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Category: Algebra

Compare-8-and-8-

Question Number 209059 by hardmath last updated on 01/Jul/24 $$\mathrm{Compare}: \\ $$$$\mathrm{8}!\:\:\:\mathrm{and}\:\:\:\mathrm{8}!! \\ $$ Commented by mr W last updated on 01/Jul/24 $${if}\:\frac{{a}}{{b}}=\mathrm{1}\:\Rightarrow{a}={b} \\ $$$$\frac{\cancel{\mathrm{8}!}!}{\cancel{\mathrm{8}!}}=\mathrm{1}!=\mathrm{1}\:\Rightarrow\mathrm{8}!!=\mathrm{8}!\:\:\:\:\:\underset{\mid<>\mid}…

If-z-1-2-3-2-i-Find-z-4-2z-z-3-z-

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 $$ \\…

2-2024-2024-Remainder-

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}} \\…

4-cos-2-x-4-cos-2-3x-cos-x-cos-2-3x-0-0-pi-2-Find-x-

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…

Find-16-9-

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-208880

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-208852

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}} \\…