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Question-191009

Question Number 191009 by pascal889 last updated on 16/Apr/23 Answered by Frix last updated on 16/Apr/23 $$\int\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}{dt}=\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}\int{dt}=\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}{t}+{C} \\ $$ Commented by ARUNG_Brandon_MBU…

Question-190972

Question Number 190972 by pascal889 last updated on 15/Apr/23 Answered by Frix last updated on 15/Apr/23 $$\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}=\mathrm{2}+\sqrt{\mathrm{3}}\wedge\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$${a}=\mathrm{2}\wedge{b}=\sqrt{\mathrm{3}} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} +\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\mathrm{14}…

Question-190961

Question Number 190961 by pascal889 last updated on 15/Apr/23 Answered by cortano12 last updated on 16/Apr/23 $$\:\Rightarrow\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{3x}^{\mathrm{2}} +\mathrm{8}\right)=\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{5x}+\mathrm{10}\right) \\ $$$$\:\Rightarrow\mathrm{3x}^{\mathrm{2}} −\mathrm{5x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)=\mathrm{0}\:…

Question-190960

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-59787

Question Number 59787 by Khairun Nisa last updated on 14/May/19 Answered by MJS last updated on 14/May/19 $$\mathrm{both}\:\mathrm{should}\:\mathrm{be}\:\mathrm{square}\:\mathrm{numbers},\:\mathrm{trying}\:\mathrm{I}\:\mathrm{get} \\ $$$${x}=\mathrm{9} \\ $$$${y}=\mathrm{4} \\ $$$$ \\…

An-earth-based-observer-sees-rocket-A-moving-at-0-70c-directly-towards-rocket-B-which-is-moving-towards-A-at-0-80c-How-fast-does-rocket-A-sees-rocket-B-approaching-

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-59752

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}}…