Question Number 208814 by Ismoiljon_008 last updated on 23/Jun/24 Commented by Ismoiljon_008 last updated on 23/Jun/24 $${Help}\:{me}\:{please} \\ $$ Commented by Ismoiljon_008 last updated on…
Question Number 208809 by kgmxdd last updated on 23/Jun/24 Commented by Frix last updated on 23/Jun/24 $$\mathrm{As}\:\mathrm{always}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{42}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 208676 by RoseAli last updated on 20/Jun/24 Answered by Berbere last updated on 20/Jun/24 $$\sqrt[{\mathrm{4}}]{{a}}+\sqrt[{\mathrm{4}}]{{b}}=\sqrt{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}=\sqrt[{\mathrm{4}}]{\mathrm{2}}.\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\mathrm{2}{a}}+\sqrt[{\mathrm{4}}]{\mathrm{2}{b}}=\sqrt{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}+\sqrt[{\mathrm{4}}]{\frac{{b}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$${b}=\mathrm{0}\:\Rightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1}\:{no}\:{solution}\:\mathbb{N} \\…
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Question Number 208591 by RoseAli last updated on 18/Jun/24 Answered by A5T last updated on 18/Jun/24 $$\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…
Question Number 208594 by Jimenez000 last updated on 18/Jun/24 $${Find} \\ $$$${x}+\mathrm{3}^{{x}} <\mathrm{4} \\ $$ Answered by mr W last updated on 18/Jun/24 $$\mathrm{3}^{{x}} <\mathrm{4}−{x}…
Question Number 208617 by abdomath last updated on 18/Jun/24 Answered by efronzo1 last updated on 19/Jun/24 $$\:\:\begin{cases}{\:\downharpoonleft}\end{cases} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 208555 by liuxinnan last updated on 18/Jun/24 $${can}\:{you}\:{find}\:{any}\:{arrangement}\:{of}\:{nine}\:{digits}\:{of}\:\mathrm{1}−\mathrm{9} \\ $$$${such}\:{as}\:\mathrm{967854312} \\ $$$${and}\:{the}\:{first}\:{digit}\:{should}\:{be}\:{divisible}\:{by}\:\mathrm{1} \\ $$$${thefirst}\:{two}\:{digitds}\:{should}\:{be}\:{divisible}\:{by}\:\mathrm{2} \\ $$$${the}\:{first}\:{three}\:{digitds}\:{should}\:{be}\:{divisible}\:{by}\:\mathrm{3}\: \\ $$$$…… \\ $$$${the}\:{number}\:{should}\:{be}\:{fivisible}\:{by}\:\mathrm{9} \\ $$$${if}\:{is}\:{such}\:{a}\:{number}\:{available} \\…
Question Number 208540 by otchereabdullai@gmail.com last updated on 17/Jun/24 Answered by Sutrisno last updated on 18/Jun/24 $${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 \\…