Question Number 199844 by jlewis last updated on 10/Nov/23 $$\mathrm{solve}\:\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dx}^{\mathrm{2}} }\:\mathrm{x} \\ $$ Answered by AST last updated on 10/Nov/23 $$\frac{{d}}{{dx}}\left(\frac{{d}}{{dx}}{x}\right)=\frac{{d}}{{dx}}\left(\mathrm{1}\right)=\mathrm{0} \\ $$ Answered…
Question Number 199847 by cortano12 last updated on 10/Nov/23 $$\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{value}\: \\ $$$$\:\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}+\mathrm{d}\:}\:+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\:+\:\frac{\mathrm{c}}{\mathrm{b}+\mathrm{c}+\mathrm{d}}+\frac{\mathrm{d}}{\mathrm{a}+\mathrm{c}+\mathrm{d}}\: \\ $$$$\:\mathrm{when}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{vary}\:\mathrm{over}\:\mathrm{positive} \\ $$$$\:\mathrm{reals}\: \\ $$ Commented by Frix last updated on 10/Nov/23…
Question Number 199840 by Calculusboy last updated on 10/Nov/23 Answered by AST last updated on 10/Nov/23 $${Let}\:\frac{{y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{{y}}=\mathrm{1}+\sqrt{\frac{\mathrm{1}}{{y}^{\mathrm{2}} }+\mathrm{1}}={x} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}=\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\Rightarrow{y}=\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{x}}} \\…
Question Number 199843 by universe last updated on 10/Nov/23 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{1}+{x}\right)\:}−\frac{\mathrm{1}}{\mathrm{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)}\right)\:=\:?? \\ $$ Answered by witcher3 last updated on 10/Nov/23 $$=\frac{\mathrm{ln}\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)−\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{ln}\left(\left(\mathrm{1}+\mathrm{x}\right)\mathrm{ln}\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)\right.}= \\…
Question Number 199836 by Calculusboy last updated on 10/Nov/23 Commented by Rasheed.Sindhi last updated on 10/Nov/23 $$\mathcal{D}{o}\:{you}\:{mean}\:\mathrm{2}^{{x}^{\mathrm{2}} } =\mathrm{16}\:? \\ $$$${If}\:{so}: \\ $$$$\mathrm{2}^{{x}^{\mathrm{2}} } =\mathrm{2}^{\mathrm{4}}…
Question Number 199837 by Calculusboy last updated on 10/Nov/23 Answered by AST last updated on 10/Nov/23 $$\sqrt{{a}+\sqrt{{b}}}={p};\sqrt{{a}−\sqrt{{b}}}={q}\Rightarrow{p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{2}{a};{p}^{\mathrm{2}} {q}^{\mathrm{2}} ={a}^{\mathrm{2}} −{b} \\ $$$$\Rightarrow{pq}=\sqrt{{a}^{\mathrm{2}} −{b}}\Rightarrow\left({p}+{q}\right)^{\mathrm{2}}…
Question Number 199838 by Calculusboy last updated on 10/Nov/23 Answered by Frix last updated on 10/Nov/23 $${y}=\sqrt[{{x}}]{{xy}} \\ $$$${y}={x}^{\frac{\mathrm{1}}{{x}−\mathrm{1}}} \\ $$$$\left({x}+\mathrm{5}\right)^{{x}} =\mathrm{23}{x}+\mathrm{1}+{x}^{\frac{\mathrm{1}}{{x}−\mathrm{1}}} \\ $$$$\mathrm{Trying}\:{x}=\mathrm{1},\:\mathrm{2},\:\mathrm{3} \\…
Question Number 199834 by Calculusboy last updated on 10/Nov/23 $$\boldsymbol{{Let}}\:\boldsymbol{{C}}\:\boldsymbol{{be}}\:\boldsymbol{{the}}\:\boldsymbol{{circle}}\:\boldsymbol{{with}}\:\boldsymbol{{the}}\:\boldsymbol{{center}}\:\left(\mathrm{2},\mathrm{3}\right)\:\boldsymbol{{and}}\:\boldsymbol{{radius}}\:\mathrm{5} \\ $$$$\left.\boldsymbol{{a}}\right)\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{P}}\left(\mathrm{5},\mathrm{7}\right)\:\boldsymbol{{lies}}\:\boldsymbol{{on}}\:\boldsymbol{{C}}\:\boldsymbol{{and}}\:\boldsymbol{{find}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{equation}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{tangent}}\:\boldsymbol{{at}}\:\boldsymbol{{P}} \\ $$$$\left.\boldsymbol{{b}}\right)\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{line}}\:\mathrm{3}\boldsymbol{{x}}−\mathrm{4}\boldsymbol{{y}}+\mathrm{31}=\mathrm{0}\:\boldsymbol{{is}}\:\boldsymbol{{a}}\:\boldsymbol{{tangent}}\:\boldsymbol{{to}}\:\boldsymbol{{C}} \\ $$ Commented by Calculusboy last updated on 10/Nov/23…
Question Number 199893 by hardmath last updated on 10/Nov/23 Commented by mr W last updated on 10/Nov/23 $${do}\:{you}\:{know}\:{what}\:{the}\:{question}\:{is}? \\ $$$${if}\:{you}\:{don}'{t}\:{know},\:{then}\:{other}\:{people} \\ $$$${don}'{t}\:{know}\:{either}. \\ $$$${badly}\:{prepared}\:{questions}\:{don}'{t} \\…
Question Number 199830 by cortano12 last updated on 10/Nov/23 $$\:\:\mathrm{Si}\:\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\:, \\ $$$$\:\mathrm{halle}\:\mathrm{el}\:\mathrm{valor}\:\mathrm{de}\:\mathrm{la}\:\mathrm{expresion}\: \\ $$$$\:\mathrm{R}=\:\mathrm{16}\left(\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}\right)+\mathrm{3}\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}+\mathrm{csc}^{\mathrm{2}} \:\mathrm{x}\right) \\ $$ Answered by Frix last updated…