Question Number 225954 by fantastic2 last updated on 17/Nov/25 $$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x} \\ $$$$ \\ $$ Answered by mr W last updated on 17/Nov/25 Commented by mr…
Question Number 226003 by ajfour last updated on 17/Nov/25 $${If}\:\:{r}^{\mathrm{2}} +{r}\left(\sqrt{\mathrm{3}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\mathrm{sin}\:\theta=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${find}\:{A}=\int_{\pi/\mathrm{6}} ^{\:\pi/\mathrm{2}} \left(\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\right){d}\theta \\ $$$$\: \\ $$ Commented by mr W last…
Question Number 225955 by fantastic2 last updated on 17/Nov/25 $${can}\:{we}\:{find}\:{the}\:{perimeter} \\ $$$${of}\:{an}\:{ellipse}? \\ $$ Commented by mr W last updated on 17/Nov/25 $${certainly}\:{we}\:{can}\:{find},\:{but}\:{not}\:{in} \\ $$$${a}\:{simple}\:{formula}\:{as}\:{for}\:{perimeter}…
Question Number 225932 by Rojarani last updated on 16/Nov/25 $$\:{If},\:\frac{{by}+{cz}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }=\frac{{cz}+{ax}}{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{{ax}+{by}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\:{then}\:{prove}\:{that},\:\frac{{x}}{{a}}=\frac{{y}}{{b}}=\frac{{z}}{{c}} \\ $$ Answered by som(math1967) last updated…
Question Number 225934 by gregori last updated on 16/Nov/25 $$\:\: \begin{cases}{\lceil\:\frac{\mathrm{8}−\mathrm{2}{x}}{\mathrm{3}}\:\rceil\:;\:{x}\geqslant\:\mathrm{0}}\\{\lfloor\:\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{4}}\:\rfloor\:;\:{x}<\mathrm{0}}\end{cases}. \\ $$$$\left.\:\: − −\mathrm{1}\right)+\: \\ $$$$ \\ $$ Answered by efronzo1 last updated on…
Question Number 225941 by mr W last updated on 16/Nov/25 Commented by mr W last updated on 16/Nov/25 $${Q}\mathrm{222520} \\ $$$${at}\:{least}\:{dropped}\:{from}\:{some}\:{points}, \\ $$$${the}\:{ball}\:{will}\:{return}\:{back}\:{to}\:{its} \\ $$$${starting}\:{position}\:{A}.…
Question Number 225938 by Spillover last updated on 16/Nov/25 Answered by Spillover last updated on 17/Nov/25 Answered by Spillover last updated on 17/Nov/25 Answered by…
Question Number 225939 by Spillover last updated on 16/Nov/25 Answered by Ghisom_ last updated on 16/Nov/25 $$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{cot}\:{x}}\:\rightarrow\:{dx}=−\mathrm{2sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cot}\:{x}}\right] \\ $$$$=−\mathrm{2}\underset{\infty}…
Question Number 225885 by Mingma last updated on 15/Nov/25 Commented by Mingma last updated on 15/Nov/25 Can you show workings, Prof? Commented by fantastic2 last updated on 15/Nov/25 $$\angle{DCH}+\angle{DBH}=\mathrm{45}^{\mathrm{0}}…
Question Number 225837 by Rojarani last updated on 14/Nov/25 $${Show}\:{that},\:{log}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}….\alpha}}}}\:=\mathrm{1} \\ $$ Answered by fantastic last updated on 14/Nov/25 $${let}\:\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}}={x} \\ $$$${x}^{\mathrm{2}} =\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}} \\ $$$$\mathrm{7}{x}=\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}}…