Question Number 200175 by mokys last updated on 15/Nov/23 $${find}\:{all}\:{values}\:{of}\:{x}\:{if}\:{x}^{\mathrm{2}} \equiv\mathrm{4}{mod}\left(\mathrm{5}\right)\:{and}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{11}\:? \\ $$ Commented by AST last updated on 15/Nov/23 $$\mathrm{2},\mathrm{3},\mathrm{7},\mathrm{8} \\ $$ Commented by…
Question Number 200122 by Blackpanther last updated on 14/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200109 by sonukgindia last updated on 14/Nov/23 Answered by mr W last updated on 14/Nov/23 $$\mathrm{2}^{{x}} =\mathrm{5}−{x} \\ $$$$\mathrm{2}^{{x}−\mathrm{5}} =\frac{\mathrm{5}−{x}}{\mathrm{32}} \\ $$$${e}^{\left({x}−\mathrm{5}\right)\mathrm{ln}\:\mathrm{2}} =\frac{\mathrm{5}−{x}}{\mathrm{32}}…
Question Number 200110 by sonukgindia last updated on 14/Nov/23 Answered by mr W last updated on 14/Nov/23 $${x}^{{x}^{{x}} } ={x}^{\mathrm{3}} \\ $$$$\Rightarrow{x}=\pm\mathrm{1} \\ $$$$\Rightarrow{x}^{{x}} =\mathrm{3}\Rightarrow{e}^{\mathrm{ln}\:{x}}…
Question Number 200133 by sonukgindia last updated on 14/Nov/23 Answered by Frix last updated on 14/Nov/23 $$\mathrm{You}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${x}\approx\mathrm{1}.\mathrm{68421311} \\ $$ Terms of Service Privacy…
Question Number 200060 by sonukgindia last updated on 13/Nov/23 Answered by cortano12 last updated on 13/Nov/23 $$\:\mathrm{L}=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\mathrm{2}{ab}−{a}\sqrt{{ab}}−{ab}}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)} \\ $$$$\:\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{{ab}−{a}\sqrt{{ab}}}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)} \\ $$$$\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\sqrt{{ab}}\:\left(\sqrt{{ab}}−{a}\right)}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)}\: \\…
Question Number 200053 by sonukgindia last updated on 13/Nov/23 Commented by MathematicalUser2357 last updated on 15/Nov/23 $${The}\:{red}\:{one} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 200075 by sonukgindia last updated on 13/Nov/23 Answered by ajfour last updated on 13/Nov/23 $${R}\mathrm{cos}\:\theta=\frac{{a}+{b}}{\mathrm{2}} \\ $$$${R}={b}\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\:\mathrm{2}{b}\mathrm{cos}\:^{\mathrm{2}} \theta={a}+{b} \\ $$$$\frac{{a}}{{b}}=\mathrm{2cos}\:^{\mathrm{2}} \theta−\mathrm{1}…
Question Number 200102 by Blackpanther last updated on 13/Nov/23 Answered by som(math1967) last updated on 14/Nov/23 $$\:{E}\:{is}\:{mid}\:{pt}\:{of}\:{AC}\:{and}\:{AB}\parallel{DE} \\ $$$$\therefore\:{D}\:{is}\:{mid}\:{pt}\:{of}\:{BC} \\ $$$$\mathrm{2}{ar}\bigtriangleup{ABD}={ar}\bigtriangleup{ABC} \\ $$$$\:{AB}\parallel{DE}\:\therefore\:{ar}\bigtriangleup{ADE}={ar}\bigtriangleup{ABD} \\ $$$${ar}\:\bigtriangleup{ADE}=\bigtriangleup{AEF}+\bigtriangleup{EFD}…
Question Number 200007 by Hummayoun last updated on 12/Nov/23 $${f}\left({x}\right)=\sqrt{{x}} \\ $$$${g}\left({x}\right)=\mathrm{2}{x}+\mathrm{1} \\ $$$${fog}\left({x}\right)={f}\left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$${fog}\left({x}\right)=\sqrt{\mathrm{2}{x}+\mathrm{1}} \\ $$$${gof}\left({x}\right)={g}\left(\sqrt{\left.{x}\right)}\right. \\ $$$${gof}\left({x}\right)=\mathrm{2}\sqrt{{x}} \\ $$ Terms of Service…