Question Number 138275 by mathmax by abdo last updated on 11/Apr/21 $$\left.\mathrm{1}\right)\mathrm{calculate}\:\mathrm{U}_{\mathrm{n}} =\int\int_{\left[\frac{\mathrm{1}}{\mathrm{n}},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\left(\mathrm{2x}+\mathrm{3y}\right)\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} }\mathrm{dxdy} \\ $$$$\left.\mathrm{2}\right)\mathrm{find}\:\int\int_{\left.\right]\left.\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \left(\mathrm{2x}+\mathrm{3y}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\mathrm{dxdy} \\ $$…
Question Number 7200 by Tawakalitu. last updated on 16/Aug/16 $${Prove}\:{that}\: \\ $$$$\frac{{sin}\left(\frac{\Theta}{\mathrm{2}}\right)}{\left(\frac{\Theta}{\mathrm{2}}\right)}\:\:=\:\:\mathrm{1} \\ $$ Commented by Rasheed Soomro last updated on 16/Aug/16 $${Do}\:{you}\:{mean}\:\underset{\Theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\Theta}{\mathrm{2}}\right)\:}{\frac{\Theta}{\mathrm{2}}}=\mathrm{1}\:? \\…
Question Number 72734 by Mr Jor last updated on 01/Nov/19 $${A}\:{triangle}\:{ABC}\:{is}\:{inscribed}\:{in}\:{a} \\ $$$${circle}.{AC}=\mathrm{10}{cm},{BC}=\mathrm{7}{cm}\:{and}\: \\ $$$${AB}=\mathrm{10}{cm}.{Find}\:{the}\:{radius}\:{of}\:{the} \\ $$$${circle}. \\ $$ Answered by $@ty@m123 last updated on…
Question Number 7196 by Tawakalitu. last updated on 15/Aug/16 Answered by Rasheed Soomro last updated on 18/Aug/16 $$\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{4}}&{\mathrm{5}}&{\mathrm{6}}&{\mathrm{7}}&{\mathrm{8}}&{\mathrm{9}}&{\mathrm{10}}\\{\mathrm{10}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{4}}&{\mathrm{5}}&{\mathrm{6}}&{\mathrm{7}}&{\mathrm{8}}&{\mathrm{9}}\\{\mathrm{9}}&{\mathrm{10}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{4}}&{\mathrm{5}}&{\mathrm{6}}&{\mathrm{7}}&{\mathrm{8}}\\{\mathrm{8}}&{\mathrm{9}}&{\mathrm{10}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{4}}&{\mathrm{5}}&{\mathrm{6}}&{\mathrm{7}}\\{\mathrm{7}}&{\mathrm{8}}&{\mathrm{9}}&{\mathrm{10}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{4}}&{\mathrm{5}}&{\mathrm{6}}\\{\mathrm{6}}&{\mathrm{7}}&{\mathrm{8}}&{\mathrm{9}}&{\mathrm{10}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{4}}&{\mathrm{5}}\\{\mathrm{5}}&{\mathrm{6}}&{\mathrm{7}}&{\mathrm{8}}&{\mathrm{9}}&{\mathrm{10}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{4}}\\{\mathrm{4}}&{\mathrm{5}}&{\mathrm{6}}&{\mathrm{7}}&{\mathrm{8}}&{\mathrm{9}}&{\mathrm{10}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{3}}&{\mathrm{4}}&{\mathrm{5}}&{\mathrm{6}}&{\mathrm{7}}&{\mathrm{8}}&{\mathrm{9}}&{\mathrm{10}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{4}}&{\mathrm{5}}&{\mathrm{6}}&{\mathrm{7}}&{\mathrm{8}}&{\mathrm{9}}&{\mathrm{10}}&{\mathrm{1}}\end{bmatrix}\:\:\:\: \\ $$$${Subtracting}\:{each}\:{row}\left({start}\:{from}\:\mathrm{2}{nd}\:{row}\right)\:\:{from}\:{previous}\:{row} \\ $$$$\begin{bmatrix}{-\mathrm{9}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{-\mathrm{9}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{-\mathrm{9}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{-\mathrm{9}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{-\mathrm{9}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{-\mathrm{9}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{-\mathrm{9}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{-\mathrm{9}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{-\mathrm{9}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{4}}&{\mathrm{5}}&{\mathrm{6}}&{\mathrm{7}}&{\mathrm{8}}&{\mathrm{9}}&{\mathrm{10}}&{\mathrm{1}}\end{bmatrix}\:\:\:\:\:\:\:\: \\ $$$${Again}\:{subtracting}\:{each}\:{row}\:{from}\:{previous}\:{row} \\…
Question Number 7193 by peter james last updated on 15/Aug/16 Commented by Yozzia last updated on 15/Aug/16 $${S}\left({n}\right)=\mathrm{1}+\mathrm{2}^{\mathrm{2}} −\mathrm{3}^{\mathrm{3}} +\mathrm{4}+\mathrm{5}^{\mathrm{2}} −\mathrm{6}^{\mathrm{3}} +\mathrm{7}+\mathrm{8}^{\mathrm{2}} −\mathrm{9}^{\mathrm{3}} +… \\…
Question Number 7191 by Tawakalitu. last updated on 15/Aug/16 $${Evaluate}\:\:\:\:\:\Sigma\:\frac{{sin}\left(\mathrm{3}{n}\right)}{{n}}\:\:\:\:\:{from}\:\:\mathrm{1}\:\:{to}\:\:{infinity}\: \\ $$ Answered by Yozzia last updated on 15/Aug/16 $${Define}\:{the}\:{function}\: \\ $$$${f}\left({x}\right)={x}\:{for}\:\mathrm{0}<{x}<\mathrm{1}\:,\:{period}=\mathrm{2}. \\ $$$${For}\:{Fourier}\:{series}\:{of}\:{f}\:{having}\:{the}\:{form} \\…
Question Number 72724 by rajesh4661kumar@gmail.com last updated on 01/Nov/19 Commented by kaivan.ahmadi last updated on 01/Nov/19 $$\frac{\pi}{\mathrm{2}}<{x}_{\mathrm{1}} <{x}_{\mathrm{2}} <\pi\Rightarrow\mathrm{0}<\mid{cosx}_{\mathrm{2}} \mid<\mid{cosx}_{\mathrm{1}} \mid<\mathrm{1}\Rightarrow \\ $$$${log}\mid{cosx}_{\mathrm{2}} \mid<{log}\mid{cosx}_{\mathrm{1}} \mid…
Question Number 7189 by Tawakalitu. last updated on 15/Aug/16 $${Evaluate}\:\:\::\:\:\:\Sigma\:\frac{{sin}\left({n}\right)}{{n}}\:\:,\:\:\:{From}\:\:\:\mathrm{1}\:{to}\:\:{infinity}. \\ $$ Commented by Yozzia last updated on 15/Aug/16 $${f}\left({x}\right)=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{{a}_{{n}} {cos}\frac{{n}\pi{x}}{{L}}+{b}_{{n}} {sin}\frac{{n}\pi{x}}{{L}}\right\}…
Question Number 138257 by greg_ed last updated on 11/Apr/21 $$\boldsymbol{\mathrm{hi}}\:! \\ $$$$\boldsymbol{\mathrm{calculate}}\::\: \\ $$$$\int\int_{\mathrm{A}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdy}\:{with}\:\mathrm{A}=\left\{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\leqslant\:\mathrm{1}\right\} \\ $$ Answered by…
Question Number 7187 by 314159 last updated on 15/Aug/16 $${Prove}\:{that}\:\frac{\sqrt{\mathrm{4}{sin}^{\mathrm{2}} \mathrm{36}−\mathrm{1}}}{\mathrm{2}}={cos}\mathrm{72}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com