Question Number 222801 by hardmath last updated on 07/Jul/25 Answered by ajfour last updated on 08/Jul/25 $$\mathrm{256}\:\:{becuz}\:\mid{x}\mid+\mid{y}\mid=\mathrm{1}\:\:{got}\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2} \\ $$ Commented by hardmath last updated…
Question Number 222743 by hardmath last updated on 06/Jul/25 $$\mathrm{Compare}: \\ $$$$\boldsymbol{\mathrm{a}}\:=\:\mathrm{arcctg}\:\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{b}}\:=\:\mathrm{arccos}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{c}}\:=\:\mathrm{arctg}\:\sqrt{\mathrm{2}} \\ $$ Answered by mr W last updated on…
Question Number 222754 by hardmath last updated on 06/Jul/25 Commented by hardmath last updated on 06/Jul/25 $$\int_{\mathrm{0}} ^{\:\mathrm{8}} \left(\mathrm{f}^{\:−\mathrm{1}} \left(\mathrm{x}\right)\:−\:\mathrm{f}\left(\mathrm{x}\right)\right)\:\mathrm{dx}\:=\:\mathrm{12} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{8}} \:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\…
Question Number 222616 by ajfour last updated on 01/Jul/25 Answered by ajfour last updated on 01/Jul/25 Commented by MathematicalUser2357 last updated on 02/Jul/25 $$\mathrm{I}\:\mathrm{had}\:\mathrm{grapher}\:\mathrm{free} \\…
Question Number 222584 by hu last updated on 30/Jun/25 $${find}\:{the}\:{correct}\:{const}.\:{to}\:{preconst}.\:“\:{x}^{{n}} −\mathrm{2}=−{x}\:'' \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222585 by Rojarani last updated on 30/Jun/25 Commented by AntonCWX8 last updated on 01/Jul/25 $${Now}\:{I}'{ll}\:{give}\:{you}\:{guys}\:{another}\:{challenge}. \\ $$$${What}\:{if}\:{the}\:{numbers}\:{beside}\:{the}\:\sqrt{\:\:}\:{symbol}\:{represent}\:{the}\:{nth}\:{root}? \\ $$$${Say}\:\mathrm{5}\sqrt{\:\:\:}\:=\:\sqrt[{\mathrm{5}}]{\:\:\:\:} \\ $$ Commented by…
Question Number 222582 by hardmath last updated on 30/Jun/25 $$\mathrm{If}:\:\:\:\mathrm{a}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{b}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{i}\:=\:\overline {\mathrm{1},…,\mathrm{n}} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} }\:+\:\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} }\:+…+\:\sqrt{\mathrm{a}_{\boldsymbol{\mathrm{n}}}…
Question Number 222583 by ajfour last updated on 30/Jun/25 $${solve}\:{for}\:{p},{q},{s}\:{in}\:{terms}\:{of}\:{c}. \\ $$$$\bullet\:\left(\frac{{qs}}{{q}−{sp}}\right)^{\mathrm{2}} −{s}\left(\frac{{qs}}{{q}−{sp}}\right)+{p}=\mathrm{0} \\ $$$$\bullet\:\left(\frac{{q}+{c}}{{p}+\mathrm{1}}\right)^{\mathrm{2}} ={sp}−{q} \\ $$$$\bullet\:\left({q}−{cp}\right)\left({p}+\mathrm{1}\right)^{\mathrm{2}} =\left({q}+{c}\right)^{\mathrm{3}} \\ $$$${I}\:{have}\:{to}\:{find}\:{non}\:{zero}\:{real}\:{x}=−\left(\frac{{q}+{c}}{{p}+\mathrm{1}}\right)\:. \\ $$ Terms of…
Question Number 222568 by mathlove last updated on 30/Jun/25 $${f}\left({x}\right)=\left(\sqrt{{x}−\mathrm{2}}\right)^{\mathrm{0}} \:\:{and}\:{g}\left({x}\right)=\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{0}} } \\ $$$${dom}\:{f}\left({x}\right)=?\:,\:{dom}\:{g}\left({x}\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222552 by mathlove last updated on 29/Jun/25 $${x}^{{x}} =\mathrm{64}\:\:\:\:\:\:,\:\:{x}=? \\ $$ Answered by mr W last updated on 29/Jun/25 $${x}\mathrm{ln}\:{x}=\mathrm{ln}\:\mathrm{64} \\ $$$$\mathrm{ln}\:{xe}^{\mathrm{ln}\:{x}} =\mathrm{6ln}\:\mathrm{2}…