Question Number 76077 by vishalbhardwaj last updated on 23/Dec/19 $$\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{maximum}}\:\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{isosceles}}\:\boldsymbol{\mathrm{triangle}} \\ $$$$\boldsymbol{\mathrm{inscribed}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{ellipse}}\:\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} }\:+\:\frac{\boldsymbol{{y}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} }\:=\:\mathrm{1}\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{its}}\:\boldsymbol{\mathrm{vetrex}}\: \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{end}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{major}}\:\boldsymbol{\mathrm{axis}}\:?\:? \\ $$ Commented by MJS last updated…
Question Number 141585 by ZiYangLee last updated on 20/May/21 $$\mathrm{If}\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{4}\:\mathrm{and}\:{i}=\sqrt{−\mathrm{1}}\:,\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series} \\ $$$$\:\:\:{S}=\mathrm{1}+\mathrm{2}{i}+\mathrm{3}{i}^{\mathrm{2}} +……+\left({n}+\mathrm{1}\right){i}^{{n}} \: \\ $$ Answered by mathmax by abdo last updated…
Question Number 141586 by otchereabdullai@gmail.com last updated on 20/May/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{functions}\:\mathrm{are}\:\mathrm{undefined}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{g}:\mathrm{x}\rightarrow\frac{\mathrm{5x}+\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{f}:\mathrm{x}\rightarrow\mathrm{3x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{1}\: \\ $$ Answered by physicstutes last updated on…
Question Number 141580 by learner001 last updated on 20/May/21 $${i}\:{am}\:{finding}\:{it}\:{difficult}\:{to}\:{form}\:{a}\:{pde} \\ $$$${from}\:{the}\:{function}\:{f}\left({x}+{y}+{z},\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right). \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141569 by SOMEDAVONG last updated on 20/May/21 $$\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\left(\frac{\mathrm{110}^{\mathrm{2}} }{\left(\mathrm{11}−\mathrm{10}\right)\left(\mathrm{11}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} \right)}\:+\:\frac{\mathrm{110}^{\mathrm{2}} }{\left(\mathrm{11}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} \right)\left(\mathrm{11}^{\mathrm{3}} −\mathrm{10}^{\mathrm{3}} \right)}\:+\:….+\:\frac{\mathrm{110}^{\mathrm{2}} }{\left(\mathrm{11}^{\mathrm{n}} −\mathrm{10}^{\mathrm{n}} \right)\left(\mathrm{11}^{\mathrm{n}+\mathrm{1}} −\mathrm{10}^{\mathrm{n}+\mathrm{1}} \right)}\right) \\…
Question Number 76013 by vishalbhardwaj last updated on 22/Dec/19 Answered by som(math1967) last updated on 22/Dec/19 $$\left({y}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}.\mathrm{3}\left({x}−\mathrm{9}\right) \\ $$$$\therefore{Vertex}\:\left(\mathrm{9},\mathrm{4}\right)\:\:{focus}\left(\mathrm{3}+\mathrm{9}\:,\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\:\mathrm{12},\mathrm{4}\right) \\ $$$${equation}\:{of}\:{directrix}\:{x}+\mathrm{3}=\mathrm{9} \\…
Question Number 10474 by FilupS last updated on 13/Feb/17 $$\mathrm{Just}\:\mathrm{so}\:\mathrm{you}\:\mathrm{all}\:\mathrm{know},\:\mathrm{I}\:\mathrm{had}\:\mathrm{to}\:\mathrm{make} \\ $$$$\mathrm{a}\:\mathrm{new}\:\mathrm{username}.\:\mathrm{I}\:\mathrm{am}\:\mathrm{FilupSmith} \\ $$ Commented by sandy_suhendra last updated on 13/Feb/17 $$\mathrm{ok}\:\mathrm{Filup} \\ $$ Terms…
Question Number 10468 by FilupSmith last updated on 11/Feb/17 $$\mathrm{Show}\:\mathrm{why}: \\ $$$$\Gamma\left({x}+\mathrm{1}\right)\approx\sqrt{\mathrm{2}\pi}{e}^{−{x}} {x}^{{x}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141535 by mohammad17 last updated on 20/May/21 $${how}\:{can}\:{solve}\:{this}\:{with}\:{by}\:{steps}\:{please} \\ $$$$\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\mathrm{5}\right)}\:{i}\:{want}\:{solution}\:{step}\:{by}\:{step} \\ $$$${because}\:{i}\:{understand}\:{this}? \\ $$$$ \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 141525 by hgrocks last updated on 20/May/21 $$\mathrm{Let}\:<\mathrm{x}_{\mathrm{n}} >\:\mathrm{be}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{defined}\:\mathrm{by} \\ $$$$\mathrm{x}_{\mathrm{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{k}}\left(\mathrm{x}_{\mathrm{n}} +\frac{\mathrm{k}}{\mathrm{x}_{\mathrm{n}} }\right)\:\forall\:\mathrm{n}\:\in\:\mathbb{N} \\ $$$$\mathrm{Show}\:\mathrm{that}\:<\mathrm{x}_{\mathrm{n}} >\:\mathrm{converges}\:\mathrm{to}\:\sqrt{\frac{\mathrm{k}}{\mathrm{k}−\mathrm{1}}} \\ $$$$\mathrm{x}_{\mathrm{1}} >\mathrm{0}\:,\:\mathrm{k}>\mathrm{1} \\ $$ Answered…