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Question Number 222175 Answers: 1 Comments: 0

solve (e^(2y) −y)cosx(dy/dx)=e^y sin2x klipto−quanta

$$\boldsymbol{\mathrm{solve}} \\ $$$$\left(\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\mathrm{y}}} −\boldsymbol{\mathrm{y}}\right)\boldsymbol{\mathrm{cosx}}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{y}}} \boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}} \\ $$

Question Number 222172 Answers: 2 Comments: 0

(4/5) > (8/(3x − 6)) > (2/9) find: x = ?

$$\frac{\mathrm{4}}{\mathrm{5}}\:>\:\frac{\mathrm{8}}{\mathrm{3x}\:−\:\mathrm{6}}\:>\:\frac{\mathrm{2}}{\mathrm{9}} \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Question Number 222151 Answers: 3 Comments: 8

Question Number 222153 Answers: 1 Comments: 1

Question Number 222141 Answers: 2 Comments: 0

x^2 +((x/(2x−1)))^2 =12

$${x}^{\mathrm{2}} +\left(\frac{{x}}{\mathrm{2}{x}−\mathrm{1}}\right)^{\mathrm{2}} =\mathrm{12} \\ $$

Question Number 222142 Answers: 2 Comments: 0

a=3(√2) ,b=(1/(5^(1/6) (√6))) and x,yεR such that 3x +2y=log _a (18)^(5/4) 2x−y=log _b ((√(1080))) then find the value of 4x+5y

$${a}=\mathrm{3}\sqrt{\mathrm{2}}\:,{b}=\frac{\mathrm{1}}{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{6}}} \sqrt{\mathrm{6}}}\:{and}\:{x},{y}\epsilon\mathbb{R}\:{such}\:{that} \\ $$$$\mathrm{3}{x}\:+\mathrm{2}{y}=\mathrm{log}\:_{{a}} \left(\mathrm{18}\right)^{\frac{\mathrm{5}}{\mathrm{4}}} \\ $$$$\mathrm{2}{x}−{y}=\mathrm{log}\:_{{b}} \left(\sqrt{\mathrm{1080}}\right) \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\: \\ $$$$\mathrm{4}{x}+\mathrm{5}{y} \\ $$

Question Number 222127 Answers: 0 Comments: 0

∫_0 ^( ∞) ((ln(tan(tan^(−1) (e^((1/π) tan^(−1) u) ))) )/(u^2 + 2πu + 2π^2 )) du

$$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{ln}\left(\mathrm{tan}\left(\mathrm{tan}^{−\mathrm{1}} \left({e}^{\frac{\mathrm{1}}{\pi}\:\mathrm{tan}^{−\mathrm{1}} \:{u}} \right)\right)\right)\:}{{u}^{\mathrm{2}} \:+\:\mathrm{2}\pi{u}\:+\:\mathrm{2}\pi^{\mathrm{2}} }\:{du} \\ $$$$ \\ $$

Question Number 222125 Answers: 4 Comments: 2

If: log_3 (5^x + (1/5^x ) + 7) ⇒ min = ?

$$\mathrm{If}:\:\:\:\mathrm{log}_{\mathrm{3}} \left(\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:+\:\frac{\mathrm{1}}{\mathrm{5}^{\boldsymbol{\mathrm{x}}} }\:+\:\mathrm{7}\right)\:\:\:\Rightarrow\:\:\:\mathrm{min}\:=\:? \\ $$

Question Number 222123 Answers: 1 Comments: 1

Question Number 222121 Answers: 2 Comments: 0

∫ acos(((cos(ϱ))/(1+2cos(ϱ)))) dϱ

$$\int\:\mathrm{acos}\left(\frac{\mathrm{cos}\left(\varrho\right)}{\mathrm{1}+\mathrm{2cos}\left(\varrho\right)}\right)\:\mathrm{d}\varrho \\ $$

Question Number 222117 Answers: 0 Comments: 0

problem 1. for a given positive integer m, find all triples(n,x,y) of positive integers,with n relatively prime to m,which satisfy (x^2 +y^2 )^m =(xy)^n hint:utilize AM&GM,diophantine eqn KLIPTO−QUANTA−OOZY

