Question Number 190960 by pascal889 last updated on 15/Apr/23 Commented by Rasheed.Sindhi last updated on 15/Apr/23 $$\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{28}}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{8}}\:=\sqrt{\mathrm{6}{x}^{\mathrm{2}} −\mathrm{11}{x}−\mathrm{52}}\:\:\:\:\:? \\ $$ Commented by pascal889…
Question Number 125411 by Dwaipayan Shikari last updated on 11/Dec/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\sqrt{{n}}}{{n}^{\mathrm{2}} +\mathrm{1}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 59787 by Khairun Nisa last updated on 14/May/19 Answered by MJS last updated on 14/May/19 $$\mathrm{both}\:\mathrm{should}\:\mathrm{be}\:\mathrm{square}\:\mathrm{numbers},\:\mathrm{trying}\:\mathrm{I}\:\mathrm{get} \\ $$$${x}=\mathrm{9} \\ $$$${y}=\mathrm{4} \\ $$$$ \\…
Question Number 59780 by necx1 last updated on 14/May/19 $${An}\:{earth}-{based}\:{observer}\:{sees}\:{rocket}\:{A} \\ $$$${moving}\:{at}\:\mathrm{0}.\mathrm{70}{c}\:{directly}\:{towards}\:{rocket} \\ $$$${B},{which}\:{is}\:{moving}\:{towards}\:{A}\:{at}\:\mathrm{0}.\mathrm{80}{c}. \\ $$$${How}\:{fast}\:{does}\:{rocket}\:{A}\:{sees}\:{rocket}\:{B} \\ $$$${approaching}? \\ $$$$ \\ $$ Answered by ajfour…
Question Number 59752 by Khairun Nisa last updated on 14/May/19 Commented by Smail last updated on 14/May/19 $$\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} +\left(\sqrt{\mathrm{7}−\sqrt{\mathrm{48}}}\right)^{{x}} =\mathrm{14} \\ $$$$\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} \left(\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} +\left(\sqrt{\mathrm{7}−\sqrt{\mathrm{48}}}\right)^{{x}} \right)=\mathrm{14}\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}}…
Question Number 125283 by Dwaipayan Shikari last updated on 09/Dec/20 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({log}\left({x}+\mathrm{1}\right)\right)}{\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}}\right)^{\mathrm{2}} }{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 59685 by Tawa1 last updated on 13/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 59675 by Mr X pcx last updated on 13/May/19 $${you}\:{are}\:{welcome}\:{sir}\:{ali}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 59655 by aliesam last updated on 13/May/19 Commented by Mr X pcx last updated on 13/May/19 $${r}=\xi\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:}\:\Rightarrow\frac{\partial{r}}{\partial{x}}\:=\frac{\xi{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }}\:=\frac{\xi{x}}{{r}} \\ $$$$\frac{\partial\theta}{\partial{y}}\:=\xi\frac{{y}}{\:\sqrt{{x}^{\mathrm{2}}…
Question Number 59639 by Tawa1 last updated on 12/May/19 Commented by Tawa1 last updated on 12/May/19 $$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{continue}.\:\:\mathrm{My}\:\mathrm{final}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{wrong}. \\ $$$$\:\:\:\mathrm{I}\:\mathrm{got}:\:\:\:\mathrm{S}\:\:=\:\:\frac{\mathrm{2n}^{\mathrm{3}} \:+\:\mathrm{3n}^{\mathrm{2}} \:+\:\mathrm{n}}{\mathrm{6}}\:\:\:\:\:\:\mathrm{but}\:\:\mathrm{answer}\:\mathrm{is}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{S}\:\:=\:\:\frac{\mathrm{n}}{\mathrm{6}}\:\left(\mathrm{14n}^{\mathrm{2}} \:−\:\mathrm{9n}\:+\:\mathrm{1}\right) \\ $$ Commented…