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AB = 100Let AC = xLet BC = 100-xLet CD = hInΔEAD,EA = DCED = ACtan 45° = EA/ED1 = h/xx = hIn ΔDFBFB = CDDF = CBtan 30° = FB/DF1/√3 = h/100-x1/√3 = h/100-h100-h =√3h100-h = 1.732h2.732h = 100h = 100/2.732h = 36.603mh = 36.6m

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2)tan∅ = AB/BC

tan60° = 90/BC

√3 = 90/BC

BC×√3 = 90

BC = 90/√3

BC. = 90/1.73

BC = 52.023 m (approximate)___________________________________so ship 52.023 meter far from the light house.

79 - 16 = 63

sum of 5.39 and 8.065.39+8.06 = 13.45ow subtraction16-13.45 = 2.55

thank you mam

1. -402. 29 is the correct answer

thsnks

Given that, mth term=1/n and nth term=1/m.then ,let a and d be the first term and the common difference of the A.P.so a+(m-1)d=1/n...........(1) and a+(n-1)d=1/m...........(2).subtracting equation (1) by (2) we get,md-d-nd+d=1/n-1/m=>d(m-n)=m-n/mn=>d=1/mn. again if we put this value in equation (1) or (2) we get, a=1/mn.then, let A be the mnth term of the APa+(mn-1)d=1/mn+1+(-1/mn)=1 hence proved

1

2

at last why we put sin tita

or tan tita is wrong?

why please tell me the reason

ohh ok sir I understood

tnq

Sin(90-3theta)= sin(2theta+15°)90-3theta= 2 theta+155 theta = 75theta= 15°

3pi/4 since cosinv(-x) = pi - cosinv(x)

If the principal value of cos−1 x is θ then we know, 0 ≤ θ ≤ π.

Therefore, If the principal value of cos−1 (-1/√2) be θ then cos−1 (-1/√2) = θ

⇒ cos θ = (-1/√2) = cos π/4 = cos (π - π/4) [Since, 0 ≤ θ ≤ π]

Therefore, the principal value of cos−1 (-1/√2) is π - π/4 = 3π/4

Given that, mth term=1/n and nth term=1/m.then ,let a and d be the first term and the common difference of the A.P.so a+(m-1)d=1/n...........(1) and a+(n-1)d=1/m...........(2).subtracting equation (1) by (2) we get,md-d-nd+d=1/n-1/m=>d(m-n)=m-n/mn=>d=1/mn. again if we put this value in equation (1) or (2) we get, a=1/mn.then, let A be the mnth term of the APa+(mn-1)d=1/mn+1+(-1/mn)=1 hence proved.

The process ofcentrifugationis used in dairies to separate cream from milk .Centrifugationis a method for separating the suspended particles of a substance from a liquid in which the mixture is rotated at a high speed in a centrifuge machine.

Chromatography is the separation technique which can be use to separate dyes in the black ink . This could be possible with the whatman’s filter paper.

There are way in separating mixtures of chlorides salts such as that of sodium chloride and ammonium chloride. It can be done by crystallization, filtration orsublimation. The easiest method among the three is separation bysublimation. Ammonium chloride sublimes uponheating.

There are way in separating mixtures of chlorides salts such as that of sodium chloride and ammonium chloride. It can be done by crystallization, filtration orsublimation. The easiest method among the three is separation bysublimation. Ammonium chloride sublimes uponheating.

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a cyclist travel 0.9 km in 3 minute . find his speed in km per hour

इस समीकरण में एक्स की घात 2 का गुणांक माइनस वन है

3x²-8x+2k+1 a=3, b=-8 ,c=2k+1suppose the zeroes of the equation are α and β . by the given condition α=7βas we know , α+β=-b/a7β+β=-(-8)/38β=8/3β=8/3×1/8=1/3αβ=c/a7β×β=2k+1/37×1/3×1/3=2k+1/37/9=2k+1/321/9=2k+121/9-1=2k12/9=2k4/3=2k4/6=k2/3=k

Out of two friends, one girl say, Savita's bday can be any day of the year. Now, Hamida's bday can also be on any day out of 365 days in the Year.

We assume that these 365 outcomes are equally likely.(i) If Hamida's bday is different from savita's, the no. Of favorable outcomes for her bday is 365-1= 364

So, P(Hamida' bday is diffrent from Savita's bday)=364/365

(ii) P(Savita and Harmida have the same bday)= 1-P(Both have diffrent bdays)=1-(364/365) [ using P(not event)=1-P(E)]=1/365

l= 140 cm b=80 cm therefore Area =l×B=140 ×80 =11200 Sq.cmHence ,11200 Sq. . cm is the area of required hardboard.

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sorry i didn't understand math in hidi language. please like

SinA/secA+tanA-1+cosA/cosecA+cotA-1=sinA/(1/cosA+sinA/cosA-1)+cosA/(1/sinA+cosA/sinA-1)=sinA/{(1+sinA-cosA)/cosA}+cosA/{(1+cosA-sinA)/sinA}=sinAcosA/(1+sinA-cosA)+sinAcosA/(1+cosA-sinA)=sinAcosA[(1+cosA-sinA+1+sinA-cosA)/(1+sinA-cosA)(1+cosA-sinA)]=2sinAcosA/(1+sinA-cosA+cosA+sinAcosA-cos²A-sinA-sin²A+sinAcosA)=2sinAcosA/{1+2sinAcosA-(sin²A+cos²A)}=2sinAcosA/(1+2sinAcosA-1)=2sinAcosA/2sinAcosA=1 (Proved)

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suppose 4000 at rate of x so rate for 5000 is x-2 both are for 3 years total interest is 1050 (4000×x×3)/100 + (5000(x+2)3)/100 = 1050 120x + 150x + 300 = 1050 270x = 750 x = 27/75 = 9/25 = 0.36% and second rate is 2.36%

Cosxcos2xcos3x=or, (2cosxcos3x)(2cos2x)=1or, (cos4x+cos2x)(2cos2x)=1or, 2cos4xcos2x+(2cos²2x-1)=0or, 2cos4xcos2x+cos4x=0or, cos4x(2cos2x+1)=0Either, cos4x=0or, 4x=(2n+1)π/2or, x=(2n+1)π/8Or, 2cos2x+1=0or, 2cos2x=-1or, cos2x=-1/2or, cos2x=cos(-π/3)or, cos2x=cosπ/3 [neglecting the negative sign for cos]or, 2x=2nπ⁺₋π/3or, x=nπ⁺₋π/6(⁺₋ means plus-minus)

here the amount is 882 the principle is 800 and time is 2 years , let r be the rate.

so, A = p(1+r)ⁿ=> 882 = 800(1+r)²=> (1+r)² = 88/80 = 11/10 = 1.1=> 1+r = √1.1 = 1.048

so, r = 0.048 or 4.8%