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first lcm you write any multiply

correct answer is (-13/70)

first take lcm then multiply so you will get -13/70 is your answer

-13/70 is the correct answer of the given question is

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area of path=pi. {R^2-r^2}R=17.5/2+2=8.75+2=10.75r=8.75now area of path =22/7 x ({10.75)^2-(8.75)^2=22/7 x 2 x 19.5=122.460 m^2nowcost of constructing path =122.46 x 253061.5 Rs

Side of a square ABCD= 56 mAC = BD (diagonals of a square are equal in length)Diagonal of a square (AC) =√2×side of a square.Diagonal of a square (AC) =√2 × 56 = 56√2 m.OA= OB = 1/2AC = ½(56√2)= 28√2 m.[Diagonals of a square bisect each other]Let OA = OB = r m (ràdius of sector)Area of sector OAB = (90°/360°) πr² Area of sector OAB =(1/4)πr²= (1/4)(22/7)(28√2)² m²= (1/4)(22/7)(28×28 ×2) m²= 22 × 4 × 7 ×2= 22× 56= 1232 m²Area of sector OAB = 1232 m²Area of ΔOAB = ½ × base ×height= 1/2×OB × OA= ½(28√2)(28√2) = ½(28×28×2)= 28×28=784 m²Area of flower bed AB =area of sector OAB - area of ∆OAB= 1232 - 784 = 448 m²Area of flower bed AB = 448 m²Similarly, area of the other flower bed CD = 448 m²Therefore, total area = Area of square ABCD + area of flower bed AB + area of flower bed CD=(56× 56) + 448 +448= 3136 + 896= 4032 m²Hence, the sum of the areas of the lawns and the flower beds are 4032 m².

The equation x-2=0 on number line is represented by a line

14t+8 is the velocity and 14 is the acceleration of the given question

14 is the correct answer

Sol:Given that ax + by = c ---------(1)bx + ay =1+c ------------(2)Multiply equation (1) by a and equation (2) by b and substractx(a2 - b2) = ac - b - bcx = ( c( a - b) - b) / (a2 - b2).Multiply equation (1) by b and equation (2) by a and substracty(b2 - a2) = bc - ac - ay = ( c( b - a) - a) / (b2 - a2).∴ x = ( c( a - b) - b) / (a2 - b2), y = ( c( b - a) - a) / (b2 - a2).

For this first find area for both the shapesHence,Area of rectangular field is l*b=127*48=6096m^2Area of triangular flower beds=using herons formula √s(s-a)(s-b)(s-c)s=10+10+12/2=16mhencearea will be √16(16-10)(16-10)(16-12)=√16*6*6*4=4*6*2=48m^2henceNo of flower beds=6096/48=127beds

Area of right triangle=1/2*b*h=1/2*12*13=6*13=78m^2Hence,area of rectangular field is l*b=91*24=2184m^2Hence number of flower beds=2184/78=28

thanks

Given:Side of a square ABCD= 56 mAC = BD (diagonals of a square are equal in length)Diagonal of a square (AC) =√2×side of a square.Diagonal of a square (AC) =√2 × 56 = 56√2 m.OA= OB = 1/2AC = ½(56√2)= 28√2 m.[Diagonals of a square bisect each other]Let OA = OB = r m (ràdius of sector)Area of sector OAB = (90°/360°) πr² Area of sector OAB =(1/4)πr²= (1/4)(22/7)(28√2)² m²= (1/4)(22/7)(28×28 ×2) m²= 22 × 4 × 7 ×2= 22× 56= 1232 m²Area of sector OAB = 1232 m²Area of ΔOAB = ½ × base ×height= 1/2×OB × OA= ½(28√2)(28√2) = ½(28×28×2)= 28×28=784 m²Area of flower bed AB =area of sector OAB - area of ∆OAB= 1232 - 784 = 448 m²Area of flower bed AB = 448 m²Similarly, area of the other flower bed CD = 448 m²Therefore, total area = Area of square ABCD + area of flower bed AB + area of flower bed CD=(56× 56) + 448 +448= 3136 + 896= 4032 m²Hence, the sum of the areas of the lawns and the flower beds are 4032 m².

AREA OF RECTANGLE=LENGTH×BREATH LENGTH=48 cm BREATH=20×100cm=2000cm AREA=48×200 =96000cm²AREA OF TRIANGLE=HEIGHT×BASE×1/2HERE, BASE=12×100cm=1200cm HEIGHT=5CMAREA=1200×5×1/2 =6000/2 =3000cm²AREA OF RECTANGLE/AREA OF TRIANGLE=NO. OF FLOWER BED =96000/3000 =96/3=32 FLOWER BED

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given x = 5n and n < 3 => n = 1,2,

so. the value of x = 5*1 = 5 and 5*2 = 10

set A = {5,10}

a.if A is a subset of B, then whole A is in Bhenceanswer is A

1. x-3=0

x=0+3

x=3

thanx didi

simple,x-2=7;shift constant side on R.H.S. sign would be oppositex=7-5;x=2 Ans.

shift constant term on R.H.Sx=7+2x=9 ans

y = 5 or y = -5 this answer

x+y=5; x-y=5; y=5/-5; x=0

x+y=5= (0)+y=5; y=5; x-5=5; x=5+5=10

If lines kx + 3y = k - 3 and12x + ky = k are coincident lines

K/12 = 3/k = k-3/k

Then,K/12 = 3/kk^2 = 36k = 6, - 6

Therefore value of k = 6, - 6

lines are3x-y+8=0..... (1)6x-ky=-166x-ky+16=0.... (2)for lines to coincide 3/6 = 1/k =8/161/k=1/2k=2

Given:A ∆ XYZ in which ∠XYZ = 90°.

To prove:XZ^2= XY^2+ YZ^2

Construction:Draw YO ⊥ XZ

Proof:In ∆XOY and ∆XYZ, we have,

∠X = ∠X → common

∠XOY = ∠XYZ → each equal to 90°

Therefore, ∆ XOY ~ ∆ XYZ → by AA-similarity

⇒XO/XY = XY/XZ

⇒ XO × XZ = XY^2----------------- (i)

In ∆YOZ and ∆XYZ, we have,

∠Z = ∠Z → common

∠YOZ = ∠XYZ → each equal to 90°

Therefore, ∆ YOZ ~ ∆ XYZ → by AA-similarity

⇒ OZ/YZ = YZ/XZ

⇒ OZ × XZ = YZ^2----------------- (ii)

From (i) and (ii) we get,

XO × XZ + OZ × XZ = (XY^2+ YZ^2)

⇒ (XO + OZ) × XZ = (XY^2+ YZ^2)

⇒ XZ × XZ = (XY^2+ YZ^2)

⇒ XZ^2= (XY^2+ YZ^2)

12*15 = 3*x

=> 3x = 180=>x = 180/3 = 60

2x-12 = -3x+28=> 5x = 28+12=> x = 50/5 = 10

In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.

a² + b² = c²

x/2+y+2/5=0......(1)4x+8y+5/16=0....(2)

multiply (1) by 84x+8y+16/5=0

so both the equation has same coefficients,that means they will not collideThat's why they are coincident lines

No, these equations represent parallel lines because slope of lines are same but c for both equations are different so they are not coincident lines