Question Number 208959 by hardmath last updated on 28/Jun/24 $$\mathrm{If}\:\:\:\boldsymbol{\mathrm{z}}\:=\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\:+\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\boldsymbol{\mathrm{i}} \\ $$$$\mathrm{Find}\:\:\:\left(\mathrm{z}^{\mathrm{4}} \:+\:\mathrm{2z}\right)\centerdot\left(\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{z}\right)\:=\:? \\ $$ Answered by grigoriy last updated on 29/Jun/24 $$ \\…
Question Number 208951 by hardmath last updated on 28/Jun/24 $$\mathrm{2}^{\mathrm{2024}} \::\:\mathrm{2024}\:=\:…\:\left(\mathrm{Remainder}\:=\:?\right) \\ $$ Answered by A5T last updated on 28/Jun/24 $$\left(\mathrm{2}^{\mathrm{11}} \right)^{\mathrm{184}} \:\:\overset{\mathrm{2024}} {\equiv}\:\mathrm{24}^{\mathrm{184}} \\…
Question Number 208945 by hardmath last updated on 28/Jun/24 $$\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{x}\:−\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{3x}\:\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{3x}\:=\:\mathrm{0} \\ $$$$\left[\:\mathrm{0}\:;\:\frac{\pi}{\mathrm{2}}\:\right] \\ $$$$\mathrm{Find}:\:\:\mathrm{x}\:=\:? \\ $$ Answered by Berbere last updated on…
Question Number 208892 by hardmath last updated on 26/Jun/24 $$\mathrm{Find}: \\ $$$$\sqrt{−\mathrm{16}}\:\:\centerdot\:\:\sqrt{−\mathrm{9}}\:\:=\:\:? \\ $$ Commented by Adeyemi889 last updated on 26/Jun/24 $$ \\ $$$$\sqrt{−\mathrm{16}}\:=\:\sqrt{\:\left(\mathrm{16}\right)\left(−\mathrm{1}\right)}\:=\sqrt{−\mathrm{1}}\:×\sqrt{\mathrm{16}\:} \\…
Question Number 208891 by efronzo1 last updated on 26/Jun/24 $$\:\:\:\underline{\kappa} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208880 by efronzo1 last updated on 26/Jun/24 Answered by Frix last updated on 26/Jun/24 $${a},\:{b},\:{c}\:>\mathrm{0}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{{a}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{4}}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}>\mathrm{0} \\ $$$$\mathrm{The}\:\mathrm{question}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}. \\ $$ Terms of Service Privacy…
Question Number 208852 by Tawa11 last updated on 25/Jun/24 Answered by Frix last updated on 25/Jun/24 $$\mathrm{3}^{−\mathrm{3}\left({x}^{\mathrm{2}} −{x}\right)} ={x}^{\mathrm{2}} −{x}\:\Rightarrow\:{x}^{\mathrm{2}} −{x}>\mathrm{0}\:\Leftrightarrow\:{x}<\mathrm{0}\vee{x}>\mathrm{1} \\ $$$${t}={x}^{\mathrm{2}} −{x}\:\Leftrightarrow\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}{t}+\mathrm{1}}}{\mathrm{2}} \\…
Question Number 208783 by Tawa11 last updated on 22/Jun/24 $$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x}\:\:\mathrm{and}\:\:\mathrm{y}. \\ $$$$\:\:\:\:\mathrm{x}^{\mathrm{y}} \:\:−\:\:\mathrm{y}^{\mathrm{x}} \:\:=\:\:\mathrm{17} \\ $$ Commented by Tawa11 last updated on 22/Jun/24 $$\mathrm{x}\:=\:\mathrm{3}\:\mathrm{and}\:\mathrm{y}\:=\:\mathrm{4}. \\…
Question Number 208738 by Tawa11 last updated on 22/Jun/24 $$\mathrm{sum}: \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{4}.\mathrm{7}}\:\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{7}.\mathrm{10}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{10}.\mathrm{13}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{10}.\mathrm{13}.\mathrm{16}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{13}.\mathrm{16}.\mathrm{19}}\:\:+\:\:…\:\:+\:\:\frac{\mathrm{1}}{\mathrm{94}.\mathrm{97}.\mathrm{100}} \\ $$ Answered by BaliramKumar last updated on 22/Jun/24 $$\frac{\mathrm{1}}{\mathrm{6}}\left[\frac{\mathrm{1}}{\mathrm{1}×\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{97}×\mathrm{100}}\right]\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{d}=\:\mathrm{7}−\mathrm{1}\:=\:\mathrm{10}−\mathrm{4}\:=\:\mathrm{6} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}×\frac{\left(\mathrm{97}×\mathrm{25}−\mathrm{1}\right)}{\mathrm{97}×\mathrm{100}}\:=\:\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{2424}}{\mathrm{9700}}\: \\…
Question Number 208685 by hardmath last updated on 21/Jun/24 $$\mathrm{25}\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{3} \\ $$$$\mathrm{x}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 21/Jun/24 $$\mathrm{tan}\:\mathrm{x}=\frac{\mathrm{3}}{\mathrm{25}} \\ $$$$\mathrm{x}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{25}}\right)…