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(a + b)2= a2+ b2+ 2ab

(a+b)2=a2+b2+2ab then (a+b)(a+b)=a(a+b)+b(a+b)=a2+ab+ba+b2=a2+b2+2ab

(a+b)2=a2 +b2 +2ab is formula.

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∴ (a³+4b³+c³)/[b(a²+c²)] = 3

a + b + c = 11 and ab+bc+ca = 20 (a+b+c)(a+b+c) = 121 a*a+b*b+c*c+2ab+2bc+2ca = 121 a*a+b*b+c*c+40 = 121 a*a+b*b+c*c = 81 a*a*a+b*b*b+c*c*c-3abc = (a+b+c)(a*a+b*b+c*c-ab-bc-ca) = 11(81-20) = 11*61 = 671

(a+b)(a^2-ab+b^2) =(a-b)(a^2+ab+b^2)

Let the unknown number be X. Then, 16*12 -672/12 = X -211=> 192 - 56 = X -211=> X = 347.

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16*12 - 672 + 21 = ? - 211192 - 672 + 21 = ? - 211213 + 211 - 672 = ? 424 - 672 =? ? = - 248

Ans = - 248

Let the angle be xother angle (it's supplement) =( 180-x)x+18= (180-x)2x = 162x = 162/2 = 81other angle(its supplement) = 180- x => 180-81 => 99

(a + b)(a - b) = a^2 - b^2(3 + √3)(3 - √3) = 3^2 - (√3)^2 = 9 - 3 = 6

(a +b)( a - b) =a^2 - b 2(3 + √3) (3 - √3) =3^2 - (√3)^2 =9-3 = 6

Yes, 0 is a rational number (provided that you understand 0 as the symbol of the natural or integer number zero). The definition of a rational number is 'A number that can be written in the form a/b, where a and b are integers with no common factors and b is NOT zero'. 0 can be written as (for example) 0/2.

yes because it can be written in the form of p/Q such as 0/9 /;0/1;0/8000 etc.

961 is it largest 3 digits which is a perfect square

Largest 2 digit =81Largest 3 digit =961

largest 3 digit square=961

a)circumference= 154 cmcircumference= 2πr= 154r=(154)/(2π)=24.5 cmb) Area= 9.625cm²area= π r²=9.625r²=(9.625)/(2π)=1.531r=1.237

H.C.F.×L.C.M.=multiplication of two numbers105×L.C.M.=525×945so L.C.M.=4725

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Let the 4 numbers are a , a + d , a + 2d , a + 3d.Sum of 4 numbers AP = 50

a + a + d + a + 2d + a + 3d = 50

⇒ 4a + 6d = 50

⇒ 2a + 3d = 25 --------------(1)

Also given the greatest number is 4 times the least.

4(a) = a + 3d

4a - a = 3d

a = d

putting a = d in (1) , we obtain

5d = 25

d = 5

a = 5 but d = a.

∴ First four terms are 5 , 10 ,15 , 20.

1) Given radius of circle is 5cm If point P is such that OP = 6cm. Then point P lies outside the circle at dis of 1 cm

2) No it is not possible to draw chord of 11 cm in circle of radius 5 cm. As we know longest chord of circle is diameter which is of length twice of radius. For given radius diameter length is 10 cm.

1026is correct answer of this question

The perpendicular bisector of any chordpasses through thecenter of the circle.

So when you construct the diagram, remember to mark the right angles (∠OAZ and∠OBX in the diagram).

Also, you need to identify what you are supposed to prove (joining mid point passes through the center). So you need to prove that AOB is a straight line.To do that, we can construct another line OC//WX//YZ.

The proof

∠BOC =∠OAZ = 90° (corresponding angles, OC//YZ)∠COA =∠XBO = 90°(corresponding angles, OC//WX)∠AOB= ∠BOC + ∠COA=180°

So AOB is a straight line. Therefore AB passes through O.

∠BOC =∠OAZ = 90° (corresponding angles, OC//YZ)∠COA =∠XBO = 90°(corresponding angles, OC//WX)∠AOB= ∠BOC + ∠COA=180°

So AOB is a straight line. Therefore AB passes through O.

sin(π/9)*cos(π/9)*cos(2π/9)*cos(3π/9)*cos(4π/9)/sin(π/9)=sin(2π/9)cos(2π/9)*cos(3π/9)*cos(4π/9)/[2*sin(π/9)]=sin(4π/9)*cos(3π/9)*cos(4π/9)/[4*sin(π/9)]=sin(8π/9)*cos(3π/9)/[8*sin(π/9)]=sin(π/9)*cos(3π/9)/[8*sin(π/9)]=cos(3π/9)/8=1/16

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Let the first chord=24 cmSecond chord=x cmRadius=Diameter/2 =30cm/2 =15cm Note that half of first chord,half of second chord and the radius form a right angle triangle with the radius as the hypotenuse

Using Pythagorean theorem[Half of first chord]^2+[Half of second chord]^2=Radius^212^2+[Half of second chord]^2=15^2144+[Half of second chord]^2=225[Half of second chord]^2=225-144[Half of second chord]^2=121[Half of second chord] =121^(1/2)[Half of second chord] =11

So Answer Is 11 cm.