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Question Number 54767 Answers: 1 Comments: 1

A stone is thrown vertically upwards from a cliff 20m high. After a time of 3 s it passes the edge of the cliff on its way down. Calculate a) the speed of projection b) the speed when it hits the ground c) the times when it is 10m above the top of the cliff d) the time it is 15m above the ground (take g=10ms^(−2) ).

Question Number 54765 Answers: 1 Comments: 0

Question Number 54756 Answers: 1 Comments: 0

solve ∫tan(x−θ)tan(x+θ)tan 2x dx

$${solve} \\ $$$$\int{tan}\left({x}−\theta\right){tan}\left({x}+\theta\right){tan}\:\mathrm{2}{x}\:{dx} \\ $$

Question Number 54755 Answers: 1 Comments: 0

solve for x sin[2cos^(−1) {cot(2tan^(−1) x)}]=0 (x = 1∓(√(2 )) , ∓1 ,−1∓(√2) )

$${solve}\:{for}\:{x} \\ $$$${sin}\left[\mathrm{2}{cos}^{−\mathrm{1}} \left\{{cot}\left(\mathrm{2}{tan}^{−\mathrm{1}} {x}\right)\right\}\right]=\mathrm{0} \\ $$$$ \\ $$$$\left({x}\:=\:\mathrm{1}\mp\sqrt{\mathrm{2}\:}\:,\:\mp\mathrm{1}\:,−\mathrm{1}\mp\sqrt{\mathrm{2}}\:\:\right) \\ $$

Question Number 54745 Answers: 0 Comments: 0

(√(1−x^2 +))(√(1−y^2 ))=a(x−y) (dy/dx)=(√((1−y^2 )/(1−x^2 ))) without solve put x=sinθand y=sinφ direct solve

$$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} +}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }={a}\left({x}−{y}\right) \\ $$$$\frac{{dy}}{{dx}}=\sqrt{\frac{\mathrm{1}−{y}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${without}\:{solve}\:{put}\:{x}={sin}\theta{and} \\ $$$${y}={sin}\phi \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${direct}\:{solve} \\ $$

Question Number 54741 Answers: 1 Comments: 1

such that 1. ((n),(0) )^2 + ((n),(1) )^2 +...+ ((n),(n) )^2 =(((2n)!)/((n!)^2 )) 2. ((n),(0) )+(1/2) ((n),(1) )+...+(1/(n+1)) ((n),(n) )^2 =((2^(n+1) −1)/(n+1))

$$\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{1}.\begin{pmatrix}{\mathrm{n}}\\{\mathrm{0}}\end{pmatrix}^{\mathrm{2}} +\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}^{\mathrm{2}} +...+\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}^{\mathrm{2}} =\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}.\:\begin{pmatrix}{\mathrm{n}}\\{\mathrm{0}}\end{pmatrix}+\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}+...+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}^{\mathrm{2}} =\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}} \\ $$

Question Number 54740 Answers: 3 Comments: 0

Prove that 1. ((n),(r) ) = (((n−1)),(( r)) ) + (((n−1)),(( r−1)) ) 2. ((n),(r) ) + ((( n)),((r−1)) ) = (((n+1)),(( r)) ) 3. ((n),(0) )+ ((n),(1) )+ ((n),(2) )+..+ ((n),(n) )=2^n

$${Prove}\:{that} \\ $$$$\mathrm{1}.\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:=\:\begin{pmatrix}{{n}−\mathrm{1}}\\{\:\:\:\:{r}}\end{pmatrix}\:+\:\begin{pmatrix}{{n}−\mathrm{1}}\\{\:{r}−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{2}.\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:+\begin{pmatrix}{\:\:\:{n}}\\{{r}−\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{{n}+\mathrm{1}}\\{\:\:\:\:{r}}\end{pmatrix}\: \\ $$$$\mathrm{3}.\:\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}+..+\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}=\mathrm{2}^{{n}} \\ $$

Question Number 54737 Answers: 0 Comments: 0

Question Number 54730 Answers: 0 Comments: 3

prove lnx≤x

$${prove}\:{lnx}\leqslant{x} \\ $$

Question Number 54716 Answers: 0 Comments: 6

A light horizontal meter rule PQR has the end P fixed to vertical wall, while a weight of 5N is suspended from the end R. A light string QS of length 50cm fixed to the wall at S is used to maintain the meter rule in equilibrium. If PQ = 40 cm, the tension in the string is ?

