Question Number 139476 by aliibrahim1 last updated on 27/Apr/21 Answered by qaz last updated on 27/Apr/21 $${I}=\int\frac{{dx}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)…\left({x}+{n}\right)} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)…\left({x}+{n}\right)}=\frac{{A}_{\mathrm{0}} }{{x}}+\frac{{A}_{\mathrm{1}} }{{x}+\mathrm{1}}+\frac{{A}_{\mathrm{2}} }{{x}+\mathrm{2}}+…+\frac{{A}_{{n}} }{{x}+{n}} \\ $$$${A}_{\mathrm{0}}…
Question Number 139479 by mnjuly1970 last updated on 27/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{advanced}\:….\bigstar\bigstar\bigstar…..{calculus}….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\: :=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{sin}\left({n}\right)}{{n}}\right)^{\mathrm{3}} =?\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$ Answered by Dwaipayan Shikari last…
Question Number 73940 by smartsmith459@gmail.com last updated on 16/Nov/19 Answered by Rio Michael last updated on 16/Nov/19 $$\left.{Q}\mathrm{4}\right)\:{is}\:{equivalent}\:{to}\:{solving}\: \\ $$$$\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:{M}^{−\mathrm{1}} \begin{pmatrix}{\mathrm{1}}\\{\mathrm{23}}\end{pmatrix} \\ $$$${where}\:{M}\:=\:\begin{pmatrix}{\mathrm{3}}&{−\mathrm{4}}\\{\mathrm{7}}&{\mathrm{1}}\end{pmatrix} \\ $$$${M}^{−\mathrm{1}}…
Question Number 139478 by mnjuly1970 last updated on 27/Apr/21 $$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:………\:{nice}\:…\:…\:…\:{calculus}…….. \\ $$$$\:\:\:\:\:\:\Phi:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$\:\:\:\:\:\:\:{NOTE}\:::\:{li}_{\mathrm{2}} \left({z}\right)+{li}_{\mathrm{2}} \left(\mathrm{1}−{z}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−{ln}\left({z}\right){ln}\left(\mathrm{1}−{z}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{Hence}\:::\:\:{li}_{\mathrm{2}}…
Question Number 8403 by rhm last updated on 10/Oct/16 $$\left({Q}.\mathrm{1}\right)\:{cos}\:\mathrm{4}{A}=\mathrm{1}−\mathrm{8}{cos}^{\mathrm{2}} {A}\:+\:\mathrm{8}{cos}^{\mathrm{2}} \:{A} \\ $$$$\left({Q}.\mathrm{2}\right)\:\:\frac{{sec}\mathrm{8}\:{A}\:−\mathrm{1}}{{sec}\mathrm{4}\:{A}\:−\mathrm{1}}\:=\:\frac{{tan}\:\mathrm{8}{A}}{{tan}\:\mathrm{2}{A}} \\ $$$$\left({Q}.\mathrm{3}\right)\:\:\:{tanA}+{tan}\left(\mathrm{60}^{\mathrm{0}} +{A}\right)+{tan}\left(\mathrm{120}^{\mathrm{0}} \right. \\ $$$$\left.+\mathrm{4}\right)\:=\:\mathrm{3}{tan}\:\mathrm{3}{A}\: \\ $$$$\left({Q}.\mathrm{4}\right)\:\:\:{sinA}\:{sin}\:\left(\mathrm{60}^{\mathrm{0}} −{A}\right)\:\:{sin}\left(\mathrm{60}^{\mathrm{0}} +{A}\right)\: \\…
Question Number 8402 by rhm last updated on 10/Oct/16 $${Q}.\:\left({cos}^{\mathrm{2}} \mathrm{66}^{\mathrm{0}} −{sin}^{\mathrm{2}} \mathrm{6}^{\mathrm{0}} \right)\left({cos}^{\mathrm{2}} \mathrm{48}^{\mathrm{0}} −{sin}^{\mathrm{2}} \mathrm{12}^{\mathrm{0}} \right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$ Answered by sandy_suhendra…
Question Number 8401 by rhm last updated on 10/Oct/16 $${Q}\:.\mathrm{1}\:\:\:\mathrm{1}+{cos}^{\mathrm{2}} \mathrm{2}{A}=\mathrm{2}\left({cos}^{\mathrm{4}} \:{A}+{sin}^{\mathrm{4}} \:{A}\right) \\ $$$$\mathrm{2}.\:\:{sin}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} \left(\mathrm{120}^{\mathrm{0}} +{A}\right)+{sin}^{\mathrm{2}} \left(\mathrm{120}^{\mathrm{0}} −{A}\right) \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\…
Question Number 139474 by mathdanisur last updated on 27/Apr/21 $${let}:\:\:\Omega_{\boldsymbol{{n}}} =\underset{\:\mathrm{0}} {\overset{\:\mathrm{2}\pi} {\int}}{cos}\left({x}\right)\centerdot{cos}\left(\mathrm{2}{x}\right)\centerdot…\centerdot{cos}\left({nx}\right)\:{dx} \\ $$$${for}\:{which}\:{integers}\:{n},\:\mathrm{1}\leqslant{n}\leqslant\mathrm{10},\:{is}\:\Omega_{\boldsymbol{{n}}} \neq\mathrm{0}? \\ $$ Answered by mathmax by abdo last updated…
Question Number 139468 by mathdanisur last updated on 27/Apr/21 $${prove}:\:\:\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{9}+\mathrm{4}\sqrt{\mathrm{2}}}}\:=\:\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}} \\ $$ Answered by OlafThorendsen last updated on 27/Apr/21 $${x}\:=\:\:\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{9}+\mathrm{4}\sqrt{\mathrm{2}}}} \\ $$$${x}\:=\:\:\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}−\sqrt{\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$${x}\:=\:\:\sqrt{\mathrm{3}+\sqrt{\mathrm{2}}−\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right)}…
Question Number 8390 by tawakalitu last updated on 09/Oct/16 Answered by fernandodantas1996 last updated on 12/Oct/16 $$ \\ $$$$\: \\ $$$$\left.\mathrm{i}\right)\:\overset{\rightarrow} {\mathrm{F}}+\overset{\rightarrow} {\mathrm{P}}\:=\:\mathrm{6i}\:+\:\mathrm{4j}\:−\:\mathrm{k}\:\Rightarrow\: \\ $$$$\Rightarrow\:\parallel\overset{\rightarrow}…