Question Number 200205 by cherokeesay last updated on 15/Nov/23 Commented by Frix last updated on 15/Nov/23 $$\mathrm{min}\:=\mathrm{4} \\ $$$$\mathrm{max}=+\infty \\ $$ Terms of Service Privacy…
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Question Number 200005 by Mingma last updated on 12/Nov/23 Answered by AST last updated on 12/Nov/23 $$\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \geqslant\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)\Rightarrow{a}+{b}+{c}\leqslant\mathrm{3} \\ $$$$\Sigma\frac{\mathrm{1}}{\mathrm{2}{a}+{b}}\geqslant\frac{\mathrm{9}}{\mathrm{3}\left({a}+{b}+{c}\right)}\geqslant\frac{\mathrm{9}}{\mathrm{3}×\mathrm{3}}=\mathrm{1} \\ $$$${Equality}\:{holds}\:{when}\:{a}={b}={c}=\mathrm{1}…
Question Number 200004 by Mingma last updated on 12/Nov/23 Answered by aleks041103 last updated on 12/Nov/23 $$−\mathrm{1}<{x}<\mathrm{1} \\ $$$${f}\left({x}\right)=\frac{{x}+\mathrm{2}^{{k}} }{\mathrm{2}^{{k}+\mathrm{1}} }=\frac{{x}}{\mathrm{2}^{{k}+\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }<{f}\left({x}\right)<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} } \\…
Question Number 199913 by WaqasShabbir last updated on 11/Nov/23 $$ \\ $$To calculate the mean and median of the distribution, you can follow these steps:…
Question Number 199863 by Karanji last updated on 10/Nov/23 $$−\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} \\ $$$$−\mathrm{230}=−\mathrm{5}\:+\left({n}−\mathrm{1}\right)×−\mathrm{3} \\ $$$$\Rightarrow\left({n}−\mathrm{1}\right)=\mathrm{75} \\…
Question Number 199771 by Rupesh123 last updated on 09/Nov/23 Answered by AST last updated on 09/Nov/23 $$\frac{{sin}\left(\mathrm{4}{x}\right)=\mathrm{2}{sin}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{2}{x}\right)}{{a}}=\frac{{sin}\left(\mathrm{2}{x}\right)}{{b}} \\ $$$$\Rightarrow{cos}\left(\mathrm{2}{x}\right)=\frac{{a}}{\mathrm{2}{b}}=\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\Rightarrow{cos}^{\mathrm{2}} {x}=\frac{{a}+\mathrm{2}{b}}{\mathrm{4}{b}} \\ $$$$\frac{{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)}{{b}}=\frac{{sin}\left({x}\right)}{{c}}\Rightarrow{cos}\left({x}\right)=\frac{{b}}{\mathrm{2}{c}} \\ $$$$\Rightarrow\frac{{a}+\mathrm{2}{b}}{{b}}=\frac{{b}^{\mathrm{2}}…
Question Number 199729 by Mingma last updated on 08/Nov/23 Answered by mathfreak01 last updated on 08/Nov/23 $${N}\:=\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{450}} {\sum}}\left(\mathrm{10}^{{n}} \:−\:\mathrm{1}\right) \\ $$$${N}\:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{450}} {\sum}}\mathrm{10}^{{n}} \:−\:\mathrm{450}…
Question Number 199659 by cherokeesay last updated on 06/Nov/23 Answered by Frix last updated on 07/Nov/23 $$\mathrm{If}\:\mathrm{we}\:\mathrm{stay}\:\mathrm{in}\:\mathbb{R}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}} \\ $$$$\mathrm{Obvious}\:\mathrm{solutions} \\ $$$${x}=−\mathrm{1}\vee{x}=\mathrm{0} \\ $$$$\mathrm{Not}\:\mathrm{so}\:\mathrm{obvious}\:\mathrm{solutions} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}…
Question Number 199621 by Mingma last updated on 06/Nov/23 Answered by AST last updated on 06/Nov/23 $${x}+{y}\overset{\mathrm{3}} {\equiv}\mathrm{2}…\left({i}\right) \\ $$$$\left(\mathrm{6}+\mathrm{6}+{y}\right)−\left(\mathrm{7}+{x}\right)=\mathrm{5}+{y}−{x}\overset{\mathrm{11}} {\equiv}\mathrm{0}\Rightarrow{x}−{y}\overset{\mathrm{11}} {\equiv}\mathrm{5}…\left({ii}\right) \\ $$$$\left({i}\right)\wedge\left({ii}\right)\Rightarrow\left({x},{y}\right)=\left(\mathrm{8},\mathrm{3}\right);\left(\mathrm{5},\mathrm{0}\right);\left(\mathrm{1},\mathrm{7}\right) \\…