Question Number 192367 by Rupesh123 last updated on 15/May/23 Answered by mehdee42 last updated on 15/May/23 $${t}_{{r}} ={r}−\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow{t}_{{r}+\mathrm{1}} ={r}−\frac{\mathrm{1}}{\mathrm{2}}\:\&\:{t}_{{r}+\mathrm{2}} ={r}+\frac{\mathrm{1}}{\mathrm{2}}\&\:{t}_{{r}+\mathrm{3}} ={r}+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{t}_{{r}} {t}_{{r}+\mathrm{1}} {t}_{{r}+\mathrm{2}}…
Question Number 126827 by harckinwunmy last updated on 24/Dec/20 Answered by AST last updated on 24/Dec/20 $$\mathrm{Q33}. \\ $$$$\frac{\mathrm{4}+\mathrm{7}{i}}{\mathrm{1}−\mathrm{3}{i}} \\ $$$$\mathrm{Rationalising} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{4}+\mathrm{7}{i}\right)\left(\mathrm{1}+\mathrm{3}{i}\right)}{\left(\mathrm{1}−\mathrm{3}{i}\right)\left(\mathrm{1}+\mathrm{3}{i}\right)}=\frac{\mathrm{4}+\mathrm{12}{i}+\mathrm{7}{i}+\left(\mathrm{21}{i}^{\mathrm{2}} \right)}{\mathrm{1}^{\mathrm{2}} −\left(\mathrm{3}{i}\right)^{\mathrm{2}}…
Question Number 192363 by Red1ight last updated on 15/May/23 $$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}_{{i}} −{x}+\left(\mathrm{2}{cx}−{cb}\right)\left({y}_{{i}} +{cx}^{\mathrm{2}} −{cbx}\right)=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{this}\:\mathrm{equaition} \\ $$$$\left.{i}\right)\mathrm{c}>\mathrm{0} \\ $$$$\left.{ii}\right)\mathrm{b}>\mathrm{0} \\ $$$$\left.{iii}\right)\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:\mathrm{real}\:\mathrm{solution} \\ $$…
Question Number 126824 by MathSh last updated on 24/Dec/20 $$\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{{f}\left({x}+\mathrm{2}\right)−\mathrm{6}}{{x}−\mathrm{2}}=\mathrm{4}\:\:{if},\:\:\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{\mathrm{4}\centerdot{f}\left({x}\right)−\mathrm{24}}{{x}−\mathrm{4}}=? \\ $$ Answered by mahdipoor last updated on 24/Dec/20 $$\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{{f}\left({x}+\mathrm{2}\right)−\mathrm{6}}{{x}−\mathrm{2}}=\frac{{f}\left(\mathrm{4}\right)−\mathrm{6}}{\mathrm{0}}=\mathrm{4}\Rightarrow{f}\left(\mathrm{4}\right)−\mathrm{6}=\mathrm{0} \\ $$$$\overset{{L}'{hopital}\:}…
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Question Number 61284 by naka3546 last updated on 31/May/19 $$\left(\mathrm{998}^{\mathrm{999}} \:×\:\mathrm{999}^{\mathrm{998}} \:×\:\mathrm{2019}^{\mathrm{2019}} \right)\:\:{mod}\:\left(\mathrm{1000}\right)\:\:=\:\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 61283 by naka3546 last updated on 01/Jun/19 $$\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \right)\:\:=\:\:\mathrm{2} \\ $$$$\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)\left({x}^{\mathrm{6}} +{y}^{\mathrm{6}} \right)\left({x}^{\mathrm{8}} \:+\:{y}^{\mathrm{8}} \right)\:\:=\:\:\mathrm{4} \\ $$$$\left({x}^{\mathrm{3}} +\:{y}^{\mathrm{3}}…
Question Number 192349 by 073 last updated on 15/May/23 Answered by mehdee42 last updated on 15/May/23 $${I}=\int\frac{{cos}^{\mathrm{4}} {x}\sqrt{\mathrm{1}+{sinx}}}{{cosx}}\:{dx}=\int\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)\sqrt{\mathrm{1}+{sinx}}\:{cosxdx} \\ $$$${let}\::\:\mathrm{1}+{sinx}={u}^{\mathrm{2}} \:\Rightarrow{cosxdx}=\mathrm{2}{udu} \\ $$$$\Rightarrow{I}=\mathrm{2}\int\left(−{u}^{\mathrm{2}} +\mathrm{2}{u}\right){u}^{\mathrm{2}}…
Question Number 192348 by leandrosriv02 last updated on 15/May/23 Answered by mehdee42 last updated on 15/May/23 $${AB}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${M}\:\:{is}\:{midpoint}\:<{AB}>\Rightarrow{M}\left(\frac{{a}}{\mathrm{2}}\:,\:\frac{{b}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{0}{M}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\Rightarrow{OM}={BM}={AM}…
Question Number 192350 by Abdullahrussell last updated on 15/May/23 Answered by Frix last updated on 15/May/23 $$\mathrm{Just}\:\mathrm{type}\:\mathrm{into}\:\mathrm{a}\:\mathrm{calculator}. \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\frac{\mathrm{5}}{\mathrm{197}} \\ $$$$\mathrm{But}\:\mathrm{can}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{this}? \\ $$$$\frac{\left(\left({x}−\mathrm{5}\right)^{\mathrm{4}} +\mathrm{4}\right)\left(\left({x}−\mathrm{1}\right)^{\mathrm{4}} +\mathrm{4}\right)\left(\left({x}+\mathrm{3}\right)^{\mathrm{4}}…