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Question Number 16053 Answers: 0 Comments: 1

number of positive integers a and b and c satisfying a^b^c b^c^a c^a^b =5abc

$${number}\:{of}\:{positive}\:{integers}\:\boldsymbol{{a}}\:\boldsymbol{{and}}\:\boldsymbol{{b}}\:\boldsymbol{{and}}\:\boldsymbol{{c}}\:\boldsymbol{{satisfying}} \\ $$$$\boldsymbol{{a}}^{\boldsymbol{{b}}^{\boldsymbol{{c}}} } \boldsymbol{{b}}^{\boldsymbol{{c}}^{\boldsymbol{{a}}} } \boldsymbol{{c}}^{\boldsymbol{{a}}^{\boldsymbol{{b}}} } =\mathrm{5}\boldsymbol{{abc}} \\ $$

Question Number 16047 Answers: 0 Comments: 1

number of real solution of equation a^(2008) + b^(2008) = 2008ab−2006

$${number}\:{of}\:{real}\:{solution}\:{of}\:{equation}\:\:\:\:{a}^{\mathrm{2008}} \:+\:\:{b}^{\mathrm{2008}} \:\:=\:\mathrm{2008}{ab}−\mathrm{2006}\:\:\:\:\:\:\:\:\: \\ $$

Question Number 16078 Answers: 0 Comments: 0

Question Number 16044 Answers: 1 Comments: 1

A particle is projected horizontally with speed u from point A, which is 10 m above the ground. If the particle hits the inclined plane perpendicularly at point B. [g = 10 m/s^2 ] Find horizontal speed with which the particle was projected.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{horizontally} \\ $$$$\mathrm{with}\:\mathrm{speed}\:{u}\:\mathrm{from}\:\mathrm{point}\:{A},\:\mathrm{which}\:\mathrm{is}\:\mathrm{10} \\ $$$$\mathrm{m}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{hits} \\ $$$$\mathrm{the}\:\mathrm{inclined}\:\mathrm{plane}\:\mathrm{perpendicularly}\:\mathrm{at} \\ $$$$\mathrm{point}\:{B}.\:\left[{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right] \\ $$$$\mathrm{Find}\:\mathrm{horizontal}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{the} \\ $$$$\mathrm{particle}\:\mathrm{was}\:\mathrm{projected}. \\ $$

Question Number 16079 Answers: 1 Comments: 1

An open flask contains air at 27°C. To what temperature it must be heated to expel one-fourth of the air?

$$\mathrm{An}\:\mathrm{open}\:\mathrm{flask}\:\mathrm{contains}\:\mathrm{air}\:\mathrm{at}\:\mathrm{27}°\mathrm{C}.\:\mathrm{To} \\ $$$$\mathrm{what}\:\mathrm{temperature}\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:\mathrm{heated}\:\mathrm{to} \\ $$$$\mathrm{expel}\:\mathrm{one}-\mathrm{fourth}\:\mathrm{of}\:\mathrm{the}\:\mathrm{air}? \\ $$

Question Number 16041 Answers: 1 Comments: 0

If sides a, b, c of ΔABC are in H.P., prove that sin^2 ((A/2)), sin^2 ((B/2)), sin^2 ((C/2)) are in H.P.

$$\mathrm{If}\:\mathrm{sides}\:{a},\:{b},\:{c}\:\mathrm{of}\:\Delta{ABC}\:\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}., \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{A}}{\mathrm{2}}\right),\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{B}}{\mathrm{2}}\right),\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}. \\ $$

Question Number 16039 Answers: 0 Comments: 4

Question Number 16037 Answers: 0 Comments: 1

Question Number 16032 Answers: 0 Comments: 0

Question Number 16029 Answers: 0 Comments: 1

Question Number 16027 Answers: 0 Comments: 0

Question Number 16015 Answers: 1 Comments: 1

Two horses pull horizontally on ropes attached to a stump.The two forces F and T that they applied to the stump are such that the resultant R has a magnitude equal to F and makes an angle of 90° with F.Let F=1300N and R=1300N.Find the value of T.

$$\mathrm{Two}\:\mathrm{horses}\:\mathrm{pull}\:\mathrm{horizontally}\:\mathrm{on} \\ $$$$\mathrm{ropes}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{a}\:\mathrm{stump}.\mathrm{The} \\ $$$$\mathrm{two}\:\mathrm{forces}\:\mathrm{F}\:\mathrm{and}\:\mathrm{T}\:\mathrm{that}\:\mathrm{they} \\ $$$$\mathrm{applied}\:\mathrm{to}\:\mathrm{the}\:\mathrm{stump}\:\mathrm{are}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{resultant}\:\mathrm{R}\:\mathrm{has}\:\mathrm{a}\:\mathrm{magnitude} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{F}\:\mathrm{and}\:\mathrm{makes}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of} \\ $$$$\mathrm{90}°\:\mathrm{with}\:\mathrm{F}.\mathrm{Let}\:\mathrm{F}=\mathrm{1300N}\:\mathrm{and}\: \\ $$$$\mathrm{R}=\mathrm{1300N}.\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{T}. \\ $$

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

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

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