Question Number 215715 by Hanuda354 last updated on 15/Jan/25 Commented by Hanuda354 last updated on 15/Jan/25 $$\mathrm{Find}\:{x} \\ $$ Answered by som(math1967) last updated on…
Question Number 215679 by BaliramKumar last updated on 14/Jan/25 Answered by MATHEMATICSAM last updated on 14/Jan/25 $$\mathrm{sec}^{\mathrm{2}} \theta\:=\:\frac{\mathrm{4}{xy}}{\left({x}\:+\:{y}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{cos}^{\mathrm{2}} \theta\:=\:\frac{\left({x}\:+\:{y}\right)^{\mathrm{2}} }{\mathrm{4}{xy}} \\ $$$$\left({x}\:−\:{y}\right)^{\mathrm{2}}…
Question Number 215667 by mathlove last updated on 14/Jan/25 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{sec}\left(\frac{\pi}{\mathrm{2}}{x}\right)\left({arctanx}−\frac{\pi}{\mathrm{4}}\right)=? \\ $$ Answered by issac last updated on 14/Jan/25 $$\mathrm{sec}\left(\frac{\pi}{\mathrm{2}}\right)\left(\mathrm{arctan}\left(\mathrm{1}\right)−\frac{\pi}{\mathrm{4}}\right) \\ $$ Commented by…
Question Number 215656 by MathematicalUser2357 last updated on 13/Jan/25 $$\mathrm{3}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{8}=\mathrm{0} \\ $$ Answered by Wuji last updated on 13/Jan/25 $$\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{2}\:\:,\:\mathrm{3x}^{\mathrm{2}}…
Question Number 215659 by Ari last updated on 13/Jan/25 Answered by Rasheed.Sindhi last updated on 13/Jan/25 $$\mathrm{2}\left(\overline {{abc}}\right)=\overline {{ab}}+\overline {{ba}}+\overline {{bc}}+\overline {{cb}}+\overline {{ac}}+\overline {{ca}} \\…
Question Number 215650 by mr W last updated on 13/Jan/25 Commented by mr W last updated on 14/Jan/25 $${a}\:{rotating}\:{disc}\:{with}\:{angular}\:{velocity} \\ $$$$\boldsymbol{\omega}\:{is}\:{put}\:{on}\:{an}\:{incline}\:{with}\:{rough}\: \\ $$$${surface}\:{as}\:{shown}.\: \\ $$$${find}\:{the}\:{maximum}\:{distance}\:{the}\:…
Question Number 215640 by BaliramKumar last updated on 12/Jan/25 $${If}\:\:\mathrm{2025}^{{sin}^{\mathrm{2}} {x}} \:−\:\mathrm{2025}^{{cos}^{\mathrm{2}} {x}} \:=\:\sqrt{\mathrm{2025}} \\ $$$${then}\:\mathrm{2025}^{{cos}\mathrm{2}{x}} \:+\:\frac{\mathrm{1}}{\mathrm{2025}^{{cos}\mathrm{2}{x}} }\:=\:? \\ $$ Answered by Frix last updated…
Question Number 215641 by cherokeesay last updated on 12/Jan/25 Answered by mr W last updated on 12/Jan/25 $${AC}=\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{5}} \\ $$$${DC}×{AC}=\mathrm{1}×\left(\mathrm{1}+\mathrm{1}+\mathrm{1}\right)\: \\ $$$$\Rightarrow{DC}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{5}}}=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{5}} \\…
Question Number 215625 by MathematicalUser2357 last updated on 12/Jan/25 $${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}=\mathrm{0} \\ $$$${x}=\mathrm{1}\:\vee\:{x}=\mathrm{2}\:\vee\:{x}=\mathrm{2}\pm\sqrt{\mathrm{2}} \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{right}?\:\mathrm{I}\:\mathrm{have}\:\mathrm{not}\:\mathrm{enough}\:\mathrm{time}\:\mathrm{to}\:\mathrm{edit}\:\mathrm{my}\:\mathrm{solution} \\ $$ Commented by MathematicalUser2357 last updated on…
Question Number 215639 by Wuji last updated on 12/Jan/25 $$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{z}}\right)^{\left(\mathrm{1}+\mathrm{z}\right)} =\mathrm{2} \\ $$ Answered by mr W last updated on 12/Jan/25 $$\left(\mathrm{1}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{1}+{z}} =\mathrm{2}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{1}} =\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}−\mathrm{2}} \\…