Question and Answers Forum
All Questions Topic List
AllQuestion and Answers: Page 213
Question Number 194896 Answers: 1 Comments: 0
$$\:\:\:\:\:\:\:\cancel{ } \\ $$
Question Number 194891 Answers: 1 Comments: 0
$$\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{3}^{+} } {\mathrm{lim}}\:\left(\frac{\sqrt{{x}}−\sqrt{{x}−\mathrm{3}}−\sqrt{\mathrm{3}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}\:\right)=? \\ $$
Question Number 194888 Answers: 1 Comments: 0
Question Number 194887 Answers: 1 Comments: 0
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{Inverse}\:\mathrm{laplace}\:\mathrm{transform}\:\mathrm{of} \\ $$$$\frac{\mathrm{S}\:+\:\mathrm{2}}{\mathrm{S}^{\mathrm{2}} \:+\mathrm{4S}\:+\:\mathrm{7}} \\ $$$$ \\ $$$$\mathrm{Urgent}! \\ $$
Question Number 194881 Answers: 1 Comments: 0
$${Give}\:\bigtriangleup{ABC}\: \\ $$$${Proof}:\:{sin}\:{A}\:+\:{sin}\:{B}\:+\:{sin}\:{C}\:>\:\mathrm{2} \\ $$
Question Number 194884 Answers: 1 Comments: 0
$$\:\:\:\:\:\:{x}!\:=\:\mathrm{6}!.\:\mathrm{7}!\: \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} \:=?\: \\ $$
Question Number 194879 Answers: 1 Comments: 0
Question Number 194878 Answers: 2 Comments: 0
Question Number 194877 Answers: 0 Comments: 1
$${without}\:{calculator}\:{Prove}:\:\sqrt[{\mathrm{6}}]{\pi^{\mathrm{5}} +\pi^{\mathrm{4}} }<{e} \\ $$
Question Number 194869 Answers: 1 Comments: 0
Question Number 194868 Answers: 1 Comments: 0
$$\mathrm{Prove}\:\mathrm{that}\:\forall{n}\in\mathrm{IN} \\ $$$$\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} {t}\:{sin}^{\mathrm{2}{n}} \left({lnt}\right){dt}=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{2}\pi} }\:\underset{\:\mathrm{0}} {\int}^{\:\pi} {e}^{−\mathrm{2}{t}} {sin}^{\mathrm{2}{n}} \left({t}\right){dt} \\ $$
Question Number 194861 Answers: 1 Comments: 0
Question Number 194853 Answers: 3 Comments: 0
Question Number 194852 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cosx}\right)^{\mathrm{log}\left(\mathrm{x}\right)} =? \\ $$$$ \\ $$
Question Number 194847 Answers: 0 Comments: 2
$$\:\:\:\:\:\:\underbrace{\:} \\ $$
Question Number 194846 Answers: 1 Comments: 0
Question Number 194844 Answers: 1 Comments: 0
$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +\left({x}+\mathrm{3}\right)^{\mathrm{2}} =\left({x}^{\mathrm{2}} +{ax}+{b}\right)\left({x}^{\mathrm{2}} +{cx}+{d}\right) \\ $$$${Find}\:{a},{b},{c},{d}. \\ $$
Question Number 194837 Answers: 2 Comments: 0
$${for}\:{x}>\mathrm{0}\:{find}\:{the}\:{minimum}\:{of}\:{the} \\ $$$${function}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\frac{\mathrm{5}}{{x}}. \\ $$
Question Number 194836 Answers: 0 Comments: 1
Question Number 194831 Answers: 1 Comments: 1
Question Number 194851 Answers: 0 Comments: 1
