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 $${Q}\mathrm{222520} \\ $$$${at}\:{least}\:{dropped}\:{from}\:{some}\:{points}, \\ $$$${the}\:{ball}\:{will}\:{return}\:{back}\:{to}\:{its} \\ $$$${starting}\:{position}\:{A}.…
Question Number 225938 by Spillover last updated on 16/Nov/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 225939 by Spillover last updated on 16/Nov/25 Answered by Ghisom_ last updated on 16/Nov/25 $$\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 $${let}\:\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}}={x} \\ $$$${x}^{\mathrm{2}} =\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}} \\ $$$$\mathrm{7}{x}=\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}}…
Question Number 225866 by BHOOPENDRA last updated on 15/Nov/25 $${Question}\:\mathrm{222520} \\ $$ Answered by BHOOPENDRA last updated on 15/Nov/25 $$\frac{−{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({y}+{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\…
Question Number 225892 by mnjuly1970 last updated on 15/Nov/25 Answered by A5T last updated on 15/Nov/25 $$\mathrm{Let}\:\mathrm{O}\:\mathrm{be}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{and}\:\mathrm{OX}=\mathrm{d} \\ $$$$\mathrm{Let}\:\mathrm{OB}=\mathrm{r}\Rightarrow\mathrm{XB}=\mathrm{r}−\mathrm{d};\:\mathrm{AX}=\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{e} \\ $$$$\mathrm{AO}=\sqrt{\mathrm{e}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} }\: \\ $$$$\angle\mathrm{BYC}=\mathrm{90}°\:\Rightarrow\:\mathrm{XHYC}\:\mathrm{is}\:\mathrm{cyclic}…
Question Number 225861 by fantastic2 last updated on 15/Nov/25 $$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{{ln}^{\mathrm{12}} \left(\mathrm{1}−{x}\right)}{\:\sqrt{\frac{{ln}^{\mathrm{12}} \left(\mathrm{1}−{x}\right)}{\:\sqrt{\frac{{ln}^{\mathrm{12}} \left(\mathrm{1}−{x}\right)}{\:\sqrt{\frac{{ln}^{\mathrm{12}} \left(\mathrm{1}−{x}\right)}{…}}}}}}}}{dx}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com