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According to this theory, the caste system is of divine origin. It says caste system is an extension of the varna system, where the 4 varnas originated from the body of Bramha.

At the top of the hierarchy were the Brahmins who were mainly teachers and intellectuals and came from Brahma’s head. Kshatriyas, or the warriors and rulers, came from his arms. Vaishyas, or the traders, were created from his thighs. At the bottom were the Shudras, who came from Brahma’s feet. The mouth signifies its use for preaching, learning etc, the arms – protections, thighs – to cultivate or business, feet – helps the whole body, so the duty of the Shudras is to serve all the others. The sub castes emerged later due to inter marriages between the 4 varnas.

The proponents of this theory cite Purushasuktaof Rigveda, Manusmriti etc to support their stand.

Answer: Thesumof interioranglesof a15 sided polygonis 2,340° and each interiorangleis 156°

the sum of the interior angles of a polygon is.

where n is the number of sides of the polygon.

here n=15.

sum =180∘×13=2340∘

one interior angle =2340∘15=156∘

Secularismrefers to the separation of religion from the state. It means that the state should not discriminate among its citizens on the basis of religion. It should neither encourage nor discourage the followers of any religion.

products 21/8 × 5/16=21×5/16×8=105/56=1.875; 21/8+5/16=105/56=0

x+y=21/8; xy=5/16; x^2+21/8x-5/16=16x^2+42x-5

16x^²-42x+5 = 0 is the correct answer of the given question

part 2

9,3,1

Let the numbers are x and (24 - x)

As per given conditionx*(24 - x) = 143x^2 - 24x + 143 = 0x^2 - 13x - 11x +143 = 0x(x - 13) - 11(x - 13) = 0(x - 13)(x - 11) = 0x = 13, 11

Therefore, numbers are 13 and 11

Sum of the exterior angles of regular polygon = 360°But each exterior angle = 40°number of sides of regular polygon = 360° / 40° = 9

8 sides is the answer

Nyc

P(x) = 8x^3 - ax^2 - x + 2

x = - 1/2 is a zero of polynomial

Therefore,For x = - 1/2, given polynomial equal to zero

8*(-1/2)^3 - a*(-1/2)^2 - (-1/2) + 2 = 08*(-1/8) - a*(1/4) + 1/2 + 2 = 0- 1 - a/4 + 5/2 = 05/2 - 1 = a/4(5 - 2)/2 = a/43/2 = a/4a = 4*3/2 = 6

Given,Let p, q and r are three zeroes of 3x^3 + 5x^2 - 7x - 27

p*q = 3.......(1)

We know,Product of three zeroes of polynomial = - d/a

Then, p*q*r = - (-27)/3 = 9......(2)

Put value of p*q from eq(1) in eq(2)r = 9/3r = 3

Therefore, third zero is 3

there is a stock of 1071 dhotis, 595 sarees and 357 dress. let's us calculate, what will be the maximum number of families among which these can be distributed equally. haow many of these things each, the families will receive

We need to find HCF in this question.HCF(1071,595,357)=119So maximum 119 families can get them.Each family will get=1071/119 = 9 dhotis595/119=5 sarees357/119=3 dresses

3/4 = 0.75 6/8 = 3/4 = 0.75 so both are equivalent

Given polynomialx^3 - 4x^2 + x + 6 = 0

As - 1 is one of the zero, then (x+1) should be one of factor of polynomial

x^3 + x^2 - 5x^2 - 5x + 6x + 6 = 0x^2(x + 1) - 5x(x + 1) + 6(x + 1) = 0(x + 1)(x^2 - 5x + 6) = 0(x + 1)(x^2 - 3x - 2x + 6) = 0(x + 1)(x(x - 3) - 2(x - 3)) = 0(x + 1)(x - 2)(x - 3) = 0x = - 1, 2, 3

Therefore, other two zeroes are 2 and 3

2/3 of the tank is emptied in 64 buckets

Volume of tank = (1.2)³ = 1728 cm³

Volume of 64 buckets = ⅔ × 1.728

Volume of 1 bucket = (⅔ × 1.728)/64 = 0.018 m³ = 0.018 × 1000 litres = 18 litres

9+(-17)=-8 this is answer

x-1/x = 7x^2 + 1/x^2 = 51x^3 - 1/x^3 = (x-1/x)(x^2+1/x^2+1)7*(51+1)=52*7=364

5/7 - 2/14Find LCM of 7 and 14 = (10 - 2)/14= 8/14= 4/7

Consider ∆ABC in which angle ABC=90°,AB=a,BC=b and AC=c.Let AM=x be altidude of AC from B.Take angle BAC=θ,angleBCA=90-θ.sinθ=x/a and sin(90-θ)=x/b=cosθusing (sinθ)^2+(cosθ)^2=1 we getx^2/a^2=1-x^2/b^2=>x^2(1/a^2+1/b^2)=1=>x^2=(a^2b^2)/(a^2+b^2) = (a^2b^2)/c^2 =>x=(ab)/c.

