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area= side × side = 5m × 5m = 25 m^2

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Area of floor Af = length*breadth = 4x3 = 12 m^2 = 120000 cm^2

Area of each tile At = side*side = 20*20 = 400 cm^2

Let no. of tiles required to cover floor = n

Then, n*At = Afn = 120000/400 = 300

300 tiles required to cover floor

Given:Area of floor=192sq mlength=16mNowArea=l×b192=16×bb=192/16b=12m

Now perimeter=2(l+b)=2(16+12)=2(28)=56m

the. correct answer is 56m.....

area = length x breadth=67.5 sq.m

volume=270m cube

volume of a cuboid=lxbxh

67.5xh=270

h=270/67.5

=4m

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1/2is athe greatest power among them

1/2 is a the greatest power among them

1/3 power 1/3 is the greater

225^1/2

= √225

= 15

Length of the room = 4 m

Width of the room is 3 m 25 cm = 3.25 m

To carpet the room, we need to find the area of the floor.

So, Area of the room = length × breadth = 4 × 3.25 = 13 sq. m

length of floor = 4mbreadth= 3m 25cm =3.25marea of floor=4*3.25 =13 square metres13 square metres of carpet is required to cover the floor

A) 3 power 4 B) 5 cubeC) 6 cube D) 7 cube

An Arc is a curved line that is a part of a circle.

thestraight linethat "just touches" the curve at that point is called the tangent.

Secant is a straight line that intersects a curve at two or more points.

We know that Volume of cone = πr²h/3

=π*5²*8/3

=200/3π

Now

volume of liquid displaced=volume of balls dropped

1/4(200/3)π=100*4/3πr³

50=100*4r³

50/4=100r³

50/400=r³

1/8=r³

r=1/2=0.5cm

please complete the question

49/343=49×1/49×3=1/3

cube root will be 70 (-8)*(-343)*(125) 3√(-8)*(-343)*(125)3√(-2*-2*-2)*(-7*-7*-7)*(5*5*5)-2*-7*5=70

(7)^(6/x) =7²So x=6/2=3

Given :- Weight of the heavier sphere = 5920 g

weight of the lighter sphere = 740 g

Diameter of lighter sphere = 5 cm or radius = 2.5 cm

Let the volume of the heavier sphere be 'V1' and the volume of the lighter sphere be 'V2'.

Radius of the heavier sphere be 'r1' because the diameter of the lighter sphere is given so the radius is 2.5 cm.

Weight of an object = density× volume of that object

So,

(Weight of the heavier sphere/weight of the lighter sphere) = (Density× V1/density× V2)

As both the spheres are made up of same metal, therefore, the ratio of their weights will be equal to the ratio oftheir volumes.

(weight of the heavier sphere/weight of the lighter sphere) = (V1/V2)

⇒ (5920/740) = (4/3πr₁³)/(4/3πr³)

⇒ (5920/740) = (4/3×22/7×r₁³)/(4/3× 22/7× 2.5× 2.5× 2.5)

⇒ 8 = r₁³× 1/2.5× 1/2.5× 1/2.5

⇒ 8 = r₁³× 1/15.625

⇒ r₁³ = 8× 15.625

⇒ r₁³ = 125

⇒r₁ = 5 cm

So, the radius of the heavier sphereis 5 cm.

Given :- Weight of the heavier sphere = 5920 g and weight of the lighter sphere = 740 g, and diameter of lighter sphere = 5 cm or radius = 2.5 cmLet the volume of the heavier sphere be 'V1' and the volume of the lighter sphere be 'V2'. Radius of the heavier sphere be 'r1' because the diameter of the lighter sphere is given so the radius is 2.5 cm.Weight of an object = density× volume of that objectSo,(Weight of the heavier sphere/weight of the lighter sphere) = (Density× V1/density× V2)As both the spheres are made up of same metal, therefore, the ratio of their weights will be equal to the ratio oftheir volumes.(weight of the heavier sphere/weight of the lighter sphere) = (V1/V2)⇒ (5920/740) = (4/3πr₁³)/(4/3πr³)⇒ (5920/740) = (4/3×22/7×r₁³)/(4/3× 22/7× 2.5× 2.5× 2.5)⇒ 8 = r₁³× 1/2.5× 1/2.5× 1/2.5⇒ 8 = r₁³× 1/15.625⇒ r₁³ = 8× 15.625⇒ r₁³ = 125⇒r₁ = 5 cmSo, the radius of the heavier sphereis 5 cm.Answer.

Big sphere:

r = 18cm

Vol = 4/3πr³

= 4/3*22/7*18*18*18

= 24438.8cm³

Other 3 spheres:

r₁ = 2cm

Vol = 4/3πr³ = 33.52cm³

r₂ = 12cm

Vol = 4/3πr₃ = 7241.1cm³

Vol of 3rd sphere = Vol of big sphere - vol of both small spheres

4/3πr₃³ = 24438.8 - (33.52+7241.6)

4/3πr³ = 17163

r³ = 4095.7

r = ∛4095.7

r = 34.4cm

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Milk in each can =20.41L/13 = 1.57L

Ans: Each can will have 1.57L =1570 mL.

Volume of sphere = (4/3)Πr³

(4/3) Π (18)³ =(4/3) Π (2)³ +(4/3) Π (12)³ +(4/3) Π r³

(4/3) Π (18)³ =(4/3) Π [(2)³ +(12)³ +r³ ]

(18)³ =[(2)³ +(12)³ +r³ ]

5832 = 8 + 1728 +r³

5832 = 1736 +r³

5832 - 1736 =r³

4096 =r³

r =∛16x16x16

r = 16 cm

Radius of third sphere = 16 cm

it's not correct

N term sum = 3n²/2 + 13n/2as we know that nth term = (Sum of nth term ) -( sum of (n-1)th term)

so 25th term = (sum of 25terms) - ( sum of 24 terms) =(3×(25)²/2 +(13×25)/2) - ( 3×(24)²/2 + (13×24)/2) = 1100 - 1020 =80

Sol:Coordinates of the opposite vertices of the rhombus are (3, 0) and (-1, 4); (4, 5) and (-2, -1).

Length of the diagonals:Length of first diagonal joining the vertices (3, 0) and (-1, 4) = √[(-1-3)2+ (4 - 0)2] = √[16 + 16] = √32 = 4√2Length of first diagonal joining the vertices (4, 5) and (-2, -1) = √[(-2-4)2+ (-1 - 5)2] = √[36 + 36] = √72 = 6√2Area of the rhombus = 1/2 x product of the diagonals= 1/2 x 4√2 x 6√2= 24 sq units.

thanks sir.... it was helpful

1)The given statement is true as the set contains a composite number that is 92)The given statement is true because 125 is a perfect cube but not a perfect square.

thanks you

LCM×HCF= product of no12×512=product of no6144= product of no hcf= product of no/LCM = 6144/512=12LCM= product of no/HCF = 6144/12=512

no they can't have 512 as least common multiple

1[b( a+b)+a(1(a+b)-c+