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Question Number 51700 Answers: 1 Comments: 1

Question Number 51694 Answers: 1 Comments: 0

Question Number 51691 Answers: 1 Comments: 0

Question Number 51710 Answers: 1 Comments: 0

solve the equation y = xy′+ (y′)^2

$${solve}\:{the}\:{equation} \\ $$$${y}\:=\:{xy}'+\:\left({y}'\right)^{\mathrm{2}} \\ $$

Question Number 51709 Answers: 2 Comments: 0

y′′=2y(y′) solve the equation

$${y}''=\mathrm{2}{y}\left({y}'\right) \\ $$$$ \\ $$$${solve}\:{the}\:{equation} \\ $$

Question Number 51680 Answers: 1 Comments: 2

Solve the equation: (1) z^3 + 1 − 10i = 0 (2) z^4 − i + 2 = 0

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}:\:\:\:\:\: \\ $$$$\:\left(\mathrm{1}\right)\:\:\:\:\:\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{1}\:−\:\mathrm{10i}\:\:=\:\:\mathrm{0} \\ $$$$\:\left(\mathrm{2}\right)\:\:\:\:\:\mathrm{z}^{\mathrm{4}} \:−\:\mathrm{i}\:+\:\mathrm{2}\:\:=\:\:\mathrm{0} \\ $$

Question Number 51673 Answers: 0 Comments: 1

Question Number 51658 Answers: 3 Comments: 7

Question Number 51649 Answers: 2 Comments: 0

It takes man A 6 days to a piece of work. It takes man B 4 days to do the same piece of work. How many days will A and B take to do the same piece of work?

$$\mathrm{It}\:\mathrm{takes}\:\mathrm{man}\:\mathrm{A}\:\mathrm{6}\:\mathrm{days}\:\mathrm{to}\:\mathrm{a}\:\mathrm{piece}\:\mathrm{of} \\ $$$$\mathrm{work}.\:\mathrm{It}\:\mathrm{takes}\:\mathrm{man}\:\mathrm{B}\:\mathrm{4}\:\mathrm{days}\:\mathrm{to}\:\mathrm{do}\: \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{piece}\:\mathrm{of}\:\mathrm{work}.\:\mathrm{How}\:\mathrm{many} \\ $$$$\mathrm{days}\:\mathrm{will}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{take}\:\mathrm{to}\:\mathrm{do}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{piece}\:\mathrm{of}\:\mathrm{work}? \\ $$

Question Number 51643 Answers: 2 Comments: 0

If ∣z∣ = 1, prove that ((z − 1)/(z^− − 1)) (z ≠ 1) is a pure imaginary

$$\mathrm{If}\:\:\:\mid\mathrm{z}\mid\:=\:\mathrm{1},\:\:\:\:\mathrm{prove}\:\mathrm{that}\:\:\:\:\:\frac{\mathrm{z}\:−\:\mathrm{1}}{\overset{−} {\mathrm{z}}\:−\:\mathrm{1}}\:\:\:\:\:\:\left(\mathrm{z}\:\neq\:\mathrm{1}\right)\:\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{pure}\:\mathrm{imaginary} \\ $$

Question Number 51637 Answers: 2 Comments: 0

An object 60cm tall is placed on the principal axis of a convex lens and its 30cm tall image is formed on the screen placed at a distance of 40cm from the object.What is the focal length of the lens?

$${An}\:{object}\:\mathrm{60}{cm}\:{tall}\:{is}\:{placed}\:{on} \\ $$$${the}\:{principal}\:{axis}\:{of}\:{a}\:{convex}\:{lens} \\ $$$${and}\:{its}\:\mathrm{30}{cm}\:{tall}\:{image}\:{is}\:{formed} \\ $$$${on}\:{the}\:{screen}\:{placed}\:{at}\:{a}\:{distance} \\ $$$${of}\:\mathrm{40}{cm}\:{from}\:{the}\:{object}.{What}\:{is} \\ $$$${the}\:{focal}\:{length}\:{of}\:{the}\:{lens}? \\ $$

