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Question Number 49763    Answers: 0   Comments: 0

Question Number 49767    Answers: 1   Comments: 0

Question Number 49768    Answers: 0   Comments: 1

help me sir plz

$$\mathrm{help}\:\mathrm{me}\:\mathrm{sir}\:\mathrm{plz} \\ $$$$ \\ $$

Question Number 49761    Answers: 1   Comments: 0

Calculate : ∫(( sin^2 x cos^2 x)/((sin^3 x+cos^3 x)^2 )) dx

$${Calculate}\:: \\ $$$$\:\int\frac{\:\:\mathrm{sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}}{\left(\mathrm{sin}^{\mathrm{3}} {x}+\mathrm{cos}^{\mathrm{3}} {x}\right)^{\mathrm{2}} }\:{dx} \\ $$

Question Number 49760    Answers: 0   Comments: 3

Question Number 49755    Answers: 1   Comments: 0

Solve simultaneously for s in terms of a and b. h^2 +(b−k)^2 = s^2 .....(i) (h^2 /a^2 )+(k^2 /b^2 ) = 1 .....(ii) (h−(s/2))^2 +(k+b(√(1−(s^2 /(4a^2 )))) )= s^2 ..(iii).

$${Solve}\:{simultaneously}\:{for}\:\boldsymbol{{s}}\:{in}\:{terms} \\ $$$${of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}. \\ $$$${h}^{\mathrm{2}} +\left({b}−{k}\right)^{\mathrm{2}} =\:{s}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:.....\left({i}\right) \\ $$$$\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....\left({ii}\right) \\ $$$$\left({h}−\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({k}+{b}\sqrt{\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }}\:\right)=\:{s}^{\mathrm{2}} \:\:\:..\left({iii}\right). \\ $$

Question Number 49751    Answers: 1   Comments: 1

The value of tan^(−1) (1/2) + tan^(−1) (1/3) is

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\:\:\mathrm{is} \\ $$

Question Number 49746    Answers: 1   Comments: 0

∫((sin^8 x−cos^8 x)/(1−2sin^2 x.cos^2 x)) = ? a) ((−1)/2)sin 2x b)(1/2)sin 2x c)None.

$$\int\frac{\mathrm{sin}^{\mathrm{8}} {x}−\mathrm{cos}^{\mathrm{8}} {x}}{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} {x}.\mathrm{cos}^{\mathrm{2}} {x}}\:=\:? \\ $$$$\left.{a}\left.\right)\left.\:\frac{−\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\:\:\:{b}\right)\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\:\:\:{c}\right){None}. \\ $$

Question Number 49740    Answers: 1   Comments: 1

Question Number 49737    Answers: 1   Comments: 1

Question Number 49736    Answers: 1   Comments: 1

there is two small and one grater circles that [two]are tangent to [one]and all three circles are inscribed in an ellipse with: [(a/b)=2(√2)]and tangent to it at two points such that center of circles are on major axe of ellipse. find: ((radi of great circle)/(radi of small circle)) .

$$\boldsymbol{\mathrm{there}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{small}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{grater}}\:\boldsymbol{\mathrm{circles}} \\ $$$$\boldsymbol{\mathrm{that}}\:\left[\boldsymbol{\mathrm{two}}\right]\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{to}}\:\left[\boldsymbol{\mathrm{one}}\right]\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{three}} \\ $$$$\:\boldsymbol{\mathrm{circles}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{inscribed}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{ellipse}}\:\boldsymbol{\mathrm{with}}: \\ $$$$\left[\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}=\mathrm{2}\sqrt{\mathrm{2}}\right]\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{points}}\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{center}}\: \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{circles}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{major}}\:\boldsymbol{\mathrm{axe}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{ellipse}}. \\ $$$$\boldsymbol{\mathrm{find}}:\:\:\:\:\frac{\boldsymbol{\mathrm{radi}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{great}}\:\boldsymbol{\mathrm{circle}}}{\boldsymbol{\mathrm{radi}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{small}}\:\boldsymbol{\mathrm{circle}}}\:\:. \\ $$

