Question Number 199940 by cortano12 last updated on 11/Nov/23 Commented by Frix last updated on 11/Nov/23 $$\mid{AD}\mid=\mathrm{3} \\ $$$$\mid{BD}\mid=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${r}=\frac{\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$ Commented by…
Question Number 199942 by cortano12 last updated on 11/Nov/23 Answered by MM42 last updated on 11/Nov/23 $$\mathrm{1}−\sqrt{{x}}={u}^{\mathrm{2}} \Rightarrow{x}=\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} ={x}\Rightarrow{dx}=−\mathrm{4}{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right){du} \\ $$$$\left.\Rightarrow{I}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \left({u}^{\mathrm{2}}…
Question Number 199864 by cortano12 last updated on 10/Nov/23 $$\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{2019}} {\sum}}\mathrm{n}^{\mathrm{4}} \:\mathrm{when} \\ $$$$\:\:\mathrm{divide}\:\mathrm{by}\:\mathrm{53}\: \\ $$ Answered by AST last updated on 10/Nov/23 $$\underset{{n}=\mathrm{1}}…
Question Number 199862 by universe last updated on 10/Nov/23 $$\mathrm{consider}\:\mathrm{the}\:\mathrm{taylor}\:\mathrm{expansion}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{3}} } \\ $$$$\mathrm{centered}\:\mathrm{at}\:\mathrm{x}\:=\:\mathrm{1}/\mathrm{2}\:\mathrm{then}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{convergence} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{power}\:\mathrm{series}\:\mathrm{repersentation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{is} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 199863 by Karanji last updated on 10/Nov/23 $$−\mathrm{5}+\left(−\mathrm{8}\right)+\left(−\mathrm{11}\right)+…+\left(−\mathrm{230}\right) \\ $$ Answered by som(math1967) last updated on 10/Nov/23 $$\:{c}.{d}=\left(−\mathrm{8}\right)−\left(−\mathrm{5}\right)=−\mathrm{3} \\ $$$$−\mathrm{230}=−\mathrm{5}\:+\left({n}−\mathrm{1}\right)×−\mathrm{3} \\ $$$$\Rightarrow\left({n}−\mathrm{1}\right)=\mathrm{75} \\…
Question Number 199858 by sonukgindia last updated on 10/Nov/23 Commented by AST last updated on 10/Nov/23 $$\mathrm{2} \\ $$ Answered by cortano12 last updated on…
Question Number 199854 by sonukgindia last updated on 10/Nov/23 Answered by aleks041103 last updated on 10/Nov/23 $${every}\:{fold}\:{is}\:{done}\:{at}\:{the}\:{angle}\:{bisector} \\ $$$$\Rightarrow{final}\:{accute}\:{angle}=\mathrm{90}°/\mathrm{4}=\mathrm{22}.\mathrm{5}° \\ $$$${S}=\mathrm{8}×\mathrm{6}−\mathrm{2}×\frac{\mathrm{3}×\left(\frac{\mathrm{3}}{{tan}\left(\mathrm{22}.\mathrm{5}°\right)}\right)}{\mathrm{2}}= \\ $$$$=\mathrm{48}−\mathrm{9}{cot}\left(\frac{\mathrm{45}°}{\mathrm{2}}\right)=\mathrm{48}−\mathrm{9}{cot}\left({x}\right)=\mathrm{48}−\mathrm{9}{z} \\ $$$$…
Question Number 199844 by jlewis last updated on 10/Nov/23 $$\mathrm{solve}\:\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dx}^{\mathrm{2}} }\:\mathrm{x} \\ $$ Answered by AST last updated on 10/Nov/23 $$\frac{{d}}{{dx}}\left(\frac{{d}}{{dx}}{x}\right)=\frac{{d}}{{dx}}\left(\mathrm{1}\right)=\mathrm{0} \\ $$ Answered…
Question Number 199847 by cortano12 last updated on 10/Nov/23 $$\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{value}\: \\ $$$$\:\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}+\mathrm{d}\:}\:+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\:+\:\frac{\mathrm{c}}{\mathrm{b}+\mathrm{c}+\mathrm{d}}+\frac{\mathrm{d}}{\mathrm{a}+\mathrm{c}+\mathrm{d}}\: \\ $$$$\:\mathrm{when}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{vary}\:\mathrm{over}\:\mathrm{positive} \\ $$$$\:\mathrm{reals}\: \\ $$ Commented by Frix last updated on 10/Nov/23…
Question Number 199840 by Calculusboy last updated on 10/Nov/23 Answered by AST last updated on 10/Nov/23 $${Let}\:\frac{{y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{{y}}=\mathrm{1}+\sqrt{\frac{\mathrm{1}}{{y}^{\mathrm{2}} }+\mathrm{1}}={x} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}=\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\Rightarrow{y}=\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{x}}} \\…