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Question-208676

Question Number 208676 by RoseAli last updated on 20/Jun/24 Answered by Berbere last updated on 20/Jun/24 $$\sqrt[{\mathrm{4}}]{{a}}+\sqrt[{\mathrm{4}}]{{b}}=\sqrt{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}=\sqrt[{\mathrm{4}}]{\mathrm{2}}.\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\mathrm{2}{a}}+\sqrt[{\mathrm{4}}]{\mathrm{2}{b}}=\sqrt{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}+\sqrt[{\mathrm{4}}]{\frac{{b}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$${b}=\mathrm{0}\:\Rightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1}\:{no}\:{solution}\:\mathbb{N} \\…

Question-208591

Question Number 208591 by RoseAli last updated on 18/Jun/24 Answered by A5T last updated on 18/Jun/24 $$\mathrm{0}\leqslant\mathrm{3}\mid{sinx}\mid\leqslant\mathrm{3};\:−\mathrm{4}\leqslant−\mathrm{4}\mid{cosx}\mid\leqslant\mathrm{0} \\ $$$$\Rightarrow−\mathrm{4}\leqslant\mathrm{3}\mid{sinx}\mid−\mathrm{4}\mid{cosx}\mid\leqslant\mathrm{3} \\ $$$$\Rightarrow{y}\in\left[−\mathrm{4},\mathrm{3}\right] \\ $$ Terms of…

can-you-find-any-arrangement-of-nine-digits-of-1-9-such-as-967854312-and-the-first-digit-should-be-divisible-by-1-thefirst-two-digitds-should-be-divisible-by-2-the-first-three-digitds-should-be-divisi

Question Number 208555 by liuxinnan last updated on 18/Jun/24 $${can}\:{you}\:{find}\:{any}\:{arrangement}\:{of}\:{nine}\:{digits}\:{of}\:\mathrm{1}−\mathrm{9} \\ $$$${such}\:{as}\:\mathrm{967854312} \\ $$$${and}\:{the}\:{first}\:{digit}\:{should}\:{be}\:{divisible}\:{by}\:\mathrm{1} \\ $$$${thefirst}\:{two}\:{digitds}\:{should}\:{be}\:{divisible}\:{by}\:\mathrm{2} \\ $$$${the}\:{first}\:{three}\:{digitds}\:{should}\:{be}\:{divisible}\:{by}\:\mathrm{3}\: \\ $$$$…… \\ $$$${the}\:{number}\:{should}\:{be}\:{fivisible}\:{by}\:\mathrm{9} \\ $$$${if}\:{is}\:{such}\:{a}\:{number}\:{available} \\…

Question-208540

Question Number 208540 by otchereabdullai@gmail.com last updated on 17/Jun/24 Answered by Sutrisno last updated on 18/Jun/24 $${misal}\::\:\mathrm{3}{x}=\mathrm{2}{tan}\theta \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{3}}{tan}\theta\rightarrow{dx}=\frac{\mathrm{2}}{\mathrm{3}}{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\int\frac{\frac{\mathrm{2}}{\mathrm{3}}{tan}\theta−\mathrm{3}}{\:\sqrt{\left(\mathrm{2}{tan}\theta\right)^{\mathrm{2}} +\mathrm{4}}}.\frac{\mathrm{2}}{\mathrm{3}}{sec}^{\mathrm{2}} \theta{d}\theta \\…