Question Number 49731    Answers: 1   Comments: 0

one vertex of a equilateral triangle lies on one vertex of a square and two anothers lie on opposite sides of square such that triangle have the maximum area. find: 1.ratio of: ((square side)/(triangle side)) 2.angle between square side and triangle side.[need additional data?]

$$\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{vertex}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{equilateral}}\:\boldsymbol{\mathrm{triangle}}\:\boldsymbol{\mathrm{lies}} \\ $$$$\boldsymbol{\mathrm{on}}\:\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{vertex}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{square}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{two}} \\ $$$$\boldsymbol{\mathrm{anothers}}\:\boldsymbol{\mathrm{lie}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{opposite}}\:\boldsymbol{\mathrm{sides}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{square}} \\ $$$$\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{triangle}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{maximum}} \\ $$$$\boldsymbol{\mathrm{area}}. \\ $$$$\boldsymbol{\mathrm{find}}: \\ $$$$\mathrm{1}.\boldsymbol{\mathrm{ratio}}\:\boldsymbol{\mathrm{of}}:\:\:\:\:\:\frac{\boldsymbol{\mathrm{square}}\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{side}}}{\boldsymbol{\mathrm{triangle}}\:\:\:\:\:\:\:\boldsymbol{\mathrm{side}}} \\ $$$$\mathrm{2}.\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{between}}\:\boldsymbol{\mathrm{square}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{triangle}} \\ $$$$\boldsymbol{\mathrm{side}}.\left[\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{additional}}\:\boldsymbol{\mathrm{data}}?\right] \\ $$

Question Number 49730    Answers: 0   Comments: 1

find the largest ellipse inscribed in a given rectangle and its major axe of:ellipse lies on rectangle diagonal.

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{largest}}\:\boldsymbol{\mathrm{ellipse}}\:\boldsymbol{\mathrm{inscribed}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{a}} \\ $$$$\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{rectangle}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{its}}\:\boldsymbol{\mathrm{major}}\:\boldsymbol{\mathrm{axe}}\:\boldsymbol{\mathrm{of}}:\boldsymbol{\mathrm{ellipse}} \\ $$$$\boldsymbol{\mathrm{lies}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{rectangle}}\:\boldsymbol{\mathrm{diagonal}}. \\ $$

Question Number 49725    Answers: 2   Comments: 1

Question Number 49708    Answers: 0   Comments: 0

Please integrate ∫(((e^(cos x) sin x)/(1−x^2 )))dx

$${Please}\:{integrate} \\ $$$$\int\left(\frac{\mathrm{e}^{\mathrm{cos}\:{x}} \mathrm{sin}\:{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$

Question Number 49703    Answers: 0   Comments: 0

Given f(x)= sin2x+2cos2x find f′((π/4)) hence given g(x) = { ((x, 0≤x≤2)),((3x−x^2 , 2≤x≤3)) :} for a total range of 0≤x≤3 sketch the graph for y=g(x) and find the area with makes with the x−axis otherwise, find the composite function gf in the range 2≤x≤3.

$${Given}\:{f}\left({x}\right)=\:{sin}\mathrm{2}{x}+\mathrm{2}{cos}\mathrm{2}{x} \\ $$$${find}\:{f}'\left(\frac{\pi}{\mathrm{4}}\right) \\ $$$${hence}\:{given}\:{g}\left({x}\right)\:=\:\begin{cases}{{x},\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}}\\{\mathrm{3}{x}−{x}^{\mathrm{2}} ,\:\:\mathrm{2}\leqslant{x}\leqslant\mathrm{3}}\end{cases} \\ $$$${for}\:{a}\:{total}\:{range}\:{of}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{3}\:\:{sketch}\:{the}\:{graph}\:{for}\:{y}={g}\left({x}\right) \\ $$$${and}\:{find}\:{the}\:{area}\:{with}\:{makes}\:{with}\:{the}\:{x}−{axis} \\ $$$${otherwise},\:{find}\:\:\:{the}\:{composite}\:{function}\:\:{gf}\:\:{in}\:{the}\:{range} \\ $$$$\mathrm{2}\leqslant{x}\leqslant\mathrm{3}. \\ $$

