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Question Number 181426    Answers: 1   Comments: 1

Question Number 181424    Answers: 1   Comments: 1

Question Number 173661    Answers: 0   Comments: 0

Question Number 173652    Answers: 1   Comments: 3

Question Number 173645    Answers: 0   Comments: 3

Question Number 173644    Answers: 1   Comments: 0

Question Number 173643    Answers: 0   Comments: 0

Question Number 181581    Answers: 1   Comments: 0

Question Number 181579    Answers: 1   Comments: 1

((x)^(1/3) )^(−2+log_x 11) =11 x=?

$$\left(\sqrt[{\mathrm{3}}]{{x}}\right)^{−\mathrm{2}+{log}_{{x}} \mathrm{11}} =\mathrm{11} \\ $$$${x}=? \\ $$

Question Number 176816    Answers: 3   Comments: 4

Question Number 173630    Answers: 0   Comments: 2

(x+4)^2 =x^((x+2)) Please Help...

$$\left({x}+\mathrm{4}\right)^{\mathrm{2}} ={x}^{\left({x}+\mathrm{2}\right)} \\ $$$${Please}\:\:{Help}... \\ $$

Question Number 173629    Answers: 2   Comments: 2

Question Number 173624    Answers: 0   Comments: 6

find f(x) such that f′(x)=f^(−1) (x).

$${find}\:{f}\left({x}\right)\:{such}\:{that}\:{f}'\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right). \\ $$

Question Number 173620    Answers: 2   Comments: 0

Question Number 173610    Answers: 2   Comments: 0

Question Number 173609    Answers: 2   Comments: 2

Question Number 173606    Answers: 0   Comments: 0

Question Number 173613    Answers: 1   Comments: 0

if f(x)=x^2 +4x+2, solve f(f(f(f(...f(x)))))_(n times) =0

$${if}\:{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}, \\ $$$${solve}\:\underset{{n}\:{times}} {{f}\left({f}\left({f}\left({f}\left(...{f}\left({x}\right)\right)\right)\right)\right)}=\mathrm{0} \\ $$

Question Number 173599    Answers: 0   Comments: 1

Find the area bounded by the graph y=x^2 and y=2−x^2 for 0≤x≤2.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{graph} \\ $$$$\mathrm{y}=\mathrm{x}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{y}=\mathrm{2}−\mathrm{x}^{\mathrm{2}} \:\mathrm{for}\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{2}. \\ $$

Question Number 173603    Answers: 0   Comments: 0

∫_0 ^1 ((log(x)log(1+x^2 ))/(1+x))dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{log}}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)}{\mathrm{1}+\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$

Question Number 173582    Answers: 0   Comments: 0

Question Number 173574    Answers: 5   Comments: 4

solve for x∈R ((5x−3)/(3x−5))=x^4

$${solve}\:{for}\:{x}\in{R} \\ $$$$\frac{\mathrm{5}{x}−\mathrm{3}}{\mathrm{3}{x}−\mathrm{5}}={x}^{\mathrm{4}} \\ $$

Question Number 173572    Answers: 2   Comments: 0

Solve equation 2⌊x⌋ = (x^( 2) /2) (x ∈ R )

$$\: \\ $$$$\:\:\mathrm{S}{olve}\:\:\:{equation} \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{2}\lfloor{x}\rfloor\:=\:\frac{{x}^{\:\mathrm{2}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\left({x}\:\in\:\mathbb{R}\:\right) \\ $$$$ \\ $$

Question Number 173566    Answers: 3   Comments: 0

Question Number 173558    Answers: 1   Comments: 1

A metal sphere has radius R and mass m. A spherical hollow of diameter R is made in this sphere such that its surface passes through the centre of the metal sphere and touches the outside surface of the metal sphere. A unit mass is placed at a distance from the centre of the metal sphere. The gravitational field at that point is (a) ((GM)/R^2 ) (1−(1/(8(1−((2R)/a))^2 ))) (b) ((GM)/a^2 ) (1−(1/(8(1−(R/(2a)))^2 ))) (c) ((GM)/((R+a)^2 )) (1−(1/(8(1−(R/(2a)))^2 ))) (d) ((GM)/((R−a)^2 )) (1−(1/(8(1−((2a)/R))^2 )))

$$\mathrm{A}\:\mathrm{metal}\:\mathrm{sphere}\:\mathrm{has}\:\mathrm{radius}\:{R}\:\mathrm{and}\:\mathrm{mass}\:{m}.\:\mathrm{A}\:\mathrm{spherical} \\ $$$$\mathrm{hollow}\:\mathrm{of}\:\mathrm{diameter}\:{R}\:\mathrm{is}\:\mathrm{made}\:\mathrm{in}\:\mathrm{this}\:\mathrm{sphere}\:\mathrm{such}\:\mathrm{that}\:\mathrm{its} \\ $$$$\mathrm{surface}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{metal}\:\mathrm{sphere} \\ $$$$\mathrm{and}\:\mathrm{touches}\:\mathrm{the}\:\mathrm{outside}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{the}\:\mathrm{metal}\:\mathrm{sphere}. \\ $$$$\mathrm{A}\:\mathrm{unit}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{placed}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{metal} \\ $$$$\mathrm{sphere}.\:\mathrm{The}\:\mathrm{gravitational}\:\mathrm{field}\:\mathrm{at}\:\mathrm{that}\:\mathrm{point}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\frac{{GM}}{{R}^{\mathrm{2}} }\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{2}{R}}{{a}}\right)^{\mathrm{2}} }\right) \\ $$$$\left(\mathrm{b}\right)\:\frac{{GM}}{{a}^{\mathrm{2}} }\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}−\frac{{R}}{\mathrm{2}{a}}\right)^{\mathrm{2}} }\right) \\ $$$$\left(\mathrm{c}\right)\:\frac{{GM}}{\left({R}+{a}\right)^{\mathrm{2}} }\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}−\frac{{R}}{\mathrm{2}{a}}\right)^{\mathrm{2}} }\right) \\ $$$$\left(\mathrm{d}\right)\:\frac{{GM}}{\left({R}−{a}\right)^{\mathrm{2}} }\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{2}{a}}{{R}}\right)^{\mathrm{2}} }\right) \\ $$

Question Number 173557    Answers: 0   Comments: 2

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