Question Number 58154 by behi83417@gmail.com last updated on 18/Apr/19 $$\boldsymbol{\mathrm{A}}\left(\mathrm{1},\mathrm{1}+\boldsymbol{\mathrm{i}}\right),\boldsymbol{\mathrm{B}}\left(\sqrt{\mathrm{2}}+\boldsymbol{\mathrm{i}},\mathrm{2}\right),\boldsymbol{\mathrm{C}}\left(\mathrm{1}−\mathrm{3}\boldsymbol{\mathrm{i}},\mathrm{1}−\boldsymbol{\mathrm{i}}\right) \\ $$$$\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{given}}. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{between}}:\:\:\boldsymbol{\mathrm{AB}}\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{AC}}\:. \\ $$ Answered by MJS last updated on 18/Apr/19 $$\mathrm{if}\:\mathrm{i}=\sqrt{−\mathrm{1}}\:\mathrm{then}\:{A},\:{B},\:{C}\:\in\mathbb{R}^{\mathrm{4}} \:\mathrm{with}\:\mathrm{coordinates}…
Question Number 58156 by hhghg last updated on 18/Apr/19 $$\mathrm{Jaiden}\:\mathrm{buys}\:\mathrm{334}\:\mathrm{cupcakes}.\mathrm{He}\:\mathrm{got}\:\mathrm{14}\:\mathrm{more}\:\mathrm{cupcakes}.\mathrm{How}\:\mathrm{many}\:\mathrm{cupcakes}\:\mathrm{did}\:\mathrm{he}\:\mathrm{got}\:\mathrm{altogether}? \\ $$ Answered by tanmay last updated on 19/Apr/19 $${suppose}\:{you}\:{have}\:\mathrm{3}\:{pen}\:{and}\:{you}\:{get}\:\mathrm{2}\:{more}\:{pen} \\ $$$${then}\:{how}\:{many}\:{do}\:{you}\:{have} \\ $$$$\mathrm{3}+\mathrm{2}=\mathrm{5}\:{pen} \\…
Question Number 58153 by behi83417@gmail.com last updated on 18/Apr/19 $$\boldsymbol{\mathrm{arctan}}\left(\sqrt{\mathrm{2}}−\boldsymbol{\mathrm{i}}\right)=?\:\:\:\:\:\:\:\:\:\:\left[\boldsymbol{\mathrm{i}}=\sqrt{−\mathrm{1}}\right] \\ $$ Answered by MJS last updated on 18/Apr/19 $$\mathrm{found}\:\mathrm{these}: \\ $$$$\mathrm{tan}\:\left({a}+{b}\mathrm{i}\right)\:=\frac{\mathrm{2e}^{\mathrm{2}{b}} \mathrm{sin}\:\mathrm{2}{a}}{\mathrm{2e}^{\mathrm{2}{b}} \mathrm{cos}\:\mathrm{2}{a}\:+\mathrm{e}^{\mathrm{4}{b}} +\mathrm{1}}+\frac{\mathrm{e}^{\mathrm{4}{b}}…
Question Number 58145 by Mikael_Marshall last updated on 18/Apr/19 $${how}\:{to}\:{factorize} \\ $$$${a}^{\mathrm{3}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{3}} \: \\ $$ Answered by tanmay last updated on 18/Apr/19…
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Question Number 189208 by mathlove last updated on 13/Mar/23 Answered by MJS_new last updated on 13/Mar/23 $$\mathrm{easy}\:\mathrm{to}\:\mathrm{see} \\ $$$$\mathrm{9}+\mathrm{2}=\mathrm{11} \\ $$$$\mathrm{3}+\mathrm{4}=\mathrm{7} \\ $$$${x}=\mathrm{9}\wedge{y}=\mathrm{4} \\ $$…
Question Number 58135 by rahul 19 last updated on 18/Apr/19 Answered by tanmay last updated on 18/Apr/19 $${e}^{{i}\alpha} ={cos}\alpha+{isin}\alpha \\ $$$${e}^{{i}\theta} ×{e}^{{i}\mathrm{2}\theta} ×{e}^{{i}\mathrm{3}\theta} ×…{e}^{{in}\theta} =\mathrm{1}={cos}\left(\mathrm{2}{m}\pi+{isin}\mathrm{2}{m}\pi\right)={e}^{{i}\left(\mathrm{2}{m}\pi\right)}…
Question Number 189201 by Shrinava last updated on 13/Mar/23 $$\mathrm{In}\:\:\:\bigtriangleup\mathrm{ABC}\:\:\:\mathrm{holds}: \\ $$$$\sqrt{\mathrm{2}}\:\mathrm{a}\:\mathrm{cos}\:\frac{\mathrm{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{\mathrm{C}}{\mathrm{2}}\:=\:\mathrm{s} \\ $$$$\Rightarrow\:\mathrm{sec}\:\left(\mathrm{2B}\right)\:+\:\mathrm{tan}\:\left(\mathrm{2B}\right)\:=\:\frac{\mathrm{c}\:+\:\mathrm{b}}{\mathrm{c}\:−\:\mathrm{b}} \\ $$ Answered by som(math1967) last updated on 13/Mar/23 $$\sqrt{\mathrm{2}}{acos}\frac{{B}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}={s} \\…
Question Number 123638 by 676597498 last updated on 26/Nov/20 $$\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{x}\in\mathbb{R}\:\mathrm{then}\:\mathrm{x}\:\mathrm{and}\:−\mathrm{x}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{positive} \\ $$ Commented by talminator2856791 last updated on 27/Nov/20 $$\:\mathrm{this}\:\mathrm{is}\:\mathrm{just}\:\mathrm{a}\:\mathrm{rule} \\ $$ Terms of Service…
Question Number 58092 by smiak8742 last updated on 17/Apr/19 $$\mathrm{6}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{5}=\mathrm{0} \\ $$ Answered by tanmay last updated on 17/Apr/19 $$\mathrm{6}{x}^{\mathrm{3}} −\mathrm{6}{x}+\mathrm{5}{x}^{\mathrm{2}} −\mathrm{5}=\mathrm{0} \\…