Question Number 212790 by RojaTaniya last updated on 24/Oct/24 Answered by A5T last updated on 24/Oct/24 $${x}−{y}=\left({y}−\mathrm{1}+{x}−\mathrm{1}\right)\left({y}−\mathrm{1}−{x}+\mathrm{1}\right) \\ $$$$\Rightarrow{x}={y}\:{or}\:−\mathrm{1}={x}+{y}−\mathrm{2}\Rightarrow{x}+{y}=\mathrm{1} \\ $$$$\Rightarrow{x}+\mathrm{1}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\:{or}\:{x}+\mathrm{1}={x}^{\mathrm{2}} \Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={x}+{y}+\mathrm{2}=\mathrm{3}…
Question Number 212816 by Akayx last updated on 24/Oct/24 Answered by mahdipoor last updated on 24/Oct/24 $${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}\left({t}−\mathrm{4}{k}\right)\left({u}_{\mathrm{4}{k}} −{u}_{\mathrm{4}{k}+\mathrm{2}} \right)= \\ $$$$\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left({t}−\mathrm{4}{k}\right){u}_{\mathrm{4}{k}}…
Question Number 212808 by RoseAli last updated on 24/Oct/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\left({x}^{.\mathrm{1}} −{x}^{.\mathrm{9}} \right)^{\mathrm{1}{o}} −{x}\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212809 by MrGaster last updated on 24/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{certificate}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{csc}\frac{{x}}{\mathrm{2}^{{k}} }=\mathrm{cot}\frac{{x}}{\mathrm{2}^{{n}+\mathrm{1}} }−\mathrm{cot}\:{x} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212801 by Spillover last updated on 24/Oct/24 Answered by som(math1967) last updated on 24/Oct/24 Commented by som(math1967) last updated on 24/Oct/24 $$\frac{{OL}}{{PL}}=\mathrm{tan}\:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}} \\…
Question Number 212802 by Spillover last updated on 24/Oct/24 Answered by mr W last updated on 24/Oct/24 Commented by mr W last updated on 24/Oct/24…
Question Number 212803 by Spillover last updated on 24/Oct/24 Answered by Spillover last updated on 24/Oct/24 Answered by Spillover last updated on 24/Oct/24 Answered by…
Question Number 212765 by issac last updated on 23/Oct/24 $$\mathrm{i}\:\:\mathrm{generalized}\:\boldsymbol{\mathrm{Bessel}}\:\boldsymbol{\mathrm{function}}'\mathrm{s} \\ $$$$\mathrm{Laplace}\:\mathrm{Transform} \\ $$$$\mathrm{L}.\mathrm{T}{J}_{\nu} \left({z}\right)=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}\:,\:{s}\in\left[\mathrm{0},\infty\right)\:,\:\nu\in\mathbb{R} \\ $$$$\mathrm{L}.\mathrm{T}\:{Y}_{\nu} \left({z}\right)=\frac{\mathrm{cot}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}−\frac{\mathrm{csc}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{\nu}…
Question Number 212760 by Spillover last updated on 23/Oct/24 Answered by mr W last updated on 23/Oct/24 Commented by mr W last updated on 23/Oct/24…
Question Number 212761 by Spillover last updated on 23/Oct/24 Commented by mr W last updated on 23/Oct/24 $${AB}=\mathrm{2}×{GE}=\mathrm{8}\sqrt{\mathrm{7}} \\ $$ Answered by som(math1967) last updated…