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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.

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

Question Number 51095    Answers: 0   Comments: 5

Question Number 51088    Answers: 2   Comments: 1

Question Number 51077    Answers: 0   Comments: 0

$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 51075    Answers: 0   Comments: 0

$$ \\ $$

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.

$${Two}\:{similar}\:{ball}\:{of}\:{mass} \\ $$$${m}\:{attached}\:{by}\:\:{silk}\:{thread} \\ $$$${of}\:{length}\:{a}\:\:{and}\:{carry} \\ $$$${similar}\:{charge}\:\:{q}.{assume}\:\theta\:{is} \\ $$$${small}\:{enough}\:{that} \\ $$$${tan}\theta\approx{sin}\theta\:{to}\:{this} \\ $$$${approximation},{show}\: \\ $$$${that}\:\:\:\:{X}=\left(\frac{{qa}}{\mathrm{2}\pi\varepsilon_{\mathrm{0}} {mg}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \: \\ $$$${where}\:\:{X}\:{is}\:{distance}\:{of} \\ $$$${separation}. \\ $$

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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