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Question-200109

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-200060

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-200075

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-200102

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}…

f-x-x-g-x-2x-1-fog-x-f-2x-1-fog-x-2x-1-gof-x-g-x-gof-x-2-x-

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…