Question Number 222113 Answers: 0 Comments: 2

name the following compound

$$\mathrm{name}\:\mathrm{the}\:\mathrm{following}\:\mathrm{compound} \\ $$

Question Number 222105 Answers: 1 Comments: 1

Question Number 222104 Answers: 2 Comments: 0

log _4 x − log _x^2 8 = 1 x =?

$$\:\:\:\:\mathrm{log}\:_{\mathrm{4}} \:\mathrm{x}\:−\:\mathrm{log}\:_{\mathrm{x}^{\mathrm{2}} } \:\mathrm{8}\:=\:\mathrm{1} \\ $$$$\:\:\:\:\mathrm{x}\:=?\: \\ $$

Question Number 222100 Answers: 0 Comments: 0

Question Number 222097 Answers: 0 Comments: 2

Question Number 222095 Answers: 0 Comments: 0

could I consider Y_ν (z)=cot(νπ)J_ν (z)−csc(νπ)J_(−ν) (z) as ∞−∞ form limit when ν∈Z and How can i calculate Y_ν (z)=cot(νπ)J_ν (z)−csc(νπ)J_(−ν) (z)...?? lim_(α→ν) ((cot^2 (απ)J_α ^2 (z)−csc^2 (απ)J_(−α) ^( 2) (z))/(cot(απ)J_α (z)+csc(απ)J_(−α) (z)))..... lim_(α→ν) ((((∂ )/∂α)(cot^2 (απ)J_α ^2 (z)−csc^2 (απ)J_(−α) ^2 (z)))/(((∂ )/∂α)(cot(απ)J_α ^ (z)+csc(απ)J_(−α) (z))))....??.... :(

$$\mathrm{could}\:\:\mathrm{I}\:\mathrm{consider}\:\:{Y}_{\nu} \left({z}\right)=\mathrm{cot}\left(\nu\pi\right){J}_{\nu} \left({z}\right)−\mathrm{csc}\left(\nu\pi\right){J}_{−\nu} \left({z}\right) \\ $$$$\mathrm{as}\:\infty−\infty\:\mathrm{form}\:\mathrm{limit}\:\mathrm{when}\:\nu\in\mathbb{Z} \\ $$$$\mathrm{and}\:\mathrm{How}\:\mathrm{can}\:\mathrm{i}\:\mathrm{calculate} \\ $$$${Y}_{\nu} \left({z}\right)=\mathrm{cot}\left(\nu\pi\right){J}_{\nu} \left({z}\right)−\mathrm{csc}\left(\nu\pi\right){J}_{−\nu} \left({z}\right)...?? \\ $$$$\underset{\alpha\rightarrow\nu} {\mathrm{lim}}\:\frac{\mathrm{cot}^{\mathrm{2}} \left(\alpha\pi\right){J}_{\alpha} ^{\mathrm{2}} \left({z}\right)−\mathrm{csc}^{\mathrm{2}} \left(\alpha\pi\right){J}_{−\alpha} ^{\:\mathrm{2}} \left({z}\right)}{\mathrm{cot}\left(\alpha\pi\right){J}_{\alpha} \left({z}\right)+\mathrm{csc}\left(\alpha\pi\right){J}_{−\alpha} \left({z}\right)}..... \\ $$$$\underset{\alpha\rightarrow\nu} {\mathrm{lim}}\frac{\frac{\partial\:\:}{\partial\alpha}\left(\mathrm{cot}^{\mathrm{2}} \left(\alpha\pi\right){J}_{\alpha} ^{\mathrm{2}} \left({z}\right)−\mathrm{csc}^{\mathrm{2}} \left(\alpha\pi\right){J}_{−\alpha} ^{\mathrm{2}} \left({z}\right)\right)}{\frac{\partial\:\:}{\partial\alpha}\left(\mathrm{cot}\left(\alpha\pi\right){J}_{\alpha} ^{\:} \left({z}\right)+\mathrm{csc}\left(\alpha\pi\right){J}_{−\alpha} \left({z}\right)\right)}....??.... \\ $$$$:\left(\right. \\ $$

Question Number 222076 Answers: 2 Comments: 0