Question Number 208637 by efronzo1 last updated on 20/Jun/24 $$\:\:\:{s} \\ $$ Answered by Berbere last updated on 20/Jun/24 $${find}\:{fractin}\:{that}\:{one}\:{man}\:{can}\:{compete}\:{per}\:{day}\:{V}_{{m}} \\ $$$${and}\:{women}\:{V}_{{w}} \\ $$$${V}_{{m}} =\frac{\mathrm{1}}{\mathrm{16}.\mathrm{24}}…
Question Number 208670 by hardmath last updated on 20/Jun/24 $$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{5}} \:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{2x}\:+\:\mathrm{1}\right)}\:\mathrm{dx}\:\:=\:\:? \\ $$ Commented by essaad last updated on 20/Jun/24 Answered by Berbere…
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Question Number 208639 by Tawa11 last updated on 20/Jun/24 Commented by mr W last updated on 20/Jun/24 $${is}\:{it}\:{a}\:{red}\:{square}? \\ $$$${does}\:{one}\:{corner}\:{of}\:{the}\:{square}\:{lie} \\ $$$${on}\:{the}\:{center}\:{of}\:{semicircle}? \\ $$ Commented…
Question Number 208661 by efronzo1 last updated on 20/Jun/24 $$\:\:\downharpoonleft\underline{\:} \\ $$ Answered by Berbere last updated on 20/Jun/24 $$\mathrm{3}{x}+\mathrm{4}={u}\Rightarrow{dx}=\frac{{du}}{\mathrm{3}} \\ $$$$\int_{\mathrm{10}} ^{\mathrm{25}} {f}\left({u}\right).\frac{{du}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}}\left\{.\int_{\mathrm{10}} ^{\mathrm{15}}…
Question Number 208662 by efronzo1 last updated on 20/Jun/24 $$\:\:\frac{\begin{pmatrix}{\mathrm{n}}\\{\mathrm{0}}\end{pmatrix}\:+\mathrm{3}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}\:+\mathrm{5}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{2}}\end{pmatrix}\:+…+\left(\mathrm{2n}+\mathrm{1}\right)\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}}{\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}\:+\mathrm{2}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{2}}\end{pmatrix}\:+\:\mathrm{3}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{3}}\end{pmatrix}\:+…+\mathrm{n}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}}\:=\frac{\mathrm{23}}{\mathrm{11}} \\ $$$$\:\mathrm{n}=? \\ $$ Answered by Berbere last updated on 20/Jun/24 $${A}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}+\mathrm{1}\right)\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix};\underset{{k}=\mathrm{0}} {\overset{{n}}…
Question Number 208652 by efronzo1 last updated on 20/Jun/24 Answered by Berbere last updated on 20/Jun/24 $${a},{b}\:{solution}\:{of}\:−\mathrm{3}{x}^{\mathrm{3}} +\mathrm{2}{x}={c} \\ $$$${S}_{\mathrm{1}} =\int_{\mathrm{0}} ^{{a}} {c}−\left(\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{3}} \right)=\int_{{a}} ^{{b}}…
Question Number 208681 by Noorzai last updated on 20/Jun/24 Answered by Rasheed.Sindhi last updated on 21/Jun/24 $$\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\:=\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}}\:={a}+{b}\sqrt{\mathrm{3}}\:\left(>\mathrm{0}\right)\left({let}\right) \\ $$$$\:{Where}\:{a},{b}\in\mathbb{Z} \\ $$$$\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\:={a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{\mathrm{3}}\: \\ $$$${a}^{\mathrm{2}}…
Question Number 208676 by RoseAli last updated on 20/Jun/24 Answered by Berbere last updated on 20/Jun/24 $$\sqrt[{\mathrm{4}}]{{a}}+\sqrt[{\mathrm{4}}]{{b}}=\sqrt{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}=\sqrt[{\mathrm{4}}]{\mathrm{2}}.\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\mathrm{2}{a}}+\sqrt[{\mathrm{4}}]{\mathrm{2}{b}}=\sqrt{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}+\sqrt[{\mathrm{4}}]{\frac{{b}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$${b}=\mathrm{0}\:\Rightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1}\:{no}\:{solution}\:\mathbb{N} \\…
Question Number 208645 by Mastermind last updated on 20/Jun/24 $$\mathrm{Solve}\:: \\ $$$$\mathrm{2x}_{\mathrm{1}} \:−\:\lambda_{\mathrm{1}} \:−\:\mathrm{5}\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{2x}_{\mathrm{2}} \:−\:\lambda_{\mathrm{1}} \:−\:\mathrm{2}\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{2x}_{\mathrm{3}} \:−\:\mathrm{3}\lambda_{\mathrm{1}} \:−\:\lambda_{\mathrm{2}} \:=\:\mathrm{0}…