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Question Number 208493 by Tawa11 last updated on 17/Jun/24 Answered by A5T last updated on 17/Jun/24 $${a}\left(\mathrm{3}{a}\right)={x}^{\mathrm{2}} \Rightarrow{x}={a}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{4}{a}\right)^{\mathrm{2}} =\mathrm{2}\left({a}\sqrt{\mathrm{3}}+{y}\right)^{\mathrm{2}} =\mathrm{16}{a}^{\mathrm{2}} \Rightarrow{r}={y}=\mathrm{2}{a}\sqrt{\mathrm{2}}−{a}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{r}={y}={a}\left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)…
Question Number 208526 by Tawa11 last updated on 17/Jun/24 Commented by Tawa11 last updated on 17/Jun/24 $$\mathrm{Find}\:\:\mid\mathrm{SP}\mid \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{question}\:\mathrm{correct}\:\mathrm{like}\:\mathrm{this}? \\ $$ Commented by Tawa11 last…
Question Number 208520 by Tawa11 last updated on 17/Jun/24 Answered by mr W last updated on 17/Jun/24 Commented by mr W last updated on 17/Jun/24…
Question Number 208489 by Tawa11 last updated on 17/Jun/24 Answered by Berbere last updated on 17/Jun/24 $${x}^{\mathrm{4}} +\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}}−\mathrm{4} \\ $$$$\geqslant\mathrm{2}\sqrt{\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}}}−\mathrm{4}=−\mathrm{2} \\ $$$${equaliti}\:\:{x}=\mathrm{0}…
Question Number 208519 by Tawa11 last updated on 17/Jun/24 $$\mathrm{If}\:\:\:\:\:\mathrm{a}^{\mathrm{x}} \:\:=\:\:\mathrm{b}^{\mathrm{y}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{a}}{\mathrm{x}}\:\:+\:\:\frac{\mathrm{b}}{\mathrm{y}}\:\:=\:\:\mathrm{1} \\ $$$$\mathrm{then},\:\:\:\:\mathrm{a}^{\mathrm{x}} \:\:+\:\:\mathrm{b}^{\mathrm{y}} \:\:=\:\:? \\ $$ Answered by mr W last updated…
Question Number 208513 by CrispyXYZ last updated on 17/Jun/24 $${a},\:{b}>\mathrm{0}.\:\mathrm{2}{a}+{b}=\mathrm{1},\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}{a}}+\frac{\mathrm{2}{a}}{{b}^{\mathrm{2}} }\:>\:\mathrm{2}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 208515 by mokys last updated on 17/Jun/24 $$\int_{\mathrm{0}} ^{\:\infty} \:\frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208445 by alcohol last updated on 16/Jun/24 $$\left.{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} ,\:{u}_{\mathrm{0}} \in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$${v}_{{n}} \:=\:\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} }\:=\:{f}\left({u}_{{n}} ^{\mathrm{2}} \right)\:;\:{f}\left({x}\right)\:=\:\frac{\mathrm{2}−{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }…