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Question Number 16014 Answers: 1 Comments: 0

A cirlce is drawn to touch the sides of a triangle whose sides are 12cm,10cm,and 9cm. Find the radius of the circle.

$$\mathrm{A}\:\mathrm{cirlce}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to}\:\mathrm{touch}\:\mathrm{the} \\ $$$$\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{whose}\:\mathrm{sides}\:\mathrm{are} \\ $$$$\mathrm{12cm},\mathrm{10cm},\mathrm{and}\:\mathrm{9cm}.\:\mathrm{Find}\:\mathrm{the}\: \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}. \\ $$

Question Number 16012 Answers: 0 Comments: 2

prove that when light travels through a triangular glass prism of angle A the refractive index (n) is given by n=sin (((A+D_(min) )/2))/sin ((A/2))

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{when}\:\mathrm{light}\:\mathrm{travels}\: \\ $$$$\mathrm{through}\:\mathrm{a}\:\mathrm{triangular}\:\mathrm{glass}\:\mathrm{prism}\: \\ $$$$\mathrm{of}\:\mathrm{angle}\:\mathrm{A} \\ $$$$\mathrm{the}\:\mathrm{refractive}\:\mathrm{index}\:\left(\mathrm{n}\right)\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{by} \\ $$$$\mathrm{n}=\mathrm{sin}\:\left(\frac{\mathrm{A}+\mathrm{D}_{\mathrm{min}} }{\mathrm{2}}\right)/\mathrm{sin}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right) \\ $$

Question Number 16000 Answers: 2 Comments: 0

Question Number 15999 Answers: 1 Comments: 0

Solve simultaneously 2xy = x + y 5xz = 6z − 2x 3yz = 3y + 4z

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{2xy}\:=\:\mathrm{x}\:+\:\mathrm{y} \\ $$$$\mathrm{5xz}\:=\:\mathrm{6z}\:−\:\mathrm{2x} \\ $$$$\mathrm{3yz}\:=\:\mathrm{3y}\:+\:\mathrm{4z} \\ $$

Question Number 15993 Answers: 1 Comments: 1

Question Number 15987 Answers: 1 Comments: 1

Question Number 15982 Answers: 0 Comments: 3

Question Number 15975 Answers: 1 Comments: 1

Solve sin 2x + cos 2x =0, for −π<x<π.

$${Solve}\:{sin}\:\mathrm{2}{x}\:+\:{cos}\:\mathrm{2}{x}\:=\mathrm{0}, \\ $$$${for}\:−\pi<{x}<\pi. \\ $$

Question Number 15972 Answers: 0 Comments: 1

∫tan^2 x sec x dx

$$\int\mathrm{tan}^{\mathrm{2}} \mathrm{x}\:\mathrm{sec}\:\mathrm{x}\:\mathrm{dx}\: \\ $$

Question Number 15971 Answers: 0 Comments: 1

Question Number 15969 Answers: 0 Comments: 17

Question Number 15967 Answers: 1 Comments: 1

prove that (4cos9° − 3)(4cos27° − 3) = tan9°

$$\mathrm{prove}\:\mathrm{that} \\ $$$$\left(\mathrm{4cos9}°\:−\:\mathrm{3}\right)\left(\mathrm{4cos27}°\:−\:\mathrm{3}\right)\:=\:\mathrm{tan9}° \\ $$

Question Number 15961 Answers: 0 Comments: 5

Path of a projectile as seen from another projectile is a (1) Straight line (2) Parabola (3) Ellipse (4) Hyperbola

$$\mathrm{Path}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile}\:\mathrm{as}\:\mathrm{seen}\:\mathrm{from}\:\mathrm{another} \\ $$$$\mathrm{projectile}\:\mathrm{is}\:\mathrm{a} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Straight}\:\mathrm{line} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Parabola} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Ellipse} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Hyperbola} \\ $$

