Question Number 226053 by Linton last updated on 18/Nov/25 $${The}\:{coefficient}\:{of}\:{x}^{\mathrm{2}} \:{in}\:{the}\:{expansion} \\ $$$${of}\:\left(\mathrm{1}+\:\left(\mathrm{2}/{p}\right){x}\right)^{\mathrm{5}} \:+\:\left(\mathrm{1}+{px}\right)^{\mathrm{6}} \:{is}\:\mathrm{70}. \\ $$$${Find}\:{the}\:{possible}\:{values}\:{of}\:{the} \\ $$$${constant}\:{p}. \\ $$ Answered by mr W…
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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 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 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 $$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{cot}\:{x}}\:\rightarrow\:{dx}=−\mathrm{2sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cot}\:{x}}\right] \\ $$$$=−\mathrm{2}\underset{\infty}…
Question Number 225837 by Rojarani last updated on 14/Nov/25 $${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} \\ $$$${x}^{\mathrm{2}} =\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}} \\ $$$$\mathrm{7}{x}=\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}}…
Question Number 225840 by hardmath last updated on 14/Nov/25 Commented by hardmath last updated on 14/Nov/25 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{MA}^{\mathrm{2}} +\mathrm{MB}^{\mathrm{2}} +\mathrm{MC}^{\mathrm{2}} +\mathrm{MD}^{\mathrm{2}} \:=\:\mathrm{4R}^{\mathrm{2}} \\ $$…
Question Number 225810 by hardmath last updated on 12/Nov/25 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{in}\:\mathrm{any}\:\mathrm{triangle}: \\ $$$$\frac{\mathrm{4R}}{\mathrm{r}}\:\geqslant\:\frac{\mathrm{w}_{\boldsymbol{\mathrm{a}}} \:\mathrm{w}_{\boldsymbol{\mathrm{b}}} \:\mathrm{w}_{\boldsymbol{\mathrm{c}}} }{\mathrm{h}_{\boldsymbol{\mathrm{a}}} \:\mathrm{h}_{\boldsymbol{\mathrm{b}}} \:\mathrm{h}_{\boldsymbol{\mathrm{c}}} }\:\centerdot\:\left(\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\right)\centerdot\left(\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\right)^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy…
Question Number 225613 by Jubr last updated on 04/Nov/25 Answered by mahdipoor last updated on 04/Nov/25 $$\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}^{\mathrm{2}} −\mathrm{4}.\mathrm{1}.\mathrm{1}}}{\mathrm{2}.\mathrm{1}}=\frac{−\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$=\mathrm{cos}\left(\pm\mathrm{120}\right)+\mathrm{i}.\mathrm{sin}\left(\pm\mathrm{120}\right)=\mathrm{e}^{\mathrm{i}\left(\pm\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} \\ $$$$\mathrm{v}_{\mathrm{n}} =\mathrm{e}^{\mathrm{i}\theta\mathrm{n}} +\mathrm{e}^{\mathrm{i}\left(−\theta\right)\mathrm{n}}…