Question Number 200930 by Spillover last updated on 26/Nov/23 $${If}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{{n}} {dx}\:\:{and}\:\:\frac{{I}_{{n}} }{{I}_{{n}−\mathrm{1}} }=\frac{\lambda{n}}{\lambda{n}+\mathrm{1}} \\ $$$${then}\:{find}\:\:\lambda \\ $$ Commented by mr W…
Question Number 200931 by Spillover last updated on 26/Nov/23 $$\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{18}}+\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{45}}+\frac{\mathrm{1}}{\mathrm{63}}+\frac{\mathrm{1}}{\mathrm{84}}+…\infty=? \\ $$$$ \\ $$ Answered by MM42 last updated on 26/Nov/23 $${s}_{{n}} =\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right)=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)\:\: \\ $$$$\Rightarrow{s}_{{n}}…
Question Number 200925 by akolade last updated on 26/Nov/23 Answered by mr W last updated on 26/Nov/23 $$\boldsymbol{{p}}=\overset{\rightarrow} {{AB}}=\left(\mathrm{6},\:−\mathrm{14}\right) \\ $$$${C}=\left(\mathrm{1},\mathrm{5}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{6},\:−\mathrm{14}\right)=\left(\mathrm{2}.\mathrm{5},\:\mathrm{1}.\mathrm{5}\right) \\ $$$${D}=\left(\mathrm{1},\mathrm{5}\right)+\frac{\mathrm{2}}{\mathrm{4}}\left(\mathrm{6},\:−\mathrm{14}\right)=\left(\mathrm{4},\:−\mathrm{2}\right) \\ $$$${E}=\left(\mathrm{1},\mathrm{5}\right)+\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{6},\:−\mathrm{14}\right)=\left(\mathrm{5}.\mathrm{5},\:−\mathrm{5}.\mathrm{5}\right)…
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Question Number 200915 by Rupesh123 last updated on 26/Nov/23 Answered by Frix last updated on 26/Nov/23 $$\mathrm{Assume} \\ $$$$\int\frac{\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} }{{x}^{\frac{\mathrm{8}}{\mathrm{5}}} }{dx}=\frac{{p}\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{{q}}{\mathrm{5}}} }{{x}^{\frac{{r}}{\mathrm{5}}} }…
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Question Number 200902 by Rupesh123 last updated on 26/Nov/23 Answered by Rasheed.Sindhi last updated on 26/Nov/23 $${f}\left({x}\right){f}\left({y}\right)+\mathrm{1}=\mathrm{2}{f}\left({xy}\right)+\mathrm{2}\left({x}+{y}\right) \\ $$$${x}={y}=\mathrm{1}: \\ $$$$\left[{f}\left(\mathrm{1}\right)\right]^{\mathrm{2}} +\mathrm{1}=\mathrm{2}{f}\left(\mathrm{1}\right)+\mathrm{4} \\ $$$$\left[{f}\left(\mathrm{1}\right)\right]^{\mathrm{2}} −\mathrm{2}{f}\left(\mathrm{1}\right)−\mathrm{3}=\mathrm{0}…