Question Number 208533 by alcohol last updated on 18/Jun/24 $${z}'\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({z}+\frac{\mathrm{1}}{{z}}\right) \\ $$$${z}\:{and}\:{z}'\:{are}\:{complex}\:{numbers} \\ $$$${show}\:{that}\:{z}\:=\:\mathrm{2}{e}^{{i}\theta} \\ $$$${show}\:{that}\:{M}'\:{describes}\:{a}\:{conic}\:{section} \\ $$ Answered by Berbere last updated on 18/Jun/24…
Question Number 208502 by Kalebwizeman last updated on 17/Jun/24 Answered by A5T last updated on 17/Jun/24 $$=\sqrt{\mathrm{16}}+\sqrt{\mathrm{15}}−\sqrt{\mathrm{15}}−\sqrt{\mathrm{14}}+…−\sqrt{\mathrm{9}}−\sqrt{\mathrm{8}}=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$ Commented by Kalebwizeman last updated on…
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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