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Question Number 51151 Answers: 1 Comments: 0

Find the equation of tangent to the ellipse x^2 +4y^2 =4 which are perpendicular to the line 2x−3y=1 ∗merry X−mas and happy new year∗

$${Find}\:{the}\:{equation}\:{of} \\ $$$${tangent}\:{to}\:{the}\:\:{ellipse} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} =\mathrm{4}\:{which}\:{are} \\ $$$${perpendicular}\:{to}\:{the}\: \\ $$$${line}\:\mathrm{2}{x}−\mathrm{3}{y}=\mathrm{1} \\ $$$$ \\ $$$$\ast{merry}\:{X}−{mas}\:{and}\:{happy}\:{new}\:{year}\ast \\ $$

Question Number 51150 Answers: 2 Comments: 0

from left hand sides prove that ((sinαsin β)/(cos α+cos β))=((2tan(α/2) tan (β/2))/(1−tan^2 (α/2)tan^2 (β/2)))

$${from}\:{left}\:{hand}\:{sides} \\ $$$${prove}\:{that} \\ $$$$\frac{{sin}\alpha\mathrm{sin}\:\beta}{\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta}=\frac{\mathrm{2tan}\frac{\alpha}{\mathrm{2}}\:\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}\mathrm{tan}^{\mathrm{2}} \:\frac{\beta}{\mathrm{2}}} \\ $$

Question Number 51148 Answers: 1 Comments: 0

Two hot cubes are at same temperature and both are cooled by forced convetion the cube are made from the same material but one has side of L metre and other one 2L metre. which cube is at quicker rate of cooling

$${Two}\:{hot}\:{cubes}\:{are}\:\:{at}\:{same}\: \\ $$$${temperature}\:{and} \\ $$$${both}\:{are}\:{cooled}\:{by}\:{forced}\: \\ $$$${convetion}\:{the}\:{cube}\:{are}\:{made} \\ $$$${from}\:{the}\:{same}\:{material} \\ $$$${but}\:{one}\:{has}\:{side}\:{of}\:{L}\:{metre} \\ $$$${and}\:{other}\:{one}\:\mathrm{2}{L}\:{metre}. \\ $$$${which}\:{cube}\:{is}\:{at}\:{quicker}\: \\ $$$${rate}\:{of}\:{cooling} \\ $$$$ \\ $$

Question Number 51141 Answers: 1 Comments: 0

Question Number 51134 Answers: 1 Comments: 1

Question Number 51122 Answers: 1 Comments: 1

∫ (dx/(tgx−(√(tgx))))=?

$$\int\:\:\:\:\frac{\boldsymbol{\mathrm{dx}}}{\boldsymbol{\mathrm{tgx}}−\sqrt{\boldsymbol{\mathrm{tgx}}}}=? \\ $$

Question Number 51107 Answers: 2 Comments: 1

A ball is bouncing down a flight of stairs.The coefficiate of restitution is e.The height of each step d and the ball descends one step at each bounce. After each bounce it rebounds to heigt h above the next lower step.The height h is/ large enough compare with width of a step that the empacts are effectively head on.show that h=(d/(1−e^2 ))

$${A}\:{ball}\:{is}\:{bouncing}\:{down} \\ $$$${a}\:{flight}\:{of}\:{stairs}.{The}\: \\ $$$${coefficiate}\:{of}\:{restitution} \\ $$$${is}\:{e}.{The}\:{height}\:{of}\:{each}\:{step} \\ $$$${d}\:{and}\:{the}\:{ball}\:{descends} \\ $$$${one}\:{step}\:{at}\:{each}\:{bounce}. \\ $$$${After}\:{each}\:{bounce}\:{it}\:{rebounds} \\ $$$${to}\:{heigt}\:\:{h}\:{above}\:{the}\:{next} \\ $$$${lower}\:{step}.{The}\:{height}\:{h}\:{is}/ \\ $$$${large}\:{enough}\:{compare}\:{with} \\ $$$${width}\:{of}\:{a}\:{step}\:{that} \\ $$$${the}\:{empacts}\:{are}\:{effectively}\: \\ $$$${head}\:{on}.{show}\:{that} \\ $$$${h}=\frac{{d}}{\mathrm{1}−{e}^{\mathrm{2}} } \\ $$$$ \\ $$

Question Number 51109 Answers: 0 Comments: 1

Question Number 51098 Answers: 1 Comments: 0

Two similar spherical bodies of radius R and 2R are initially are at same temperature.if they are kept to cool under the same condition.show qualitatively which of the two spherical body will cool faster.

