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'r' is the radius of circle which is given 28cmand formula to find the area of circle is (pi)r^2value of pi is 22/7 and value of 'r' is givenputting the values we'll get answer 2464

cube root of 8000= 20cube root of 729=9cube root of 343=7cube root of -512=-8

a 512000000b 387420489c 40353607d 134217728e 20661046784f 35184372088832

cube root of 8000=20cube root of 729=9cube root of 343=7cube root of -512=-8

V of cone = 216πr= 9 cmh=?

NOW,v of cone = 1/3 ×π× r ^2 × h= > 216 π= 1/3 × π× ( 9×9) × h

=> 216× 3/ 9^2= h= 648/81 = h= 8 cm.

Given The cost of fencing a square field at the rate of ₹36 per m is ₹5040

perimeter of the square field = 5040/36

= 140mperimeter =4×side

hence, Length of each side = 140/4

= 35mLike if you find it useful

Area of Garden= (Area of quarter circle with radius 7 m) - (Area of semicircle with radius 7/2 m)

Then,Area of Garden= 1/4*pi*7*7 - 1/2*pi*7/2*7/2= 1/2*pi(49/2 - 49/4)= 1/2*22/7(49/4)= 11*7/2= 77/2= 38.5 m^2

(B) is correct option

Let the interior angle of the regular polygon be x.Therefore, the exterior angle is x/4.

Exterior angle + adjacent interior angle = 180°x/4 + x. = 180°5x /4. = 180°x. = 180° * 4/5= 144°

The interior angle is 144°.The exterior angle is 36°.

Let n be the number of sides.

n = 360°/ exterior angle= 360° / 36°= 10

part 1

part 2

first define the relation or function .. then only it can be answered

although domain will have set from A = (1,2,3,4)and co-domain is always the full set of B = (1,2,3)and range will be either set B or subset of B .

Let the sides of a triangular plot is 3x , 5x and 7x 3x + 5x + 7x = 300 m15x = 300x = 300/15 x = 20then the sides of triangular plot is 3x = (3 × 20) = 605x = (5 × 20) = 1007x = (7 × 20) =140now the area of triangular plot is

area = √s(s-a)(s-b)(s-c)=√150(150-60)(150-100)(150-140)=1500√3m^2

Length of a field = 114mBreadth = 114/6 m= 19m

Perimeter of the rectangle= 2( Length + Breadth)= 2( 114 + 19)= 2 x 133= 266 m

C) -3 is the correct answer

option B ) 1/4 is the correct answer

c) is the right answer of the following

option B is correct answer

option (a) is the right answer

d is the correct answer of this question

c) is the right answer of the following

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Given- ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

To Prove-PQRS is a rectangle

Construction,AC and BD are joined.

Proof,In ΔDRS and ΔBPQ,DS = BQ (Halves of the opposite sides of the rhombus)∠SDR = ∠QBP (Opposite angles of the rhombus)DR = BP (Halves of the opposite sides of the rhombus)Thus, ΔDRS ≅ ΔBPQ by SAS congruence condition.RS = PQ by CPCT --- (i)In ΔQCR and ΔSAP,RC = PA (Halves of the opposite sides of the rhombus)∠RCQ = ∠PAS (Opposite angles of the rhombus)CQ = AS (Halves of the opposite sides of the rhombus)Thus, ΔQCR ≅ ΔSAP by SAS congruence condition.RQ = SP by CPCT --- (ii)Now,In ΔCDB,R and Q are the mid points of CD and BC respectively.⇒ QR || BDalso,P and S are the mid points of AD and AB respectively.⇒ PS || BD⇒ QR || PSThus, PQRS is a parallelogram.also, ∠PQR = 90°Now,In PQRS,RS = PQ and RQ = SP from (i) and (ii)∠Q = 90°Thus, PQRS is a rectangle.

x+|x|>0

=>|x|>-x which is possible only when x>0

Domain: (0,∞)

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Number of apples which should have been sold to avoid either loss or profit = 21+4 = 25. Loss percent = (4/25)*100 = 16%

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Length of Reactangular field= 1.2 km = 1200 m

Breadth of Reactangular field= 400 m

Ratio of Length to Breadth= 1200/400= 3/1= 3:1

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