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Question Number 187029 by yaslm last updated on 12/Feb/23 Answered by cortano12 last updated on 12/Feb/23 $$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow−\mathrm{2}^{−} } {\mathrm{lim}}\:\left({ax}+{b}\right)=\:\underset{{x}\rightarrow−\mathrm{2}^{+} } {\mathrm{lim}}\left({x}^{\mathrm{2}} +\mathrm{2}{b}−\mathrm{17}\right) \\ $$$$\:\:\:\:\:\:\:−\mathrm{2}{a}+{b}\:=\:\mathrm{4}+\mathrm{2}{b}−\mathrm{17} \\…
Question Number 187027 by yaslm last updated on 12/Feb/23 Answered by Farhadazizi last updated on 12/Feb/23 $${fog}\left({x}\right)={f}\left({g}\left({x}\right)\right)=\frac{\mathrm{2}{g}\left({x}\right)−\mathrm{4}}{{g}\left({x}\right)}=\frac{\mathrm{2}\sqrt{{x}−\mathrm{1}}−\mathrm{4}}{\:\sqrt{{x}−\mathrm{1}}} \\ $$$${gof}\left({x}\right)={g}\left({f}\left({x}\right)\right)=\sqrt{{f}\left({x}\right)−\mathrm{1}}=\sqrt{\frac{{x}−\mathrm{4}}{{x}}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}{x}−\mathrm{4}}{{x}}=\mathrm{2}−\frac{\mathrm{4}}{{x}}\Rrightarrow{f}\left({x}\right)−\mathrm{2}=−\frac{\mathrm{4}}{{x}} \\ $$$$−\frac{\mathrm{1}}{{f}\left({x}\right)−\mathrm{2}}=\frac{{x}}{\mathrm{4}}\Rrightarrow{x}=\frac{−\mathrm{4}}{{f}\left({x}\right)−\mathrm{2}}\Rrightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{−\mathrm{4}}{{x}−\mathrm{2}} \\…
Question Number 187020 by mustafazaheen last updated on 12/Feb/23 $$\frac{{a}}{{x}}=\frac{{b}}{{y}}=\frac{{c}}{{z}}=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:,{a}−\mathrm{2}{b}+{c}=\mathrm{2}\:\:{and}\:\:\mathrm{2}{y}−\mathrm{3}{z}=\mathrm{1}\:\:\:\:{x}=? \\ $$$${how}\:{is}\:{solution} \\ $$ Commented by mr W last updated on 12/Feb/23 $${you}\:{can}\:{not}\:{determine}\:{x}\:{and}\:{other} \\ $$$${unknowns}!…
Question Number 121474 by faysal last updated on 08/Nov/20 $${a}\:{rectangle}'{s}\:{a}\:{side}'{s}\:{terminal} \\ $$$${points}\:{coordinate}\:{are}\:\left(\mathrm{2},−\mathrm{1}\right),\:\left(\mathrm{6},\mathrm{5}\right)\:{and}\:{the} \\ $$$${length}\:{of}\:{the}\:{diagonal}\:{is}\:\mathrm{8}\:{unit}. \\ $$$${what}\:{are}\:{coordinate}\:{of}\:{terminal}\:{point} \\ $$$${of}\:{paralel}\:{side}\:{that}\:{side} \\ $$ Answered by TANMAY PANACEA last…
Question Number 187011 by aba last updated on 16/Feb/23 $$\mathrm{a}\in\mathbb{C} \\ $$$$\begin{vmatrix}{\mathrm{a}\:\:\:\mathrm{1}\:\:\:\:\ldots\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\ddots\:\:\ddots\:\:\vdots}\\{\vdots\:\:\ddots\:\:\ddots\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\ldots\:\:\:\:\mathrm{1}\:\:\:\mathrm{a}}\end{vmatrix}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 187010 by aba last updated on 16/Feb/23 $$\mathrm{a},\mathrm{b}\in\mathbb{C} \\ $$$$\:\begin{vmatrix}{\mathrm{a}+\mathrm{b}\:\:\:\:\:\:\:\:\:\mathrm{a}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\ldots\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\mathrm{a}+\mathrm{b}\:\:\:\:\:\:\ddots\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\vdots}\\{\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\:\:\vdots\:\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}}\\{\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\ldots\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}+\mathrm{b}}\end{vmatrix}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 187007 by Humble last updated on 12/Feb/23 $${prove}\:\:{that} \\ $$$$\mathrm{7}^{{k}+\mathrm{1}} \mid\:\mathrm{13}^{\mathrm{7}{k}} \:+\:\mathrm{1}\:,\:\forall{k}\in\mathbb{N} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Commented by Frix…
Question Number 121471 by Myrnalyn last updated on 08/Nov/20 $${f}/{g} \\ $$$$\left({f}\:/\:{g}\right)\left({x}\right)=\:{f}\left({x}\right)\:/\:{g}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:+\:\mathrm{1}\right)\frac{\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:+\:\mathrm{1}}{} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{6}{x}^{\mathrm{3}} \:−\:\mathrm{9}{x}^{\mathrm{2}} \:+\:\mathrm{8}{x}^{\mathrm{2}} \:−\mathrm{12}{x}\:+\:\mathrm{2}{x}\:−\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{6}{x}^{\mathrm{3}} \:−\:{x}^{\mathrm{2}} \:−\:\mathrm{10}{x}\:−\:\mathrm{3}…
Question Number 55933 by Gulay last updated on 06/Mar/19 Commented by Gulay last updated on 06/Mar/19 $$\mathrm{sir}\:\mathrm{could}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…