Question Number 51095 Answers: 0 Comments: 5

Question Number 51088 Answers: 2 Comments: 1

Question Number 51077 Answers: 0 Comments: 0

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Question Number 51075 Answers: 0 Comments: 0

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Question Number 51115 Answers: 0 Comments: 0

Question Number 51069 Answers: 3 Comments: 0

If sinA+cos2A =(1/2) and cosA+sin2A=(1/3) , then find the value of sin3A.

$${If}\:\mathrm{sin}{A}+\mathrm{cos2}{A}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{and} \\ $$$$\mathrm{cos}{A}+\mathrm{sin2}{A}=\frac{\mathrm{1}}{\mathrm{3}}\:,\:{then}\:{find}\:{the}\:{value} \\ $$$${of}\:\mathrm{sin3}{A}. \\ $$

Question Number 52893 Answers: 1 Comments: 0

∫sin x×cos x dx

$$\int\mathrm{sin}\:{x}×\mathrm{cos}\:{x}\:{dx} \\ $$

Question Number 52891 Answers: 1 Comments: 10

Question Number 51047 Answers: 0 Comments: 5

Question Number 51028 Answers: 2 Comments: 0

f(z)=(√r) e^(i(θ/2)) f′(z)=...?

$${f}\left({z}\right)=\sqrt{{r}}\:{e}^{{i}\frac{\theta}{\mathrm{2}}} \\ $$$${f}'\left({z}\right)=...? \\ $$

Question Number 51005 Answers: 3 Comments: 3

Question Number 50996 Answers: 1 Comments: 8

Find the minimum value of f(x)= 9tan^2 θ+4cot^2 θ ?

$${Find}\:{the}\:{minimum}\:{value}\:{of} \\ $$$${f}\left({x}\right)=\:\mathrm{9tan}^{\mathrm{2}} \theta+\mathrm{4cot}^{\mathrm{2}} \theta\:? \\ $$

Question Number 50994 Answers: 2 Comments: 0

Two similar ball of mass m attached by silk thread of length a and carry similar charge q.assume θ is small enough that tanθ≈sinθ to this approximation,show that X=(((qa)/(2πε_0 mg)))^(1/3) where X is distance of separation.

Question Number 50993 Answers: 2 Comments: 0

Prove newtons law of cooling by stefan law

$${Prove}\:{newtons}\:{law}\:{of} \\ $$$${cooling}\:{by}\:{stefan}\:{law} \\ $$

Question Number 50991 Answers: 0 Comments: 1

gururaja chitra

$${gururaja}\:{chitra} \\ $$

Question Number 50987 Answers: 0 Comments: 2

(√(x + y + 9)) + (√(x − y + 8)) = 33^2 x > y x, y ∈ Z^+ (x^y + y^x ) mod (1000) = ?

$$\sqrt{{x}\:+\:{y}\:+\:\mathrm{9}}\:\:+\:\:\sqrt{{x}\:−\:{y}\:+\:\mathrm{8}}\:\:=\:\:\mathrm{33}^{\mathrm{2}} \\ $$$${x}\:>\:{y} \\ $$$${x},\:{y}\:\:\in\:\:\mathbb{Z}^{+} \\ $$$$\left({x}^{{y}} \:+\:{y}^{{x}} \right)\:\:{mod}\:\:\left(\mathrm{1000}\right)\:\:=\:\:? \\ $$

Question Number 50979 Answers: 3 Comments: 1

a)Normal to any point on the hyperbola XY=C meet the x−axis at A and tangents meets the y−axis at B.find the locus of the mid point of AB b)find the equation of assymptotes of (i)(x^2 /4)−(y^2 /5)=1 (ii)(((x−1)^2 )/(16))−(((y−3)^2 )/9)=1

$$\left.{a}\right){Normal}\:{to}\:{any}\:{point}\:{on} \\ $$$${the}\:{hyperbola}\:{XY}={C} \\ $$$${meet}\:{the}\:{x}−{axis}\:{at}\:{A} \\ $$$${and}\:{tangents}\:{meets} \\ $$$${the}\:{y}−{axis}\:{at}\:{B}.{find}\:{the} \\ $$$${locus}\:{of}\:{the}\:{mid}\:{point}\:{of}\:{AB} \\ $$$$\left.{b}\right){find}\:\:{the}\:{equation}\:{of}\: \\ $$$${assymptotes}\:{of} \\ $$$$\left({i}\right)\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{{y}^{\mathrm{2}} }{\mathrm{5}}=\mathrm{1} \\ $$$$\left({ii}\right)\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{16}}−\frac{\left({y}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1} \\ $$

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