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Question Number 22612 Answers: 2 Comments: 0

In the binomial expasion of (a − b)^5 , the sum of 2^(nd) and 3^(rd) term is zero, then (a/b) is

$${In}\:{the}\:{binomial}\:{expasion}\:{of}\:\left({a}\:−\:{b}\right)^{\mathrm{5}} , \\ $$$${the}\:{sum}\:{of}\:\mathrm{2}^{{nd}} \:{and}\:\mathrm{3}^{{rd}} \:{term}\:{is}\:{zero}, \\ $$$${then}\:\frac{{a}}{{b}}\:{is} \\ $$

Question Number 22582 Answers: 1 Comments: 2

A chain consisting of 5 links each of mass 0.1 kg is lifted vertically with a constant acceleration of 2 m/s^2 . The force of interaction (in newton) between the top link and the link immediately below it will be : Take g = 10 m/s^2 .

$$\mathrm{A}\:\mathrm{chain}\:\mathrm{consisting}\:\mathrm{of}\:\mathrm{5}\:\mathrm{links}\:\mathrm{each}\:\mathrm{of} \\ $$$$\mathrm{mass}\:\mathrm{0}.\mathrm{1}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{lifted}\:\mathrm{vertically}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{constant}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{2}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{The} \\ $$$$\mathrm{force}\:\mathrm{of}\:\mathrm{interaction}\:\left(\mathrm{in}\:\mathrm{newton}\right)\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{top}\:\mathrm{link}\:\mathrm{and}\:\mathrm{the}\:\mathrm{link}\:\mathrm{immediately} \\ $$$$\mathrm{below}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:: \\ $$$$\mathrm{Take}\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} . \\ $$

Question Number 22576 Answers: 1 Comments: 1

Question Number 22574 Answers: 1 Comments: 1

The system is initially at origin and is moving with a velocity (3 m/s) k^∧ . A force (120t newton) i^∧ acts on mass m_2 , where t is time in seconds. The man throws a light ball, at the instant when m_1 starts slipping on m_2 , with a velocity 5 m/s vertically up w.r.t himself. Taking the masses of blocks and man as 60 kg each and assuming that the man never slips on m_2 , find (a) The time at which man throws the ball and (b) The coordinates of the point where ball lands. Neglect the dimensions of the system. (g = 10 m/s^2 )

$$\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{initially}\:\mathrm{at}\:\mathrm{origin}\:\mathrm{and}\:\mathrm{is} \\ $$$$\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\left(\mathrm{3}\:\mathrm{m}/\mathrm{s}\right)\:\overset{\wedge} {{k}}.\:\mathrm{A} \\ $$$$\mathrm{force}\:\left(\mathrm{120}{t}\:\mathrm{newton}\right)\:\overset{\wedge} {{i}}\:\mathrm{acts}\:\mathrm{on}\:\mathrm{mass}\:{m}_{\mathrm{2}} , \\ $$$$\mathrm{where}\:{t}\:\mathrm{is}\:\mathrm{time}\:\mathrm{in}\:\mathrm{seconds}. \\ $$$$\mathrm{The}\:\mathrm{man}\:\mathrm{throws}\:\mathrm{a}\:\mathrm{light}\:\mathrm{ball},\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{instant}\:\mathrm{when}\:{m}_{\mathrm{1}} \:\mathrm{starts}\:\mathrm{slipping}\:\mathrm{on}\:{m}_{\mathrm{2}} , \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{5}\:\mathrm{m}/\mathrm{s}\:\mathrm{vertically}\:\mathrm{up}\:\mathrm{w}.\mathrm{r}.\mathrm{t} \\ $$$$\mathrm{himself}.\:\mathrm{Taking}\:\mathrm{the}\:\mathrm{masses}\:\mathrm{of}\:\mathrm{blocks} \\ $$$$\mathrm{and}\:\mathrm{man}\:\mathrm{as}\:\mathrm{60}\:\mathrm{kg}\:\mathrm{each}\:\mathrm{and}\:\mathrm{assuming} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{man}\:\mathrm{never}\:\mathrm{slips}\:\mathrm{on}\:{m}_{\mathrm{2}} ,\:\mathrm{find} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{The}\:\mathrm{time}\:\mathrm{at}\:\mathrm{which}\:\mathrm{man}\:\mathrm{throws}\:\mathrm{the} \\ $$$$\mathrm{ball}\:\mathrm{and} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{coordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{where} \\ $$$$\mathrm{ball}\:\mathrm{lands}.\:\mathrm{Neglect}\:\mathrm{the}\:\mathrm{dimensions}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{system}.\:\left({g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$

Question Number 22562 Answers: 0 Comments: 0

solve∫_0 ^1 ln(x)dx/x^2 −x−1 dx

$${solve}\overset{\mathrm{1}} {\int}_{\mathrm{0}} {ln}\left({x}\right){dx}/{x}^{\mathrm{2}} −{x}−\mathrm{1}\:{dx} \\ $$

