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LHS

( cosecA - cotA )²

= ( 1/ sinA - cosA/sinA )²

= ( 1 - cosA/ sinA )²

= ( 1 + cosA )²/ sin²A

= ( 1 - cosA )²/ 1 - cos²A

= ( 1 - cosA ) ( 1 - cosA )/ ( 1 + cosA ) ( 1 - cosA )

= 1 - cosA / 1 + cosA

.°. RHS

800*40/100320will be the result

herein the longest side is BC hence it is hypotenusehencesinA=opposite /hypotenuse=12/13cosA=adjacent/hypotenuse=5/13tanA=sinA/cosA=12/5cosecA=1/sinA=13/12secA=1/cosA=13/5cotA=5/12=1/tanA

Given,Length l = ?, Breadth b = 6 m, Height = 5 m

Total surface area of cuboid => 2(l*b + l*h + b*h) = 5002(l*6 + l*5 + 6*5) = 5006l + 5l + 30 = 25011l = 220l = 220/11 = 20

Therefore, length of cuboid = 20 m

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Face of cuboid is rectangle

one of face of cuboid is rectangular and other is a square

rectangular is the right answer

The face of cubiod has shape of recatngle

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X will be the no. Of 10 rs. NotesY will be the no. Of 20 rs. Notes10x + 20y = 800x + y = 50So, 10x + 10y = 500Now,10x + 20y = 800- 10x + 10y = 50010y = 300y = 30x = 20

x= 20y= 30 is the correct answer of the given question

20, 30 is the correct answer of the given question

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D of hemisphere = 14cmr = 14/2 = 7cmCSA of hemisphere = 2πr²= 2 x 22/7 x 7 x 7 = 44x7= 308cm²r = 7cmh(total) = 13cmh(cylinder) = 13 - radius of heisphere = 13-7 = 6cmCSA of Cylinder = 2πrh = 2 x 22/7 x 7 x 6= 44x6 = 264cm²Surface area of the vessel = CSA of Hemisphere + CSA of Cyclinder= 308+264 = 572cm²

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Let side length of cube is s

Given,Volume of cube = 64s^3 = 64s = 4

When two cubes of side length 4 cm joined end to end then cuboid form is of length = 8 cm, breadth b = 4 cm, height = 4 cm

Surface area of cuboid= 2(l*b + b*h + l*h)= 2(8*4 + 4*4 + 4*4)= 2(32 + 16 + 16)= 2(64)= 128 cm^2

option (A)

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= Let Radha be x years and her brother be y years.

Given,= x = y + 5 ...... 1

= x + y = 25Substitute eq 1.= (y+5) + y = 25= y + 5 + y = 25= 2 y = 20= y = 10 years.

x = y + 5= x = 10 + 5 = 15 years.

We know that formulae:

SI = P× R × T /100

We have Principal P = 800

T = 3 years

⇒ SI = Amount – P

⇒ SI = 956 – 800 = 156

Hence,

⇒ 156 = 800 × R × 3/100

⇒156 = 24R

⇒ R = 156/24

Hence we have

R = 6.5 % p.a.

Now, we have the rate of interest increased by 4%

Hence, new rate of interest as 6.5 + 4 = 10% p.a.

New SI = 800 × 10.5 × 3/100

⇒ SI = 252

Amount after 3 years at new Rate of interest = P + SI = 800 + 252

= 1052

Answer:161.5 cm^2

Explanation:AngleQPR=90 (Traingle formed in the semi circle is of 90°IN Traingle QPRBy pythagoras theoremH^2=P^2+B^2H^2=(7)^2+(24)^2H^2=49+576H^2=625H=√625H=25cmDiameter of circle =25cmRadius of circle= 25/2Area of Semi circle=pie(r)^2=1/2×22/7×25/2×25/2=245.5Area of traingle= 1/2×b×h=1/2×24×7= 12×7=84

Area of shaded region = Area of semicircle -Area of traingle=245.5-84=161.5 cm^2

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64/40 = 16/10 = 1.6after 1 decimal

Whatever mass of 2 balls are and distance between them is anything.

The Gravitational constant is given by 6.67×10^-11N

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800(1+5/100)²

= 800*21*21/20*20

= 882 rupees

Given Volume of both cubes = 64cm^3find one side = (64)^1/3 = 4

The cubes are joined end to endLength = 8cmwidth = 4cmheight= 4 cm

Surface area,2(lb+bh+hl)2(32+16+32)160 cm^2

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First term = a₁ = 1

Second term = a₂ = 3

Common difference = d =a₂- a₁ = 3 - 1 = 2

Last term = an = 199

an = a+ (n-1)d199 = 1 + (n-1)2198 = (n-1)2198/2 = n - 199 = n - 1n = 100

No: of terms = n = 100

Sum = Sn

Sn = n/2 [a + an] = 100/2 [ 1 + 199] = 50 [ 200] = 10000

Sum = 10000

x³ -7x +6= 0=> x³ - x - 6x +6 = 0=> X(x²-1) -6(x-1)=0=> X(X+1)(x-1) -6(x-1)=0=> (x-1)(x²+x-6)=0=> (x-1)(x²+3x-2x+6)=0=> (x-1)(X(X+3)-2(X+3))=0=> (x-1)(X+3)(x-2)=0=> X = 1,2,-3.

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area of the shaded square

1rs=100paise5rs=5*100=500paise