Question Number 5561 by Rasheed Soomro last updated on 20/May/16 $$\mathrm{A}\:\mathrm{chord}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{that} \\ $$$$\mathrm{divides}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{in}\:\mathrm{1}:\mathrm{2}\:\left(\mathrm{ratio}\right). \\ $$$$\mathrm{In}\:\mathrm{what}\:\mathrm{ratio}\:\mathrm{the}\:\mathrm{chord}\:\mathrm{divides} \\ $$$$\mathrm{the}\:\mathrm{diameter}\:\mathrm{perpendicular}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{chord}? \\ $$ Commented by Rasheed Soomro…
Question Number 5556 by Rasheed Soomro last updated on 20/May/16 $$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{diretly}\: \\ $$$$\mathrm{proportional}\:\mathrm{to}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of} \\ $$$$\mathrm{its}\:\mathrm{diameter}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{of}\:\mathrm{proportionality}? \\ $$ Commented by Yozzii last updated on…
Question Number 71091 by Henri Boucatchou last updated on 11/Oct/19 Commented by Prithwish sen last updated on 11/Oct/19 $$\boldsymbol{\mathrm{It}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{same}}\:\boldsymbol{\mathrm{as}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}}\:\boldsymbol{\mathrm{dx}}\:−\int_{\mathrm{0}}…
Question Number 136626 by mnjuly1970 last updated on 24/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……….{nice}\:\:\:\:{calculus}……… \\ $$$$\:\:\:\:{suppose}\:{that}::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\varphi\left({p}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\left({p}+{x}\right)^{\mathrm{2}} }\:…\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}:: \\ $$$$\:\:\:\:\:\:\:\underset{\mathrm{0}} {\int}^{\:\mathrm{1}} \:\frac{\varphi\left({p}\right)}{\mathrm{1}+{p}}{dp}=?… \\ $$$$\:\:\:\:\:\:…
Question Number 71089 by 20190927 last updated on 11/Oct/19 $$\left(\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\right)^{\mathrm{x}} +\left(\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}\right)^{\mathrm{x}} =\mathrm{4} \\ $$ Answered by Henri Boucatchou last updated on 11/Oct/19 $$\:\sqrt{\mathrm{2}}\:+\:\mathrm{1}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:−\:\mathrm{1}},\:\boldsymbol{{let}}\:\:\boldsymbol{{a}}\:=\:\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$\:\left(\sqrt{\sqrt{\mathrm{2}}\:−\mathrm{1}}\right)^{\boldsymbol{{x}}}…
Question Number 71082 by aliesam last updated on 11/Oct/19 $${prove}\:{that}\: \\ $$$$ \\ $$$$\mid\:\sqrt{\mid{x}\mid}\:−\:\sqrt{\mid{y}\mid}\:\mid\:\leqslant\:\sqrt{\mid{x}−{y}\mid}\: \\ $$$$ \\ $$ Answered by Henri Boucatchou last updated on…
Question Number 5546 by FilupSmith last updated on 19/May/16 $$\mathrm{I}\:\mathrm{have}\:\mathrm{two}\:\mathrm{circles}\:\mathrm{side}\:\mathrm{by}\:\mathrm{side}. \\ $$$$\mathrm{Circle}\:\mathrm{1}\:\mathrm{has}\:\mathrm{radius}\:{r}. \\ $$$$\mathrm{Circle}\:\mathrm{2}\:\mathrm{has}\:\mathrm{radius}\:\frac{\mathrm{1}}{\mathrm{3}}{r}. \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{I}\:\mathrm{roll}\:\mathrm{circle}\:\mathrm{2}\:\mathrm{around}\:\mathrm{circle}\:\mathrm{1}\:\mathrm{until} \\ $$$$\mathrm{it}\:\mathrm{reaches}\:\mathrm{the}\:\mathrm{begining}, \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{times}\:\mathrm{will}\:\mathrm{it}\:\mathrm{roll}? \\ $$ Commented…
Question Number 5543 by Rasheed Soomro last updated on 19/May/16 $$\bullet\mathrm{What}\:\boldsymbol{\mathrm{plane}}\:\boldsymbol{\mathrm{geometrical}}\:\boldsymbol{\mathrm{figures}} \\ $$$$\mathrm{could}\:\mathrm{be}\:\mathrm{produced}\:\mathrm{by}\:\mathrm{joining} \\ $$$$\boldsymbol{\mathrm{vertices}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{cube}}? \\ $$$$\left(\mathrm{for}\:\mathrm{example}\::\:\mathrm{square}\right) \\ $$$$\bullet\mathrm{What}\:\mathrm{largest}\:\mathrm{area}\left(\mathrm{2}\:\mathrm{dimensional}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{obtained} \\ $$$$\mathrm{by}\:\mathrm{joining}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{cube}\:\mathrm{when}\:\mathrm{its}\:\mathrm{side}\:\mathrm{is}\:\mathrm{x}? \\ $$ Terms of…
Question Number 5542 by FilupSmith last updated on 19/May/16 $$\mathrm{The}\:\mathrm{surface}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{cylinder}\:\mathrm{is}: \\ $$$${A}=\mathrm{2}\pi{r}^{\mathrm{2}} +\mathrm{2}\pi{rh} \\ $$$$ \\ $$$$\mathrm{As}\:{h}\rightarrow\mathrm{0},\:\mathrm{the}\:\mathrm{shape}\:\mathrm{becomes}\:\mathrm{a}\:\mathrm{2D}\:\mathrm{circle}, \\ $$$$\mathrm{so}\:\mathrm{should}\:\:\:\:\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{A}\:=\:\pi{r}^{\mathrm{2}} \:\:\:??? \\ $$$$\mathrm{Or}\:\mathrm{is}\:\mathrm{it}\:\mathrm{that}\:\mathrm{as}\:{h}\rightarrow\mathrm{0},\:\mathrm{it}\:\mathrm{creates}\:\mathrm{a}\:\mathrm{circle} \\ $$$$\mathrm{of}\:\mathrm{3}\:\mathrm{dimensions}\:\mathrm{with}\:\mathrm{infintesimally}\:\mathrm{small}…
Question Number 136614 by bemath last updated on 24/Mar/21 $${If}\:{a}+\frac{\mathrm{1}}{{a}}\:=\:\mathrm{23}\:{then}\:\sqrt[{\mathrm{4}}]{{a}}\:+\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{a}}}\:=? \\ $$ Answered by mr W last updated on 24/Mar/21 $${a}+\frac{\mathrm{1}}{{a}}=\left(\sqrt{{a}}+\frac{\mathrm{1}}{\:\sqrt{{a}}}\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{23} \\ $$$$\sqrt{{a}}+\frac{\mathrm{1}}{\:\sqrt{{a}}}=\mathrm{5} \\…