Question Number 179058 by sciencestudent last updated on 24/Oct/22 $${f}\left({x}\right)=\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}\:\:\:\:\:\:\:{Dom}_{{f}\left({x}\right)} =? \\ $$ Answered by Rasheed.Sindhi last updated on 24/Oct/22 $${x}\in\mathbb{R}\:\wedge\:{x}+\mathrm{1}\neq\mathrm{0} \\ $$$${x}\in\mathbb{R}\:\wedge\:{x}\neq−\mathrm{1} \\ $$$${x}\in\mathbb{R}−\left\{−\mathrm{1}\right\}…
Question Number 113510 by Lekhraj last updated on 13/Sep/20 Answered by MJS_new last updated on 13/Sep/20 $${x}^{\mathrm{3}} −{y}^{\mathrm{3}} +\mathrm{9}{xy}+\mathrm{127}=\mathrm{0} \\ $$$${x},\:{y},\:{p}\in\mathbb{Z} \\ $$$$\mathrm{let}\:{y}={x}+{p} \\ $$$$\left(\mathrm{9}−\mathrm{3}{p}\right){x}^{\mathrm{2}}…
Question Number 113508 by Lekhraj last updated on 13/Sep/20 Commented by mr W last updated on 13/Sep/20 $${i}\:{got}\:{N}={LCM}\left(\mathrm{5},\mathrm{7},\mathrm{8},\mathrm{9}\right)=\mathrm{2520} \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 47966 by ajfour last updated on 17/Nov/18 $${a}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+{b}\left({x}+{y}\right)=\:{c} \\ $$$$\:\&\:\:\:\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:=\:{R}^{\mathrm{2}} \\ $$$${Solve}\:{for}\:{x}\:{or}\:{y}\:. \\ $$ Commented by behi83417@gmail.com last updated…
Question Number 179031 by sciencestudent last updated on 23/Oct/22 $${prove}\:{that}\:{sgn}\left(\mathrm{0}\right)=\mathrm{0} \\ $$ Commented by MJS_new last updated on 23/Oct/22 $$\mathrm{no}\:\mathrm{proof}\:\mathrm{possible},\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{definition} \\ $$ Terms of Service…
Question Number 179025 by jlewis last updated on 23/Oct/22 $$\mathrm{2}^{\mathrm{10}{x}} −{x}^{\mathrm{5}} −\mathrm{4}=\mathrm{0} \\ $$ Commented by MJS_new last updated on 23/Oct/22 $$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{1}.\mathrm{31950087530}…
Question Number 179023 by Spillover last updated on 23/Oct/22 $$\mathrm{Draw}\:\mathrm{an}\:\mathrm{electrical}\:\mathrm{network} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{p}\wedge\left(\mathrm{q}\vee\mathrm{r}\right) \\ $$$$\left(\mathrm{b}\right)\left(\sim\mathrm{p}\wedge\sim\mathrm{q}\right)\vee\left(\sim\mathrm{p}\wedge\mathrm{q}\right)\vee\left(\mathrm{p}\wedge\sim\mathrm{q}\right) \\ $$$$\left(\mathrm{c}\right)\:\mathrm{p}\leftrightarrow\mathrm{q} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 179018 by Spillover last updated on 23/Oct/22 Answered by Tokugami last updated on 26/Oct/22 $$\begin{array}{|c|c|c|c|c|c|c|c|c|}{{x}}&\hline{{y}}&\hline{{z}}&\hline{{p}}\\{\mathrm{1}}&\hline{\mathrm{1}}&\hline{\mathrm{1}}&\hline{\mathrm{0}}\\{\mathrm{1}}&\hline{\mathrm{1}}&\hline{\mathrm{0}}&\hline{\mathrm{0}}\\{\mathrm{1}}&\hline{\mathrm{0}}&\hline{\mathrm{1}}&\hline{\mathrm{0}}\\{\mathrm{1}}&\hline{\mathrm{0}}&\hline{\mathrm{0}}&\hline{\mathrm{1}}\\{\mathrm{0}}&\hline{\mathrm{1}}&\hline{\mathrm{1}}&\hline{\mathrm{0}}\\{\mathrm{0}}&\hline{\mathrm{1}}&\hline{\mathrm{0}}&\hline{\mathrm{1}}\\{\mathrm{0}}&\hline{\mathrm{0}}&\hline{\mathrm{1}}&\hline{\mathrm{1}}\\{\mathrm{0}}&\hline{\mathrm{0}}&\hline{\mathrm{0}}&\hline{\mathrm{1}}\\\hline\end{array} \\ $$$${p}=\left({x}\:\mathrm{nor}\:{y}\right)\:\mathrm{or}\:\left({y}\:\mathrm{nor}\:{z}\right)\:\mathrm{or}\:\left({z}\:\mathrm{nor}\:{x}\right) \\ $$$$\mathrm{just}\:\mathrm{draw}\:\mathrm{a}\:\mathrm{diagram}\:\mathrm{from}\:\mathrm{that}. \\ $$ Answered by…
Question Number 113468 by Algoritm last updated on 13/Sep/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 179001 by Agnibhoo98 last updated on 23/Oct/22 $$\mathrm{In}\:\bigtriangleup\mathrm{ABC}\:\angle\mathrm{BAC}\:=\:\mathrm{90}°\:\mathrm{and}\:\mathrm{AB}\:=\:\frac{\mathrm{BC}}{\mathrm{2}}. \\ $$$$\angle\mathrm{ACB}\:=\:? \\ $$ Commented by Rasheed.Sindhi last updated on 23/Oct/22 $$\mathrm{30}° \\ $$ Terms…