$$\mathrm{A}\:\mathrm{light}\:\mathrm{horizontal}\:\mathrm{meter}\:\mathrm{rule}\:\:\mathrm{PQR}\:\:\mathrm{has}\:\mathrm{the}\:\mathrm{end}\:\:\mathrm{P}\:\mathrm{fixed}\:\mathrm{to}\: \\ $$$$\mathrm{vertical}\:\mathrm{wall},\:\mathrm{while}\:\mathrm{a}\:\mathrm{weight}\:\mathrm{of}\:\:\mathrm{5N}\:\mathrm{is}\:\mathrm{suspended}\:\mathrm{from}\:\mathrm{the}\:\mathrm{end} \\ $$$$\mathrm{R}.\:\mathrm{A}\:\mathrm{light}\:\mathrm{string}\:\mathrm{QS}\:\:\mathrm{of}\:\mathrm{length}\:\:\mathrm{50cm}\:\mathrm{fixed}\:\mathrm{to}\:\mathrm{the}\:\mathrm{wall}\:\mathrm{at}\:\mathrm{S}\:\mathrm{is}\: \\ $$$$\mathrm{used}\:\mathrm{to}\:\mathrm{maintain}\:\mathrm{the}\:\mathrm{meter}\:\mathrm{rule}\:\mathrm{in}\:\mathrm{equilibrium}. \\ $$$$\mathrm{If}\:\:\mathrm{PQ}\:=\:\mathrm{40}\:\mathrm{cm},\:\:\:\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{string}\:\mathrm{is}\:? \\ $$

Question Number 54721 Answers: 0 Comments: 0

Show that ((Σ{sin 2xtan (y−z)cos^2 (y+z)})/(Σ{cos 2xtan (y−z)sin^2 (y+z)} )) = tan 2xtan 2ytan 2z

$$\mathrm{Show}\:\mathrm{that} \\ $$$$\frac{\Sigma\left\{\mathrm{sin}\:\mathrm{2xtan}\:\left(\mathrm{y}−\mathrm{z}\right)\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{y}+\mathrm{z}\right)\right\}}{\Sigma\left\{\mathrm{cos}\:\mathrm{2xtan}\:\left(\mathrm{y}−\mathrm{z}\right)\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{y}+\mathrm{z}\right)\right\}\:\:\:}\:\:=\:\mathrm{tan}\:\mathrm{2xtan}\:\mathrm{2ytan}\:\mathrm{2z} \\ $$

Question Number 54701 Answers: 2 Comments: 0

Question Number 54700 Answers: 1 Comments: 1

Find the sum to infinity 1−(1/(2 ))cos θ+(1/4)cos 2θ−(1/8)cos 3θ+..........

$$\mathrm{F}{i}\mathrm{nd}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinity} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\:}\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{3}\theta+.......... \\ $$$$ \\ $$

Question Number 54699 Answers: 2 Comments: 3

∫ e^(2x) ((1/x) −(1/(2x^2 )))dx ∫e^(2x) (((1+sin 2x)/(1+cos 2x)))dx. Solve above Questions by using the formulae : ∫e^(kx) {f(kx)+f ′(kx)}dx= e^(kx) f(kx)+c.

$$\int\:{e}^{\mathrm{2}{x}} \:\left(\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right){dx} \\ $$$$\int{e}^{\mathrm{2}{x}} \:\left(\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}\right){dx}. \\ $$$${Solve}\:{above}\:{Questions}\:{by}\:{using}\:{the} \\ $$$${formulae}\:: \\ $$$$\int{e}^{{kx}} \left\{{f}\left({kx}\right)+{f}\:'\left({kx}\right)\right\}{dx}=\:{e}^{{kx}} \:{f}\left({kx}\right)+{c}. \\ $$

Question Number 54698 Answers: 2 Comments: 0

How many words with at least 2 letters can be formed using the letters from TINKUTARA?

$${How}\:{many}\:{words}\:{with}\:{at}\:{least}\:\mathrm{2}\:{letters} \\ $$$${can}\:{be}\:{formed}\:{using}\:{the}\:{letters}\:{from} \\ $$$${TINKUTARA}? \\ $$

Question Number 54679 Answers: 2 Comments: 1

Question Number 54675 Answers: 1 Comments: 1

simplify A_n =Σ_(k=0) ^n k^2 C_n ^k cos(kθ) and B_n =Σ_(k=0) ^n k^2 C_n ^k sin(kθ) .

$${simplify}\:\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}^{\mathrm{2}} \:{C}_{{n}} ^{{k}} \:{cos}\left({k}\theta\right)\:{and}\:{B}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}^{\mathrm{2}} \:{C}_{{n}} ^{{k}} \:{sin}\left({k}\theta\right)\:. \\ $$

