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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

Question Number 173556 Answers: 1 Comments: 0

Solve over integer a^3 +2a^2 +3a+4=b!+c!

$$ \\ $$$$\mathrm{Solve}\:\mathrm{over}\:\mathrm{integer} \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{2a}^{\mathrm{2}} +\mathrm{3a}+\mathrm{4}=\mathrm{b}!+\mathrm{c}! \\ $$

Question Number 173535 Answers: 1 Comments: 0

Question Number 173532 Answers: 1 Comments: 0

Question Number 173531 Answers: 1 Comments: 0

Q: what is the greatest coefficient of the following expansion. P:= ( a+ b +c +d )^( 10) −−−−−−−

$$ \\ $$$$\:\:\:\:{Q}: \\ $$$$\:\:{what}\:{is}\:{the}\:{greatest}\:{coefficient} \\ $$$$\:\:{of}\:{the}\:{following}\:\:{expansion}. \\ $$$$\:\:\:\:\:\:\:\mathrm{P}:=\:\left(\:{a}+\:{b}\:+{c}\:+{d}\:\right)^{\:\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:−−−−−−− \\ $$

Question Number 173529 Answers: 1 Comments: 0

Find without softs: Ω = ∫_(𝛑/5) ^( ((3𝛑)/(10))) (x/(sin2x)) dx

$$\mathrm{Find}\:\mathrm{without}\:\mathrm{softs}:\:\:\:\Omega\:=\:\int_{\frac{\boldsymbol{\pi}}{\mathrm{5}}} ^{\:\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{10}}} \:\frac{\mathrm{x}}{\mathrm{sin2x}}\:\mathrm{dx} \\ $$

Question Number 173523 Answers: 2 Comments: 0

Question Number 173522 Answers: 0 Comments: 0

∫_0 ^1 ((ln(1−x)ln(1+x^2 ))/x)dx evaluate

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

Question Number 173515 Answers: 1 Comments: 0

Question Number 173516 Answers: 3 Comments: 0

calcul ∫_o ^(+oo) ((tlnt)/((t^2 +1)^2 ))dt

$${calcul} \\ $$$$\int_{{o}} ^{+{oo}} \frac{{tlnt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$

Question Number 173504 Answers: 2 Comments: 0

Question Number 173503 Answers: 0 Comments: 0

Question Number 173500 Answers: 2 Comments: 0

Find without any software: Ω = ∫ (x + (5/x))(1 − (5/x^2 ))sin(ln(x + (5/x)))dx

$$\mathrm{Find}\:\mathrm{without}\:\mathrm{any}\:\mathrm{software}: \\ $$$$\Omega\:=\:\int\:\left(\mathrm{x}\:+\:\frac{\mathrm{5}}{\mathrm{x}}\right)\left(\mathrm{1}\:−\:\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} }\right)\mathrm{sin}\left(\mathrm{ln}\left(\mathrm{x}\:+\:\frac{\mathrm{5}}{\mathrm{x}}\right)\right)\mathrm{dx} \\ $$

Question Number 173499 Answers: 0 Comments: 2

Question Number 173498 Answers: 1 Comments: 1

lim_(x→0) ((8cot (x)+9 sin ((1/x)))/(12 csc (x)−4sin ((1/x)))) =?

$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{8cot}\:\left({x}\right)+\mathrm{9}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}{\mathrm{12}\:\mathrm{csc}\:\left({x}\right)−\mathrm{4sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}\:=? \\ $$

Question Number 173497 Answers: 0 Comments: 0

Bonus du Mardi 12/07/2022 I= ∫_0 ^(Π/2) ((sin^2 (x))/(1+cos^2 (x)))dx = ? J = ∫_0 ^(Π/2) (dx/(1+cos^2 (x))) I = ∫_0 ^(Π/2) ((2−(1+cos^2 (x))/(1+cos^2 (x)))dx = 2∫_0 ^(Π/2) (dx/(1+cos^2 (x)))−∫_0 ^(Π/2) dx . = 2J−(Π/2) determinant (((J=∫_0 ^(Π/2) (dx/(1+cos^2 (x))))),((I=2J−(Π/2)))) e

$$\:\:\: \\ $$$$\:\:{Bonus}\:{du}\:{Mardi}\:\mathrm{12}/\mathrm{07}/\mathrm{2022} \\ $$$$\:\:\:\:{I}=\:\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \:\frac{{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)}{dx}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:{J}\:\:=\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)} \\ $$$$\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{\mathrm{2}−\left(\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)\right.}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)}−\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \:{dx}\:.\:\: \\ $$$$=\:\mathrm{2}{J}−\frac{\Pi}{\mathrm{2}} \\ $$$$ \\ $$$$\begin{array}{|c|c|}{{J}=\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)}}\\{{I}=\mathrm{2}{J}−\frac{\Pi}{\mathrm{2}}}\\\hline\end{array} \\ $$$$\:{e}\: \\ $$

Question Number 173493 Answers: 1 Comments: 0

find lim_(x→0) tan(tanhx)−tanh(tanx)

$$\boldsymbol{{find}}\:\boldsymbol{{lim}}_{\boldsymbol{{x}}\rightarrow\mathrm{0}} \:\boldsymbol{{tan}}\left(\boldsymbol{{tanhx}}\right)−\boldsymbol{{tanh}}\left(\boldsymbol{{tanx}}\right) \\ $$

