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New-version-of-the-app-with-usability-enhacements-is-now-available-on-playstore-Pleasd-update-app-Comment-on-this-post-for-problem-feedback-suggestions-

Question Number 93371 by Tinku Tara last updated on 12/May/20 $$\mathrm{New}\:\mathrm{version}\:\mathrm{of}\:\mathrm{the}\:\mathrm{app}\:\mathrm{with}\:\mathrm{usability} \\ $$$$\mathrm{enhacements}\:\mathrm{is}\:\mathrm{now}\:\mathrm{available}\:\mathrm{on} \\ $$$$\mathrm{playstore}. \\ $$$$\mathrm{Pleasd}\:\mathrm{update}\:\mathrm{app}.\: \\ $$$$\mathrm{Comment}\:\mathrm{on}\:\mathrm{this}\:\mathrm{post}\:\mathrm{for} \\ $$$$\mathrm{problem}/\mathrm{feedback}\:\mathrm{suggestions}. \\ $$ Commented by…

1-2-3-4-

Question Number 158899 by ehab last updated on 10/Nov/21 $$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}= \\ $$ Answered by EbrimaDanjo last updated on 10/Nov/21 $$\mathrm{the}\:\mathrm{LCM}\:\mathrm{of}\:\mathrm{denominators}\:=\mathrm{4} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\frac{\mathrm{2}\:+\:\mathrm{3}}{\mathrm{4}}\:\:\:\Rightarrow\:\frac{\mathrm{5}}{\mathrm{4}}\:=\mathrm{1}\frac{\mathrm{1}}{\mathrm{4}} \\…

Question-158884

Question Number 158884 by akolade last updated on 10/Nov/21 Answered by ghimisi last updated on 10/Nov/21 $${log}_{\mathrm{3}} \left({a}+\mathrm{1}\right)={log}_{\mathrm{4}} \left({a}+\mathrm{8}\right)={t}\Rightarrow \\ $$$${a}+\mathrm{1}=\mathrm{3}^{{t}} \\ $$$${a}+\mathrm{8}=\mathrm{4}^{{t}} \Rightarrow\mathrm{4}^{{t}} −\mathrm{3}^{{t}}…

Question-158883

Question Number 158883 by akolade last updated on 10/Nov/21 Commented by Rasheed.Sindhi last updated on 10/Nov/21 $$\left(\underset{{a}} {\underbrace{\sqrt{\mathrm{3}}\:+{x}}}\right)^{\mathrm{12}} +\left(\underset{{b}} {\underbrace{\sqrt{\mathrm{3}}\:−{x}}}\right)^{\mathrm{12}} =\mathrm{172928};{x}=? \\ $$$${a}+{b}=\mathrm{2}\sqrt{\mathrm{3}}\:;{ab}=\mathrm{3}−{x}^{\mathrm{2}} ={y}\:\left({say}\right) \\…