Question Number 5278 by sara last updated on 04/May/16 $$\int\left\{\left(\mathrm{3}{x}\right){i}+\left(\mathrm{2}{x}\right){y}\right\}{dx}= \\ $$$$ \\ $$ Answered by FilupSmith last updated on 04/May/16 $$\int\left(\mathrm{3}{xi}+\mathrm{2}{xy}\right){dx} \\ $$$$=\mathrm{3}{i}\int{xdx}+\mathrm{2}{y}\int{xdx} \\…
Question Number 5275 by sanusihammed last updated on 03/May/16 Answered by Rasheed Soomro last updated on 04/May/16 $$\mathrm{Three}\:\mathrm{members}\:\mathrm{of}\:\mathrm{the}\:\mathrm{commetee}\:\mathrm{of}\:\mathrm{six} \\ $$$$\mathrm{are}\:\mathrm{chairman},\:\mathrm{secratery}\:\mathrm{and}\:\mathrm{treasurer}. \\ $$$$\mathrm{Remaining}\:\mathrm{3}\:\mathrm{members}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{out}\:\mathrm{of} \\ $$$$\mathrm{7}\:\mathrm{members}: \\…
Question Number 136344 by Gideon last updated on 21/Mar/21 Commented by mindispower last updated on 21/Mar/21 $${x}^{\mathrm{2}} +\mathrm{6}{x}=\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{9} \\ $$$${t}=\left({x}+\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{5}}{\:\sqrt{{t}−\mathrm{1}}}+\frac{\mathrm{1}}{\:\sqrt{{t}−\mathrm{4}}}=\frac{\mathrm{4}}{{t}} \\ $$$${we}\:{have}\:{t}>\mathrm{4},{t}−\mathrm{1}>\mathrm{3}\Rightarrow\sqrt{{t}−\mathrm{1}}<{t}\Leftrightarrow…
Question Number 5274 by sanusihammed last updated on 03/May/16 $${Please}\:{help}\: \\ $$$$ \\ $$$${The}\:{stock}\:\:{of}\:{umar}\:{pharmacy}\:{in}\:{january}\:{was}\:\mathrm{20240}\:{items}. \\ $$$${its}\:{stock}\:{decline}\:{by}\:\mathrm{20}\:{percent}\:{in}\:{may}\:{due}\:{to}\:{increase}\:{sales}. \\ $$$${What}\:{is}\:{the}\:{amount}\:{of}\:{stock}\:{in}\:{may}\:.\: \\ $$ Commented by Rasheed Soomro last…
Question Number 70808 by rajesh4661kumar@gmail.com last updated on 08/Oct/19 Answered by MJS last updated on 08/Oct/19 $$\int\left(\mathrm{sec}\:{x}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{csc}\:{x}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} {dx}=\int\frac{{dx}}{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:\left(\mathrm{tan}\:{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\left(\mathrm{cos}\:{x}\right)^{\mathrm{2}} {dt}\right] \\ $$$$=\int\frac{{dt}}{{t}^{\frac{\mathrm{4}}{\mathrm{3}}}…
Question Number 136346 by ZiYangLee last updated on 21/Mar/21 $$\mathrm{If}\:\mathrm{1},\omega\:\mathrm{and}\:\omega^{\mathrm{2}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}, \\ $$$$\mathrm{prove}\:\mathrm{that}\:\left(\mathrm{1}−\omega\right)\left(\mathrm{1}−\omega^{\mathrm{2}} \right)\left(\mathrm{1}−\omega^{\mathrm{4}} \right)\left(\mathrm{1}−\omega^{\mathrm{5}} \right)=\mathrm{9} \\ $$ Answered by Rasheed.Sindhi last updated on 21/Mar/21…
Question Number 136340 by liberty last updated on 21/Mar/21 $${If}\:\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}}\:=\:\sqrt{\frac{{a}+\sqrt{{b}}}{{c}}}\:+\:\sqrt{\frac{{a}−\sqrt{{b}}}{{c}}} \\ $$$${then}\:{a}+{bc}\:=\:? \\ $$ Answered by Ñï= last updated on 21/Mar/21 $$\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}}=\sqrt{{x}}+\sqrt{{y}} \\ $$$$\mathrm{3}+\sqrt{\mathrm{2}}={x}+{y}+\mathrm{2}\sqrt{{xy}} \\…
Question Number 136343 by snipers237 last updated on 21/Mar/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{x}^{\mathrm{2021}} }{{x}−{ln}\left(\underset{{k}=\mathrm{0}} {\overset{\mathrm{2020}} {\sum}}\frac{{x}^{{k}} }{{k}!}\right)}\:\overset{?} {=}\:\mathrm{2021}!\: \\ $$ Answered by mindispower last updated on 21/Mar/21…
Question Number 5266 by Kasih last updated on 03/May/16 $$\int\:\frac{\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} +\:\mathrm{2}{x}+\:\mathrm{5}}}\:{dx} \\ $$ Commented by prakash jain last updated on 03/May/16 $$\mathrm{You}\:\mathrm{can}\:\mathrm{integrate}\:\mathrm{as}\:\mathrm{following} \\ $$$$\frac{\mathrm{3}{x}+\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}}−\frac{\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}}…
Question Number 5265 by Kasih last updated on 03/May/16 $$\int\:\frac{{x}}{\mathrm{1}+{x}}\:{dx} \\ $$ Answered by Yozzii last updated on 03/May/16 $$\frac{{x}}{\mathrm{1}+{x}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}}=\frac{\mathrm{1}+{x}−\mathrm{1}}{\mathrm{1}+{x}}=\frac{{x}}{\mathrm{1}+{x}} \\ $$$$\Rightarrow\int\frac{{x}}{\mathrm{1}+{x}}{dx}=\int\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}}\right){dx}={x}−{ln}\mid{x}+\mathrm{1}\mid+{C} \\ $$$$ \\…