Question Number 49696    Answers: 1   Comments: 1

Question Number 49680    Answers: 1   Comments: 0

Question Number 49678    Answers: 1   Comments: 1

Question Number 50924    Answers: 1   Comments: 0

factor the expression: E=x^5 +x^4 +1

$$\mathrm{factor}\:\mathrm{the}\:\mathrm{expression}: \\ $$$$\mathrm{E}={x}^{\mathrm{5}} +{x}^{\mathrm{4}} +\mathrm{1} \\ $$

Question Number 49692    Answers: 1   Comments: 0

The vectors a=xi+(x+1)j+(x+2)k, b=(x+3)i+(x+4)j+(x+5)k and c=(x+6)i+(x+7)j+(x+8)k are coplanar for

$$\mathrm{The}\:\mathrm{vectors}\:\boldsymbol{\mathrm{a}}={x}\boldsymbol{\mathrm{i}}+\left({x}+\mathrm{1}\right)\boldsymbol{\mathrm{j}}+\left({x}+\mathrm{2}\right)\boldsymbol{\mathrm{k}}, \\ $$$$\boldsymbol{\mathrm{b}}=\left({x}+\mathrm{3}\right)\boldsymbol{\mathrm{i}}+\left({x}+\mathrm{4}\right)\boldsymbol{\mathrm{j}}+\left({x}+\mathrm{5}\right)\boldsymbol{\mathrm{k}}\:\:\:\mathrm{and} \\ $$$$\boldsymbol{\mathrm{c}}=\left({x}+\mathrm{6}\right)\boldsymbol{\mathrm{i}}+\left({x}+\mathrm{7}\right)\boldsymbol{\mathrm{j}}+\left({x}+\mathrm{8}\right)\boldsymbol{\mathrm{k}}\:\mathrm{are}\:\mathrm{coplanar} \\ $$$$\mathrm{for} \\ $$

Question Number 49691    Answers: 1   Comments: 0

simplify ((log_3 64 × log_4 243)/(log_2 16))

$${simplify}\:\frac{{log}_{\mathrm{3}} \mathrm{64}\:×\:{log}_{\mathrm{4}} \mathrm{243}}{{log}_{\mathrm{2}} \mathrm{16}} \\ $$

Question Number 49748    Answers: 1   Comments: 0

Question Number 49661    Answers: 1   Comments: 2

calculateA_n =(1/(2i)) ∫_0 ^1 {(1+ix)^n −(1−ix)^n }dx

$${calculateA}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}{i}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\left(\mathrm{1}+{ix}\right)^{{n}} −\left(\mathrm{1}−{ix}\right)^{{n}} \right\}{dx} \\ $$

Question Number 49660    Answers: 2   Comments: 1

Question Number 49647    Answers: 1   Comments: 3

let p(x) =x^(2n) −x^n +1 1) determine the roots of p(x) 2) factorize inside C[x] the polynom p(x) . 3)solve p(x)=0 and p(x) =2

$${let}\:{p}\left({x}\right)\:={x}^{\mathrm{2}{n}} \:−{x}^{{n}} \:+\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{inside}\:{C}\left[{x}\right]\:{the}\:{polynom}\:{p}\left({x}\right)\:. \\ $$$$\left.\mathrm{3}\right){solve}\:{p}\left({x}\right)=\mathrm{0}\:\:{and}\:{p}\left({x}\right)\:=\mathrm{2} \\ $$

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