Question Number 138916 by mathsuji last updated on 20/Apr/21 $$\underset{\:\mathrm{1}} {\overset{\:\mathrm{4}} {\int}}\:\sqrt{\frac{\sqrt{{z}}−\mathrm{1}}{\:\sqrt{{z}}}}\:{dz}\:=? \\ $$ Answered by bramlexs22 last updated on 20/Apr/21 $$\sqrt{\frac{\sqrt{\mathrm{z}}−\mathrm{1}}{\:\sqrt{\mathrm{z}}}}\:=\:\mathrm{r}\:;\:\frac{\sqrt{\mathrm{z}}\:−\mathrm{1}}{\:\sqrt{\mathrm{z}}}\:=\:\mathrm{r}^{\mathrm{2}} \rightarrow\begin{cases}{\mathrm{z}=\mathrm{1}\rightarrow\mathrm{r}=\mathrm{0}}\\{\mathrm{z}=\mathrm{4}\rightarrow\mathrm{r}=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}\end{cases} \\ $$$$\Rightarrow\mathrm{r}^{\mathrm{2}}…
Question Number 73378 by mind is power last updated on 10/Nov/19 $${Hello}\:,{i}\:{shar}\:{withe}\:{you}\:{nice}\:{problem}\: \\ $$$${show}\:{that}\:\forall{k}\in\mathbb{N}^{\ast} \:\exists{n}\in\mathbb{N}\:{such}\:{that} \\ $$$${k}\leqslant\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{j}}<{k}+\mathrm{1} \\ $$$${have}\:{a}\:{very}\:{Nice}\:{day} \\ $$$$ \\ $$…
Question Number 73356 by behi83417@gmail.com last updated on 10/Nov/19 Commented by ajfour last updated on 11/Nov/19 $${I}\:{had}\:{tried}\:{Q}.\mathrm{3}\:.\:{It}\:{did}\:{not}\:{yield} \\ $$$${any}\:{solution},\:{sir}. \\ $$ Commented by behi83417@gmail.com last…
Question Number 7820 by ridwan balatif last updated on 17/Sep/16 $${if}\:\mathrm{2}{x}+\mathrm{1}=\sqrt{\mathrm{111}},\:\mathrm{2}{x}^{\mathrm{5}} +\mathrm{2}{x}^{\mathrm{4}} −\mathrm{53}{x}^{\mathrm{3}} −\mathrm{57}{x}+\mathrm{54}=…? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 138888 by mathdanisur last updated on 19/Apr/21 $${Calculus}: \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\:\left[\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}^{{x}} }\:+\:\frac{\mathrm{2}^{{x}} }{{x}!}\:+\:\frac{{x}!}{{x}^{{x}} }\:\right]\:=\:? \\ $$ Answered by mathmax by abdo last…
Question Number 7815 by Rohit 57 last updated on 17/Sep/16 $${now}\:{wright}\:{the}\:{of}\:{A}.{P}\:{and}\:{G}.{P}\: \\ $$$${wiyh}\:{example}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 7809 by Tawakalitu. last updated on 16/Sep/16 $${x}\:+\:{y}\:+\:{z}\:=\:\mathrm{1}\:\:\:\:\:\:………\:\left({i}\right) \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{37}\:\:\:\:\:……..\:\left({ii}\right) \\ $$$${x}^{\mathrm{3}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{3}} \:=\:\mathrm{91}\:\:\:\:\:……..\:\left({iii}\right) \\ $$$$ \\ $$$${Solve}\:{simultaneously}.\: \\…
Question Number 138879 by peter frank last updated on 19/Apr/21 Answered by Dwaipayan Shikari last updated on 19/Apr/21 $${e}^{{i}\theta\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+..{n}\right)} =\mathrm{1}\Rightarrow{e}^{{i}\theta\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} ={e}^{\mathrm{2}\pi{im}} \:\:\left({m}\in\mathbb{Z}\right) \\ $$$$\Rightarrow\theta=\frac{\mathrm{4}\pi{m}}{{n}\left({n}+\mathrm{1}\right)} \\…
Question Number 138872 by peter frank last updated on 19/Apr/21 Answered by bramlexs22 last updated on 19/Apr/21 $$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} +\left(\mathrm{A}−\mathrm{7}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{Ax}−\mathrm{8} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{x}^{\mathrm{3}} +\left(\mathrm{A}−\mathrm{7}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{Ax}−\mathrm{8}=\mathrm{0} \\…
Question Number 7803 by Tawakalitu. last updated on 16/Sep/16 Answered by Rasheed Soomro last updated on 16/Sep/16 $$\left(\frac{\mathrm{1}×\mathrm{3}×\mathrm{5}×…×\mathrm{2015}}{\mathrm{2}×\mathrm{4}×\mathrm{6}×…×\mathrm{2016}}\right)^{\mathrm{1}/\mathrm{4}} \\ $$$$\left(\frac{\mathrm{2015}!/\left(\mathrm{2}.\mathrm{4}.\mathrm{6}…..\mathrm{2014}\right.}{\mathrm{2}.\mathrm{4}.\mathrm{6}…..\mathrm{2016}}\right)^{\mathrm{1}/\mathrm{4}} \\ $$$$\left(\frac{\mathrm{2015}!}{\left(\mathrm{2}.\mathrm{4}.\mathrm{6}….\mathrm{2014}\right)\left(\mathrm{2}.\mathrm{4}.\mathrm{6}…..\mathrm{2016}\right)}\right)^{\mathrm{1}/\mathrm{4}} \\ $$$$\left(\frac{\mathrm{2015}!}{\left(\mathrm{2}.\mathrm{4}.\mathrm{6}….\mathrm{2014}\right)^{\mathrm{2}} \left(\mathrm{2015}.\mathrm{2016}\right)}\right)^{\mathrm{1}/\mathrm{4}}…