Question Number 107690 by bemath last updated on 12/Aug/20 $$\:\:\:\:\:\:\:\:“\mathcal{B}{e}\mathcal{M}{ath}“ \\ $$$${Let}\:{the}\:{complex}\:{number}\:{z}\:{satisfies}\:{the} \\ $$$${equation}\:\mathrm{3}\left({z}−\mathrm{1}\right)=\:{i}\left({z}+\mathrm{1}\right)\: \\ $$$$\left(\mathrm{1}\right)\:{find}\:{z}\:{in}\:{the}\:{form}\:{a}+{bi}\:{where}\:{a},{b}\:\in\mathbb{R}\: \\ $$$$\left(\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\mid{z}\mid\:{and}\:\mid{z}−{z}^{\ast} \mid\: \\ $$$$ \\ $$ Answered by…
Question Number 173226 by AgniMath last updated on 08/Jul/22 $$\:\:\:\:\boldsymbol{\mathrm{Factorize}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{equation}}: \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\:\left({a}\:+\:{b}\right)^{\mathrm{2}} \:−\:\frac{\mathrm{9}}{\mathrm{16}}\:\left(\mathrm{2}{a}\:−\:{b}\right)^{\mathrm{2}} \\ $$ Commented by mr W last updated on 08/Jul/22 $$=\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} −\left[\frac{\mathrm{3}\left(\mathrm{2}{a}−{b}\right)}{\mathrm{4}}\right]^{\mathrm{2}}…
Question Number 173227 by mnjuly1970 last updated on 08/Jul/22 Commented by mr W last updated on 08/Jul/22 $${not}\:\mathrm{1234567900}? \\ $$ Answered by floor(10²Eta[1]) last updated…
Question Number 173216 by Shrinava last updated on 08/Jul/22 $$\mathrm{O}-\mathrm{circumcenter}\:,\:\mathrm{I}-\mathrm{incenter},\:\mathrm{R}-\mathrm{circumradii}, \\ $$$$\mathrm{r}-\mathrm{radii},\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}-\mathrm{sides}\:\mathrm{in}\:\mathrm{a}\:\mathrm{bicenteric} \\ $$$$\mathrm{quadrilateral}.\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{20I}^{\mathrm{2}} \:+\:\mathrm{r}\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\sqrt{\mathrm{4R}^{\mathrm{2}} \:−\:\mathrm{a}^{\mathrm{2}} }\:=\:\mathrm{2}\left(\mathrm{R}^{\mathrm{2}} \:+\:\mathrm{2r}^{\mathrm{2}} \right) \\ $$ Terms…
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Question Number 173212 by peter frank last updated on 08/Jul/22 Answered by Frix last updated on 08/Jul/22 $$\mathrm{3}{x}^{\mathrm{3}} −\mathrm{2}{x}−\mathrm{6}=\mathrm{3}\left({x}−{r}\right)\left({x}^{\mathrm{2}} +{rx}+{s}\right) \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{useable}\:\mathrm{solution} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{\left({x}−{r}\right)\left({x}^{\mathrm{2}} +{rx}+{s}\right)}=…
Question Number 107674 by Study last updated on 12/Aug/20 $${f}\left({x}\right)=\sqrt{{x}}\sqrt{{x}}\:\:\:\:\:\:\:{D}_{{f}} =??? \\ $$ Answered by Her_Majesty last updated on 12/Aug/20 $${x}={re}^{{i}\theta} ;\:{r}\in\mathbb{R}^{+} \wedge\theta\in\mathbb{R} \\ $$$$\Rightarrow…
Question Number 42133 by rahul 19 last updated on 18/Aug/18 $$\mathrm{Let}\:\mathrm{a},\mathrm{b},\mathrm{c}\epsilon\:\mathrm{R}\:\mathrm{such}\:\mathrm{that}\:\mathrm{a}+\mathrm{2b}+\mathrm{c}=\mathrm{4}. \\ $$$$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right). \\ $$ Answered by MrW3 last updated on 18/Aug/18 $${b}=\frac{\mathrm{4}−{a}−{c}}{\mathrm{2}}=\mathrm{2}−\frac{{a}+{c}}{\mathrm{2}} \\ $$$${F}\left({a},{c}\right)={ab}+{bc}+{ca}=\left({a}+{c}\right){b}+{ac}=\mathrm{2}\left({a}+{c}\right)−\frac{\left({a}+{c}\right)^{\mathrm{2}}…
Question Number 107673 by qwerty111 last updated on 12/Aug/20 Answered by Her_Majesty last updated on 12/Aug/20 $$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}^{\mathrm{2}{k}} +\mathrm{2}^{−\mathrm{2}{k}} +\mathrm{2}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{4}^{{k}} +\underset{{k}=\mathrm{1}} {\overset{{n}}…
Question Number 173196 by AgniMath last updated on 08/Jul/22 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{{y}^{\mathrm{2}} \:+\:{yz}\:+\:{z}^{\mathrm{2}} }{\left({x}\:−\:{y}\right)\left({x}\:−\:{z}\right)}+\frac{{z}^{\mathrm{2}} \:+\:{zx}\:+\:{x}^{\mathrm{2}} }{\left({y}\:−\:{z}\right)\left({z}\:−\:{y}\right)}+\frac{{x}^{\mathrm{2}} \:+\:{xy}\:+\:{y}^{\mathrm{2}} }{\left({z}\:−\:{x}\right)\left({z}−{y}\right)} \\ $$ Commented by som(math1967) last updated…