Question Number 90100 by ar247 last updated on 21/Apr/20 Answered by john santu last updated on 21/Apr/20 $${let}\:{w}\:=\:\underset{{y}\:=\:\mathrm{2}} {\overset{\mathrm{2020}} {\prod}}\:\frac{\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right)}{{y}^{\mathrm{2}} } \\ $$$${w}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{3}}\right)\left(\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{4}}\right)×…× \\ $$$$\left(\frac{\mathrm{2018}}{\mathrm{2019}}×\frac{\mathrm{2020}}{\mathrm{2019}}\right)\left(\frac{\mathrm{2019}}{\mathrm{2020}}×\frac{\mathrm{2021}}{\mathrm{2020}}\right)…
Question Number 24565 by ajfour last updated on 21/Nov/17 $$\:\:\boldsymbol{{y}}=\boldsymbol{{ax}}^{\mathrm{3}} +\boldsymbol{{bx}}^{\mathrm{2}} +\boldsymbol{{cx}}+\boldsymbol{{d}}\:,\:{then} \\ $$$${prove}\:{that}\:{the}\:{equation}\:{y}=\mathrm{0} \\ $$$${has}\:{only}\:{one}\:{real}\:{root}\:{if} \\ $$$$\:\boldsymbol{{a}}\left[\left(\mathrm{9}\boldsymbol{{ad}}−\boldsymbol{{bc}}\right)^{\mathrm{2}} −\mathrm{4}\left(\boldsymbol{{b}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{ac}}\right)\left(\boldsymbol{{c}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{bd}}\right)\right] \\ $$$$\:\:\:\:>\:\mathrm{0}\:\:\:\:\:{provided}\:\:\:\boldsymbol{{b}}^{\mathrm{2}} \:>\:\mathrm{3}\boldsymbol{{ac}}\:. \\…
Question Number 155638 by cortano last updated on 03/Oct/21 Commented by Rasheed.Sindhi last updated on 03/Oct/21 $${x}=\mathrm{3},{y}=\mathrm{2} \\ $$ Commented by cortano last updated on…
Question Number 90099 by Rio Michael last updated on 21/Apr/20 $$\:\mathrm{given}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{equation} \\ $$$$\:{r}\:=\:{a}^{\mathrm{2}} \:\mathrm{sin2}\theta\:\:\mathrm{show}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{at}\: \\ $$$$\mathrm{the}\:\mathrm{poles}\:\mathrm{of}\:\mathrm{this}\:\mathrm{polar}\:\mathrm{equation}\:\mathrm{is}. \\ $$$$\:\theta\:=\:\left\{\frac{\pi}{\mathrm{4}},\frac{\mathrm{3}\pi}{\mathrm{4}},\frac{\mathrm{5}\pi}{\mathrm{4}},\frac{\mathrm{7}\pi}{\mathrm{4}}\right\} \\ $$ Commented by jagoll last updated…
Question Number 90097 by jagoll last updated on 21/Apr/20 $$\left(\sqrt{\mathrm{3}+\sqrt{\mathrm{8}}}\right)^{\mathrm{x}} \:+\left(\sqrt{\mathrm{3}−\sqrt{\mathrm{8}}}\right)^{\mathrm{x}} \:=\:\mathrm{6} \\ $$ Commented by john santu last updated on 21/Apr/20 $$\left(\mathrm{3}+\sqrt{\mathrm{8}}\right)\left(\mathrm{3}−\sqrt{\mathrm{8}}\right)=\mathrm{1} \\ $$$$\mathrm{3}−\sqrt{\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{3}+\sqrt{\mathrm{8}}}\:…
Question Number 155628 by amin96 last updated on 03/Oct/21 Answered by mr W last updated on 03/Oct/21 $${BM}={MC}=\mathrm{1} \\ $$$${AC}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:\left(\mathrm{45}+{x}\right)\:\mathrm{sin}\:\mathrm{30}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{sin}\:\left(\mathrm{30}+{x}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}=\frac{\mathrm{2}}{\mathrm{cos}\:{x}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}} \\…
Question Number 155631 by aunzo last updated on 03/Oct/21 $$\left({x}+\mathrm{3}\left({x}−\mathrm{2}\right)={x}+\mathrm{10}\right. \\ $$ Commented by aunzo last updated on 03/Oct/21 $${plss} \\ $$ Commented by Ar…
Question Number 90092 by jagoll last updated on 21/Apr/20 $$\mathrm{G}\left(\sqrt{\mathrm{x}+\mathrm{5}}\right)\:=\:\mathrm{x} \\ $$$$\mathrm{G}\left(\mathrm{x}^{\mathrm{2}} \right)\:=\:\mathrm{x}^{\mathrm{a}} −\mathrm{b} \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{b}\: \\ $$ Commented by john santu last updated on…
Question Number 155630 by SANOGO last updated on 03/Oct/21 Answered by Ar Brandon last updated on 03/Oct/21 $$\mathscr{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{exp}\left(−\frac{{n}}{{k}}\right)}{{k}^{\mathrm{2}} }=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}^{\mathrm{2}}…
Question Number 155625 by puissant last updated on 02/Oct/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{tanx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} =\:? \\ $$ Answered by yeti123 last updated on 03/Oct/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{tan}\:{x}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}…