Question Number 51636 Answers: 1 Comments: 4

Question Number 51644 Answers: 0 Comments: 0

Question Number 51630 Answers: 1 Comments: 4

Question Number 51628 Answers: 0 Comments: 2

Question Number 51621 Answers: 0 Comments: 6

Question Number 51617 Answers: 1 Comments: 2

A=lim_(x→0) ((e^x −e^(−x) −2x)/x^3 )(without using L′hopital rule)

$$\mathrm{A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} −{e}^{−{x}} −\mathrm{2}{x}}{{x}^{\mathrm{3}} }\left({without}\:\right. \\ $$$$\left.{using}\:{L}'\mathrm{hopital}\:\mathrm{rule}\right) \\ $$$$ \\ $$

Question Number 51614 Answers: 2 Comments: 0

Prove that length of lactus rectum when directrix are parallel to y−axis is y_1 −y_2 =((2b^2 )/a)

$${Prove}\:{that}\:{length}\:{of} \\ $$$${lactus}\:{rectum}\:{when}\: \\ $$$${directrix}\:{are}\:{parallel}\:{to} \\ $$$${y}−{axis}\:{is}\: \\ $$$${y}_{\mathrm{1}} −{y}_{\mathrm{2}} =\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}} \\ $$

Question Number 51613 Answers: 2 Comments: 0

Prove that the perpendicilar tangent to the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 meets on the circle x^2 +y^2 =a^2 +b^2 .

$${Prove}\:{that}\:{the}\:{perpendicilar} \\ $$$${tangent}\:{to}\:{the}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:{meets}\:{on}\:{the} \\ $$$${circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} . \\ $$

Question Number 51612 Answers: 2 Comments: 0

Find the equation of the ellipse with ecentricity (1/2) and the focus (2,1) Does the line x=3 touches ellipse.if so at what point?if line x=5 is the line of direction.

$${Find}\:{the}\:{equation}\:{of} \\ $$$${the}\:{ellipse}\:{with}\:{ecentricity} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{the}\:{focus}\:\left(\mathrm{2},\mathrm{1}\right) \\ $$$${Does}\:{the}\:{line}\:{x}=\mathrm{3}\:{touches} \\ $$$${ellipse}.{if}\:{so}\:{at}\:{what}\: \\ $$$${point}?{if}\:{line}\:{x}=\mathrm{5}\:{is}\:{the} \\ $$$${line}\:{of}\:{direction}. \\ $$

Question Number 51600 Answers: 1 Comments: 1

solve (sinθ)Z^2 −i(cosθ)Z+(1/4) sinθ=0

$$\mathrm{solve} \\ $$$$\left(\mathrm{sin}\theta\right)\mathrm{Z}^{\mathrm{2}} −\mathrm{i}\left(\mathrm{cos}\theta\right)\mathrm{Z}+\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{sin}\theta=\mathrm{0} \\ $$

Question Number 51594 Answers: 0 Comments: 0

Question Number 51592 Answers: 0 Comments: 4

Question Number 51590 Answers: 2 Comments: 1

The line y=mx+c touches ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 prove that the foot of perpendicular from focus into this line lie on auxillary circle x^2 +y^2 =a^2

$${The}\:{line}\:{y}={mx}+{c}\:{touches} \\ $$$${ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${prove}\:{that}\:{the}\:{foot}\:{of}\: \\ $$$${perpendicular}\:{from} \\ $$$${focus}\:{into}\:{this}\:{line}\:{lie}\:{on} \\ $$$${auxillary}\:{circle}\: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$

Question Number 51588 Answers: 1 Comments: 0

The tangent at P to an ellipse meets directrix at Q prove that the line joining the corresponding focus to P and Q are perpendicular

$${The}\:{tangent}\:{at}\:{P}\:\:{to}\:{an}\:{ellipse} \\ $$$${meets}\:{directrix}\:{at}\:{Q} \\ $$$${prove}\:{that}\:{the}\:{line} \\ $$$${joining}\:{the}\:{corresponding} \\ $$$${focus}\:{to}\:{P}\:{and}\:{Q}\:{are} \\ $$$${perpendicular} \\ $$

Question Number 51605 Answers: 0 Comments: 0

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