Question Number 160487 by HongKing last updated on 30/Nov/21 Answered by Rasheed.Sindhi last updated on 30/Nov/21 $$\left(\mathrm{10}{a}+{b}\right)+\left(\mathrm{10}{b}+{a}\right)=\mathrm{10}{c}+\mathrm{3} \\ $$$$\mathrm{11}{a}+\mathrm{11}{b}=\mathrm{10}{c}+\mathrm{3} \\ $$$$\mathrm{11}\left({a}+{b}\right)=\mathrm{10}{c}+\mathrm{3}…………..\left({i}\right) \\ $$$$\mathrm{11}\mid\:\mathrm{10}{c}+\mathrm{3}\Rightarrow\mathrm{10}{c}+\mathrm{3}=\mathrm{33}\Rightarrow{c}=\mathrm{3} \\ $$$$\left(\mathrm{2}-{digit}\:{multiple}\:{of}\:\mathrm{11}\:{with}\:{unit}\:\mathrm{3}\right.…
Question Number 160482 by mr W last updated on 30/Nov/21 Commented by mr W last updated on 30/Nov/21 $${solve}\:{for}\:\mathbb{R} \\ $$ Answered by mr W…
Question Number 160473 by quvonnn last updated on 30/Nov/21 Answered by Rasheed.Sindhi last updated on 30/Nov/21 $$ \\ $$$${a}+{b}>{c}\:\wedge\:{b}+{c}>{a}\:\wedge\:{c}+{a}>{b} \\ $$$${a}+{b}−{c}>\mathrm{0}\:\wedge\:{b}+{c}−{a}>\mathrm{0}\:\wedge\:{c}+{a}−{b}>\mathrm{0} \\ $$$$\sqrt{{a}+{b}−{c}}\:\leqslant\sqrt{{a}}\:+\sqrt{{b}}\:−\sqrt{{c}} \\ $$$$\frac{\sqrt{{a}+{b}−{c}}\:}{\:\sqrt{{a}}\:+\sqrt{{b}}\:−\sqrt{{c}}}\leqslant\mathrm{1}……………..\left({i}\right)…
Question Number 160457 by HongKing last updated on 29/Nov/21 $$\mathrm{Simplfy}: \\ $$$$\frac{\mathrm{1}\:+\:\mathrm{cos}\boldsymbol{\alpha}}{\mathrm{sin}^{\mathrm{2}} \boldsymbol{\alpha}}\::\:\left(\mathrm{1}\:+\:\left(\frac{\mathrm{1}\:+\:\mathrm{cos}\boldsymbol{\alpha}}{\mathrm{sin}\boldsymbol{\alpha}}\right)^{\mathrm{2}} \right) \\ $$ Answered by MJS_new last updated on 29/Nov/21 $$\frac{\frac{\mathrm{1}+{c}}{{s}^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{\mathrm{1}+{c}}{{s}}\right)^{\mathrm{2}}…
Question Number 160445 by HongKing last updated on 29/Nov/21 $$\mathrm{Find}: \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\centerdot\:\frac{\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{2}}}}{\mathrm{2}}\:\centerdot\:\frac{\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{2}}}}}{\mathrm{2}}\:\centerdot\:…\:=\:? \\ $$ Commented by HongKing last updated on 29/Nov/21 $$\mathrm{Yes}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{but}\:\mathrm{how}\:\mathrm{please} \\ $$ Commented…
Question Number 160444 by HongKing last updated on 29/Nov/21 $$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{x}}\:+\:\mathrm{1}\right)\:-\:\frac{\sqrt{\pi}}{\mathrm{x}}}{\mathrm{x}^{\mathrm{3}} \:-\:\mathrm{8}}\:=\:? \\ $$ Answered by Ar Brandon last updated on 29/Nov/21 $$\mathscr{L}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\Gamma\left(\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)−\frac{\sqrt{\pi}}{{x}}}{{x}^{\mathrm{3}} −\mathrm{8}}…
Question Number 160436 by HongKing last updated on 29/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 29362 by puneet1789 last updated on 08/Feb/18 Commented by puneet1789 last updated on 08/Feb/18 how..? Commented by math solver last updated on 08/Feb/18…
Question Number 160424 by cortano last updated on 29/Nov/21 Answered by MJS_new last updated on 29/Nov/21 $$\mathrm{let}\:{y}={px}\:\Rightarrow \\ $$$$\begin{cases}{{x}\left(\left({p}^{\mathrm{4}} −\mathrm{1}\right){x}^{\mathrm{3}} −{p}+\mathrm{2}\right)=\mathrm{0}}\\{\left({p}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} {x}^{\mathrm{6}} +\mathrm{3}=\mathrm{0}}\end{cases} \\…
Question Number 29345 by Victor31926 last updated on 07/Feb/18 $$\mathrm{solve}\:\mathrm{simultanrodly} \\ $$$$ \\ $$$$\mathrm{x}+\mathrm{y}=\mathrm{5}…..\left(\mathrm{1}\right) \\ $$$$\mathrm{x}^{\mathrm{y}} +\mathrm{y}^{\mathrm{x}} =\mathrm{17}…..\left(\mathrm{2}\right) \\ $$ Answered by NECx last updated…