Volume of cuboid = 25*20*4= 2000cm^2volume of cube is half of 2000that is 1000cm ^3as volume of cube = a^3= 1000a= 10cmnow surface area = 6(side)^2= 6(10^2)= 600cm^2

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R = 28 cmr = 21 cmLet height of frustum be h cm

∴Volume held by bucket = 28.490 litres = 28490 cm³Also,

Volume of frustum = π(R²+Rr+r²)h/3=> 28490 = 22*(784+588+441)*h/(3*7)=> 1813h =28490*21/22=> h = 27195/1813 = 15cm

In cylinder,r=7cm=0.7ml=15km=15000m

In tank,l=50mb=44mh=0.21m

Vol.of water in tank=lbh =50*44*0.21 =462m³

Height of cylindrical pipe=Vol. /πr² =462/(0.07)²(22/7) =462/0.0154 =30000m

Time = 30000/15000 = 2 hours

plz like. and. mark

x+2y=103x+4y=3603x+4y-3x-6y=360-30=330-2y=330y=-165lx+5=-165so lx=-170now.x=10+330=340so l=-1/2

Radius of Sphere = 10.5 cmRadius of cone = 3.5 cmHeight of cone = 3 cm

The volume of the metallic sphere = (4/3)Πr³ = (4/3) × Π × (10.5)³ Volume of cone = (1/3)Π r² h = (1/3) × Π × (3.5)² × 3

No of cones = Volume of Metallic Sphere/Volume of cone = (4/3) × Π × (10.5)³/(1/3) × Π × (3.5)² × 3 = 4 × 10.5 × 10.5 × 10.5/3.5 × 3.5 × 3.5 = 126 Cones

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Let Number of Students Playing Chess be : n(Ch) = 22

Let Number of Students Playing Carrom be : n(Ca) = 21

Let Number of Students Playing Table Tennis be : n(TT) = 15

Given Number of Students Playing both Chess and Table Tennis = 10

⇒ n(Ch ∩ TT) = 10

Given Number of Students Playing both Carrom and Table Tennis = 8

⇒ n(Ca ∩ TT) = 8

Given Number of Students Playing all the three Games = 6

⇒ n(Ch ∩ Ca ∩ TT) = 6

Given that Each Student in the Class of 35 Plays atleast one Game among Chess, Carrom and Table Tennis

⇒ Total number of Students Playing atleast one Game = 35

⇒ n(Ch ∪ Ca ∪ TT) = 35

We know that :

n(Ch ∪ Ca ∪ TT) = n(Ch) + n(TT) + n(Ca) - n(Ch ∩ TT) - n(Ca ∩ TT) - n(Ch ∩ Ca) + n(Ch ∩ Ca ∩ TT)

Substituting all the Respective values we get :

⇒ 35 = 22 + 15 + 21 - 10 - 8 - n(Ch ∩ Ca) + 6

⇒ 35 = 46 - n(Ch ∩ Ca)

⇒ n(Ch ∩ Ca) = 11

⇒ Number of Students Playing both Carrom and Chess = 11

From the Venn Diagram :

We can Notice that Number of Students Playing both Carrom and Chess also includes the students who are playing all the Three Games.

⇒ Number of Students who are playing both Carrom and Chess but not Table Tennis can be found by Subtracting Number of students playing all the Three games from Number of students playing both Carrom and Chess.

(i) Number of Students Playing both Carrom and Chess but not Table Tennis = n(Ch ∩ Ca) - n(Ch ∩ Ca ∩ TT) = 11 - 6 = 5

(ii) Number of Students playing only Chess will be :n(Ch) - { n(Ch ∩ Ca) - n(Ch ∩ Ca ∩ TT) + n(Ch ∩ Ca ∩ TT) + n(Ch ∩ TT) - n(Ch ∩ Ca ∩ TT) }⇒ 22 - {11 - 6 + 6 + 10 - 6}⇒ 22 - 15 = 7Number of Students only playing Chess will be : 7

(iii) Number of Students playing only Carrom will be :n(Ca) - { n(Ch ∩ Ca) - n(Ch ∩ Ca ∩ TT) + n(Ch ∩ Ca ∩ TT) + n(Ca ∩ TT) - n(Ch ∩ Ca ∩ TT) }⇒ 21 - {11 - 6 + 6 + 8 - 6}⇒ 21 - 13 = 8Number of Students only playing Carrom will be : 8

i )49 chess,carrom but not table tennisii)22 only chess

let number of rides on giant wheel = x

and number of played hoopla games = y

so according to the question

x=2y and 3x+4y=20

by putting x value in second equation from equation first we get

⇒3*(2y)+4y=20

⇒6y+4y=20

⇒10y=20

⇒y=2

putting this value in first equation

we get x=4

it means she played hoopla 2 times and giant wheel ride is 4 times.

let number of rides on giant wheel is x

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