3 x 10
3 goes into 6, 2 times ⇒ 6 x -1/3 = -2
∴ (2⁶)^-1/3 x 10 = 2^-2 x 10
Applying, a^-n = 1 / aⁿ
⇒ 2^-2 = 1 /2²
2² = 4
∴ 2^-2 x 10 = 1/4 x 10
Using 2 to cancel each other, 2 goes into 4, 2 times and 10, 5 times
⇒ 1/4 x 10 = 1/2 x 5 = 5/2
Note 1/2 x 5 = 1/2 x 5/1 , every number is divided by 1
Applying a/b x c/d = axc / bxd
⇒ 1/2 x 5/1 = 1 x 5 / 2 x 1 = 5/2
∴ (0.064)^-1/3 = 5/2

3(y+1) - 2(2y-1) = 24

3y + 3 - 4y + 2 = 24

Group like terms

3y - 4y = 24 - 3 -2

diffence of two squards

(27.63 - 12.37)(27.63 + 12.37)

610.4

Run it to three dicimal place

since 4 is less than 5 you maintian it

T ={ 2,3,5,7}

M ={1,3,5,7,9}

U {1,2,3,4,5,6,7,8,9,10}

T¹ is called T compliment, which means numbers found in universal set but can't be found in set T

M¹ is called M compliment, which means numbers found in universal set but can't be found in set M

T¹ = {1,4,6,8,9,10}

M ¹ = { 2,4,6,8,10}

T ¹ M ¹ = { 4,6,8,10}

1st expand the equation by multipyling

number y from zero, alway number it from your right to left do same to the other side of the equatiion

(2xy¹) + (3 x y⁰) = (1x2³)+ (1x2²) + (1x2¹) + (1x2⁰)

2y + 3 = 8 + 4 + 2 + 1

Grouping like terms

2y = 8 + 4 + 2 + 1

2y⁄2= 12⁄2

y = 6

let represent initial term d represent difference

a = 6

d = common diffence

To find common diffence is the second term minus the first time that's current minus intial term

d = p - 6 = 14 -p

p - 6 = 14 - p

Group like terms

p + p = 14 + 6

2p⁄2 = 20⁄2

p = 10

look for two numbers If you add it will give you the co-efficient of x and multipy will give 60

The two numbers are -10,-6

substitute -10,-6 in place of the co-efficient of x which is -16

Gives 4x² - 10x - 6x + 15 = 0

Group like terms

(4x² - 6x ) - (10x + 15)

2x(2x - 3) - 5(2x - 3) = 0

(2x -5)(2x -3) = 0

2x -5 = 0

2x⁄2 = 5⁄2

x = 5⁄2 = 2 1⁄2

2x - 3 = 0

2x⁄2 = 3⁄2

x = 3⁄2 = 1 1⁄2

x = 1 1⁄2 or 2 1⁄2

selling price = N6,900.00

Profit percent = 15%

percentage profit if he had sold if for N6,600.00

let C.P represent cost price

S.P represnt selling Price

% represent Profit percent

C.P = S.P - P

but put P = %P X C.P

put P = 15% x C.P

P = 0.15 C.P

Gives = C.P = S.P - ( 0.15 - C.P )

but S.P = 6,900.00

C.P = 6,900 - (0.15 C.P )

Group like terms

C.P + 0.15 C.P = 6,900.00

1.5 C.P⁄1.5 = 6,900.00⁄1.5

therefore cost price of the article is 6,000

If the owner of the article had sold it N6,600.000

Profit = Selling price - Cost Price

= 6,600 - 6000

N600.00

since profit is know

percent profit = P⁄C.P x 100

= 600⁄6000 x 100

10%

3p = 4q ...... 1eqn

9p = 8p - 12 .......2eqn

multiply equation 1 by 3

3 x 3p = 3 x 4q

9p = 12q ..... 3eqn

put 3eqn into 2eqn

12q = 8q - 12

12q - 8q = -12

4q⁄4 =-12⁄4

q = 3

put q = -3 into 1

3p = 4(-3)

3P⁄3 = -12⁄3

Pq = (-4)(-3)

12

(25⁄100)^y = 32

(1⁄4)^y 32

log(1⁄4)^y = log32

ylog 0.25 = log32

y = log32⁄log0.25

y = -2.5

y = -25⁄10

y = -5⁄2

Number of Boys = 5

Number of Girls = 4

Total pupils = 8 + 4

To get the probability of boys out of lift is the number of boys over the total number of pupils

P(a boys steps out of the lift) = 8⁄12

2⁄3

diameter = 7cm

Area of cylinder = 165cm² ϖ = 22⁄7

since radius = 35

165 = 8(22⁄7)(3.5)(h) + 2(22⁄7)(3.5)²

165 = 22h + 44⁄7 (12.25)

165 =22h + 77

f=group like terms

165 - 77 = 22h

88 =22h

divide through by 22

h=4

Considering 2^a = √64

2^a = √64

what raise the power 2 will give you 64

2⁶ = √64

2^a = 2^61⁄2

since √ = 1⁄2

2a = 2^6(1⁄2)

divide 2 by 6

2^a = 2³⁽¹⁾

2^a = 2²

since the base arethe same we equate the exponent considering b⁄a = 3

b⁄a = 3

cross multiple

b⁄a = 3⁄l

b(1) = 3(a)

b =3a

since a = 3

b = 3(3)

b = 9

therefore a² + b² = 9² + 3²

81 + 9

90

∅ = 52

opposite ∣YZ∣ = 32

Adjacement ∣XZ∣ = ?

