Question Number 121111 by zakirullah last updated on 05/Nov/20 Commented by benjo_mathlover last updated on 05/Nov/20 $$\left(\mathrm{Q4}\right)\:\mathrm{n}\left(\mathrm{A}\cup\mathrm{B}\right)=\mathrm{50}+\mathrm{20}−\mathrm{10}=\mathrm{60} \\ $$ Commented by zakirullah last updated on…
Question Number 186647 by MathsFan last updated on 07/Feb/23 $$ \\ $$Due to Earth's rotation. celestial objects like the moon and the stars appear to…
Question Number 55561 by otchereabdullai@gmail.com last updated on 26/Feb/19 Commented by otchereabdullai@gmail.com last updated on 26/Feb/19 $${prof}\:{W}\:{please}\:{am}\:{learning}\:{this}\:{question} \\ $$$${solved}\:{for}\:{me}\:{on}\:{the}\:\mathrm{21}{st}\:{but}\:{i}\:{dont}\: \\ $$$${understand}\:{this}\:{part}\::\:{if}\:{n}=\mathrm{16}:\alpha_{{n}} =−\mathrm{5}<\mathrm{0} \\ $$$${and}\:{if}\:{n}=\mathrm{18}:\alpha_{{n}} =\mathrm{355}>\mathrm{180}…
Question Number 186631 by DAVONG last updated on 07/Feb/23 Commented by mr W last updated on 08/Feb/23 $${all}\:{answers}\:{given}\:{are}\:{wrong}. \\ $$$${a}_{{k}} \:\in{N}\:\Rightarrow{a}_{{k}} \geqslant\mathrm{1} \\ $$$${since} \\…
Question Number 186630 by DAVONG last updated on 07/Feb/23 Answered by mr W last updated on 07/Feb/23 $${a}_{{k}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{k}=\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}=\frac{{k}^{\mathrm{2}} +{k}}{\mathrm{2}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{k}=\mathrm{1}}…
Question Number 186622 by maqsood last updated on 07/Feb/23 $$\:\:\:\mathrm{Solve}\:\mathrm{plz} \\ $$$$\:\:\:\:\left(\sqrt{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{25}+\mathrm{4}\sqrt{\mathrm{16}}}}\right)^{\mathrm{2}} \\ $$ Commented by maqsood last updated on 07/Feb/23 $$\mathrm{Process}\:\mathrm{plz} \\ $$ Commented…
Question Number 186617 by mustafazaheen last updated on 07/Feb/23 $$\underset{{x}\rightarrow\infty} {{lim}}\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt[{\mathrm{5}}]{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}}=?\:\:\:\:\:\:\:\:{with}\:{solution} \\ $$ Commented by aba last updated on 07/Feb/23 $$ \\ $$$$=\underset{{x}\rightarrow\infty}…
Question Number 121081 by zakirullah last updated on 05/Nov/20 Commented by zakirullah last updated on 05/Nov/20 $${please}\:{help}????\upuparrows \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 121077 by zakirullah last updated on 05/Nov/20 Commented by zakirullah last updated on 05/Nov/20 $$\boldsymbol{{please}}\:\boldsymbol{{help}}??\upuparrows\Uparrow \\ $$ Answered by liberty last updated on…
Question Number 121073 by Algoritm last updated on 05/Nov/20 Answered by MJS_new last updated on 05/Nov/20 $$\mathrm{both}\:\mathrm{sides}\:\mathrm{must}\:\mathrm{be}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{zero}\:\Rightarrow\:{x}<\mathrm{0}\wedge{x}\neq−\mathrm{1}\:\vee\:{x}>\mathrm{1} \\ $$$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming}\:\mathrm{leads}\:\mathrm{to} \\ $$$${x}^{\mathrm{5}} +\frac{{x}^{\mathrm{4}} }{\mathrm{6}}+\frac{{x}^{\mathrm{3}}…