Question Number 15955 Answers: 0 Comments: 2

The motion of a particle moving along x-axis is represented by the equation (dv/dt) = 6 − 3v, where v is in m/s and t is in second. If the particle is at rest at t = 0, then (1) The speed of the particle is 2 m/s when the acceleration of particle is zero (2) After a long time the particle moves with a constant velocity of 2 m/s (3) The speed is 0.1 m/s, when the acceleration is half of its initial value (4) The magnitude of final acceleration is 6 m/s^2

$$\mathrm{The}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{along} \\ $$$${x}-\mathrm{axis}\:\mathrm{is}\:\mathrm{represented}\:\mathrm{by}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\frac{{dv}}{{dt}}\:=\:\mathrm{6}\:−\:\mathrm{3}{v},\:\mathrm{where}\:{v}\:\mathrm{is}\:\mathrm{in}\:\mathrm{m}/\mathrm{s}\:\mathrm{and}\:{t}\:\mathrm{is} \\ $$$$\mathrm{in}\:\mathrm{second}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest}\:\mathrm{at}\:{t}\:= \\ $$$$\mathrm{0},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{2}\:\mathrm{m}/\mathrm{s} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{particle}\:\mathrm{is} \\ $$$$\mathrm{zero} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{After}\:\mathrm{a}\:\mathrm{long}\:\mathrm{time}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{moves} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{2}\:\mathrm{m}/\mathrm{s} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{The}\:\mathrm{speed}\:\mathrm{is}\:\mathrm{0}.\mathrm{1}\:\mathrm{m}/\mathrm{s},\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{is}\:\mathrm{half}\:\mathrm{of}\:\mathrm{its}\:\mathrm{initial}\:\mathrm{value} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{The}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{final}\:\mathrm{acceleration} \\ $$$$\mathrm{is}\:\mathrm{6}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$

Question Number 15948 Answers: 1 Comments: 1

Question Number 15943 Answers: 0 Comments: 0

Solve: x(dy/dx) + 3y = 3x^x y^(2/3)

$$\mathrm{Solve}: \\ $$$$\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{3y}\:=\:\mathrm{3x}^{\mathrm{x}} \:\mathrm{y}^{\mathrm{2}/\mathrm{3}} \\ $$

Question Number 15942 Answers: 1 Comments: 0

Help. solve(x+2)dy/dx=3−2y/x

$$\mathrm{Help}. \\ $$$$\mathrm{solve}\left(\mathrm{x}+\mathrm{2}\right)\mathrm{dy}/\mathrm{dx}=\mathrm{3}−\mathrm{2y}/\mathrm{x} \\ $$

Question Number 15932 Answers: 1 Comments: 1

A particle is projected from the foot of an inclined plane having inclination 45°, with the velocity u at an angle θ (> 45°) with the horizontal in a vertical plane containing the line of greatest slope through the point of projection. Find the value of tan θ if the particle strikes the plane (i) Horizontally (ii) Normally

Question Number 15939 Answers: 0 Comments: 1

Question Number 15927 Answers: 0 Comments: 5

Question Number 15919 Answers: 0 Comments: 1

multiply 3x+4y+5x−8y

$${multiply}\:\mathrm{3}{x}+\mathrm{4}{y}+\mathrm{5}{x}−\mathrm{8}{y} \\ $$

Question Number 15917 Answers: 1 Comments: 1

Question Number 15916 Answers: 2 Comments: 2

Question Number 15908 Answers: 1 Comments: 7

Question Number 15906 Answers: 1 Comments: 0

If λ_1 and λ_2 are respectively the wavelengths of the series limit of Lyman and Balmer series of Hydrogen atom, then the wavelength of the first line of the Lyman series of the H-atom is (1) λ_1 − λ_2 (2) (√(λ_1 λ_2 )) (3) ((λ_2 − λ_1 )/(λ_1 λ_2 )) (4) ((λ_1 λ_2 )/(λ_2 − λ_1 ))

$$\mathrm{If}\:\lambda_{\mathrm{1}} \:\mathrm{and}\:\lambda_{\mathrm{2}} \:\mathrm{are}\:\mathrm{respectively}\:\mathrm{the} \\ $$$$\mathrm{wavelengths}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}\:\mathrm{limit}\:\mathrm{of}\:\mathrm{Lyman} \\ $$$$\mathrm{and}\:\mathrm{Balmer}\:\mathrm{series}\:\mathrm{of}\:\mathrm{Hydrogen}\:\mathrm{atom}, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{wavelength}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{line}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{Lyman}\:\mathrm{series}\:\mathrm{of}\:\mathrm{the}\:\mathrm{H}-\mathrm{atom}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\lambda_{\mathrm{1}} \:−\:\lambda_{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\sqrt{\lambda_{\mathrm{1}} \lambda_{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\frac{\lambda_{\mathrm{2}} \:−\:\lambda_{\mathrm{1}} }{\lambda_{\mathrm{1}} \lambda_{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:\frac{\lambda_{\mathrm{1}} \lambda_{\mathrm{2}} }{\lambda_{\mathrm{2}} \:−\:\lambda_{\mathrm{1}} } \\ $$

Question Number 15904 Answers: 2 Comments: 1

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