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Author: Tinku Tara

If-4-p-5-Find-2-3p-

Question Number 206473 by hardmath last updated on 15/Apr/24 $$\mathrm{If}\:\:\:\mathrm{4}^{\boldsymbol{\mathrm{p}}} \:=\:\mathrm{5} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{2}^{\mathrm{3}\boldsymbol{\mathrm{p}}} \:=\:? \\ $$ Answered by A5T last updated on 15/Apr/24 $$\mathrm{2}^{\mathrm{2}{p}} =\mathrm{5}\Rightarrow\mathrm{2}^{\mathrm{3}{p}}…

if-f-x-x-x-2-then-f-1-x-

Question Number 206458 by mathlove last updated on 15/Apr/24 $${if}\:{f}\left({x}\right)=\sqrt{{x}−{x}^{\mathrm{2}} }\:\:\:\:\:{then}\:\:{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$ Answered by Skabetix last updated on 15/Apr/24 $${f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}\pm\sqrt{−\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}} \\…

Question-206452

Question Number 206452 by langatfredrick last updated on 15/Apr/24 Answered by A5T last updated on 15/Apr/24 $${If}\:{remainder}\:{means}\:\mid{x}−{y}\mid,\:{then}: \\ $$$$\frac{{t}_{{x}} }{{x}}={x};\frac{{t}_{{y}} }{{y}}={y}\Rightarrow{t}_{\mid{x}−{y}\mid} =\mid{t}_{{x}} −{t}_{{y}} \mid=\mid{x}^{\mathrm{2}} −{y}^{\mathrm{2}}…

If-asin-bcos-2ctan-1-tan-2-then-prove-that-a-2-b-2-2-4c-2-a-2-b-2-

Question Number 206471 by MATHEMATICSAM last updated on 15/Apr/24 $$\mathrm{If}\:{a}\mathrm{sin}\theta\:=\:{b}\mathrm{cos}\theta\:=\:\frac{\mathrm{2}{c}\mathrm{tan}\theta}{\mathrm{1}\:−\:\mathrm{tan}^{\mathrm{2}} \theta}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\left({a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:=\:\mathrm{4}{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \right). \\ $$ Answered by lepuissantcedricjunior last…

Question-206442

Question Number 206442 by universe last updated on 14/Apr/24 Answered by Berbere last updated on 14/Apr/24 $${let}\:{f}\left({x}\right)={e}^{{g}\left({x}\right)} ;\:{particular}\:{Solution}\:{just}\:{to}\:{simplifie} \\ $$$${the}\:{problems};{f}\left(\mathrm{0}\right)=\mathrm{1}={e}^{{g}\left(\mathrm{0}\right)} \Rightarrow{g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}'\left({x}\right)={g}'{e}^{{g}\left({x}\right)} ;{f}''\left({x}\right)=\left({g}'^{\mathrm{2}} +{g}''\right){e}^{{g}\left({x}\right)}…