-6<-4

0>-1

6-2=4

4>5

If A={1,2,3} n(A)=3

if B={ },n(B)= 0

if C={a,b},n(C)=2

if D={1,6,8,90} n(D)=5

P={1,2,3,5,7,11,13} .it is set of prime number less than 14.

Q={4,6,8,8,9,10}its Description form is

Q={ set of composite number less than 11}

S={1,4,9,16,25,36,49,64,81,100}its description form is

S={ set of square number from 1 to 100}

C={1,8,27,64,125}it's set builder form is

C={x:x is cube number ,x < 126}

E={2,4,6,8,10}it's in Listing Method .

A set can be represented in three ways:

i)Description Method .Example:A is set of Counting numbers less than 10.

ii)Tabulation/Listing/Roster Method .Example:A={1,2,3,...9}

iii)Set builder Method. Example:A={x:x is Counting numbers less than 10}

Total number of elements in a set is the Cardinal of set .

A={2,3,4,8,9}, n(A)=5

B={x:x is a prime number less than 10} ={2,3,5,7},n(A)=4

C={Set of vowels in English alhabets}={a,e,i,o,u}. so, n(C)= 5

D={7,8,9} ,N(D)=24

$$6\in B\ ,6\notin A$$  

$$4\in A\ ,4\notin B$$  

set A and set B are overlapping sets as well as equivalent sets. $$A\sim B$$  

set A and set B are disjoint sets as well as equivalent sets

Subtract 5a+7 from 8a-6=(8a-6)-(5a+7)=8a-6-5a-7=8a-5a-6-7=3a-13

Add:(4x+5y)+(-3x-2y)=4x+5y-3x-2y=4x-3x+5y-2y=x+3y

Add:(4x+5y)+(-3x-2y)=4x+5y-3x-2y=4x-3x+5y-2y=x+3

Subtract 5a+7c from 8a-6c=(8a-6c)-(5a+7c)=8a-6c-5a-7c=8a-5a-6c-7c=3a-13c

Subtract -5a-7c from -8a-6c=(-8a-6c)-(-5a-7c)=-8a-6c+5a+7c=-8a+5a-6c+7c=-3a+c

$AB\parallel DC,AD\parallel BC,EF\parallel HG,EH\parallel HG$ AB is parallel to DC ,(AB // DC )

$AB\ \perp\ AD\ ,\ AD\ \perp DC\ ,AE\perp EH\ \ ,DH\perp EH$ AB is perpendicular to AD (AB $\perp$ AD)

above figure is rectangle.

V= l x b x h

What is H.C.F of 24 and 40 by prime factor Method .

Solution: 24= 2 x 2 x 2 x 3

40= 2 x 2 x 2 x 5

H.C.F = 2 x 2 x 2 = 8

Find H.C. F of 12 and 18 by factor method .

Solution : Factor of 12 (F₁₂)= { 1,2,3,4,6,12}

Factor of 12 (F₂₄) = { 1,2,3,4,6,8,12,24}

Common factor of 12 and 24 = {1,2,3,4,6,12}

Highest common factor of 12 and 24 = {12}

H.C.F = 12

Find H.C. F of 12 and 18 by division method

Solution : 12 ) 18 ( 1

- 12

.................

6 ) 12 ( 2

- 12 So , H.C.F = 6

...................

x

Find H.C. F of 2 and 3

Solution : H.C.F of 2 and 3 = 2 x3 =6

Find H.C. F of 2 and 3

Solution : H.C.F of 2 and 3 = 1

Find the L.C.M of 6 and 8 by factor method .

Solution : 6 = 2 x 3

8 = 2 x 2 x 2

L. C . M = 2 X 2 X 2 X 3 = 24

Find the L.C.M of 6 and 8 by multiple method .

Solution : Multiples of 6( M₆ ) = { 6,12,18,24,30,36 ,42, 48,..., }

Multiples of 8 (M₈) = { 8 ,16, 24 ,32, 40 , 48 ,....,.}

Common Multiples of 6 and 8 = { 24 , 48 ,....}

Least Common Multiple of 6 and 8 = { 24}

L.C.M = 24

Find the L.C.M of 6 and 8

Solution : 6 and 8 = 2

sum of y and 6=y+6

subtract 6 from y=y+6

Multiply 6 and y =6X6

subtract 6 from y=y - 6

Multiply 6 and y =6Xy=6y

Diameter is line segment passing through origin and joins any two points of circumference of circle .

Chord is line segment not passing through origin and joins any two points of circumference of circle .

Radious is line segment which joins origin and any point of circumference of circle .

Sector is line segment which joins origin and any point of circumference of circle .

