Question Number 144026 by mohammad17 last updated on 20/Jun/21 Commented by mohammad17 last updated on 20/Jun/21 $${help}\:{me}\:{sir} \\ $$ Answered by Olaf_Thorendsen last updated on…
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Question Number 78456 by arkanmath7@gmail.com last updated on 17/Jan/20 $${for}\:{a}>\mathrm{0}\:{and}\:{b}>{a}+\mathrm{2}\:,\:\:{verify}\:{the}\:{follwing}\: \\ $$$${claim}: \\ $$$$\:\:\:\sum_{{n}=\mathrm{1}} ^{\:\:\infty} \:{n}\:\frac{{a}\left({a}+\mathrm{1}\right)\left({a}+\mathrm{2}\right)…\left({a}+{n}−\mathrm{1}\right)}{{b}\left({b}+\mathrm{1}\right)\left({b}+\mathrm{2}\right)…\left({b}+{n}−\mathrm{1}\right)}\:=\frac{{a}\left({b}−\mathrm{1}\right)}{\left({b}−{a}−\mathrm{1}\right)\left({b}−{a}−\mathrm{2}\right)} \\ $$ Answered by mind is power last updated…
Question Number 78447 by arkanmath7@gmail.com last updated on 17/Jan/20 $${prove}\:{that}\:{the}\:{seq}\:{a}_{{n}} \:=\:\frac{{ncos}\left(\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right)}{{n}+\mathrm{1}} \\ $$$${has}\:{convergent}\:{subsequence} \\ $$ Answered by mind is power last updated on 17/Jan/20…
Question Number 78443 by Maclaurin Stickker last updated on 17/Jan/20 Answered by mr W last updated on 18/Jan/20 Commented by mr W last updated on…
Question Number 143970 by ZiYangLee last updated on 20/Jun/21 $$\mathrm{Given}\:\mathrm{that}\:\omega\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}, \\ $$$$\omega^{\mathrm{7}} =\mathrm{1},\:\omega\neq\mathrm{1},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\omega^{\mathrm{1}} +\omega^{\mathrm{2}} +\omega^{\mathrm{3}} +\omega^{\mathrm{4}} +\omega^{\mathrm{5}} +\omega^{\mathrm{6}} . \\ $$ Answered by…
Question Number 143965 by Khalmohmmad last updated on 20/Jun/21 Commented by Canebulok last updated on 20/Jun/21 $$\boldsymbol{{Solution}}: \\ $$$$\Rightarrow\:\frac{{log}\left({x}\right)}{{log}\left(\mathrm{3}\right)}\:+\:\frac{{log}\left(\mathrm{5}\right)}{{log}\left({x}\right)}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{log}\left({x}\right)^{\mathrm{2}} \:+\:{log}\left(\mathrm{5}\right){log}\left(\mathrm{3}\right)\:=\:\mathrm{0} \\ $$$$\: \\…
Question Number 143964 by naka3546 last updated on 20/Jun/21 $${x}_{\mathrm{1}} \:{and}\:\:{x}_{\mathrm{2}} \:\:{are}\:\:{solutions}\:\:{of}\:\:{equality}\:: \\ $$$$\:\:\mathrm{cos}\:\left(\frac{\pi{x}+\pi}{\mathrm{6}}\right)\:−\:\mathrm{sin}\:\left(\frac{\pi{x}−\pi}{\mathrm{6}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\mathrm{3}}\:\:\:,\:\:\:\mathrm{0}\:\leqslant\:{x}\:\leqslant\:\mathrm{12} \\ $$$${Find}\:\:{the}\:\:{value}\:\:{of}\:\:{x}_{\mathrm{1}} +\:{x}_{\mathrm{2}} \:. \\ $$ Commented by Canebulok last updated…
Question Number 143961 by SOMEDAVONG last updated on 20/Jun/21 $$\mathrm{I}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{cos}^{\mathrm{6}} \mathrm{x}}{\mathrm{1}−\mathrm{3sin}^{\mathrm{2}} \mathrm{xcos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}=? \\ $$ Answered by Dwaipayan Shikari last updated on 20/Jun/21…
Question Number 143963 by SOMEDAVONG last updated on 20/Jun/21 $$\mathrm{A}=\mathrm{x}^{\mathrm{ln}\frac{\mathrm{y}}{\mathrm{x}}} .\mathrm{y}^{\mathrm{ln}\frac{\mathrm{z}}{\mathrm{x}}} .\mathrm{z}^{\mathrm{ln}\frac{\mathrm{x}}{\mathrm{y}}} =?? \\ $$ Answered by Olaf_Thorendsen last updated on 20/Jun/21 $$\mathrm{A}\:=\:{x}^{\mathrm{ln}\frac{{y}}{{x}}} {y}^{\mathrm{ln}\frac{{z}}{{x}}} {z}^{\mathrm{ln}\frac{{x}}{{y}}}…