Question Number 161338 by akolade last updated on 16/Dec/21 Commented by cortano last updated on 16/Dec/21 $$\:\begin{cases}{\mathrm{16}−{x}^{\mathrm{2}} >\mathrm{0}\Rightarrow\left({x}−\mathrm{4}\right)\left({x}+\mathrm{4}\right)<\mathrm{0}\:;−\mathrm{4}<{x}<\mathrm{4}}\\{\mathrm{3}{x}−\mathrm{4}>\mathrm{0}\Rightarrow{x}>\frac{\mathrm{4}}{\mathrm{3}}}\\{\mathrm{3}{x}−\mathrm{4}\neq\mathrm{1}\:\Rightarrow{x}\neq\frac{\mathrm{5}}{\mathrm{3}}}\end{cases} \\ $$$$\Leftrightarrow\mathrm{16}−{x}^{\mathrm{2}} =\left(\mathrm{3}{x}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{16}−{x}^{\mathrm{2}} =\mathrm{9}{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{16}…
Question Number 161335 by kapoorshah last updated on 16/Dec/21 Commented by mr W last updated on 17/Dec/21 $${see}\:{Q}\mathrm{161366} \\ $$ Terms of Service Privacy Policy…
Question Number 30259 by mondodotto@gmail.com last updated on 19/Feb/18 Commented by prof Abdo imad last updated on 20/Feb/18 $${we}\:{have}\:\:\frac{{x}}{{x}−\mathrm{1}}\:−\frac{\mathrm{1}}{{lnx}}=\:\frac{{xlnx}\:−{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right){lnx}}=\frac{{u}\left({x}\right)}{{v}\left({x}\right)} \\ $$$${we}\:{hsve}\:{u}\left(\mathrm{1}\right)={v}\left(\mathrm{1}\right)=\mathrm{0}\:{let}\:{use}\:{hospital}\:{theorem} \\ $$$${u}^{'} \left({x}\right)=\mathrm{1}+{lnx}\:−\mathrm{1}\Rightarrow{u}^{''} \left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:{anf}\:{u}^{''}…
Question Number 30257 by mondodotto@gmail.com last updated on 19/Feb/18 Commented by rahul 19 last updated on 19/Feb/18 $$\mathrm{Sandwhich}\:\mathrm{theorem}\:! \\ $$ Commented by abdo imad last…
Question Number 30258 by mondodotto@gmail.com last updated on 19/Feb/18 Answered by $@ty@m last updated on 19/Feb/18 $${Let}\:\mathrm{sec}\:{x}−\mathrm{1}={z} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} {xdx}={dz} \\ $$$${dx}=\frac{{dz}}{\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}}=\frac{{dz}}{{z}\left({z}+\mathrm{2}\right)} \\ $$$${I}=\int\frac{{dx}}{{z}^{\mathrm{2}}…
Question Number 30256 by mondodotto@gmail.com last updated on 19/Feb/18 Commented by abdo imad last updated on 19/Feb/18 $${in}\:{this}\:{case}\:{is}\:{better}\:{to}\:{use}\:{hospital}\:{theorem}\:{let}\:{put} \\ $$$${u}\left({x}\right)={x}^{{x}} \:−{x}\:{and}\:{v}\left({x}\right)=\mathrm{1}−{x}\:+{lnx}\:{wehave}\:{u}\left(\mathrm{1}\right)={v}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${u}\left({x}\right)={e}^{{xlnx}} −{x}\Rightarrow{u}^{'} \left({x}\right)=\left(\mathrm{1}+{lnx}\right){e}^{{xlnx}}…
Question Number 95782 by john santu last updated on 27/May/20 $$\mathrm{it}\:\mathrm{looks}\:\mathrm{version}\:\mathrm{2}.\mathrm{077}\:\mathrm{has}\:\mathrm{a}\: \\ $$$$\mathrm{problem}.\:\:\mathrm{for}\:\mathrm{very}\:\mathrm{long}\:\mathrm{loading} \\ $$ Commented by Tinku Tara last updated on 27/May/20 $$\mathrm{Initial}\:\mathrm{loading}?\:\mathrm{Immediately} \\…
Question Number 161322 by kapoorshah last updated on 16/Dec/21 Commented by cortano last updated on 16/Dec/21 $$\mathrm{sin}\:\left(\mathrm{30}°+\mathrm{10}°+\alpha\right)={b} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{10}°+\alpha\right)+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\mathrm{10}°+\alpha\right)={b} \\ $$$$\mathrm{cos}\:\left(\mathrm{10}°+\alpha\right)+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\mathrm{10}°+\alpha\right)=\mathrm{2}{b} \\ $$$$\sqrt{\mathrm{3}}\:\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {t}}\:=\:\mathrm{2}{b}−\mathrm{cos}\:{t}\:;\:{t}=\mathrm{10}°+\alpha \\…
Question Number 161316 by otchereabdullai@gmail.com last updated on 16/Dec/21 $$\mathrm{Three}\:\mathrm{quarters}\:\mathrm{of}\:\mathrm{a}\:\mathrm{number}\:\mathrm{added}\:\mathrm{to} \\ $$$$\mathrm{two}\:\mathrm{and}\:\mathrm{a}\:\mathrm{half}\:\mathrm{of}\:\mathrm{that}\:\mathrm{number}\:\mathrm{gives}\: \\ $$$$\mathrm{13}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{number} \\ $$$$ \\ $$ Answered by FelipeLz last updated on 16/Dec/21…
Question Number 30245 by .none. last updated on 19/Feb/18 $$\mathrm{1}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} +\bullet\bullet\bullet+\mathrm{17}^{\mathrm{2}} −\mathrm{19}^{\mathrm{2}} \\ $$ Commented by Penguin last updated on 19/Feb/18 $$=\underset{{n}=\mathrm{1}}…