since opposite is known and we are ask to find adjacent (∣XZ∣)

tan∅ formular will be appiead.

which says tan∅ = opposite⁄Adjacement ∣YZ∣⁄∣XZ∣

tan(32) = 32⁄∣XZ∣

cross multiply

tan32⁄1 = 32⁄∣XZ∣

∣XZ⁄ tan(52) =32

On your scientific calculator tan(52) = 0.52

∣XZ (∅.28) = 32

making ∣XZ∣ the subject divide through by 0.53

∣XZ⁄0.53 = 32⁄1.28

∣XZ = 32⁄1.28

∣XZ∣ = 25cm

length of an arc = ∅⁄360 X 2Πr

Angle (Π) = 72

Π = 22⁄7

length of arc = 72;⁄360 x 2 22;⁄7 x 3.5

= 4.4

Ib = 1⁄2 (a+b)h

multiply through by 2

2 x Ib = 1⁄2(a + b)h

2b = 1(a + b)h

2b =(a + b)h

2b =ah + bh

Group like terms

factorize b out

Gives b(2 - h) = ah

Divide through by 2 - b to make b the subject

b(2 - h)⁄2-h = ah⁄2-h

b = ah⁄2-h

Commision on selling price let selling price of house = x

commision on selling price = x x 8⁄100

$0.08x

Eric share of the house = 17,760.00

Total selling price of the house = Agent commision + Eric share x = $117,760 + $0.80x

group like terms

x - 0.08x = 117,760.00

0.92x⁄0.92 = 117760⁄0.92

x = $178,000.000

The total selling price of the house is 128,000

Length of an arc = 22 cm

raduis =

π = 22⁄7

angle (∅)= ?

formular for length of an arc = ∅⁄360 x 2πr

since length of an arc =22 cm

r = 15 and π = 22⁄7

22 = π⁄360 X 2(22)⁄7 X 15

22 = π⁄360 X 44⁄7

22 = π⁄360 X 94.286

22 = 94.2860⁄360

22⁄0.262 = 0.2620⁄0.262

π =22⁄0.262

π = 84

The angle is 84 °

Length = 15cm

width = x cm

If the side are double

New length = 15 X 2

30cm

New width = 2 X x

2x

2x

Area of new rectangular box = Length(L) X (W)

30cm X 2xcm

60x cm²

∆OST = OTS (Isocicoles)

OTS = 50°

considering ∆OST

50 + 50 +∆SOT = 180

100 + ∆SOT =180

Group like terms

∆SOT = 180

∆SOT = 80°

∆SOT = ∆ROP = 80° (vertical opposite angles are equal

80 + OPQ = 180

OPQ =180 - 80

OPQ = 100

Expand the bracket

2x(x) +2y(x-y) +2x(x+y) 2y(x+y)

2x² - 2xy + 2xy + - 2y² + 2x² + 2xy - 2xy - 2y²

2x² - 2y² + 2x² + 2y²

Group like terms

2x² - 2y² - 2y²

factorize 4 out

4(x - y²)

(x - y²) = (x - y)(x + y)

4[(x - y)(x + y)]

Since the angle of the polygon are five

It means the polygon is pentagon

interior angles of polygon = n-2x180⁄n

number of side (n) = 5

Interior angles of pentagon = (5-2)X180⁄5

=3X180⁄5

=540⁄5

108

since 180 is the interior

3x which is the smalllest angle

3x⁄3 = 108⁄3

x =60°

the smallest angle is 60°

Equation of a line y = mx + c

If point (1,-5) is given and gradient 3⁄4

y = -5

x = 1

y = mx + c

Gives : -5 = 3⁄4 (1) + c

-5 = 3⁄4 + c

Group like terms

-5-3⁄4 = c

-5⁄1 -3⁄4 = c

-20-3⁄4 = c

-23⁄4

y = 3⁄4x - 23⁄4

multipy through by 4

4 x y = 4 X 3⁄4x - 23⁄4 x 4

4y =3x - 23

0 =3x - 23 - 4y = 0

3x - 4y - 23 = 0

Let label the part A,B and C

to find the ladder we use pythagories theorem

which says ∣AB∣²+∣BC∣² = ∣AC∣²

∣AB∣ =6m,

∣B∣=8m,

∣AB∣²+∣BC∣² = ∣AC∣²

6²+8² = ∣AC∣²

36 + 64 = ∣AC∣²

100 ∣AC∣²

Root both side

√100 = √∣AC∣²

√100 = AC

10 = AC

The ladder is 10 m long

∠ZXY = ∠ZOY = 70°

Length of an arc = Δ⁄2πr

since angle (Δ)π = 22⁄7 and r = 18cm

length of an arc =70⁄360 x 2 x 22⁄7 x 18

cancelation will take place here

70⁄360 x 2 x 22⁄7 x 18

1⁄18 x 22 x 18

22cm