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Let k be ANY value of n

Let Pk be proposition that

1 / ( 1•3) + 1 / (3•5) + 1 / (5•7) -----1 / [ (2k - 1)(2k + 1) ] = k / ( 2k + 1 )

Have to show that P1 is true and P ( k + 1 ) is true

Consider P1--------------------LHS = 1/3RHS = 1/3Thus P1 is true

Consider P ( k + 1 )-----------------------------LHS1/( 1•3) + 1/(3•5) + 1/(5•7) ----------1 / [ ( 2k + 1 ) ( 2k + 3 ) ]

RHSk / ( 2k + 1 ) + 1 / [ ( 2k + 1 ) ( 2k + 3 ) ]

Have to show that P ( k + 1 ) is true :-

2k^2 + 3 k + 1------------------------------------( 2 k + 3 ) ( 2 k + 1 )

( 2k + 1 ) ( k + 1 )----------------------------( 2 k + 3 ) ( 2 k + 1 )

k + 1-----------2 k + 3

P1 is true and P (k + 1 ) is trueThus true for all kThus P n is true

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64 is the right answer

64 is the right answer

hence by principal of mathematical induction. p(n) is true for a. n => N

Answer:935

The least common multiple of 38 and 57 is 114 and the multiple which is between 900 and 1000 is 912.

Now, 912 + 23 i,e 935 lies between 900 and 1000 and when divided by 38 and 57 leaves in each case 23 as the remainder.

Therefore, 935 is number required.

Area of circle=πr2πr2

Area of rectangle=l×b=2×r×xl×b=2×r×x

A=2rx+12πr2A=2rx+12πr2

=r[10−(π+2)r]+12πr2=r[10−(π+2)r]+12πr2

=10r−(12=10r−(12π+2)r2π+2)r2

For maximum areadAdrdAdr=0=0andd2Adr2d2Adr2is -ve.

⇒10−(π+4)r=0⇒10−(π+4)r=0

(π+4)r=10(π+4)r=10

r=10π+4

total water 120Lused 5/8 of the water that is (5/8)× 120= 75L is used .So remaining water is 120- 75= 45L

There is 120 L of water So, 5/8 of water is used soRemaining water = 5/8 × 120 L = 75 L

64/112= -32/56= -16/28= -8/14= -4/7

AB = AC (Sides opposite to equal angle are equal)Subtracting BM from both sides, we getAB – BM = AC – BM⟹AB – BM = AC – CN (∵BM =CN)⟹AM =AN

∴∠AMN =∠ ANM (Angles opposite to equal sides are equal)Now, in ∆ABC,∠A+ ∠B+ ∠C=180°----(1)(Angle Sum Property of triangle)

Again In ∆AMN,

∠A + ∠AMN + ∠ ANM =180°---(2)Angle Sum Property of triangle

From (1) and (2), we get

∠B+ ∠C= ∠ AMN + ∠ ANM 2∠B = 2∠ AMN∠B = ∠ AMN

Since, ∠B and ∠ AMN are corresponding angles.

∴ MN ‖ BC

[57/x+y]+[6/x-y]=5 ......(1)

[57*(1/x+y)]+[6*(1/x-y)]=5

[38/x+y]+[21/x-y]=9 ......(2)

[38*(1/x+y)]+[21*(1/x-y)]=9

Let 1/x+y = A & 1/x-y = B

therefore,

57A+6B=5 .......(3)

38A=21B=9 .......(4)

Now,

(3)*7 & (4)*2

=

399A+42B=35

76A+42B=18

(-) (-) (-)

___________

323A = 17

A=17/323

A=1/19 .....(5)

put (5)in (4)=

38*(1/19)+21B=9

2+21B=9

21B=9-2

21B=7

B=7/21

B=1/3 ....(6)

1/x+y = A & 1/x-y = B=

x+y=1/A=19

x+y=19 ....(7)

x-y=1/B=3

x-y=3 ....(8)

Add (7)+(8)=

x+y=19

x -y= 3

______

2x = 22

x=11 &

y=8

(1) 25*t^-4/5^-3*10*t^-8 = (5)^2*t^-4/5^-3*(5*2)*t^-8 = (5)^[2+2]*t^[-4+8]/2 = 625t^4/2

(2) 3^-5 * 10^-5*125/5^-7*6^-5= 3^-5*(5*2)^-5*5^3/5^-7*(3*2)^-5= 3^[-5+5]*5^[-5+3+7]*2^[-5+5]= 1*5^5*1= 3125

Here Height is 24m

Base IS 7 m

HYPOTNUSE IS LENGTH OF Ladder

BY PYTHAGOROUS Theorem

HYPOTNUSE^2 = BASE^2 + HEIGHT^2

HYPOTNUSE^2 = 7^2+24^2

HYP^2 =576+49

HYP^2 = 625

HYP = UNDER ROOT OF 625= 25

a pole 81m High broke at a point but did not seperate. its top touched the ground 27m from its base. at what height above the ground did the pole

(5×5×5×6×6×6×6×625)/(3×3×3×3×10×10×10×10×10) = (2×2×2×2×625)/(2×2×2×10×10) = (625×2)/(2×2×5×5) = 25/2

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12*(26/100)*(5/78)*(38/100)*(7/152)*10000= 12*(5/300)*(7/400)*10000= 12*(5/3)*(7/4)= 12*(35/12)= 35

(2) is correct option

a= 0.06b=1230c=0.00000078d=600e=100f=100g=0.0063h=0.0006i=10j=1000k=79.1l=3.2

500 + 30 + 26/100 + 9/1000

In decimal form

530 + 0.26 + 0.009

530.269

Let X be the common multipleHence 5x+3x=648x=64x=8litresHence water will be 3x=3*8=24litres

no

answer is 128/3

thanks

It is line segment PQ.

Distance between two succesive pole = (268.51/11) = 24.41m24metre 41centimetre

Distance between two succesive pole = (268.51/11) = 24.41m= 24m 41cm

distance between two successive poles= 268.51÷11=24.4124metres 41 centimetres

distance between two successive poles=268.51÷11=24.41

Distance between each successive two poles = 268.51/11 m = 24.41m

24.41 is the correct answer of the given question is

start with y =(x-1)/(x+1)

replace x and y

x = (y-1)/(y+1)

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1/2 = 0.51/4 = 0.251/8 = 0.1251/16 = 0.0625So their sum is 0.9375

decimal form

4/5 = 0.8

3/4= 0.7

9/1000= 0.009

7xy-8xy is the answer

-15xy is the answer when we substract -8xy from 7xy

7xy-(-8xy)7xy+8xy15xy

7xy - 8xy = -xy

15xy is the correct answer of the given question

15xy is the correct answer

7xy-(-8xy) = 15xy

3/4=0.75 is the answer

method

24 is the correct answer

24 will be correct answers

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Applying Pythagoras theorem...AC^2=AB^2+BC^225^2-20^2=BC^2sqrt(625-400)=BCsqrt(225)=BC15 m=BC-is the distance

A| | | | B---------------------CAC=length of ladder=10mAB=height of window=8maccording to pathagorous theorem(H)^2=(P)^2+(B)^2(AC)^2=(AB)^2+(BC)^2(10)^2-(8)^2=(BC)^2100-64=(BC)^236=(BC)^2√36=BC6=BCdist of the foot of ladder from the base of wall is 6 m

using Pythagoras theorem we know that a square is equal to b square + c square 10 square is equal to 8 8 square + x square so x is equal to 6 metre