Question Number 219697 by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 2 2…
Question Number 219695 by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219678 by universe last updated on 30/Apr/25 Commented by mr W last updated on 30/Apr/25 $${what}\:{do}\:{you}\:{mean}\:{with}\:\mathrm{1}\centerdot\mathrm{2}?\: \\ $$$$\mathrm{1}\centerdot\mathrm{2}=\mathrm{1}×\mathrm{2}?\:{etc}. \\ $$ Commented by universe…
Question Number 219668 by Rojarani last updated on 30/Apr/25 Answered by SdC355 last updated on 30/Apr/25 $${p}={p}\left(\mathrm{4}−{p}\right)\left(\mathrm{4}−{p}\left(\mathrm{4}−{p}\right)\right) \\ $$$$\mathrm{1}=\left(\mathrm{4}−{p}\right)\left({p}^{\mathrm{2}} −\mathrm{4}{p}+\mathrm{4}\right) \\ $$$$\mathrm{4}{p}^{\mathrm{2}} −\mathrm{16}{p}+\mathrm{16}−{p}^{\mathrm{3}} +\mathrm{4}{p}^{\mathrm{2}} −\mathrm{4}{p}=\mathrm{1}…
Question Number 219659 by Nicholas666 last updated on 30/Apr/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{prove}; \\ $$$$\:\:{cos}\:\left({B}+{C}−{A}\right)−{cos}\left({C}+{A}−{B}\right)+{cos}\left({A}+{B}−{C}\right)−{cos}\left({A}+{B}+{C}\right)\:=\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{sinAcosBsinC} \\ $$$$ \\ $$ Answered by som(math1967) last updated…
Question Number 219651 by SdC355 last updated on 30/Apr/25 $$\mathrm{Solve} \\ $$$${x}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({x}\right)+{xy}^{\left(\mathrm{1}\right)} \left({x}\right)+\left({x}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({x}\right)={e}^{−{kx}} \\ $$ Answered by MrGaster last updated on…
Question Number 219570 by Rojarani last updated on 29/Apr/25 $$\:\mathrm{8}{x}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} =\mathrm{36}\left({x}+{y}\right),\: \\ $$$$\:\:{x},{y}\in{R},\:{find}\:{maximum}\:\left({x}+{y}\right) \\ $$$$\: \\ $$ Answered by SdC355 last updated on 29/Apr/25…
Question Number 219548 by universe last updated on 28/Apr/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219564 by hardmath last updated on 28/Apr/25 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\frac{\mathrm{xy}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} \right)+\mathrm{yz}\left(\mathrm{y}^{\mathrm{3}} −\mathrm{z}^{\mathrm{3}} \right)+\mathrm{zx}\left(\mathrm{z}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{xy}\left(\mathrm{x}−\mathrm{y}\right)+\mathrm{yz}\left(\mathrm{y}−\mathrm{z}\right)+\mathrm{zx}\left(\mathrm{z}−\mathrm{x}\right)}\:=\:\mathrm{55}}\\{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:=\:\mathrm{99}}\\{\frac{\mathrm{xy}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} \right)+\mathrm{yz}\left(\mathrm{y}^{\mathrm{3}} −\mathrm{z}^{\mathrm{3}} \right)+\mathrm{zx}\left(\mathrm{z}^{\mathrm{3}}…
Question Number 219561 by hardmath last updated on 28/Apr/25 $$\mathrm{let}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sequence}\:\:\:\left(\mathrm{x}_{\boldsymbol{\mathrm{n}}} \right)\mathrm{n}\:\geqslant\:\mathrm{1} \\ $$$$\mathrm{defined}\:\mathrm{by}\:\:\:\mathrm{x}_{\mathrm{1}} =\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}_{\boldsymbol{\mathrm{n}}+\mathrm{2}} \:=\:\mathrm{3x}_{\boldsymbol{\mathrm{n}}+\mathrm{1}} −\:\mathrm{x}_{\boldsymbol{\mathrm{n}}} \\ $$$$\forall\mathrm{n}\:\in\:\mathbb{N} \\ $$$$\mathrm{find}\:\:\:\boldsymbol{\mathrm{L}}\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\mathrm{0}} {\sum}}\:\:\frac{\mathrm{x}_{\mathrm{2}\boldsymbol{\mathrm{k}}+\mathrm{1}}…