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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 \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{3}}{tan}\theta\rightarrow{dx}=\frac{\mathrm{2}}{\mathrm{3}}{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\int\frac{\frac{\mathrm{2}}{\mathrm{3}}{tan}\theta−\mathrm{3}}{\:\sqrt{\left(\mathrm{2}{tan}\theta\right)^{\mathrm{2}} +\mathrm{4}}}.\frac{\mathrm{2}}{\mathrm{3}}{sec}^{\mathrm{2}} \theta{d}\theta \\…
Question Number 208502 by Kalebwizeman last updated on 17/Jun/24 Answered by A5T last updated on 17/Jun/24 $$=\sqrt{\mathrm{16}}+\sqrt{\mathrm{15}}−\sqrt{\mathrm{15}}−\sqrt{\mathrm{14}}+…−\sqrt{\mathrm{9}}−\sqrt{\mathrm{8}}=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$ Commented by Kalebwizeman last updated on…
Question Number 208513 by CrispyXYZ last updated on 17/Jun/24 $${a},\:{b}>\mathrm{0}.\:\mathrm{2}{a}+{b}=\mathrm{1},\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}{a}}+\frac{\mathrm{2}{a}}{{b}^{\mathrm{2}} }\:>\:\mathrm{2}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208515 by mokys last updated on 17/Jun/24 $$\int_{\mathrm{0}} ^{\:\infty} \:\frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208440 by liuxinnan last updated on 16/Jun/24 $${a}_{\mathrm{1}} >{a}_{\mathrm{2}} >{a}_{\mathrm{3}} >…>{a}_{{n}} >\mathrm{0} \\ $$$${b}_{\mathrm{1}} >{b}_{\mathrm{2}} >{b}_{\mathrm{3}} >…>{b}_{{n}} >\mathrm{0} \\ $$$${prove} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}}…