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Question Number 86586 by Power last updated on 29/Mar/20 Commented by Power last updated on 29/Mar/20 $$\left(\mathrm{x}_{\mathrm{1}} ^{\mathrm{5}} −\mathrm{20}\right)\left(\mathrm{3x}_{\mathrm{2}} ^{\mathrm{4}} −\mathrm{2x}_{\mathrm{2}} −\mathrm{35}\right)=? \\ $$ Commented…
Question Number 152091 by rexford last updated on 25/Aug/21 $${Given}\:{tbat}\:{Arg}\left({z}+\mathrm{1}\right)=\frac{\Pi}{\mathrm{6}}\:{and}\: \\ $$$${Arg}\left({z}−\mathrm{1}\right)=\frac{\mathrm{2}\Pi}{\mathrm{3}}.{Find}\:{z}. \\ $$$${please}\:{help}\:{me} \\ $$ Answered by Olaf_Thorendsen last updated on 25/Aug/21 $$\mathrm{Let}\:{z}\:=\:{x}+{iy} \\…
Question Number 152082 by Tawa11 last updated on 25/Aug/21 Answered by Ar Brandon last updated on 25/Aug/21 $$=\underset{\mathrm{1}\leqslant\mathrm{r}\leqslant\mathrm{80}} {\sum}\mathrm{r}=\mathrm{40}\left(\mathrm{81}\right)=\mathrm{3240} \\ $$ Commented by Tawa11 last…
Question Number 21005 by Tinkutara last updated on 10/Sep/17 $$\mathrm{If}\:{z}_{\mathrm{1}} \:=\:{a}\:+\:{ib}\:\mathrm{and}\:{z}_{\mathrm{2}} \:=\:{c}\:+\:{id}\:\mathrm{are}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{Re}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{the}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\omega_{\mathrm{1}} \:=\:{a}\:+\:{ic}\:\mathrm{and}\:\omega_{\mathrm{2}} \:=\:{b}\:+\:{id}…
Question Number 21006 by Tinkutara last updated on 10/Sep/17 $$\mathrm{Let}\:{z}_{\mathrm{1}} \:\mathrm{and}\:{z}_{\mathrm{2}} \:\mathrm{be}\:\mathrm{two}\:\mathrm{distinct}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\mathrm{and}\:\mathrm{let}\:{z}\:=\:\left(\mathrm{1}\:−\:{t}\right){z}_{\mathrm{1}} \:+\:{tz}_{\mathrm{2}} \:\mathrm{for} \\ $$$$\mathrm{some}\:\mathrm{real}\:\mathrm{number}\:{t}\:\mathrm{with}\:\mathrm{0}\:<\:{t}\:<\:\mathrm{1}.\:\mathrm{If} \\ $$$$\mathrm{arg}\left({w}\right)\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{principal}\:\mathrm{argument} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{non}-\mathrm{zero}\:\mathrm{complex}\:\mathrm{number}\:{w},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mid{z}\:−\:{z}_{\mathrm{1}} \mid\:+\:\mid{z}\:−\:{z}_{\mathrm{2}}…
Question Number 86541 by peter frank last updated on 29/Mar/20 Commented by jagoll last updated on 29/Mar/20 $$\mathrm{Q}^{\mathrm{2}} \:=\:\mathrm{PR} \\ $$$$\mathrm{4Q}\:=\:\mathrm{P}+\mathrm{R}\:\Rightarrow\:\mathrm{4ar}\:=\:\mathrm{a}+\mathrm{ar}^{\mathrm{2}} \\ $$$$\mathrm{a}\left(\mathrm{r}^{\mathrm{2}} −\mathrm{4r}+\mathrm{1}\right)\:=\:\mathrm{0} \\…
Question Number 152063 by Tawa11 last updated on 25/Aug/21 $$\mathrm{If}\:\:\mathrm{x}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{and}\:\:\:\mathrm{y}\:\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\:\:\:\:\frac{\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{x}\:\:+\:\:\mathrm{1}}, \\ $$$$\mathrm{show}\:\mathrm{that}\:\:\:\:\:\mid\mathrm{y}\:\:\:−\:\:\:\frac{\mathrm{4}}{\mathrm{3}}\mid\:\:\:\leqslant\:\:\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$ Answered by mr W last updated on 25/Aug/21 $${y}=\frac{{x}^{\mathrm{2}}…
Question Number 20983 by Tinkutara last updated on 09/Sep/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ordered}\:\mathrm{triples} \\ $$$$\left({a},\:{b},\:{c}\right)\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{such}\:\mathrm{that} \\ $$$${abc}\:=\:\mathrm{108}. \\ $$ Answered by dioph last updated on 10/Sep/17 $$\mathrm{108}\:=\:\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{3}}…
Question Number 152049 by mathdanisur last updated on 25/Aug/21 $$\mathrm{If}\:\:\mathrm{a};\mathrm{b}\geqslant\mathrm{1}\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\left(\mathrm{a}+\mathrm{1}+\frac{\mathrm{a}+\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\right)^{\boldsymbol{\mathrm{a}}} \centerdot\:\left(\mathrm{b}+\mathrm{1}+\frac{\mathrm{b}+\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }\right)^{\boldsymbol{\mathrm{b}}} \geqslant\:\mathrm{2}^{\mathrm{2}\left(\mathrm{1}+\sqrt{\boldsymbol{\mathrm{ab}}}\right)} \: \\ $$ Terms of Service Privacy Policy Contact:…