Question Number 22561 Answers: 1 Comments: 0

Solve for all integer x,y ∈Z x^2 +y^2 =19

$$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{integer}}\:\boldsymbol{{x}},\boldsymbol{{y}}\:\in\boldsymbol{{Z}} \\ $$$$\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} =\mathrm{19} \\ $$

Question Number 22547 Answers: 1 Comments: 0

If α = (5/(2!3)) + ((5.7)/(3!3^2 )) + ((5.7.9)/(4!3^3 )) ,... then find the value of α^2 + 4α.

$$\mathrm{If}\:\alpha\:=\:\frac{\mathrm{5}}{\mathrm{2}!\mathrm{3}}\:+\:\frac{\mathrm{5}.\mathrm{7}}{\mathrm{3}!\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{5}.\mathrm{7}.\mathrm{9}}{\mathrm{4}!\mathrm{3}^{\mathrm{3}} }\:,...\:\mathrm{then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\alpha^{\mathrm{2}} \:+\:\mathrm{4}\alpha. \\ $$

Question Number 22545 Answers: 2 Comments: 0

∫(x^(1/2) /(x^(1/2) −x^(1/3) ))dx=

$$\int\frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }\mathrm{dx}= \\ $$$$ \\ $$

Question Number 22567 Answers: 3 Comments: 0

Question Number 22537 Answers: 0 Comments: 22

Mr. Ajfour, you are very good at solving difficult questions.How do you do that ? Please tell us something about yourself.

$${Mr}.\:{Ajfour},\:{you}\:{are}\:{very} \\ $$$${good}\:{at}\:{solving}\:{difficult} \\ $$$${questions}.{How}\:{do}\:{you} \\ $$$${do}\:{that}\:?\:{Please}\:{tell}\:{us}\: \\ $$$${something}\:{about}\:{yourself}. \\ $$$$ \\ $$

Question Number 22536 Answers: 1 Comments: 1

show that (((a+b+c)^2 )/(a^2 +b^2 +c^2 ))= ((cot (1/2)A+cot (1/2)B+cot (1/2)C)/(cot A+cot B+cot C)) please help

$${show}\:{that}\:\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }= \\ $$$$\frac{\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{A}+\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{B}+\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{C}}{\mathrm{cot}\:{A}+\mathrm{cot}\:{B}+\mathrm{cot}\:{C}} \\ $$$$ \\ $$$$ \\ $$$${please}\:{help} \\ $$

Question Number 22535 Answers: 0 Comments: 0

How to solve this homogeneous equation. Can u help me plz Q. solve (x sin (y/x))dy −(y sin^(−1) (y/x))dx=0

$$\mathrm{H}{ow}\:{to}\:{solve}\:{this}\:{homogeneous}\:{equation}.\:{Can}\:{u}\:{help}\:{me}\:{plz} \\ $$$${Q}.\:{solve}\:\left({x}\:{sin}\:\frac{{y}}{{x}}\right){dy}\:−\left({y}\:{sin}^{−\mathrm{1}} \:\frac{{y}}{{x}}\right){dx}=\mathrm{0} \\ $$

Question Number 22525 Answers: 0 Comments: 4

Question Number 22524 Answers: 0 Comments: 1

toutes lessolutions du systeme

$${toutes}\:{lessolutions}\:{du}\:{systeme} \\ $$

Question Number 22522 Answers: 0 Comments: 1

Happy Diwali Friends !! :)

$$\mathrm{H}{appy}\: \\ $$$${Diwali} \\ $$$$\left.{Friends}\:!!\::\right) \\ $$

Question Number 22518 Answers: 2 Comments: 1

Question Number 22517 Answers: 0 Comments: 0

Find the coefficient of x in the expansion of [(√(1 + x^2 )) − x]^(−1) in ascending power of x when ∣x∣ < 1.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$$$\mathrm{of}\:\left[\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:−\:{x}\right]^{−\mathrm{1}} \:\mathrm{in}\:\mathrm{ascending}\:\mathrm{power} \\ $$$$\mathrm{of}\:{x}\:\mathrm{when}\:\mid{x}\mid\:<\:\mathrm{1}. \\ $$

Question Number 22516 Answers: 0 Comments: 1

toutes les solutions ?

$${toutes}\:{les}\:{solutions}\:? \\ $$

Question Number 22515 Answers: 0 Comments: 0

In a quadrilateral ABCD, it is given that AB is parallel to CD and the diagonals AC and BD are perpendicular to each other. Show that (a) AD.BC ≥ AB.CD; (b) AD + BC ≥ AB + CD.

