Question Number 72641 by aliesam last updated on 31/Oct/19 Commented by mind is power last updated on 31/Oct/19 $$\mathrm{nice}\:\:\mathrm{integral}\:\:!!! \\ $$$$ \\ $$ Commented by…
Question Number 72639 by kennethalli last updated on 31/Oct/19 $${prove}:\frac{\pi}{\mathrm{5}}\left(\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\pi \\ $$ Answered by MJS last updated on 31/Oct/19 $$\mathrm{are}\:\mathrm{you}\:\mathrm{serious}? \\ $$$$\frac{\pi}{\mathrm{5}}\left(\mathrm{3}^{\mathrm{2}}…
Question Number 7102 by Tawakalitu. last updated on 10/Aug/16 Commented by 123456 last updated on 10/Aug/16 $${S}_{\mathrm{0}} =\mathrm{2} \\ $$$${S}_{\mathrm{1}} =\mathrm{2} \\ $$ Commented by…
Question Number 138175 by ajfour last updated on 10/Apr/21 Commented by ajfour last updated on 10/Apr/21 $${If}\:{the}\:{yellow}\:{and}\:{blue}\:{areas}\:{are} \\ $$$${equal},\:{find}\:{edge}\:{length}\:{of}\:{the} \\ $$$${blue}\:{square}.\:\left({radius}\:{of}\:{circle}=\mathrm{1}\right) \\ $$ Answered by…
Question Number 7101 by Tawakalitu. last updated on 10/Aug/16 Commented by Tawakalitu. last updated on 10/Aug/16 $${Working}\:{please}\:………… \\ $$ Answered by Yozzii last updated on…
Question Number 72634 by Rio Michael last updated on 30/Oct/19 $${help}\:{me}\:{with}\:{the}\:{conditions}\:{please}\: \\ $$$${for}\:{a}\:{function}\:{f}\:{to}\:{be}\:{continuous}\:{at}\:{a}\:{point}\:{a} \\ $$ Commented by Prithwish sen last updated on 31/Oct/19 $$\boldsymbol{\mathrm{If}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{able}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{draw}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{function}}\:\boldsymbol{\mathrm{f}}\:\:\boldsymbol{\mathrm{through}} \\…
Question Number 138171 by mnjuly1970 last updated on 11/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:…….{mathematical}….{analysis}……. \\ $$$$\:\:\:\:\:{if}\::\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{arctan}\left({x}\right).{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{48}}\left({a}\pi^{\mathrm{2}} −{b}\pi{ln}\left(\mathrm{2}\right)+{c}\:{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:……{then}\:\:….\:{a}^{\mathrm{2}} +\left({b}−{c}+\mathrm{1}\right)^{\mathrm{2}} =??? \\…
Question Number 72633 by Rio Michael last updated on 30/Oct/19 $${prove}\:{using}\:{th}\:{sandwich}\:{or}\:{Squeez}\:{theorem}\:{that} \\ $$$${for}\:{any}\:\:{a}\:>\:\mathrm{0} \\ $$$$\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\sqrt{{x}}\:=\:\sqrt{{a}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 7096 by Tawakalitu. last updated on 10/Aug/16 Commented by Rasheed Soomro last updated on 10/Aug/16 $${First}\:{vertex}\:{of}\:{a}\:{triangle}\:{can}\:{be} \\ $$$${chosen}\:{in}\:{n}\:{ways}\:{in}\:{a}\:{regular}\:{n}-{gon}. \\ $$$${The}\:{second}\:{vertex}\:{can}\:{be}\:{chosen}\:{in} \\ $$$${n}−\mathrm{1}\:{ways}.\:{The}\:{third}\:{vertex}\:{can}\:{be}\: \\…
Question Number 138167 by physicstutes last updated on 10/Apr/21 $$\mathrm{let}\:{f}\left({x}\right)\:=\:\begin{vmatrix}{{x}}&{{x}^{\mathrm{2}} }&{{x}^{\mathrm{3}} }\\{\mathrm{0}}&{\mathrm{2}{x}\:}&{\mathrm{3}{x}^{\mathrm{2}} }\\{\mathrm{1}}&{\mathrm{0}}&{{x}}\end{vmatrix}\:\mathrm{find} \\ $$$$\:{f}\:'\left({x}\right)\: \\ $$ Answered by Dwaipayan Shikari last updated on 10/Apr/21…