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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)

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

Question Number 54626 Answers: 2 Comments: 0

((1+sinx+cox)/(1+sinx−cosx))=((1−cosx)/(1+cosx))

$$\frac{\mathrm{1}+\mathrm{sinx}+\mathrm{cox}}{\mathrm{1}+\mathrm{sinx}−\mathrm{cosx}}=\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{1}+\mathrm{cosx}} \\ $$

Question Number 54616 Answers: 2 Comments: 0

Question Number 54607 Answers: 1 Comments: 2

If ((u^5 +v^5 )/((u+v)^5 )) = −(1/5) , find ((u^3 +v^3 )/((u+v)^3 )) = ?

$${If}\:\:\frac{{u}^{\mathrm{5}} +{v}^{\mathrm{5}} }{\left({u}+{v}\right)^{\mathrm{5}} }\:=\:−\frac{\mathrm{1}}{\mathrm{5}}\:, \\ $$$${find}\:\:\:\frac{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} }{\left({u}+{v}\right)^{\mathrm{3}} }\:\:=\:? \\ $$

Question Number 54603 Answers: 1 Comments: 0

a,b,c ,are nonnegative real numbers and: a+b+c=1 . show that: 0≤ ab+bc+ca−2abc ≤(7/(27)) .

$${a},{b},{c}\:,{are}\:{nonnegative}\:{real}\:{numbers} \\ $$$${and}:\:\:\:{a}+{b}+{c}=\mathrm{1}\:\:. \\ $$$${show}\:{that}: \\ $$$$\:\:\:\:\:\:\mathrm{0}\leqslant\:\:\boldsymbol{\mathrm{ab}}+\boldsymbol{\mathrm{bc}}+\boldsymbol{\mathrm{ca}}−\mathrm{2}\boldsymbol{\mathrm{abc}}\:\:\:\leqslant\frac{\mathrm{7}}{\mathrm{27}}\:. \\ $$

Question Number 54602 Answers: 0 Comments: 1

in a given triangle: tg(C/2)=((a.tgA+b.tgB)/(a+b)) . define the kind of triangle.

$${in}\:{a}\:{given}\:{triangle}: \\ $$$$\:\:\:\boldsymbol{\mathrm{tg}}\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{tgA}}+\boldsymbol{\mathrm{b}}.\boldsymbol{\mathrm{tgB}}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}\:. \\ $$$$\boldsymbol{\mathrm{define}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{kind}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{triangle}}. \\ $$

Question Number 54600 Answers: 2 Comments: 2

solve for: x 1) (√(3−x))+(√(x+1))>(1/2) 2) cos^2 x+cos^2 2x+cos^2 3x=1 3)(√(x^2 −p))+2(√(x^2 −1))=x [p∈R]

Question Number 54595 Answers: 0 Comments: 3

x+y=a z+bx=c bz+xy=d Find yz in terms of a,b,c,d.

$$\:\:\:\:\:{x}+{y}={a} \\ $$$$\:\:\:\:\:{z}+{bx}={c} \\ $$$$\:\:\:\:\:{bz}+{xy}={d} \\ $$$$\:\:\:\:\:{Find}\:{yz}\:{in}\:{terms}\:{of}\:{a},{b},{c},{d}. \\ $$

Question Number 54588 Answers: 1 Comments: 0

What is : (d/dx) [ u(x) × v(x) × w(x) ] = ... and more generally, what is : (d/dx) [ Π_(i=1) ^n u_i (x)] = ... Thank you

$$\mathrm{What}\:\mathrm{is}\:: \\ $$$$\:\:\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\:{u}\left({x}\right)\:×\:{v}\left({x}\right)\:×\:{w}\left({x}\right)\:\right]\:=\:... \\ $$$$ \\ $$$$\mathrm{and}\:\mathrm{more}\:\mathrm{generally},\:\mathrm{what}\:\mathrm{is}\:: \\ $$$$\:\:\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\:{u}_{{i}} \left({x}\right)\right]\:=\:... \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you} \\ $$

Question Number 54586 Answers: 1 Comments: 0

(√(1−x^2 )) + (√(1−y^2 )) = a(x−y) prove that (dy/dx)=(√((1−y^2 )/(1−x^2 )))

$$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\:=\:{a}\left({x}−{y}\right) \\ $$$${prove}\:{that} \\ $$$$\frac{{dy}}{{dx}}=\sqrt{\frac{\mathrm{1}−{y}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$

Question Number 54585 Answers: 1 Comments: 1

show that ∫_0 ^∞ (x/(1+x^6 ))dx=(π/(3(√3)))

$${show}\:{that}\:\int_{\mathrm{0}} ^{\infty} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{6}} }{dx}=\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

Question Number 54584 Answers: 2 Comments: 0

Question Number 54582 Answers: 0 Comments: 1

Solve for x: (1/(√(x + (√x) + 1))) + (2/(√(x + (√x) − 2))) = (√(x + 1))

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{x}\:+\:\sqrt{\mathrm{x}}\:+\:\mathrm{1}}}\:+\:\frac{\mathrm{2}}{\sqrt{\mathrm{x}\:+\:\sqrt{\mathrm{x}}\:−\:\mathrm{2}}}\:\:=\:\sqrt{\mathrm{x}\:+\:\mathrm{1}} \\ $$

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