Volume of cuboid = l x b x h

The formula to find total surface area of cuboid = 2(lb+lh+bh)

Volume of cube = l x l x l = l³

रूढ सङ्ख्याहरू(P) ={2,3,5,7,.... }

संयुक्त सङ्ख्याहरू(C)=

{4,6,8,9,10,12,14,15.....}

वर्ग सङ्ख्याहरू(S)=

{1,4,9,16,25.......}.

24 का गुणन खन्डहरू

$$\left(F_{24}\right)$$ ={0,1,2,3,4,6,8,12,24}

24 का गुणन खन्डहरू

$$\left(F_{24}\right)$$ ={1,2,3,4,6,8,12,24}

Solution:

length of cuboid(l)= 6a m

breadth of cuboid(b)= 3am

height of cuboid(h)= 4a m

volume of cuboid(V)=?

volume of cuboid(V)= l x b x h = 6a m x 3a m x 4a m

= $$72\ a^3m^3$$  

Now answer in exponential form,

$$72a^{3\ }m^3=2X2X2X3X3\ a^3m^3$$ $$2^3\ X\ \ 3^2\ X\ \ a^3X\ \ m^3$$    

$$72a^3m^3\ =$$   $$2^2\ X\ \ 3^3\ X\ \ a^3X\ \ m^3$$

Solution: $$a^3b^4$$ in expanded form=aXaXaXbXbXbXb

Solution: $$a^3b^4$$ in expanded form=aXaXaXaXbXbXb

Solution: $$a^3b^4$$ in expanded form= $$a^{3+4}$$  = $$a^7$$  

Solution:area of rectangle(A)= $$64x^2y^2\ cm^2$$  

length of rectangle(l)=16xy

breadth of rectangle(b)=?

$$\because$$   l X b= A

$$\therefore\ b=\frac{A}{l}=\frac{64x^2y^{2\ ^{ }}cm^2}{16xy\ cm^{ }}$$  

= $$4xy\ cm$$   

area of rectangle(b)=4 xy cm= 4 X 2 X 3 cm=24 cm

Area of rectangle=

64x $$64x^2y^2cm^2=64\left(2\right)^2\left(3\right)^{2^{ }}cm^2$$  =64 X 4 X 9 $$cm^2$$  

= $$2304cm^2$$  

Solution:sum of -4 and 5=(-4)+(5)

=-4+5=1

Solution: product of 14 and 5 =(14)×(5)

=70

Solution:Difference of 14 and 5 =(14)-(5)

=9

Solution:Difference of 14 and 5 =(14)+(5)

=19

No,It is not a multiple Bar Diagram.

It is a simple Bar Diagram.

I)Solution, in 1 meter there is 100 cm.

II) Solution, The length of 3 foot long White board is 36 inch .

Because, [1 foot = 12 inch

3 foot =12 ×3 inch =36 inch]

I)Solution, in 1 meter there is 1000 cm.

II) Solution, The length of 3 foot long White board is 3.6 inch .

Because, [1 foot = 1.2 inch

3 foot =1.2 ×3 inch =3.6 inch]

III)If you walked 500m towards school then you walked 500x3.28 =1640 foot .

because, [1m=3.28foot]

[5m=3.28foot]

Iv) If length of Bench is 250cm its length is 8.2 foot long.

[30.48cm=1foot]

or,
$$1\ cm=\frac{1\ fooot}{30.48}$$
$$250\ cm=\frac{250foot}{30.48}$$

=8.2 foot

V)Solution :

Length of class room is 500cm then it's length is 196.85 inch long.

Because, [2.54cm=1 inch]

Or, $$1cm=\frac{1\ inch}{2.54}$$

Or $$500cm=\frac{500\ inch}{2.54}$$

=196.85 inch

III)If you walked 500m towards school then you walked $$\frac{500}{3.28}$$ foot =152.43 cm

because, [1m=3.28foot]

[so, 500m=500x3.28foot=152.43foot]

I) Solution:3m 60 cm

=3m+ 60 cm

=3x100cm+ 60 cm

=300cm+60 cm

=360cm.

II) Solution:6 ft

= 6x30.48 centimeter

=182.88 cm.

iii) Solution:

Convert 8 ft 6inch

=8 ft+6 inch

=8x12 inch+6 inch

=96 inch+ 6 inch.

IV)Solution:

11 ft 10 inch

=11 ft +10 inch

= 11 x 30.48 cm+10x 2.54 cm

= (335.28+25.40)cm

=360.68cm

.

IV)Solution:

11 ft 10 inch

=11 ft +10 inch

= 11 x 30.48 cm+10x 2.54 cm

= (33528+2540)cm

=36068cm