Question Number 54688 Answers: 1 Comments: 1

f(x)=((2[x])/(3x−[x])) examine its continuity at x=((−1)/2) where [x] is greatest integer function

$${f}\left({x}\right)=\frac{\mathrm{2}\left[{x}\right]}{\mathrm{3}{x}−\left[{x}\right]}\:{examine}\:{its}\:{continuity}\:{at}\:{x}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${where}\:\left[{x}\right]\:{is}\:{greatest}\:{integer}\:{function} \\ $$

Question Number 54663 Answers: 1 Comments: 1

y = cos2t×sen2t y′ = ?

$${y}\:=\:{cos}\mathrm{2}{t}×{sen}\mathrm{2}{t} \\ $$$$ \\ $$$${y}'\:=\:? \\ $$

Question Number 54661 Answers: 1 Comments: 3

Question Number 54659 Answers: 1 Comments: 2

Question Number 54658 Answers: 0 Comments: 0

Question Number 54647 Answers: 0 Comments: 3

show that a. Σ_(r=1) ^(n) r^3 ._n C_r =n^2 (n+3).2^(n−3) b. _n C_0 ._n C_1 +_n C_1 ._n C_2 +...+_n C_(n−1) ._n C_n =(((2n)!)/((n−1)!.(n+1)!))

$$\mathrm{show}\:\mathrm{that} \\ $$$${a}.\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\Sigma}}\:{r}^{\mathrm{3}} ._{{n}} {C}_{{r}} ={n}^{\mathrm{2}} \left({n}+\mathrm{3}\right).\mathrm{2}^{{n}−\mathrm{3}} \\ $$$${b}.\:_{{n}} {C}_{\mathrm{0}} ._{{n}} {C}_{\mathrm{1}} +_{{n}} {C}_{\mathrm{1}} ._{{n}} {C}_{\mathrm{2}} +...+_{{n}} {C}_{{n}−\mathrm{1}} ._{{n}} {C}_{{n}} =\frac{\left(\mathrm{2}{n}\right)!}{\left({n}−\mathrm{1}\right)!.\left({n}+\mathrm{1}\right)!} \\ $$

Question Number 54646 Answers: 2 Comments: 0

Such That a. _(n+1) C_r =(((n+1). _n C_r )/((n−r+1))) b. _n C_0 +_n C_2 +_n C_(4...) =_n C_1 +_n C_3 +_n C_(5...) =2^(n−1)

$$\mathrm{Such}\:\mathrm{That} \\ $$$$\mathrm{a}.\:_{{n}+\mathrm{1}} {C}_{{r}} =\frac{\left({n}+\mathrm{1}\right).\:_{{n}} {C}_{{r}} }{\left({n}−{r}+\mathrm{1}\right)} \\ $$$$\mathrm{b}.\:_{{n}} {C}_{\mathrm{0}} +_{{n}} {C}_{\mathrm{2}} +_{{n}} {C}_{\mathrm{4}...} =_{{n}} {C}_{\mathrm{1}} +_{{n}} {C}_{\mathrm{3}} +_{{n}} {C}_{\mathrm{5}...} =\mathrm{2}^{{n}−\mathrm{1}} \\ $$$$ \\ $$

Question Number 54644 Answers: 1 Comments: 0

A= [(3,7),((−1),(−2)) ] A^(27) +A^(31) +A^(40) =...

$$\mathrm{A}=\begin{bmatrix}{\mathrm{3}}&{\mathrm{7}}\\{−\mathrm{1}}&{−\mathrm{2}}\end{bmatrix} \\ $$$${A}^{\mathrm{27}} +{A}^{\mathrm{31}} +{A}^{\mathrm{40}} =... \\ $$

Question Number 54641 Answers: 0 Comments: 1

Given f(x) = ((4x + (√(4x^2 − 1)))/((√(2x + 1)) − (√(2x − 1)))) Find the value of f(13) + f(14) + f(15) + ... + f(112)

$$\mathrm{Given} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{4}{x}\:+\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:−\:\mathrm{1}}}{\sqrt{\mathrm{2}{x}\:+\:\mathrm{1}}\:−\:\sqrt{\mathrm{2}{x}\:−\:\mathrm{1}}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$${f}\left(\mathrm{13}\right)\:+\:{f}\left(\mathrm{14}\right)\:+\:{f}\left(\mathrm{15}\right)\:+\:...\:+\:{f}\left(\mathrm{112}\right) \\ $$

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