Explore topic-wise InterviewSolutions in .

For this you should know the formula of inverse trigonometric functions .

So sin^-1(2x/1+x^2)

Is the formula for Tan^-1(x)

Diff.

Dy/Dx= d(Tan^-1(x))/dx

So .. 1/1+x^2 is the answer

Hope it helps

Let roots be X=-1 and y=-2hencesum of roots=X+y=-1-2=-3=-b/a=-3/phence p=1andproduct of roots =-1*-2=2=c/a=q/pas p=1hence2=q/1hence q=2

If A and G be A.M and G.m, respectively between 2 positive numbers, prove that one of the numbers is

=> A ± √[ A² - G² ]

Soln : Let x and y be the two numbers. Then --

....... x + y = 2A .......... (1) and xy = G² ......... (2)

From (2) we get --

=> x = G²/y . Substitute this value of x in eqn (1). We get ---

( G²/y ) + y = 2A

=> G² + y² - 2 A y = 0

=> y² - 2 A y + G² = 0 . this is a quadratic eqn in y . Hence

2 A ± √[ 4 A² - 4 G² ]

=> y = A ± √[ A² - G² ]

=> y = A ± √[ ( A + G ) ( A - G ) ]

Like my answer if you find it useful!

Integrate each term henceintegral (x^2)dx-5 integral(X)dx+7 integral(dx)=x^3/3-5x^2/2+7x+C

fencing means we have to find perimeter of rectangleso 2×(44+22)= 2× 66 = 132therefore cost of fencing is 132× 45= Rs 5940

it's not 45..it's 4.5

BTW thnx bro

sum will be double when I=PI=PRT/100I=I×12.5×T/100T=100/12.5T=8yrs

Please hit the like button if this helped you

thanks sir

ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC .

Now, ∆ABE and ∆AEC ∠AEB = ∠ACE = 90° AE is common side of both triangles , AB = AC [ all sides of equilateral triangle are equal ]From R - H - S congruence rule , ∆ABE ≡ ∆ACE ∴ BE = EC = BC/2

Now, from Pythagoras theorem , ∆ADE is right angle triangle ∴ AD² = AE² + DE² ------(1)∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ------(2)

From equation (1) and (2) AB² - AD² = BE² - DE² = (BC/2)² - (BE - BD)² = BC²/4 - {(BC/2) - (BC/3)}² = BC²/4 - (BC/6)² = BC²/4 - BC²/36 = 8BC²/36 = 2BC²/9 ∵AB = BC = CA So, AB² = AD² + 2AB²/9 9AB² - 2AB² = 9AD²Hence, 9AD² = 7AB²

Ty☺️

Given r = 4.2 m, h = 38 mVolume of well = pi*r*r*h= 22/7*4.2*4.2*38= 22*.6*4.2*38= 21067. 2 m^3

Let rise in level of field = d

Then, Volume of Reactangular field= Volume of welll*b*d = 21067.2130*115*d = 21067.2d = 21067.2/(130*115)d = 162/115d = 1.4 m

Rise in level of field = 1.4 m

Interest = Rs. (81-72) = Rs. 9Let the time be t years.9 = (72×25×t)/(4×100)t=(9×400)/(72×25)=2 years

Given:ST || RQPS= 3 cmSR = 4cm

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

ar(∆PST) /ar(∆PRQ)= (PS)²/(PR)²

ar(∆PST) /ar(∆PRQ)= 3²/(PS+SR)²

ar(∆PST) /ar(∆PRQ)= 9/(3+4)²= 9/7²=9/49

Hence, the required ratio ar(∆PST) :ar(∆PRQ)= 9:49

bhoot hard bhai bhoot hard which is your class

b^2-4ac=5roots are unequal b^2-4ac= -5roots are irrational

only you solve and write the answer

Given: P is a point on the chord BC. such that AB = AP

To prove: CP = CQ

Solution:

in the triangle ABP, AB = AP

therefore ∠ABP = ∠APB........(1)

Angle subtended by the chord at the circumference are equal.

∠ABC = ∠AQC [angles subtended by the chord AC at circumference ]

∠ABP = ∠AQC ..........(2) [∠ABC = ∠ABP same angle]

∠APB = ∠CPQ ......(3)

from (1), (2) and (3)

∠CPQ = ∠AQC i.e. ∠CPQ = ∠PQC

Hence CP = CQ [sides opposite to equal angles are equal]

Like if you find it useful

Given :ABCD is a parallelogram. P is the mid point of BC and ∠BAP = ∠DAP

To prove :AD = 2 CD

Proof : Given, ∠BAP = ∠DAP∴ ∠1 = ∠BAP = 1/2 ∠A ...(1)

ABCD is a parallelogram,∴ AD || BC (Opposite sides of the parallelogram are equal)∠A + ∠B = 180° (Sum of adjacent interior angles is 180°)∴ ∠B = 180° – ∠A ...(2)

In ΔABP,∠1 + ∠2 + ∠B = 180°(Angle sum property)=> 1/2∠A + ∠2 + 180 - ∠A = 180 [Using equations (1) and (2)]=> ∠A - 1/2 ∠A = 0=> ∠A = 1/2 ∠A ...(3)From (1) and (2), we have∠1 = ∠2In ΔABP,∠1 = ∠2∴ BP = AB (In a triangle, equal angles have equal sides opposite to them)=> 1/2 BC = AB (P is the midpoint on BC)=> BC = 2AB⇒ AD = 2CD (Opposite sides of the parallelogram are equal)

(V3/2)^3 +(i/2)^3+3(V3/2)(i/2)[V3/2 + i/2) = = 3V3/8 +1i/8+3V3/8 +'i/8= 8i/8 =i

thank you so much, can you explain it step by step

can u help me with more come questions:

180-x+65+50= 180X= 115

180-x+65+50=180x=115

180-x+65+50=180x=115 is the right answer

90-3x = x90 = 4xx = 90/4 x = 45/2 no option is correct

wrongc option is correct

3x+x=180°4x=180°x=180/4x= 45°C option is correct

y = Acos(nx)+Bsin(nx)---(1)

differentiating wrt to x

y' = -nAsin(nx)+ nBcos(nx)

differentiating wrt to x

y" = -n²Acos(nx)-n²Bsin(nx) =-n²y { from 1}

y"+ n²y =0 hence proved

Please hit the like button if this helped you

16 articles cost = 72then, price of one article = 72÷16= 4.5

Then 30 articles will be of = 30*4.5 = 135

And at Rs. 207 the number of articles bought =207/4.5 = 46 articles

Like my answer if you find it useful!

Sin A= a/b= P/Hcosec A = b/a= H/Pas cosectheta= 1/ sin theta

what the ans?