Question Number 122850 by liberty last updated on 20/Nov/20 $$\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{4}{k}}{\mathrm{4}{k}^{\mathrm{4}} +\mathrm{1}}\:=\:? \\ $$ Answered by bemath last updated on 20/Nov/20 $$\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\left(\mathrm{2}{k}^{\mathrm{2}}…
Question Number 188375 by Rupesh123 last updated on 28/Feb/23 Commented by Frix last updated on 28/Feb/23 $${M}=\mathrm{45}\:\mathrm{at}\:{x}=−\frac{\mathrm{17}}{\mathrm{2}} \\ $$ Commented by Rupesh123 last updated on…
Question Number 122705 by john santu last updated on 19/Nov/20 Commented by liberty last updated on 19/Nov/20 $$\Rightarrow{a}\::\:{b}\:=\:\mathrm{2}\::\:\mathrm{3}\: \\ $$$$\Rightarrow{b}\::\:{c}\:=\:\mathrm{3}\::\:\mathrm{4} \\ $$$$\Rightarrow\:{c}\::\:{d}\:=\:\mathrm{5}\::\:\mathrm{6}\: \\ $$$${then}\:{a}\::\:{b}\::\:{c}\::\:{d}\:=\:\mathrm{10}\::\:\mathrm{15}\::\:\mathrm{20}\::\:\mathrm{24} \\…
Question Number 122608 by bramlexs22 last updated on 18/Nov/20 $$\:{Sum}\:{the}\:{series}\:{to}\:\mathrm{10}^{{th}} \:{terms}\: \\ $$$$\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{x}}}\:+\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:+\frac{\mathrm{1}}{\mathrm{1}−\sqrt{{x}}}\:+… \\ $$$${equal}\:{to}\:\_\_\_\: \\ $$ Answered by liberty last updated on 18/Nov/20 $$\:\mathrm{T}_{\mathrm{1}}…
Question Number 122605 by bramlexs22 last updated on 18/Nov/20 Answered by Dwaipayan Shikari last updated on 18/Nov/20 $${T}_{{n}} =\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$${S}_{{n}} +\mathrm{1}=\overset{{n}} {\sum}\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$${S}_{{n}}…
Question Number 122484 by JBocanegra last updated on 17/Nov/20 $$\:\frac{\mathrm{4}{a}}{\mathrm{3}{b}\centerdot\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \centerdot{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \centerdot{b}^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{3}\centerdot\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} }\right)\left(\frac{{a}}{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \centerdot{b}^{\frac{\mathrm{4}}{\mathrm{3}}} }\right) \\ $$$$\left(\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}\centerdot\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} }\right)\left({a}^{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}} \centerdot{b}^{−\frac{\mathrm{4}}{\mathrm{3}}} \right)=\left(\mathrm{2}^{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{3}}} \right)\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 188012 by Shlock last updated on 24/Feb/23 Commented by Rasheed.Sindhi last updated on 25/Feb/23 $${Perhaps}\:{there}'{s}\:{an}\:{error}\:{in}\:{some} \\ $$$${figures}\:{of}\:{the}\:{question}.{Have}\:{you} \\ $$$${an}\:{answer}\:{of}\:{the}\:{question}? \\ $$ Terms of…
Question Number 56857 by azam2412 last updated on 25/Mar/19 $${romi}−{romo}=\mathrm{0} \\ $$$$\mathcal{L}.{x}_{\mathrm{0}} +\mathcal{G}.{y}_{{n}} =\mathcal{L}.{x}_{{n}} +\mathcal{G}.{y}_{\mathrm{0}} \\ $$$$\mathcal{L}\left({x}_{\mathrm{0}} −{x}_{{n}} \right)=\mathcal{G}\left({y}_{\mathrm{0}} −{y}_{{n}} \right) \\ $$$${y}_{\mathrm{0}} =\frac{\mathcal{L}}{\mathcal{G}}.\left({x}_{\mathrm{0}} −{x}_{{n}}…
Question Number 56852 by peter frank last updated on 25/Mar/19 Answered by MJS last updated on 25/Mar/19 $$\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:{f}\circ{g}\:\mathrm{means}\:{f}\left({g}\right)\:\mathrm{or}\:{g}\left({f}\right) \\ $$$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{only}\:\mathrm{a}\:\mathrm{matter}\:\mathrm{of}\:\mathrm{thinking}\:\mathrm{logically} \\ $$$${f}\left({g}\left({x}\right)\right)=\begin{cases}{\mathrm{1};\:{x}<−\mathrm{1}}\\{\mathrm{3};\:−\mathrm{1}\leqslant{x}<−\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{4}{x}^{\mathrm{2}} ;\:−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{2};\:\frac{\mathrm{1}}{\mathrm{2}}<{x}\leqslant\mathrm{1}}\\{\mathrm{1};\:{x}>\mathrm{1}}\end{cases} \\ $$$${g}\left({f}\left({x}\right)\right)=\begin{cases}{\mathrm{1};\:{x}<−\mathrm{1}}\\{\mathrm{2}{x}^{\mathrm{2}}…
Question Number 56845 by Tawa1 last updated on 25/Mar/19 Answered by math1967 last updated on 25/Mar/19 $$\left(\mathrm{11}…\mathrm{108}{times}\right)×\mathrm{10}+\mathrm{1}+\left(\mathrm{22}..\mathrm{108}{times}\right)×\mathrm{10}+\mathrm{2} \\ $$$$……\left(\mathrm{77}..\mathrm{108}{times}\right)×\mathrm{10}+\mathrm{7} \\ $$$$\frac{\mathrm{11}….\mathrm{108}{times}+…..\mathrm{77}…\mathrm{108}{times}}{\mathrm{37}}+\frac{\mathrm{1}+..\mathrm{7}}{\mathrm{37}} \\ $$$${remainder}\:\mathrm{0}\:+{remainder}\mathrm{28} \\ $$$${so}\:\mathrm{28}\:{ans}…