Question Number 103888 by mr W last updated on 18/Jul/20 $${find}\:{all}\:{such}\:{numbers}: \\ $$$${if}\:{we}\:{make}\:{its}\:{last}\:{digit},\:{say}\:{k},\:{as}\:{its} \\ $$$${first}\:{digit},\:{the}\:{number}\:{becomes}\:{k} \\ $$$${times}\:{large}\:{as}\:{before}. \\ $$$$\left(\Box\Box…\Box{k}\right)\rightarrow\left({k}\Box\Box…\Box\right)={k}×\left(\Box\Box…\Box{k}\right) \\ $$ Answered by MAB last…
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Question Number 169407 by Shrinava last updated on 29/Apr/22 Answered by Rasheed.Sindhi last updated on 01/May/22 $$\mathrm{If}\:\mathbb{N}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},…\right\} \\ $$$$\mathrm{M}=\left\{\left(\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0}\right)\right\} \\ $$$$\Omega=\mathrm{0} \\ $$$$\mathrm{16}^{{x}} +\mathrm{16}^{\frac{\mathrm{1}}{{x}}} =\mathrm{8}\:{has}\:{no}\:{solution}.…
Question Number 169389 by Shrinava last updated on 29/Apr/22 $$\mathrm{1}.\:\mathrm{2}\sqrt{\mathrm{y}}\:\mathrm{dx}\:=\:\mathrm{dy} \\ $$$$\mathrm{2}.\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\mathrm{dx}\:=\:\mathrm{2xydy} \\ $$$$\mathrm{3}.\:\mathrm{xdx}\:+\:\frac{\mathrm{1}}{\mathrm{y}}\:\mathrm{dy}\:=\:\mathrm{0} \\ $$$$\mathrm{4}.\:\mathrm{dy}\:=\:\mathrm{3x}^{\mathrm{2}} \:\mathrm{dx} \\ $$$$\mathrm{5}.\:\mathrm{2y}^{\mathrm{2}} \mathrm{dx}\:+\:\mathrm{x}\left(\mathrm{1}\:+\:\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dy}\:=\:\mathrm{0} \\ $$$$\mathrm{6}.\:\mathrm{4x}^{\mathrm{3}}…
Question Number 169391 by mathlove last updated on 29/Apr/22 Commented by infinityaction last updated on 29/Apr/22 $$\mathrm{0} \\ $$ Commented by mathlove last updated on…
Question Number 169382 by infinityaction last updated on 29/Apr/22 $$\:\:\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\:\left[\boldsymbol{\mathrm{v}}\right]\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{v}}\:\boldsymbol{\mathrm{den}{o}\mathrm{tes}}\:\boldsymbol{\mathrm{maximum}}\:\: \\ $$$$\:\:\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:+\:\boldsymbol{\mathrm{y}}^{\mathrm{2}} \:,\:\boldsymbol{\mathrm{where}}\:\left(\boldsymbol{\mathrm{x}}+\mathrm{5}\right)^{\mathrm{2}} \:+\:\left(\boldsymbol{\mathrm{y}}−\mathrm{12}\right)^{\mathrm{2}} \:=\:\mathrm{14} \\ $$$$\:\:\:\:\left(\boldsymbol{\mathrm{hint}}\:\left[\bullet\right]\:\boldsymbol{\mathrm{repersent}}\:\boldsymbol{\mathrm{greatest}}\:\boldsymbol{\mathrm{integer}}\:\boldsymbol{\mathrm{function}}\:\boldsymbol{\mathrm{of}}\:“\:\bullet''\right) \\ $$$$ \\ $$$$\: \\ $$ Commented…
Question Number 169356 by Shrinava last updated on 29/Apr/22 $$\mathrm{1}.\:\mathrm{y}\:=\:\mathrm{arcsin}\left(\mathrm{sinx}\right)\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$$$\mathrm{2}.\:\mathrm{y}\:=\:\mathrm{sin}\:\sqrt{\mathrm{x}\:+\:\mathrm{1}}\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$$$\mathrm{3}.\:\mathrm{y}\:=\:\mathrm{ln}^{\mathrm{5}} \:\mathrm{sinx}\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$$$\mathrm{4}.\:\mathrm{y}\:=\:\mathrm{cos}\left(\mathrm{2x}\:+\:\mathrm{3}\right)\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$ Answered by…
Question Number 169354 by Shrinava last updated on 29/Apr/22 $$\mathrm{1}.\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{1}\:+\:\mathrm{x}} \\ $$$$\mathrm{2}.\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\mathrm{6}\:-\:\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\mathrm{3}.\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{cosx}}{\mathrm{1}\:+\:\mathrm{sinx}}\:\mathrm{dx} \\ $$ Answered by…
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Question Number 103804 by ajfour last updated on 17/Jul/20 $$\:\:\:\boldsymbol{{S}}{olve}\:{for}\:\boldsymbol{{r}} \\ $$$$\:\:\frac{\boldsymbol{{h}}−\boldsymbol{{p}}^{\mathrm{2}} }{\boldsymbol{{p}}−\boldsymbol{{r}}}=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{p}}}\:\:\:\:,\:\:\:\:\frac{\boldsymbol{{h}}−\boldsymbol{{q}}}{\boldsymbol{{q}}^{\mathrm{2}} −\boldsymbol{{r}}}=\:\mathrm{2}\boldsymbol{{q}} \\ $$$$\left(\boldsymbol{{h}}−\boldsymbol{{p}}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\boldsymbol{{p}}−\boldsymbol{{r}}\right)^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \\ $$$$\:\left(\boldsymbol{{h}}−\boldsymbol{{q}}\right)^{\mathrm{2}} +\left(\boldsymbol{{q}}^{\mathrm{2}} −\boldsymbol{{r}}\right)^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \\…