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Question-132594

Question Number 132594 by Study last updated on 15/Feb/21 Commented by mr W last updated on 15/Feb/21 $${the}\:{sum}\:{of}\:{two}\:{complex}\:{numbers} \\ $$$${can}\:{be}\:{a}\:{real}\:{number},\:{e}.{g}.\: \\ $$$$\left(\mathrm{1}+{i}\right)+\left(\mathrm{1}−{i}\right)=\mathrm{2} \\ $$$${the}\:{product}\:{of}\:{two}\:{complex}\:{numbers} \\…

let-Z-N-0-f-Z-Z-Z-f-m-n-m-n-m-n-1-2-m-prove-that-f-is-a-one-to-one-function-and-also-an-onto-function-

Question Number 67055 by Tony Lin last updated on 22/Aug/19 $${let}\:\mathbb{Z}_{+} =\mathbb{N}\cup\left\{\mathrm{0}\right\},\:{f}:\:\mathbb{Z}_{+} ×\mathbb{Z}_{+} \rightarrow\mathbb{Z}_{+} \\ $$$${f}\left({m},\:{n}\right)=\frac{\left({m}+{n}\right)\left({m}+{n}+\mathrm{1}\right)}{\mathrm{2}}+{m} \\ $$$${prove}\:{that}\:{f}\:{is}\:{a}\:{one}-{to}-{one}\:{function} \\ $$$${and}\:{also}\:{an}\:{onto}\:{function} \\ $$ Terms of Service…

Hello-Tinku-Tara-I-have-a-new-phone-and-I-am-unable-to-remember-the-password-for-my-previous-account-Is-there-anyway-to-help-me-retrieve-it-

Question Number 132575 by nadovic last updated on 15/Feb/21 $$\mathrm{Hello}\:\mathrm{Tinku}\:\mathrm{Tara},\:\mathrm{I}\:\mathrm{have}\:\mathrm{a}\:\mathrm{new}\:\mathrm{phone} \\ $$$$\mathrm{and}\:\mathrm{I}\:\mathrm{am}\:\mathrm{unable}\:\mathrm{to}\:\mathrm{remember}\:\mathrm{the}\: \\ $$$$\mathrm{password}\:\mathrm{for}\:\mathrm{my}\:\mathrm{previous}\:\mathrm{account}.\:{Is} \\ $$$${there}\:{anyway}\:{to}\:{help}\:{me}\:{retrieve}\:{it}? \\ $$ Commented by mr W last updated on…

Given-f-x-log-2020-x-and-p-p-p-2020-2020-then-the-value-of-f-p-a-2020-1-2020-c-1-2020-1-2020-b-1-2020-d-2020-e-log-10-20

Question Number 132574 by liberty last updated on 15/Feb/21 $$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{log}\:_{\mathrm{2020}} \left(\mathrm{x}\right)\:\mathrm{and}\: \\ $$$$\mathrm{p}^{\left(\mathrm{p}\right)^{\mathrm{p}^{\mathrm{2020}} } } \:=\:\mathrm{2020}\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{f}\left(\mathrm{p}\right)\:=\:… \\ $$$$\left(\mathrm{a}\right)\sqrt[{\mathrm{2020}}]{\mathrm{2020}}\:\:\:\:\:\left(\mathrm{c}\right)\:\sqrt[{\mathrm{2020}}]{\frac{\mathrm{1}}{\mathrm{2020}}} \\ $$$$\left(\mathrm{b}\right)\:\frac{\mathrm{1}}{\mathrm{2020}}\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{2020}\:\:\:\:\:\:\left(\mathrm{e}\right)\:\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2020}\right) \\ $$…