Question Number 211696 by mnjuly1970 last updated on 16/Sep/24 $$ \\ $$$$\:\:\:\:\:\underset{\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\:\:\mathrm{1}}{\left(\:\mathrm{2}\:+\mathrm{2}{x}\:+\:{x}^{\mathrm{2}} \:\right)^{\mathrm{3}} }\:{dx}=\:? \\ $$$$\:\:\:\:\:\:\underbrace{\underset{\:\:\:\:\overset{\mathrm{Improper}\:\mathrm{integral}\:} {\:}\:\:\:\:\:} {\:}} \\ $$$$\:\:\:\:\:\:\:\:−−−−−−−−− \\ $$ Answered…
Question Number 211680 by RojaTaniya last updated on 16/Sep/24 $$\:{x},\:{y}\:{are}\:{positive}\:{integer}\:{such}\: \\ $$$$\:\:{that},\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{xy}=\mathrm{911}.\:\left({x},{y}\right)=? \\ $$ Commented by AlagaIbile last updated on 16/Sep/24 $$\:{That}'{s}\:{a}\:{symmetric}\:{Function} \\…
Question Number 211672 by MATHEMATICSAM last updated on 15/Sep/24 $$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{the}\:\mathrm{side}\:{c}\:\mathrm{is} \\ $$$$\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{side}\:{b}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{2tanC}\:+\:\mathrm{tanA}\:=\:\mathrm{0}. \\ $$ Commented by Frix last updated on 16/Sep/24 $$\mathrm{If}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{a}\:{side}\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to} \\…
Question Number 211643 by MrGaster last updated on 15/Sep/24 $$\mathrm{Let}\:\boldsymbol{{A}}=\left\{\boldsymbol{{x}}\:\in\:\mathbb{R}\mid\boldsymbol{{x}}^{\mathrm{2}} <\mathrm{4}\right\}\mathrm{and} \\ $$$$\boldsymbol{{B}}=\left\{\boldsymbol{{y}}\:\in\:\mathbb{Q}\mid\boldsymbol{{y}}>−\mathrm{3}\right\}\mathrm{find}\:\boldsymbol{{A}}\cap\boldsymbol{{B}} \\ $$ Answered by Faetmaaa last updated on 15/Sep/24 $${A}=\left(−\mathrm{2},\:\mathrm{2}\right) \\ $$$${B}=\boldsymbol{\mathrm{Q}}\cap\left(−\mathrm{3},\:\infty\right)…
Question Number 211636 by BaliramKumar last updated on 15/Sep/24 Answered by Rasheed.Sindhi last updated on 15/Sep/24 $$\blacktriangleright{A}={x}^{\mathrm{8}{k}+\mathrm{3}} +{x}^{\mathrm{8}{k}+\mathrm{6}} +{x}^{\mathrm{8}{k}+\mathrm{9}} +{x}^{\mathrm{8}{k}+\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:={x}^{\mathrm{8}{k}+\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{3}} +{x}^{\mathrm{6}} +{x}^{\mathrm{9}}…
Question Number 211637 by CrispyXYZ last updated on 15/Sep/24 $${f}\left({x}\right)\:=\:\mathrm{sin}\:{x}\:−\:\mathrm{e}^{{x}} \:+\:\mathrm{1}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{f}\left({x}\right)\:\mathrm{has}\:\mathrm{only}\:\mathrm{2}\:\mathrm{zeros}\:\mathrm{in}\:−\pi\leqslant{x}\leqslant\mathrm{0}. \\ $$ Answered by mehdee1342 last updated on 15/Sep/24 $${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left(−\pi\right)=\mathrm{1}−{e}^{−\pi}…
Question Number 211632 by sonukgindia last updated on 15/Sep/24 Answered by A5T last updated on 15/Sep/24 $$\phi\left(\mathrm{1000}\right)=\mathrm{400} \\ $$$$\mathrm{2024}^{\mathrm{2024}} \equiv\mathrm{0}\left({mod}\:\mathrm{16}\right);\mathrm{2024}^{\mathrm{2024}} \equiv\mathrm{1}\left({mod}\:\mathrm{25}\right) \\ $$$$\Rightarrow\mathrm{2024}^{\mathrm{2024}} =\mathrm{25}{q}+\mathrm{1}\equiv\left(\mathrm{0}\:{mod}\:\mathrm{16}\right)\Rightarrow{q}\equiv\mathrm{7}\left({mod}\mathrm{16}\right) \\…
Question Number 211633 by sonukgindia last updated on 15/Sep/24 Commented by Ghisom last updated on 15/Sep/24 $$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\frac{\mathrm{7}}{\mathrm{30}} \\ $$ Answered by MathematicalUser2357 last updated on…
Question Number 211634 by BaliramKumar last updated on 15/Sep/24 Answered by golsendro last updated on 15/Sep/24 $$\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{64}+\mathrm{1}}{\mathrm{8}}\:=\:\mathrm{2}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} } \\ $$$$\:\mathrm{y}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{3}} }\:=\:\frac{\mathrm{729}+\mathrm{1}}{\mathrm{27}}=\:\mathrm{3}^{\mathrm{3}}…
Question Number 211635 by BaliramKumar last updated on 15/Sep/24 Answered by golsendro last updated on 15/Sep/24 $$\:\mathrm{10}^{\mathrm{31}} −\mathrm{5}−\mathrm{10}^{\mathrm{30}} −\mathrm{p}=\mathrm{10}^{\mathrm{30}} \left(\mathrm{10}−\mathrm{1}\right)−\left(\mathrm{5}+\mathrm{p}\right) \\ $$$$\:\mathrm{9}.\mathrm{10}^{\mathrm{3}} −\left(\mathrm{5}+\mathrm{p}\right)\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3} \\ $$$$\:\mathrm{so}\:\mathrm{p}=\mathrm{1}\:,\:\mathrm{4}\:,\:\mathrm{7}\:…