$$\mathrm{In}\:\mathrm{a}\:\mathrm{quadrilateral}\:{ABCD},\:\mathrm{it}\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{that}\:{AB}\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:{CD}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{diagonals}\:{AC}\:\mathrm{and}\:{BD}\:\mathrm{are}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{each}\:\mathrm{other}. \\ $$$$\mathrm{Show}\:\mathrm{that} \\ $$$$\left(\mathrm{a}\right)\:{AD}.{BC}\:\geqslant\:{AB}.{CD}; \\ $$$$\left(\mathrm{b}\right)\:{AD}\:+\:{BC}\:\geqslant\:{AB}\:+\:{CD}. \\ $$

Question Number 22503 Answers: 1 Comments: 4

If (a + bx)^(−2) = (1/4) − 3x + ..., then (a, b) =

$$\mathrm{If}\:\left({a}\:+\:{bx}\right)^{−\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\:−\:\mathrm{3}{x}\:+\:...,\:\mathrm{then}\:\left({a},\:{b}\right)\:= \\ $$

Question Number 22499 Answers: 1 Comments: 1

A cylinder of weight 200 N is supported on a smooth horizontal plane by a light cord AC and pulled with force of 400 N. The normal reaction at B is equal to

$$\mathrm{A}\:\mathrm{cylinder}\:\mathrm{of}\:\mathrm{weight}\:\mathrm{200}\:\mathrm{N}\:\mathrm{is}\:\mathrm{supported} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{plane}\:\mathrm{by}\:\mathrm{a}\:\mathrm{light} \\ $$$$\mathrm{cord}\:{AC}\:\mathrm{and}\:\mathrm{pulled}\:\mathrm{with}\:\mathrm{force}\:\mathrm{of}\:\mathrm{400}\:\mathrm{N}. \\ $$$$\mathrm{The}\:\mathrm{normal}\:\mathrm{reaction}\:\mathrm{at}\:{B}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 22492 Answers: 1 Comments: 1

A ball of mass 400 g travels horizontally along the ground and collides with a wall. The velocity-time graph below represents the motion of the ball for the first 1.2 seconds. The magnitude of average force between the ball and the wall is

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{400}\:\mathrm{g}\:\mathrm{travels}\:\mathrm{horizontally} \\ $$$$\mathrm{along}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{and}\:\mathrm{collides}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{wall}.\:\mathrm{The}\:\mathrm{velocity}-\mathrm{time}\:\mathrm{graph}\:\mathrm{below} \\ $$$$\mathrm{represents}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{1}.\mathrm{2}\:\mathrm{seconds}. \\ $$$$\mathrm{The}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{average}\:\mathrm{force}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{ball}\:\mathrm{and}\:\mathrm{the}\:\mathrm{wall}\:\mathrm{is} \\ $$

Question Number 22491 Answers: 1 Comments: 0

The coefficient of x^r in the expansion of (1 − 2x)^(−1/2) is (1) (((2r)!)/((r!)^2 )) (2) (((2r)!)/(2^r (r!)^2 )) (3) (((2r)!)/((r!)^2 2^(2r) )) (4) (((2r)!)/(2^r (r + 1)!(r − 1)!))

$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{r}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left(\mathrm{1}\:−\:\mathrm{2}{x}\right)^{−\mathrm{1}/\mathrm{2}} \:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\left({r}!\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\mathrm{2}^{{r}} \:\left({r}!\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\left({r}!\right)^{\mathrm{2}} \:\mathrm{2}^{\mathrm{2}{r}} } \\ $$$$\left(\mathrm{4}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\mathrm{2}^{{r}} \:\left({r}\:+\:\mathrm{1}\right)!\left({r}\:−\:\mathrm{1}\right)!} \\ $$

Question Number 22487 Answers: 1 Comments: 1

Question Number 22483 Answers: 0 Comments: 0

Predict the density of Cs from the density of the following elements K 0.86 g/cm^3 Ca 1.548 g/cm^3 Sc 2.991 g/cm^3 Rb 1.532 g/cm^3 Sr 2.68 g/cm^3 Y 4.34 g/cm^3 Cs ? Ba 3.51 g/cm^3 La 6.16 g/cm^3

$$\mathrm{Predict}\:\mathrm{the}\:\mathrm{density}\:\mathrm{of}\:\mathrm{Cs}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{density}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{elements} \\ $$$$\mathrm{K}\:\mathrm{0}.\mathrm{86}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \:\:\:\:\:\:\:\:\mathrm{Ca}\:\mathrm{1}.\mathrm{548}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \\ $$$$\mathrm{Sc}\:\mathrm{2}.\mathrm{991}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \:\:\:\:\:\mathrm{Rb}\:\mathrm{1}.\mathrm{532}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \\ $$$$\mathrm{Sr}\:\mathrm{2}.\mathrm{68}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \:\:\:\:\:\:\:\:\mathrm{Y}\:\mathrm{4}.\mathrm{34}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \\ $$$$\mathrm{Cs}\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Ba}\:\mathrm{3}.\mathrm{51}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{La}\:\mathrm{6}.\mathrm{16}\:\mathrm{g}/\mathrm{cm}^{\mathrm{3}} \\ $$

Question Number 22481 Answers: 1 Comments: 1

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