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…
Question Number 178996 by Spillover last updated on 23/Oct/22 Answered by qaz last updated on 23/Oct/22 $$\left({a}\right). \\ $$$$\mathrm{2}{sinhx}−{cosh}=\left({e}^{{x}} −{e}^{−{x}} \right)−\mathrm{2}\left({e}^{{x}} +{e}^{−{x}} \right) \\ $$$$=−{e}^{{x}}…
Question Number 113465 by Algoritm last updated on 13/Sep/20 Commented by Her_Majesty last updated on 13/Sep/20 $${use}\:{your}\:{brains} \\ $$$$\sqrt{\mathrm{2019}}\approx\mathrm{45} \\ $$$${how}\:{could}\:\mathrm{45}\left(\frac{\mathrm{1}}{\mathrm{45}+\mathrm{2019}}+\frac{\mathrm{1}}{\mathrm{45}+\mathrm{2019}^{\mathrm{2}} }+…\right) \\ $$$${be}\:{greater}\:{than}\:\mathrm{1}??? \\…
Question Number 47916 by peter frank last updated on 16/Nov/18 Commented by maxmathsup by imad last updated on 17/Nov/18 $$\left.{b}\right)\:{let}\:{I}\:=\int\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}}}{{x}+\mathrm{2}}\:{dx}\:\Rightarrow\:{I}\:=\:\int\:\frac{\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}}{{x}+\mathrm{2}}\:{dx}\:=_{{x}+\mathrm{1}=\mathrm{2}{cht}} \:\int\frac{\mathrm{2}\sqrt{{ch}^{\mathrm{2}} {t}−\mathrm{1}}}{\mathrm{2}{ch}\left({t}\right)−\mathrm{1}+\mathrm{2}}\mathrm{2}{sh}\left({t}\right){dt} \\…
Question Number 113451 by pete last updated on 13/Sep/20 $$\mathrm{If}\:\mathrm{2x}=\mathrm{a}^{\mathrm{n}} +\mathrm{a}^{−\mathrm{n}} \:\mathrm{and}\:\mathrm{2y}=\mathrm{a}^{\mathrm{n}} −\mathrm{a}^{−\mathrm{n}} \:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}\:} \:\mathrm{in}\:\mathrm{its}\:\mathrm{simplest}\:\mathrm{form} \\ $$ Answered by bemath last updated…
Question Number 113429 by Khanacademy last updated on 13/Sep/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 178957 by Shrinava last updated on 23/Oct/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{recurring}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{polynomial}: \\ $$$$\mathrm{1}.\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{5}} −\mathrm{6x}^{\mathrm{4}} +\mathrm{16x}^{\mathrm{3}} −\mathrm{24x}^{\mathrm{2}} +\mathrm{20x}−\mathrm{8} \\ $$$$\mathrm{2}.\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{3}} −\mathrm{30x}^{\mathrm{2}} −\mathrm{7x}−\mathrm{56} \\ $$…