Question Number 173489 Answers: 3 Comments: 0

Question Number 173486 Answers: 0 Comments: 1

𝛗 = ∫_0 ^( ∞) ((cos (ax)− sin(bx))/x^( 2) )dx =? 𝛗=∫_0 ^( ∞) ((1−2sin^( 2) (((ax)/2))−(1−2sin^( 2) (((bx)/2))))/(x^2 ))dx =2 {∫_0 ^( ∞) (((sin(((bx)/2)))/x))^2 dx=Θ_1 }−2{∫_0 ^( ∞) (((sin(((ax)/2)))/x))^2 dx=Θ_2 } Θ_( 1) =^(((bx)/2)=t) (b/2)∫_0 ^( ∞) ((sin^( 2) (t))/t^( 2) )dt= ((πb)/4) similarly : Θ_( 2) =((πa)/4) ∴ 𝛗= (π/2)(∣b∣−∣a∣) Dirichlet′s integrals: ∫_0 ^( ∞) ((sin(x))/x)dx=(π/2)=∫_0 ^( ∞) ((sin^2 (x))/x^( 2) )dx

$$ \\ $$$$\:\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\:\left({ax}\right)−\:{sin}\left({bx}\right)}{{x}^{\:\mathrm{2}} }{dx}\:\:=? \\ $$$$\:\:\: \\ $$$$\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−\mathrm{2}{sin}^{\:\mathrm{2}} \left(\frac{{ax}}{\mathrm{2}}\right)−\left(\mathrm{1}−\mathrm{2}{sin}^{\:\mathrm{2}} \left(\frac{{bx}}{\mathrm{2}}\right)\right)}{{x}\:^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:=\mathrm{2}\:\left\{\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{sin}\left(\frac{{bx}}{\mathrm{2}}\right)}{{x}}\right)^{\mathrm{2}} {dx}=\Theta_{\mathrm{1}} \right\}−\mathrm{2}\left\{\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{sin}\left(\frac{{ax}}{\mathrm{2}}\right)}{{x}}\right)^{\mathrm{2}} {dx}=\Theta_{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\Theta_{\:\mathrm{1}} \overset{\frac{{bx}}{\mathrm{2}}={t}} {=}\:\frac{{b}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\:\mathrm{2}} \left({t}\right)}{{t}^{\:\mathrm{2}} }{dt}=\:\frac{\pi{b}}{\mathrm{4}}\: \\ $$$$\:\:\:\:\:\:\:\:{similarly}\::\:\:\Theta_{\:\mathrm{2}} =\frac{\pi{a}}{\mathrm{4}}\:\:\:\:\:\therefore\:\:\:\boldsymbol{\phi}=\:\frac{\pi}{\mathrm{2}}\left(\mid{b}\mid−\mid{a}\mid\right)\: \\ $$$$\:\:\:\:\:{Dirichlet}'{s}\:{integrals}:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{2}} \left({x}\right)}{{x}^{\:\mathrm{2}} }{dx} \\ $$

Question Number 173485 Answers: 2 Comments: 0

Solve (x ∈ R ) x2^( (1/x)) +(1/x) 2^( x) = 4 −Source: L.Panaitopol

$$ \\ $$$$\:\:\:\:\:{Solve}\:\:\:\:\:\:\:\left({x}\:\in\:\mathbb{R}\:\right) \\ $$$$\:\:\:\:\:\:{x}\mathrm{2}^{\:\frac{\mathrm{1}}{{x}}} \:+\frac{\mathrm{1}}{{x}}\:\mathrm{2}^{\:{x}} =\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:−{Source}:\:{L}.{Panaitopol} \\ $$

Question Number 173484 Answers: 2 Comments: 0

1) lim_(x→0) ((x^(sinx) −(sinx)^x )/(x^(cosx) +1)) 2) lim_(x→∞) [x lnx − 2x ln(sin(1/( (√x)))) ] how can solve this proplem ?

$$\left.\mathrm{1}\right)\:\boldsymbol{{lim}}_{\boldsymbol{{x}}\rightarrow\mathrm{0}} \:\frac{\boldsymbol{{x}}^{\boldsymbol{{sinx}}} −\left(\boldsymbol{{sinx}}\right)^{\boldsymbol{{x}}} }{\boldsymbol{{x}}^{\boldsymbol{{cosx}}} +\mathrm{1}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\boldsymbol{{lim}}_{\boldsymbol{{x}}\rightarrow\infty} \:\left[\boldsymbol{{x}}\:\boldsymbol{{lnx}}\:−\:\mathrm{2}\boldsymbol{{x}}\:\boldsymbol{{ln}}\left(\boldsymbol{{sin}}\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{x}}}}\right)\:\right] \\ $$$$ \\ $$$$\boldsymbol{{how}}\:\boldsymbol{{can}}\:\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{proplem}}\:? \\ $$

Question Number 173445 Answers: 4 Comments: 0

Question Number 173431 Answers: 1 Comments: 2

Question Number 173430 Answers: 0 Comments: 3

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