Question Number 200321 by mr W last updated on 17/Nov/23 $${for}\:{x},{y},{z}\:\in{N},\:{if} \\ $$$$\mathrm{38}{x}+\mathrm{40}{y}+\mathrm{41}{z}=\mathrm{520} \\ $$$${find}\:{x}+{y}+{z}=? \\ $$ Answered by Rasheed.Sindhi last updated on 17/Nov/23 $$\mathrm{520}−\mathrm{38}{x}−\mathrm{40}{y}=\mathrm{41}{z}…
Question Number 200253 by Calculusboy last updated on 16/Nov/23 Answered by kapoorshah last updated on 16/Nov/23 $$\mathrm{1} \\ $$ Answered by kapoorshah last updated on…
Question Number 200254 by mnjuly1970 last updated on 16/Nov/23 $$ \\ $$$$\:\:\:\:\:\:{calculate}\:… \\ $$$$\:\:\Omega\:=\:\int_{\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\mathrm{ln}\left(\mathrm{tan}\left({x}\right)\right){dx}} ^{\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{sin}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} }\:{dx}} \mathrm{ln}\left(\mathrm{sin}\left({x}\right)\right){dx}=? \\ $$ Answered…
Question Number 200249 by Calculusboy last updated on 16/Nov/23 $$\boldsymbol{{Solve}}:\:\:\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{distance}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{point}}\:\boldsymbol{{P}}\left(\mathrm{3},\mathrm{4}\right)\:\boldsymbol{{from}}\:\boldsymbol{{the}}\:\boldsymbol{{line}}\:\boldsymbol{{y}}=−\mathrm{2}\boldsymbol{{x}}+\mathrm{3} \\ $$ Commented by mr W last updated on 18/Nov/23 Answered by Frix last updated…
Question Number 200250 by Calculusboy last updated on 16/Nov/23 Answered by Mathspace last updated on 16/Nov/23 $$\int_{\mathrm{0}} ^{{x}} {sin}\left({x}+{t}\right){dt} \\ $$$$=\left[{cos}\left({x}+{t}\right)\right]_{{t}=\mathrm{0}} ^{{t}={x}} ={cos}\left(\mathrm{2}{x}\right)−{cosx} \\ $$$$\Rightarrow\frac{{d}}{{dx}}\left(\int_{\mathrm{0}}…
Question Number 200251 by Calculusboy last updated on 16/Nov/23 Answered by Mathspace last updated on 16/Nov/23 $${f}\left({x},{y}\right)={sin}\left({e}^{{xy}} +{e}^{{x}} \right)\:\Rightarrow \\ $$$$\frac{\partial}{\partial{x}}{f}\left({x},{y}\right)=\frac{\partial}{\partial{x}}\left({e}^{{xy}} +{e}^{{x}} \right){cos}\left({e}^{{xy}} +{e}^{{x}} \right)…
Question Number 200304 by Calculusboy last updated on 16/Nov/23 Answered by Sutrisno last updated on 17/Nov/23 $${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty}…
Question Number 200240 by universe last updated on 16/Nov/23 $$\mathrm{the}\:\mathrm{local}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function} \\ $$$${f}\left({x},{y}\right)\:=\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{3}{y}+\mathrm{11} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200242 by cherokeesay last updated on 16/Nov/23 Answered by cortano12 last updated on 17/Nov/23 $$\:\frac{\left({ab}+{c}\right){x}}{{b}−\frac{{c}}{{a}}\:}\:−\frac{\left({ab}−{c}\right){x}}{{b}+\frac{{c}}{{a}}}\:=\:\frac{{ab}−{c}}{{b}+\frac{{c}}{{a}}}\:−\frac{{ab}+{c}}{{b}−\frac{{c}}{{a}}} \\ $$$$\:\frac{{ax}\left({ab}+{c}\right)}{{ab}−{c}}\:−\frac{{ax}\left({ab}−{c}\right)}{{ab}+{c}}\:=\:\frac{{a}\left({ab}−{c}\right)}{{ab}+{c}}\:−\frac{{a}\left({ab}+{c}\right)}{{ab}−{c}} \\ $$$$\:\:\frac{{x}\left({ab}+{c}\right)}{{ab}−{c}}\:−\frac{{x}\left({ab}−{c}\right)}{{ab}+{c}}\:=\:\frac{{ab}−{c}}{{ab}+{c}}−\frac{{ab}+{c}}{{ab}−{c}} \\ $$$$\:\frac{\left({ab}+{c}\right)\left({x}+\mathrm{1}\right)}{{ab}−{c}}\:=\:\frac{\left({ab}−{c}\right)\left({x}+\mathrm{1}\right)}{{ab}+{c}} \\ $$$$\:\left({ab}+{c}\right)^{\mathrm{2}}…
Question Number 200300 by Calculusboy last updated on 16/Nov/23 Answered by MM42 last updated on 17/Nov/23 $${lnA}={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\:\left[{ln}\left(\mathrm{1}+\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{1}} \right)+{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{n}}\right)^{\mathrm{2}} +…+{ln}\left(\mathrm{1}+\left(\frac{{n}}{{n}}\right)^{\mathrm{2}} \right)\right.\right. \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\:\underset{{i}=\mathrm{1}} {\overset{{n}}…