$$\boldsymbol{\mathrm{Name}}\:\:\:\boldsymbol{\mathrm{Zainab}}\:\boldsymbol{\mathrm{Bibi}} \\ $$$$\boldsymbol{\mathrm{BC}}\mathrm{200400692} \\ $$$$\boldsymbol{\mathrm{A}}\mathrm{ssignmeng}\:\mathrm{No}#\mathrm{2}\: \\ $$$$\boldsymbol{\mathrm{Mth}}\:\mathrm{621} \\ $$$$\mathrm{solution}.. \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\left(\mathrm{n}− \right)!}{\left(\mathrm{n}+ \right)^{\mathrm{2}} } \\ $$$$\mathrm{a}_{\mathrm{n}+ } =\frac{\left(\mathrm{n}+ − \right)!}{\left(\mathrm{n}+ + \right)^{\mathrm{2}} }=\frac{\left(\mathrm{n}\right)!}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\mathrm{by}\:\mathrm{ratio}\:\mathrm{test} \\ $$$$\frac{\mathrm{a}_{\mathrm{n}+ } }{\mathrm{a}_{\mathrm{n}} }=\frac{\frac{\left(\mathrm{n}\right)!}{\left(\mathrm{n}+\mathrm{2}\right)}}{\frac{\left(\mathrm{n}− \right)!}{\left(\mathrm{n}+ \right)^{\mathrm{2}} }}=\frac{\left(\mathrm{n}\right)!}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }×\frac{\left(\mathrm{n}+ \right)^{\mathrm{2}} }{\left(\mathrm{n}+ \right)!} \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\:\frac{\mathrm{a}_{\mathrm{n}+ } }{\mathrm{a}_{\mathrm{n}} }=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{n}\left(\mathrm{n}− \right)!}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }×\frac{\left(\mathrm{n}+ \right)^{\mathrm{2}} }{\left(\mathrm{n}− \right)!}\:\:\:\:\:\because\left(\mathrm{n}\right)!=\mathrm{n}\left(\mathrm{n}− \right)! \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{a}_{\mathrm{n}+ } }{\mathrm{a}_{\mathrm{n}} }=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{n}\left(\mathrm{n}+ \right)^{\mathrm{2}} }{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{n}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}+\mathrm{2n}\right)}{\mathrm{n}^{\mathrm{2}} +\mathrm{4}+\mathrm{4n}} \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\left(\mathrm{n}^{\mathrm{3}} +\mathrm{n}+\mathrm{2n}^{\mathrm{2}} \right)}{\mathrm{n}^{\mathrm{2}} +\mathrm{4}+\mathrm{4n}}=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{n}}\right)}{\frac{\mathrm{1}}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{n}^{\mathrm{2}} }} \\ $$$$\mathrm{Divided}\:\mathrm{by}\:\mathrm{n}^{\mathrm{3}} \:\mathrm{to}\:\mathrm{numerator}\:\mathrm{and}\:\mathrm{denominator}\: \\ $$$$\mathrm{Now}\:\mathrm{by}\:\mathrm{Applying}\:\mathrm{limit} \\ $$$$\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\infty^{\mathrm{2}\:} }+\frac{\mathrm{2}}{\infty}\right)}{\frac{\mathrm{1}}{\infty}+\frac{\mathrm{4}}{\infty^{\mathrm{3}} }+\frac{\mathrm{4}}{\infty^{\mathrm{2}} }}=\frac{\mathrm{1}+\mathrm{0}+\mathrm{0}}{\mathrm{0}+\mathrm{0}+\mathrm{0}}=\frac{\mathrm{1}}{\mathrm{0}}=\infty \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{a}_{\mathrm{n}+\mathrm{1}} }{\mathrm{a}_{\mathrm{n}} }=\infty \\ $$
Question Number 194850 Answers: 0 Comments: 0
Question Number 194826 Answers: 1 Comments: 0
$$\:\:\:\:\:\:\mathrm{tan}\:\mathrm{19}°\:=\:{p}\: \\ $$$$\:\:\:\:\:\:\mathrm{tan}\:\mathrm{7}°\:=? \\ $$
Question Number 194821 Answers: 0 Comments: 2
$${M}\:{a}\:{inside}\:{poin}\:{in}\:\:\Delta{ABC}. \\ $$$${M}\:=\:{bar}\:\left\{\left({A},\:{area}\left({MBC}\right)\right),\:\left({B},\:{area}\left({MAC}\right)\right),\left({C},{area}\left({MAB}\right)\right)\right\} \\ $$
Question Number 194819 Answers: 0 Comments: 0
Question Number 194818 Answers: 1 Comments: 0
Pg 208 Pg 209 Pg 210 Pg 211 Pg 212 Pg 213 Pg 214 Pg 215 Pg 216 Pg 217
Terms of Service
Privacy Policy
Contact: info@tinkutara.com